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Q1. A curve passing through the point and satisfying the condition that
slope of the normal at any point is equal to the
ratio of ordinate and abscissa of that point, then the curve also passes through
the point
(1)
(2)
(3)
(4)
Q2. Let be the angle in radians between and the circle
at their points of intersection. If tan
, then find the value of .
Q3. The equation of the tangent line at the point to the curve with
parametric equation given by and
where is parameter is
(1)
(2)
(3)
(4)
Q4. If the line joining the points and is a tangent to the curve
, then find the value of
Q5. The shortest distance between the line and the curve is
(1)
(2)
(3)
(4)
Q6. The number of values of for which the curves and
are orthogonal is
Q7. Find the least positive integer for which function defined as
is a decreasing function for all
Q8. Find the maximum value of the function
in the set
(1)
(2)
(3)
(4)
Q9. Let be a polynomial function. If has
extreme at and such that and , then the equation
has
(1) three distinct real roots.
(2) one positive root, if and .
(3) one negative root, if and .
(4) All of the above
Q10. Let a function be continuous, and be defined
as:
, where
Then for the function , the point is:
(1) a point of local minima
(2) not a critical point
(3) a point of local maxima
(4) a point of inflection
Q11. If has its extremum values at and
then find value of .
Q12. The point on the curve which is closest to , is
(1)
(2)
(3)
(4)
Q13. If is a critical point of the function ,
then
(1) and are local minima of
(2) and is a local maxima of
(3) is a local maxima and is a local minima of
(4) is a local minima and are local maxima of
Q14. On the interval the function takes its maximum value
at the point
(1)
(2)
(3)
(4)
Q15. The value of in order that decreases for
all real values is given by:
(1)
(2)
(3)
(4)
Q16. is monotonically decreasing in the largest
possible interval . Then find greatest value of
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(1, 2) (0, 0) (2, 2) (2, 1) (3, 2)
+ = 1 x2
> 36
> y2
> 4
x2 + y2 = 12 1 k
> 23
> k2
> 4
(4, 2)
x = t2 y = t3 3 tty = 15 x 58
y = 2
y = x 2
y = 7 9x
> 4
(0, 3) (5, 2)
y = cx+1 cy = x y2 = x 2
> 742 7811 42
2
a 4x2 + a2y2 = 4 a2
y2 = 16 xm ff(x) = sin x mx + k x Rf(x) = 3 x3 18 x2 + 27 x 40
S = {x R : x2 + 30 11 x 0 }
122 222 122 222
f(x) = ax 3 + 5 x2 + cx + 1 f(x)
x = < 0 f()f() < 0
f(x) = 0
f() < 0 f() > 0
f() > 0 f() < 0
f : [0, 5] R f(1) = 3 FF (x) = x
> 1
t2g(t)dt g(t) = t
> 1
f(u)du .
F (x) x = 1
y = a log | x| + bx 2 + x x = 1 x = 2
a + by = x2 (4, )12
(1, 1) (2, 4)
( , )2349
( , )4316 9
x = 1 f(x) = (3x2 + ax 2 a) ex
x = 1 x = 23 fx = 1 x = 23 fx = 1 x = 22 fx = 1 x = 23 f
[0, 1] x25 (1 x)75
01/4 1/2 1/3
k f(x) = sin x cos x kx + bk < 1
k > 1
k > 2
k < 2
f(x) = x3 + 4 x2 + x + 2
(2, )
> 2 3
Q17. The set of value(s) of for which the function
possess a negative point of
inflection, is
(1) empty set.
(2)
(3)
(4)
Q18. Consider the function defined by
Then is :
(1) monotonic on
(2) not monotonic on and
(3) monotonic on only
(4) monotonic on only
Q19. A spherical iron ball of radius is coated with a layer of ice of
uniform thickness that melts at a rate of .
When the thickness of ice is , then the rate (in .) at which of the
thickness of ice decreases, is:
(1)
(2)
(3)
(4)
Q20. A spherical balloon is expanding. If the radius is increasing at the rate of
the rate at which the volume increases
(in cubic centimeters per minute) when the radius is , is
(1)
(2)
(3)
(4)
Q21. Suppose that is differentiable for all and that for all If
and , then has the value equal to
(1)
(2)
(3)
(4)
Q22. If Rolle's theorem holds for the function
, at the point , then equals:
(1)
(2)
(3)
(4)
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af(x) = + ( a + 2) x2 + ( a 1) x + 2 ax 3
> 3
{ }45
(2, 0) (, 2) (0, )
f : R Rf(x) = { (2 sin ( )) |x|, x 0 0, x = 0.
> 1
> x
f
(, 0) (0, ) (, 0) (0, ) (0, ) (, 0) 10 cm 50 cm 3/min 5 cm cm/min
> 56
> 13
> 136
> 118
2 cm/min 5 cm 10
100
200
50
f x f (x) 2 x. f(1) = 2 f(4) = 8 f(2) 3468
f(x) = 2 x3 + bx 2 + cx , x [1, 1] x = 12 2 b + c 211 3 Answer Key
Q1 (3) Q2 (4) Q3 (4) Q4 (4.00)
Q5 (1) Q6 (2) Q7 (2) Q8 (1)
Q9 (4) Q10 (1) Q11 (1.50) Q12 (1)
Q13 (4) Q14 (2) Q15 (3) Q16 (4)
Q17 (4) Q18 (2) Q19 (4) Q20 (3)
Q21 (2) Q22 (3)
Application of Derivatives
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If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Q1. Given
Since, it passes through
Also, passes through
Q2.
Given ellipse equation
and circle equation
After solving equation the points of intersection we
get,
consider the point
Equation of the tangent at to the circle is
Equation of the tangent at P to the ellipse is
if is angle between these tangents, then
compare with we get ,
so,
Q3. Slope
at
Equation of tangent at is
Q4.
Given that
line joining
is tangent to curve i.e
touches
Q5.
Differentiating w.r.t. we get,
For the shortest distance, the tangent at point will be parallel to the given
line
The shortest distance between the given curve & the line
= The perpendicular distance of point from the line
Q6.
Given curves are
...(i)
and ...(ii)
If the curves intersect at , then
and
On differentiating equation (i), we get,
On differentiating equation (ii), we get,
For curves to be orthogonal,
i.e.
Hence, two values of
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=1
> ()
> dy dx
> yx
=
> dy dx xy
ydy + xdx = 0
y2 + x2 = C
(1,2) 4 + 1 = C C = 5
x2 + y2 = 5 (2,1) + = 1 . . .( i)x2
> 36
> y2
> 4
x2 + y2 = 12 . . .( ii )(i) & ( ii )
y = 3 & x = 3
P (3, 3 )
P
3x + 3 y = 12 + y = 1 x
> 12 3 4
tan = =23 423
tan 1 ( )= 423
= tan 1 ( )k
> 23
k = 4 = = 4 k2
> 442
> 4
M = =
> dy dx dy /dt dx /dt
M = 3t23 2t t = 2 M = 94
(4, 2) M = 94
(y 2)= (x 4) y = x 7. 9494
(0, 3)(5, 2)
y 3 = (x 0)
> 23 50
y 3 = xy = 3 xy = cx+1
(3 x)= cx+1
3x + 3 x2 x = cx2 + 2 x + (c 3 ) = 0
D = 0
(2 )2 4 1 (c 3 ) = 0 4 4 c + 12 = 0
c = 4
y2 = x 2
x
2yy ' = 1 y ' = 12y
P
y'p = = 1 y1 =12y1
> 12
x1 = 2 + ( )2
=1294 ( y21 = x1 2 )
P
===
> 9412
> 12+1 274
> 2 742
4x2 + a2y2 = 4 a2
y2 = 16 xP (, )+ = 1
> 2
> a2
> 2
> 4
2 = 16 .+ y = 0 2xa2
> 2y
> 4
y = 4xa2y
m1 = 4 a2
2yy ' = 16 m2 = 8
>
m1m2 = 1
( )( )= 1 4a2
> 8
>
32 = a22
2 2 = a22
a2 = 2 a = 2
aQ7.
is differentiable.
For function to be decreasing for all
The least positive integer
Q8.
for is increasing function
hence maximum value in this interval occurs at
so
Q9.
Given,
and are of opposite signs.
Let and .
It is given that has extremum at .
Therefore, and are two distinct real roots of .
But we know that between two distinct real roots of a polynomial,
there is at least one real root of its derivative.
Therefore, has three distinct real roots and (say) such
that .
Thus, first option is correct.
If has exactly one positive root, then it is evident from
the figure that and .
Therefore,
[ lies between and ]
Thus, second option is also correct.
If has exactly one negative real root, then from the
figure, we have and .
[ lies between and ]
Thus, third option is also correct.
Q10.
Given,
By Leibnitz rule we get,
Now
has a local minimum at .
Q11.
or at
or
or
solving we get
Q12.
Let any point on this parabola is
Equation of normal at this point is
It passes through
So point is
Q13.
is a critical point
maxima at minima at
Q14. Let f (x) = x 25 (1 - x) 75 , x [0,1]
f ' (x) = 25 x 24 (1 - x) 75 - 75x 25 (1- x) 74
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f(x) = sin x mx + kf f (x) = cos x mf (x) < 0 cos x m < 0 m > cos xf x m > 1 m = 2
f(x)= 3 x3 18 x2 + 27 x 40
x2 11 x + 30 0 x [5, 6]
f ' (x)= 9 x2 36 x + 27 = 9 (x2 4 x + 3 )= 9 (x 1) (x 3)
x [5, 6] f(x)
x = 6
f(6)= 648 648 + 162 40 = 122
< 0 < 0 > 0
f(x) x = , f (x)= 0
f(x) , v < < < < vf(x)= 0
v > 0 , < 0
< < 0 f()f(0)< 0 0 f()< 0 [ f(0)= 1 > 0] f()> 0 [ f()f()< 0]
f(x)= 0
< 0 , v > 0 0 < <
f(0) f()< 0 0
f()< 0 [ f(0)= 1 > 0]
F (x)= x
> 1
t2g(t)dt F (x)= x2g(x) F (1)= 1. g(1)= 0 ( g(1)= 0)
F (x)= 2 xg (x)+ x2g(x) F (x)= 2 xg (x)+ x2f(x) ( g(x)= f(x)) F (1)= 0 + 1 3 F (1)= 3
F (x) x = 1 = + 2 bx + 1
> dy dx ax
= 0
> dy dx
x = 1, 2 + 2 b(1)+1 = 0 a
> 1
a 2 b + 1 = 0 + 4 b + 1 = 0 a
> 2
a + 8 b + 2 = 0 ; a = 2, b = 12
y = x2 (t, t2)
x + 2 ty = t + 2 t3
(4, )12
4 t = t + 2 t3
2t3 + 2 t 4 = 0
t3 + t 2 = 0
t = 1 (1, 1)
f(x) =(3x2 + ax 2 a)ex
f'(x) =(3x2 + ax 2 a)ex + ex(6x + a) = ex(3x2 + (a + 6 )x 2 )
x = 1 f (1) = 0
3 + a + 6 2 = 0
a = 7
f (x) = ex(3x2 x 2 )= ex(3x2 3 x + 2 x 2 )= ex(3x + 2 )( x 1 )
x =
> 2 3
x = 1
= 25 x 24 (1 - x) 74 [(1 - x) - 3x]
= 25 x 24 (1 - x) 74 (1 - 4x)
For maximum value of f (x), put f'(x) = 0
25x 24 (1 - x) 74 (1 - 4x) = 0
Also, at x = 0 , y = 0
at x = 1, y = 0
and x = 1/4, y > 0
f (x) attains maximum at x = 1 / 4.
Q15.
Given:
For decreasing function
So,
The maximum value of is .
Q16. f(x) = 3 +
now 2 < x < 2 + < x + <
<
f(x) < 3 +
4
Greatest value of is 4
Q17.
Given
Differentiating both sides with respect to , we get
Again, differentiating both sides with respect to , we get
Since, has a point of inflection.
So,
Since, has a negative point of inflection.
So,
Q18.
is an oscillating function which is non-monotonic in .
Q19. Let thickness
Total volume
Given
At
Q20.
As, volume of sphere
Q21. Using Lagrange's mean value theorem for f in [1, 2]
for
(1)
again using Lagrange's mean value theorem in [2, 4]
for
(2)
from (1) and (2), f(2) = 4.
Q22. If Rolle's theorem is satisfied in the interval [-1, 1], then
also
Also if them
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x = 0,1, 14
f(x)= sin x cos x kx + b f (x)= cos x + sin x kf (x) < 0 cos x + sin x k < 0 cos x + sin x 2 2 k < 0 k > 2
(x2 + + )8x
> 316 9
= 3 (x + )2
+ 16 34316 32343432323 (x + )<4323 (x + )2
<43494916 3
f(x)= +( a + 2) x2 +( a 1) x + 2 ax 3
> 3
xf (x)= + 2( a + 2) x +( a 1)= ax 2 + 2( a + 2) x +( a 1)
> 3ax 2
> 3
xf (x)= 2 ax + 2( a + 2)
f(x)
f (x)= 0 2 ax + 2( a + 2)= 0 x = ( a+2 )
> a
f(x)
x = < 0
> (a+2 )
> a
> 0
> (a+2 )
> a
a (, 2)(0, )
f(x)= { x(2 sin ( )) x < 0 0 x = 0
> 1
> x
f (x)=
(2 sin )x ( cos ( )) x < 0
(2 sin )+x ( cos ( )) x > 0
> 1x1x1x2
> 1x1x1x2
f (x)= { 2 + sin cos , x < 0 2 sin + cos , x > 0
> 1
> x
> 1
> x
> 1
> x
> 1
> x
> 1
> x
> 1
> x
f (x) (, 0) (0, ) = x cm
V = (10 + x)343
= 4 (10 + x)2 . . (i)dV dt dx dt
= 50 cm 3 / min dV dt
x = 5 cm 50 = 4 (10 + 5) 2 dx dt
= cm / min dx dt
> 118
V = r 343
= 4 r 2 = 4 r 2. (2) [ = 2 ]dV dt dr dt dr dt
= 4 . 25. 2 = 200 dV dt
c (1, 2), = f (c) 2
> f ( 2 ) f ( 1 ) 21
f(2)f(1) 2 f(2) 4 d (1, 2), = f (d) 2
> f ( 4 ) f ( 2 ) 42
f(4)f(2) 4 8 f(2) 4 f(2) 4
f(1)= f(1) 2 + b c = 2 + b + cc = 2 f ' (x)= 6 x2 + 2 bx + cf ' ( )= 0 12
6 + 2 b + c = 0 1412
+ b + c = 0 32
c = 2,
b = 12
2b + c = 2 ( )+(2 )12=
=
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1 2 1. Q1. The Area of a region bounded by from to
is square unit, find the integral value of
Given that graph of lies above the axis from to
(1)
(2)
(3)
(4)
Q2. The area of closed region bounded by the parabolas
and the line is
(1) sq. unit
(2) sq. unit
(3) sq. unit
(4) None of these
Q3. If the area bounded by
and the coordinates axes
is equal to , then is equal to (where, [.] denotes the greatest
integer function
(1)
(2)
(3)
(4)
Q4. The area enclosed between the curve and the
coordinate-axes is
(1)
(2)
(3)
(4)
Q5. The area of the closed figure bounded by and
and the -axis is
(1) sq. units
(2) sq. units
(3) sq. units
(4) sq. units
Q6. The area common to and is
(1)
(2)
(3)
(4)
Q7. The area of the region bounded by from
to is
(1) sq. units
(2) sq. units
(3) sq. units
(4) sq. units
Q8. The volume of the solid formed by rotating the area enclosed
between the curve and the line about is (in cubic
units)
(1)
(2)
(3)
(4)
Q9. The area of the region is
(1)
(2)
(3)
(4)
Q10. Area enclosed between the curves and is
(1) sq. units
(2) sq. units
(3) sq. units
(4) None of these
Q11. The area of the region bounded by
and is
(1) sq. units
(2) sq. units
(3) sq. units
(4) sq. units
Q12. Let Maximum , where
. Determine the area of the region bounded by the curves
-axis,
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y = sin ax , y = 0 x =
> 3a
x = a 3
a(a > 0)
y = sin ax x = a
x = }
> 3a
3
> 131211
y = 4 x2, y = x2
> 9
y = 2
> 202 3102 3402 3
f(x) = max(sin x, cos x); 0 x , x =
> 2
>
> 2
k [k + 3] 2846
y = log e(x + e)3412
x = 1, x = 2
y = { x2 + 2, x 1 2x 1, x > 1 x
> 16 310 313 373
x2 + y2 = 64 y2 = 4 x
(4 + 3) 16 3
(8 3) 16 3
(4 3) 16 3
(8 + 3) 16 3
f(x) = sin x, g(x) = cos x
x = 0 x =
> 2
2(2 + 1) (3 1) 2(3 1) 2(2 1)
y = x2 y = 1 y = 1
> 9
> 54
> 38
> 37
> 5
{(x, y) : xy 8, 1 y x 2}
16 log e 2 14 3
8 log e 2 73
8 log e 2 14 3
16 log e 2 6 |y| = 1 x 2 x2 + y 2 = 1
> 38 3
> 8 328 3
x = 0, y = 0, x = 2, y = 2, y ex y ln x
6 4 ln 2 4 ln 2 2 2 ln 2 4 6 2 ln 2
f(x) = {x2, (1 x)2, 2 x(1 x)}
0 x 1
y = f(x), x x = 0& x = 1 (1)
(2)
(3)
(4)
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> 17 27 15 25 13 23 14 29
# Answer Key
Q1 (3) Q2 (1) Q3 (3) Q4 (3)
Q5 (1) Q6 (2) Q7 (4) Q8 (2)
Q9 (1) Q10 (1) Q11 (1) Q12 (1)
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We have,
Given that the given curve is above the axis from
Then, the area bounded by the curve from
Q2.
Q3.
Required Area
Q4.
Curve :
at x-axis
when
Required area
.
Q5.
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y = sin ax
x = to x = .a
> 3a
x = to x = = /3a
> /a
ydx a
> 3a
A = sin axdx
> a
> 3a
A = ( cos ax )/3a
> /a 1
> a
A = [ ]1
> a
> 32
A = = 3 (Given) 32a
a = 12
Area = 2
> 2
> 0
(9y ) dy
> y4
= 2 {3. (y) . (y) }
> 2023
> 321223
> 32
= 2 { . y }2053
> 32
= . 22 10 3
f(x)= { cos x for 0 x /4 sin x for /4 < x /2
= 2 /4 0 cos xdx = 2[sin x] /4 0
= 2 sq units k = 2 [k + 3]= [2 + 3 ]= 4
y = log e(x + e), x axis, y axis
y = 0 log e(x + e)= 0 x + e = 1 x = 1 ex e+ y = 01 e log( x + e) dx
= x log ( x + e) 01 e 01 e x dx 1
> x+e
= 0 11 e dx
> x+eex+e
= 0 ( x e log( x + e)) 01 e
= e +(1 e) e log(1) = 1 Q6.
The given curves are and
Solving both the equations, we get
Since the intersection point is lying in the first quadrant, so
So, the required area is
On integrating, we get
.
Q7.
We have, and
Area of the shaded region
sq. units.
Q8. Volume of the solid formed by rotating the area enclosed between
the curve and line will be
cu. units.
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A = 11 (x2 + 2 )dx + 21 (2 x 1) dx
= ( + 2 x)11 + (x2 x)2x3
> 3
= sq. units 16 3
x2 + y2 = 64. . .(i) y2 = 4 x. . .(ii)
x2 + 4 x 64 = 0
x = 2 217
x = 2 + 217 = 2 2+217 0 4 xdx + 2+217 0 64 x2dx
= (8 3 )16 3
cos x sin x, 0 x
> 4
sin x cos x, x
> 4
>
> 2
= 0 (cos x sin x)dx + (sin x cos x)dx
>
> 4
>
> 2
>
> 4
= [sinx + cosx] 0 + [cosx sinx]
>
> 4
>
> 2
>
> 4
={ + (0 + 1) }{1 ( + )} 12 12 12 12
= 2 42
= 22 2 = 2 (2 1 )
y = x2 y = 1
V = 10 2xdy = 2 10 ydy = [y ]10
=4
> 3
> 324
> 3
Q9.
To draw the inequality, let us draw the equation
and and
For point of intersection
(i) and
(ii) and
(iii) and and
Now region which contains origin
region above line
region outside the parabola
Now required area
Method I:
Using x-axis:
Method II:
Using y-axis:
Note: The question should include bounded area term as in quadrant
there exist a area which satisfy the inequality and is unbounded.
Q10.
The dotted area is
Hence, area bounded by circle and
Q11.
Required area
Q12.
coordinate of
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xy = 8 y = 1 y = x2
xy = 8 y = 1 A(8,1)
xy = 8 y = x2 x3 = 8 x = 2 B(2, 4)
y = x2 y = 1 C(1, 1) D( 1, 1)
xy 8
y 1 y = 1
y x2
A =
> 2
> 1
(x2 1 )dx +
> 8
> 2
( 1 )dx
> 8
> x
A = [ x]21
+ [8ln x x]82
> x3
> 3
A = + 16ln2 14 3
A =
> 4
> 1
( y)dy 8
> y
A = [8ln y y 3/2 ]41
A = + 16ln2 2314 3
2nd
A = 10 (1 x 2)dx = (x )10
= 1 =x3
> 31323
x2 + y 2 = 1 |y|= 1 x 2
= Lined area = Area of curcle Area bounded by |y|= 1 x 2
= 4. ( )= sq. units 2338 3
A =
> 2
> 1
ln x dx
= [ x ln x x]21
= 2 ln 2 1 = 4 2(2 ln 2 1)= 6 4 ln 2 sq.units
f(x) = Max {x2, (1 x)2, 2 x(1 x)}
A, (1 x)2 = 2 x(1 x)(1 x)(1 x 2 x) = 0 coordinate of Req. Area
Area Under Curves
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x = , A ( , 0 )1313 B
2x(1 x) = x2, 2(1 x) = x, 2 3 x = 0, x = 23
B ( , 0 )23 = 1/3 0 (1 x)2dx + 2/3 1/3 2x(1 x)dx + 1 x2dx 23
= 17 27 Q1. If is defined by
(1)
(2)
(3)
(4)
Q2. If
(1)
(2)
(3)
(4)
Q3. The number of points of discontinuity of in is/are (where
[.] denotes Greatest Integer Function)
(1)
(2)
(3)
(4)
Q4. Let where then is discontinuous only at
(1)
(2)
(3)
(4) None of these
Q5. Let and The function is
discontinuous at
(1) infinitely many points.
(2) exactly one point.
(3) exactly three points.
(4) no point.
Q6. Let . If is continuous
functions at , then is equal to
(1)
(2)
(3)
(4)
Q7. If
(1)
(2)
(3)
(4) none of these
Q8. Let be defined by where is
the greatest integer less than or equal to .
Let denote the set containing all where is discontinuous, and denote
the set containing all
where is not differentiable. Then the sum of number of elements in and is
equal to
Q9. Let and . Then
(1) is differentiable at , but is not continuous at
(2) is not differentiable at
(3) is differentiable at
(4) is continuous at but is not differentiable at
Q10. Let be a differentiable function with and
Then
(1)
(2)
(3)
(4)
Q11. Let be the set of points where the function, , is
not differentiable. Then is equal to
Q12. Let
If is differentiable at
(1)
(2)
(3)
(4) None of these
Q13. If and , then
(1) not be differentiable at every non-zero .
(2) differentiable for all .
(3) twice differentiable at .
(4) none of the above.
Q14. If , then at will be
(1) Continuous but not differentiable
(2) Neither continuous nor differentiable
(3) Continuous and differentiable
(4) Differentiable but not continuous
Q15. Let be a polynomial of degree one and be a continuous and
differentiable function defined by
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f : R Rf(x) =
, if x R {1, 2} 1, if x = 2 then f is continuous on the se 0, if x = 1
> x+2
> x2+3 x+2
RR {2}
R {1}
R {1, 2}
f(x)=
, x < 0
q , x = 0 is continuous at x = 0, then the ordered pair ( p, q) is equal to , x > 0
> sin( p+1) x+sin xx
> x+x2 xx3/2
( , )3212
( , )1232
( , )5212
( , )3212
f(x) = [x3 + 1 ] (1, 2) 1654
y = 1
> u2+u2
u = 1
> x1
y x =1, 2 1, 2 1, , 2 12
f(x) = sgn( x) g(x) = x (x2 5 x + 6 ) . f(g(x))
f(x) =
, x k , x =
> 1+cos x
> (x)2
> sin 2x
> log (1+ 22 x +x2)
f(x)
x = k
> 14121 2
14
f(x)=
x + 2, x > 0 x2 2, 0 x < 1, then the number of points of discontinuity of | f(x)| is: x, x 1 102f : [0, 3] R f(x) = min{ x [ x], 1 + [ x] x} [x]
xP x [0, 3] f Q
x (0, 3)
f P Q
f(x) = x|x|, g(x) = sin x h(x) = ( gof )( x)
h(x) x = 0 h(x) x = 0
h(x) x = 0
h(x) x = 0
h(x) x = 0 x = 0 f : (1, 1) R f(0) = 1 f (0) = 1, g(x) = {f(2f(x) + 2)} 2. g(0) = 4
> 4
02
S f(x) = |2 | x 3 , x R
xS f(f(x))
f(x) = { 3x p : 0 x 2 2x2 + qx : 2 < x 3
f(x) x = 2, ( p, q) = (8, 5)
( , 3 )52
(10, 4)
f(x + y) = f(x) + f(y) + | x|y + x2y2, x, y R f (0) = 0
xx Rx = 0
f(x) = { 3x, 1 x 1 4 x, 1 < x < 4 x = 1, f(x)
g(x) f(x)If , then
(1)
(2)
(3)
(4)
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f(x) =
g(x), x 0
( ) , x > 0 .1+ x
> 2+ x
> 1
> x
f (1) = f (1)
f (1) = ( + ln )231632
f (1) = + ( + ln )321632
f (1) = ( + ln )236123
f (1) = ( ln )231632Answer Key
Q1 (3) Q2 (4) Q3 (2) Q4 (3)
Q5 (3) Q6 (2) Q7 (1) Q8 (5)
Q9 (4) Q10 (2) Q11 (3) Q12 (1)
Q13 (2) Q14 (1) Q15 (1)
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Since, is continuous for , not sure about
Now, we have check continuity at these points.
At
It is continuous at
Now, check for
It is not continuous at
The required function is continuous in
Q2.
is continuous at
Q3.
By property of we know that is discontinuous at every integral point.
So we have to just check that for how many values of , is taking
integral values.
As , then integer lying in this range are
points are there at which is becoming
discontinuous.
Q4. is discontinuous at
is discontinuous at . If then If
then Hence the composite function is discontinuous
only at
Q5.
Let
if
if
if
Therefore, is discontinuous at three points .
Q6. The given function is
Since, f(x) continuous at x = .
Q7.
Given function
Using properties of modulus function,
For continuityof at
Hence, function is discontinuous at .
Therefore, number of discontinuity is 1.
Q8.
Given,
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f(x) R {1, 2}.
x = 2, LHL = lim
> n0 ( 2 n) +2 ( 2 n)2+3 ( 2 n) +2
= lim
> n0
= 1
> nn2+n
RHL = lim
> n0 ( 2+ n) +2 ( 2+ n)2+3 ( 2+ n) +2
= lim
> n0
= 1 nn2n
LHL = RHL = f(2)
x = 2
x = 1 LHL = lim
> n0 ( 1 n) +2 ( 1 n)2+3 ( 1 n) +2
= lim
> n0
= 1 nn2n
RHL = lim
> n0 ( 1+ n) +2 ( 1+ n)2+3 ( 1+ n) +2
= lim
> n0
=
> 1+ nn2+n
LHL = RHL f(1)
x = 1
R {1}
f(x)=
q
: x < 0 : x = 0 : x > 0
> sin ( p+1 ) x+sinx x
> x2+x xx3/2
f x = 0
L. H. L =lim
> h0
= lim
> h0 sin ( p+1 ) ( h) +sin( h)h
> sin ( p+1 ) h+sin hh
= lim
> h0
(p + 1)+ lim
> h0
= p + 1 + 1
> sin ( p+1 ) hh(p+1 ) sin hh
= p + 2
R. H. L = lim
> h0
= lim
> h0
> h2+hhh
> 32
> h2+h+ h
> h2+h+ hh2+hhhh(h+1+1 )
> 3212
lim
> h0
=11+ h+1 12
L. H. L. = R. H. L. = f(0) p + 2 = = q12
= p = & q =3212
f(x)= [x3 + 1 ]
GIF, [x]
x (1, 2) (x3 + 1 )
1 < x < 2 1 < x3 < 8 2 < x3 + 1 < 9
(x3 + 1 )(2, 9)
[x3 + 1 ]= 3, 4, 5, 6, 7, 8 6 f(x)
u = ( x)= 1
> x1
x = 1, y = f(u)= =1
> u2+u2 1(u+2 ) ( u1 )
u = 2, u = 1 u = 2 2 = x = .1
> x1 12
u = 1 1 = x = 2. 1
> x1
x = 1, , 2. 12
f(x)= sgn( x)=
1, x < 0 0, x = 0 1, x > 0
g(x)= x(x2 5 x + 6 )= x(x 2)( x 3)
f(g(x))= sgn( g(x))= sgn( x(x 2)( x 3)) =
1 , x(x 2)( x 3)< 0 0 , x(x 2)( x 3)= 0 1 , x(x 2)( x 3)> 0
h(x)= f(g(x))
h(x)= 1 x (, 0)(2, 3)
h(x)= 0, x = 0, 2, 3
h(x)= 1 x (0, 2)(3, )
f(g(x)) 0, 2, 3 f(x)=
, x
k , x=
> 1+cos x (x ) 2
> sin 2xlog (1+ 22 x+x 2)
lim x f(x)= f( ) lim
> h0
f( + h)= k lim
> h0
> 1+cos ( +h ) (h ) 2
= k
> sin 2(+h ) log [1+ 22 (+h ) + ( +h ) ]2
lim
> h0
= k
> 1cos h
> 2
> sin 2hlog (1+h 2)
lim
> h0
( )2
( )2
= k 12sin h/2 h/2 h2
> log (1+h 2)
> sin h h
= k 12
f
x
=
x + 2, x > 0
x2 2, 0 x < 1
x, x 1 .
|f(x)|=
|x + 2|, x < 0 x2 + 2 , 0 x < 1 x, x 1 .
|f(x)|=
x 2, x > 2 x + 2, 2 x < 0 x2 + 2, 0 x < 1 x, x 1 .|f(x)| x = 1, L. H. L. = f (1) = 1 2 + 2 = 3 R. H. L. = f (1+) = 1
x = 1 So, the graph of the function is
From the graph we can say that the function is continuous at all points in the
interval .
So,
Again, from the diagram, we can say that at , there are sharp
edges.
So, the function is non differentiable at
Hence,
So,
Q9.
Hence,
So
Now,
Hence, is non-derivable at .
Q10.
We have on differentiation with respect to x,
g'(x) = 2f (2f (x) + 2) f '(2f (x) + 2) 2f ' (x)
g'(0) = 2f (2f (0) + 2) f ' (2f (0) + 2 f ' (0)
= - 4.
Q11.
Check non-differentiability of function, we get,
is non differentiable at
Q12. diff. at conti. at
Now
Q13.
Given:
and
Substitute,
Clearly is differentiable at all points.
Q14. Since
and f(1) = 3 1 = 3
is continuous at x = 1
Again
and
f(x) is not differentiable at x = 1
Q15.
Let
and
As is continuous,
For ,
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f(x) = min{ x [ x], 1 + [ x] x}1 { x} = 1 x; 0 x < 1 [0, 3]
P = 0 x = , 1, , 2, 123252
x = , 1, , 2, 123252
Q = 5
P + Q = 5
h(x)= { sin x2 x 0 sin x2 x < 0
h(0+)= lim
> x0 +
= 0 sin x2
> x
h(0)= lim
> x0
= 0
> sin x2
> x
h(0)= 0
h(x)
2x cos x2 x > 0 0 x = 0 2 x cos x2 x < 0
h (0 +)= lim
> x0 +
= 2
> 2xcos x2
> x
h (0 - ) = lim
> x0
= 2
> 2 xcos x2
> x
h(x) x = 0 g(x)= { f(2 f(x)+2)} 2
f(x) x = 1, 3, 5
f(f(x))= f(f(1))+ f(f(3))+ f(f(5)) = 1 + 1 + 1 = 3
x = 2 x = 2
f(2)= f(2+) 6 p = 8 + 2 q p + 2 q = 2 . . . .(1)
f '(x) =[ 3 , 0 < x < 2 4x + q , 2 < x 3
f '(2+)= f '(2) 8 + q = 3 q = 5, p = 8
f(x + y)= f(x) + f(y) +| x|y + x2y2
f'(0) = 0
x = y = 0
f(0 + 0) = f(0) + f(0) + 0 + 0 f(0) = 0
f (x) = lim
> h0
> f(x+h) f(x)
> h
f (x) = lim
> h0
> f(x)+f(h)+ | x|h+x2h2f(x)
> h
= lim
> h0
+| x|+ x2h
> f(h)
> h
= f (0) +| x|
f (x) = 0 +| x|=| x|
f (x) ={ x x < 0
x x 0
f(x)f(1 0)= lim
> x1
3x = 3 f(1 + 0)= lim
> x1
(4 x)= 3
f(1 0)= f(1 + 0)= f(1) f(x) f(1 + 0)= lim
> x1 +
= lim
> x1
= lim
> h0
= 3lim
> h0
= 3log 3
> f ( x ) f ( 1 ) x1 3x3 x1 31+h 3 h3h1 h
f(1 + 0)= lim
> x1
= lim
> x1
= 1
> f ( x ) f ( 1 ) x1 4x3 x1
f(1 + 0) f (1 0)
g(x)= ax + b
LHL = lim
> x0
f(x)= lim
> x0
g(x)= b
RHL = lim
> x0 +
f(x)= ( ) = ( )
= 0
> 1+0 2+0
> 1012
f(x) LHL = RHL = f(0) b = 0
x > 0 f(x)= ( )1/ x1+ x
> 2+ x
ln( f(x))= (ln(1 + x) ln(2 + x)) 1
> x
f (x)= [ (log )+ ( )] 1
> f(x)1
> x2
> ( 1+ x)( 2+ x)1
> x
> 11+ x
> 12+ x
f (1)= f(1) ( ln + 1 ( )) 231213
=( )( ln + )232316
f (1)= a
f (1)= f (1) a = ( + ln )231632
f (1)= ( + ln )231632Q1. If , where , then
(1)
(2)
(3)
(4)
Q2. The value of satisfying the equation , is
(1)
(2)
(3)
(4)
Q3. The integral is equal to
(1)
(2)
(3)
(4)
Q4. Let , then
(1)
(2)
(3)
(4)
Q5. A value of such that is
(1)
(2)
(3)
(4)
Q6. The value of , where denotes the greatest integer
less than or equal to , is
(1)
(2)
(3)
(4)
Q7. Consider (where ), then the value of
is
(1)
(2)
(3)
(4)
Q8. If is a real valued function defined by ,
then the value of is equal to
(1)
(2)
(3)
(4)
Q9. If , where , then
(1)
(2)
(3)
(4) none of these
Q10. The value of , where denotes the
greatest integer function is
(1)
(2)
(3)
(4)
Q11. equals:
(1)
(2)
(3)
(4) None of these
Q12. If then is equal to
(1)
(2)
(3)
(4)
Q13. The integral value of
(1)
(2)
(3)
(4)
Q14. Statement I : If , then
Statement II : given
(1) Both Statement I and Statement II are true and the Statement II is the
correct explanation of the Statement I
(2) Both Statement I and Statement II are true but the Statement II is not
the correct explanation of the Statement I
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In = 0 xn sin xdx
>
> 2
n > 1
In + n(n 1) In2 = n( )n
> 2
In + n(n 1) In2 = n( )n1
>
> 2
In n(n 1) In2 = n( )n
> 2
In n(n 1) In2 = n( )n
> 2
x > 1 x
> 1
t log tdt = 14
ee 32
e2
2e 1
sec x cosec xdx
>
> 3
>
> 62343
3 3
> 7656
3 3
> 4313
3 3
> 5623
3 3
> 5313
l1 = 0 dx , l2 = 0 dx
> x2x
> (1+ x)6
> xx
> (1+ x)6
l1 = 2 l2
l2 = 2 l1
l1 = l22
l1 = l2
+1
>
= log e( )dx
> (x+)( x++1) 98
1212
2 2
/2 /2
> dx
> [x]+[sin x]+4
[t]
t
(4 3) 320
(4 3) 310
(7 5) 112
(7 + 5) 112
I( ) = 2
> dx x
> 0
5
> r=2
I( r) + 5
> k=2
I ( )1
> k
01ln 2 ln 4 f f(x + y) + f(x y) = f(2x)
f(2) f(2) f(x) 0123
10 +
> 0
| sin x|dx = k cos 0 < < k =101 100 201
2
> 0
[sin 2 x(1 + cos 3 x)] dx [t]
2
2
2n
> 0
{| sin x| sin x} dx 12
n
2n
2 nI = 41 ({ x}) [x]dx I24 13
1234
dx =
> 3
> 4
>
> 4
> x
> 1+sin x
(2 + 1) 2(2 1) 2(2 + 1)
>
> 2+1
10 esin xdx = 200 0 esin xdx = 200
na
> 0
fxdx = n a
> 0
f(x)dx , n I f(a + x) = f(x) (3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Q15. , where represents fractional part of Then
is
(1)
(2)
(3)
(4) None of these
Q16. Evaluate:
(1)
(2)
(3)
(4)
Q17. The value of is
(1)
(2)
(3)
(4)
Q18. , then is :
Q19. If , then the value of is
(1)
(2)
(3)
(4)
Q20. is
(1)
(2)
(3)
(4)
Q21. is equal to
(1)
(2)
(3)
(4)
Q22. The value of
(1)
(2)
(3)
(4) None of these
Q23. If and , then the
value of is
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40 {sin{ x}} dx = N {x} x.
> N
> 4
cos 1 1 cos 1 1 sin 1
199
>
( )dx 1cos 2 x
> 2
200 400 200 400
Lt { }
> x20sec 2tdt xsin x
0321
1sin x t2f(t)dt = 1 sin xx (0, /2) f ( )13
x2(1+ x)0 f(t)d t = x f(2) 1/2 1/3 1/4 1/5 lim n 1p+2 p+3 p++ np
> np+1
(p > 1)
> 1
> p+1 1
> p1
1
> p
> 1
> p1 1
> p+2
lim n ( + + + )
> (n+1) 1/3
> n4/3
> (n+2) 1/3
> n4/3
> (2 n)1/3
> n4/3
(2) 4/3 3434
(2) 3/4 43
(2) 4/3 43
(2) 4/3 3443
lim n nk=1 =
> k{na+ka}
> 1
> a1
> a1
> a
> na+1
123
f(x + y) = f(x) + 3 y2 + kxy , f(1) = 1 f(2) = 6
21 f(x)dx 110 Answer Key
Q1 (2) Q2 (1) Q3 (1) Q4 (4)
Q5 (3) Q6 (1) Q7 (1) Q8 (1)
Q9 (3) Q10 (3) Q11 (2) Q12 (2)
Q13 (4) Q14 (4) Q15 (2) Q16 (2)
Q17 (4) Q18 (3) Q19 (4) Q20 (1)
Q21 (1) Q22 (1) Q23 (1.3)
Definite Integration
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We have,
Hence,
Q2. Consider that,
(as )
Q3.
Let
.
Q4. We have,
Let then,
Q5.
Using partial fractions
or .
Q6.
We have
Q7.
Thus, and
Hence, their sum equals to zero
Q8.
Put
is an odd function.
Q9.
Let
Period of is &
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In = 0 xn sin xdx
>
> 2
In = [ xn cos x]0 + 0 nx n1 cos xdx
>
> 2
>
> 2
In = n 0 xn1 cos xdx
>
> 2
In = n[[xn1 sin x]0 0 (n 1) xn2 sin xdx ]
>
> 2
>
> 2
In = n[( )n1
( n 1) In2 ]
> 2
In + n(n 1) In2 = n( )n1
>
> 2
I =
> x
> 1
t log t dt
= [log t ]x
> 1
> x
> 1
dt t2
> 21
> tt2
> 2
= log x [ ]x
> 1
> x2
> 212
> t2
> 2
= log x [ ]x2
> 212
> x2
> 212
= log x (x2 1 )14
> x2
> 214
x2 log x x2 = 0 1214 x > 1 x2(2 log x 1)= 0 2 log x 1 = 0 log x = 12
x = e 12
x = eI = sec x cosec x dx
>
> 3
>
> 62343
= dx
>
> 3
>
> 6
> sec 2x
> tan 4/3 x
tan x = t, sec 2xdx = dt
I =
> 3
> 13
> dt t
> 43
= [ ]
> 3
> t+1 43
> +1
> 4313
= 3 ( 3 )13
> 1616
= 3 3
> 7656
l1 =
>
> 0
dx
> x2x
> ( 1+ x)6
x = 1
> t
dx = dt 1
> t2
l1 =
> 0
>
( dt )
> 1
> t2 t
> (1+ )61
> t
> 1
> t2
l1 =
>
> 0
dt = l2
> tt
> ( 1+ t)6
I =
> +1
>
=
> +1
>
( )dx dx
> (x+)( x++1) 1
> x+
> 1
> x++1
= [ln| x + |ln| x + + 1|] +1
>
= [ln ]+1
> x+x++1
= ln ln
> 2+1 2+2 2
> 2+1
= ln = ln
> ( 2 +1 ) 2
> ( 2 +1 ) 21 98
(2 + 1) 2 = 9 2 + 1 = 3 = 1 2
>
> 2
> 2
> dx
> [x]+[sin x]+4
= 0 + 0
> 2
> dx
> [x]+3
>
> 2dx
> [x]+4
= 1 + 01 + 10 + 1
> 2
> dx
> 1
> dx
> 2
> dx
> 4
>
> 2dx
> 5
= [ x]1 + [ ]01 + [ ]10 + [ ]1 =(1 + )+(0 + )+ +
> 2
> x
> 2
> x
> 4
> x
> 5
>
> 2
> 21214
>
> 10 15
= = = (4 3)
> 20+10 +10+5+2 4 20 12 9 20 320
I( )= [ln x]2
>
= ln 2 ln
= ln ( )2
>
= ln
5
> r=2
I( r)= ln 2 + ln 3 + ln 4 + ln 5
5
> r=2
I( )= ln ( )+ ln ( )+ ln ( )+ ln ( )1
> k
> 12131415
= (ln 2 + ln 3 + ln 4 + ln 5) f(x + y)+f(x y)= f(2x)= f(x + y + x y) x = 0, y = x f(x)+f(x)= 0 f(x)= f(x) f(x) f(2)= f(2)
> f ( 2 )
> f ( 2 )
f(x)dx =
> f ( 2 )
> f ( 2 )
f(x)dx = 0
I = 100 +
> 0
|sin x| dx
= 100
> 0
|sin x| dx + 100 +
> 100
|sin x| dx
= 100
> 0
sin x dx +
> 0
sin x dx
|sin x| 0 < <
= 100 ( cos x)
> 0
+ ( cos x)
> 0
Q10.
Applying
Adding and
Q11.
Let
Now,
Q12.
= =
Q13. Let ....(i)
....(ii)
By adding Equations (i) and (ii), we get
Q14.
We have given that
is a periodic function, its value will repeat as repeat its value.
Period of
But is not an integer
Statement I is wrong.
and
So Statement II is true.
Q15.
Let
Using if is a
periodic function with period
We know that has period equals to .
Also, for
Q16. Let,
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= 100 (1 + 1)+( cos + 1)= 201 cos
k = 201
I =
> 2
> 0
[sin 2 x(1 + cos 3 x)] dx (1)
I =
> 2
> 0
[sin(2 2 x)(1 + cos(2 3 x))] dx
(
> a
> 0
f(x)=
> a
> 0
f(a x)dx )
I =
> 2
> 0
[ sin 2 x(1 + cos 3 x)] dx (2) (1) (2) 2I =
> 2
> 0
([sin 2 x(1 + cos 3 x)]+[ sin 2 x(1 + cos 3 x)]) dx
2 I =
> 2
> 0
1 dx { [x]+[ x]= { 0 : x I 1 : x I
2 I = ( x)2
> 0
2 I = 2
I = I = 2n
> 0
{|sin x| sin x}dx 12
= 2n
> 0
|sin x|dx 12
= (n 2
> 0
|sin x|dx )12
I1 =
> 2
> 0
|sin x|dx I1 =
>
> 0
sin x dx
> 2
>
sin x dx
= [ cos x]
> 0
+ [cos x]2 = [1 1]+[1 + 1] = 2 + 2 = 4
I = (4 n)= 2 n12
I =
> 2
> 1
{x}+
> 3
> 2
{x}2 +
> 3
> 2
{x}3
=
> 1
> 0
x + x2 + x3 + +1213146+4+3 12
= 13 12
I = dx
> 3
> 4
>
> 4
> x
> 1+sin x
I =
> 3
> 4
>
> 4
> (+x)3
> 4
>
> 4
> 1+sin (+x)
> 3
> 4
>
> 4
=
> 3
> 4
>
> 4
> (x)dx
> 1+sin x
[ ba f(x)dx = ba f(a + b x)dx ]
2I =
> 3
> 4
>
> 4
> dx
> 1+sin x
2I = dx
> 3
> 4
>
> 4
> 1sin x
> ( 1+sin x)( 1sin x)
2 I = dx
> 3
> 4
>
> 4
> 1sin x
> cos 2x
2 I = [ sec 2 x sec x tan x]dx
> 3
> 4
>
> 4
2 I = [tan x sec x]
> 43
> 4
2 I = [1 (2 )(1 2 )]
2 I = [1 + 2 1 + 2 ]
2 I = [2 + 22 ]
I = [2 1 ]
= (2 + 1 )
> (21 )(2+1 )
=
> 2+1
10 esin xdx =
f(x) = e sinx
f(x) sinx f(x) = 2
200 0 esin xdx = 2
> 0
esin xdx
> 100
>
> 100
>
na
> 0
fxdx = n a
> 0
f(x)dx , n II = 40 {sin{ x}}dx
nT
> 0
f(x)dx = n T
> 0
f(x)dx , n Z f(x)
T .{x} 1 I = 4 10 {sin{ x}} dx x [0, 1], {sin{ x}}= sin x
I = 4 10 sin xdx
I = 4[cos x]10
I = 4(1 cos 1).
I =
> 199
>
( ) dx
> 1cos 2 x
> 2
=
> 199
>
|sin x|dx ( is periodic with period and
if is the period of the function ).
.
Q17.
Apply L'Hospital's Rule, we get
Q18.
Applying Leibnitz Rule,
Q19.
Differentiating both sides w.r.t. , we get
(Use Newton-Leibniz Rule for
differentiation)
At
Q20.
Q21. The given limit can be written as
Q22.
(applying definite integral as limit of sum)
Q23.
..(i)
In (i), put
.......(ii)
In (ii), put
Using (ii),
Replace
Now
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=(199 (1))
>
> 0
|sin x|dx
|sin x| nT mT f(x)dx =( n m) T
> 0
f(x)dx T f(x)= 200
>
> 0
sin xdx
= 200| cos x|
> 0
= 200(1 (1))= 400
Lt
> x0
{ } ( form )
> x20sec 2tdt xsin x
> 00
= Lt
> x0
( form )
> (sec 2x2)2xxcos x+sin x
> 00
= Lt
> x0
= = 1 2 sec 2 x2
> cos x+sin xx
> 21+1 1
> sin x
t2f (t) dt = 1 sin x
sin 2 x f (sin x) cos x = cos x
f (sin x) = 1sin 2 x
f (t) = 1
> t2
f ( ) = = 3 13 1
> ()
> 213
x
f(x2(1 + x))(2x + 3 x2)= 1
x = 1 f(2)= 15
Lt
> n
> 1p+2 p+...+ np
> np
> 1
> n
= Lt
> n
[( )p
+ ( )p
+. .. + ( )p]1
> n
> 1
> n
> 2
> nnn
= Lt n
> r=n
> r=1
( )p1
> nrn
=
> 1
> 0
xpdx [ ]10 =
> xp+1
> p+1 1
> p+1
lim
> n
[(1 + ) + (1 + ) + . + (1 + ) ]1
> n
> 1
> n
> 132
> n
> 13nn
> 13
= lim
> n
nr=1 (1 + )1
> nrn
> 13
=
> 1
> 0
(1 + x)1/3 dx
=
> 10( 1+ x)
> 4343
= (24/3 1 )
> 34
= . 2 4/3
> 3434
lim
> n
nk=1
> k{na+ka}
> 1
> a1
> a1
> a
> na+1
= lim
> n
nk=1 {( ) + ( )a
}1
> nkn
> 1
> akn
= 10 (x + xa)dx
> 1
> a
={ + ]10
> x()+1 1
> a
> +1 1
> a
> xa+1
> a+1
= + = 1 aa+1 1
> a+1
f(x + y)= f(x)+3 y2 + kxy x = 1, f(1 + y)= f(1)+3 y2 + ky
f(y + 1)= 3 y2 + ky + 1
y = 1, f(2)= 3(1)+ k + 1 = 6 k = 2
f(y + 1)= 3 y2 + 2 y + 1
y x 1 f(x)= 3( x 1) 2 + 2( x 1)+1 = 3( x 1) 2 + 2 x 1
21 f(x)dx = 21 (3( x 1) 2 + 2 x 1 )dx
= [(x 1) 3 + x2 x]21 =[(8 + 1)+(4 1)(2 + 1)]= 13 Q1. Given a differential equation ,
whose order and degree are respectively, then
(1)
(2)
(3)
(4)
Q2. If and are order and degree of the differential equation
, then the value of is
Q3. The degree of the differential equation, of which
is a solution, is
(1)
(2)
(3)
(4) None of these
Q4. Solution of the differential equation is
(1)
(2)
(3)
(4) None of these
Q5. If satisfies the differential equation and
given that , then
(1)
(2)
(3)
(4)
Q6. The slope at any point of a curve is given by
and it passes through The equation of the curve is
(1)
(2)
(3)
(4)
Q7. The general solution of the differential equation
is (where is an arbitrary constant)
(1)
(2)
(3)
(4)
Q8. If and , then
is equal to :
(1)
(2)
(3)
(4)
Q9. The equation of the curve satisfying the equation
and passing through the point is
(1)
(2)
(3)
(4)
Q10. The solution of is
(1)
(2)
(3)
(4) none of these
Q11. Let be the solution of the differential equation
If , then is
equal to
(1)
(2)
(3)
(4)
Q12. The solution of is
(1)
(2)
(3)
(4)
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x( ) 1 = 2 ( ) + sin x
> dy dx
> 32d2ydx 2
p&q
q = 3
p q = 1
pq + qp = 8
pq qp = 1
m n
( )5
+ 4 + = x2 1
> d2ydx 2
> ()
> 3
> d2ydx 2
> d3ydx 3
> d3ydx 3
m + ny2 = 4 a(x + a)123=
> x++x33!
> x55!
> 1+ +
> x22
> x44
> dx dy dx +dy
2ye 2x = ce 2x + 1 2ye 2x = 2 ce 2x 1
ye 2x = ce 2x + 2
f(x) = ( x y)2dy dx
y(1) = 1 ln = 2( x 1)
> 1 x+y
> 1+ xy
ln = x + y 1
> 2 y
> 2 x
ln = 2( x 1)
> 1 x+y
> 1+ xy
ln + ln | x| = 0 121 x+y
> 1+ xy
y = f(x) = 3 x2dy dx
(1, 1)
y = x3 + 2
y = x3 2
y = 3 x3 + 4
y = x3 + 2 + sin ( ) = sin ( )
> dy dx x+y
> 2
> xy
> 2
c
ln tan ( ) = c 2 sin x
> y
> 2
ln tan ( ) = c 2 sin ( )
> y
> 4
> x
> 2
ln tan ( + ) = c 2 sin x
> y
> 2
>
> 4
ln tan ( + ) = c 2 sin ( )
> y
> 4
>
> 4
> x
> 2
x3dy + xy dx = x2dy + 2 ydx ; y(2) = e x > 1 y(4)
> e
> 2
+ e12
e32
+ e32
(xy x2) = y2dy dx (1, 1)
y = (log y 1) xy = (log y + 1) xx = (log x 1) yx = (log x + 1) y
(x2 + xy ) dy = (x2 + y2) dx
log x = log( x y) + + C
> yx
log x = 2 log( x y) + + C
> yx
log x = log( x y) + + Cxy
y = y(x)sin x + y cos x = 4 x, x (0, ).
> dy dx
y ( ) = 0
> 2
y ( )
> 6
249
2493
28 93
289
+ log z = (log z)2dz dx zxzx2
( ) x = 2 x2c1log z
( ) x = 2 + x2c1log z
( ) x = x2c1log z
( ) x = + cx 21log z
> 12
Q13. A curve passes through the point . Let the slope of the
curve at each point be . Then the equation
of the curve is
(1)
(2)
(3)
(4)
Q14. The integrating factor of the differential equation
may be
(1)
(2)
(3)
(4)
Q15. The solution of differential equation
is
(1)
(2)
(3)
(4)
Q16. The solution of the initial value problem
and is given by
which of the following options?
(1)
(2)
(3)
(4)
Q17. The solution of the differential equation
is (where is the constant
of integration)
(1)
(2)
(3)
(4)
Q18. If and then find the value of
Q19. If a curve passing through satisfies the
differential equation , then which of the
following option is correct?
(1)
(2)
(3)
(4)
Q20. The solution of differential equation is
(1)
(2)
(3)
(4) None of these
Q21. The orthogonal trajectory of , where is an
arbitrary constant, is
(1) A parabola
(2) A circle
(3) An ellipse
(4) A hyperbola
Q22. The population at a time of a certain mouse species
satisfies the differential equation . If
, then the time at which the population becomes zero is
(1)
(2)
(3)
(4)
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(1, )
> 6
(x, y) + sec ( ), x > 0
> yxyx
sin ( ) = log x +
> yx
> 12
cosec ( ) = log x + 2
> yx
sec ( ) = log x + 2
> 2yx
cos ( ) = log x +
> 2yx
> 12
+ = 1 x
> dy
> dxy
> (1 x) x
> 1 x
> 1+ x
> 1+ x
> 1 x
> 1 x
> 1+ x
> x
> 1 x
(1 + y2) + (x etan 1 y) = 0
> dy dx
2xe tan 1 y = e2 tan 1 y + k
2xe tan 1 y = etan 1 y + kxe tan 1 y = etan 1 y + kxe tan 1 y = etan 1 y k
(2 ln x) + = cos x, y > 0, x > 1
> dy dx yx
> 1
> y
y ( ) = 0
> 3
> 2
y = a 1sin x
> ln x
y = a 1+sin x
> ln x
y = a 1cos x
> ln x
y = a 1+cos x
> ln x
y (sin 2 x) dy + (sin x cos x)y2dx = xdx C
sin 2 x y = x2 + C
sin 2 x y2 = x2 + C
sin x y2 = x2 + C
sin 2 x y2 = x + Cxdy = ydx + y2dy y(1) = 1 y(3)
y = f(x) (1, 2)
y(1 + xy )dx xdy = 0
f(x) = 2x
> 2 x2
f(x) = x+1
> x2+1
f(x) = x1 4 x2
f(x) = 4x
> 12 x2
=
> dy dx yf(x)y 2
> f(x)
f(x) = y(x c)
f(x) = y(c x)
f(x) = y(x + c)
x2 y2 = a2 ap(t) t
= 0.5 p(t) 450
> dp (t)
> dt
p(0) = 850 ln 18 12
ln 18 2 ln 18 ln 9 Answer Key
Q1 (3) Q2 (5.00) Q3 (2) Q4 (2)
Q5 (1) Q6 (1) Q7 (2) Q8 (3)
Q9 (1) Q10 (2) Q11 (4) Q12 (4)
Q13 (1) Q14 (2) Q15 (1) Q16 (2)
Q17 (2) Q18 (3.00) Q19 (1) Q20 (3)
Q21 (4) Q22 (3)
Differential Equations
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We have,
Square both the sides we get,
Here, Order
Degree
Then,
Q2.
The given differential equation can be written as
Q3.
We have,
On differentiating w.r.t. , we get
On substituting the value of in equation (1), we get
which is the required differential
equation.
The degree of the differential equation is
Hence, is the correct answer.
Q4.
.
Apply componendo and dividendo
.
Q5.
Hence
At and , we get
Hence
Hence
Q6.
We have,
Integrating both sides, we have
It is passing through . Therefore,
Hence, the required curve is
Q7.
Given equation
On integrating both sides, we get
Q8.
Given
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x( ) = 2 ( )+ sin x + 1
> dy dx
> 32d2ydx 2
x2( )3
= (2( )+ sin x + 1 )2
> dy dx d2ydx
= p = 2 = q = 2
pq + qp = 2 2 + 2 2 = 8
( )
> 5
+ 4 ( )
> 3
+ ( )
> 2d2ydx 2
> d3ydx 3
> d2ydx 2
> d3ydx 3
=(x2 1 ) d3ydx 3
m = 3, n = 2 y2 = 4a (x + a) (1) x2y = 4a a = .
> dy dx y2dy dx
a
y2 = 2y [x + ]
> dy dx y2dy dx
y = 2x + y ( )2dy dx dy dx
y [1 ( )2
]= 2x . ,
> dy dx dy dx
2(B)
ex = 1 + + + +. . . . . . . . . . . . . . . x
> 1!
> x2
> 2!
> x3
> 3!
ex = 1 + +. . . . . . . . . . . . . . . x
> 1!
> x2
> 2!
> x3
> 3!
=
> x++x33!
> x55!
> 1+ +x22!
> x44!
> dx dy dx +dy
=
> exex
> 2
> ex+ex
> 2
> dx dy dx +dy
= e2 xdy dx
= e2 x y = e2 x + c
> dy dx
> 12
2ye 2x = 2 ce 2x 1
x y = t
1 =
> dy dx dt dx
1 = t2dt dx
1 t2 = dt dx
= dx dt
> 1 t2
log = x + c121+ tt1
log = x + c12
> xy+1
> xy1
x = 1 y = 1 c = 1 log = x 1 12
> xy+1
> xy1
ln = 2( x 1)
> 1 x+y
> 1+ xy
= 3 x2dy dx
dy = 3 x2dx
dy = 3 x2dx
y = 3 ( )+c
> x3
> 3
y = x3 + c
(1, 1) 1 = (1) 3 + c c = 2
y = x3 + 2 + sin ( )= sin ( )
> dy dx x+y
> 2
> xy
> 2
= sin ( ) sin ( )
> dy dx xy
> 2
> x+y
> 2
= 2 sin ( )cos ( )
> dy dx y
> 2
> x
> 2
cosec ( )dy = 2 cos ( )dx
> y
> 2
> x
> 2
cosec ( )dy = 2 cos ( )dx
> y
> 2
> x
> 2
= + c
> ln (tan )
> y
> 412
> 2 sin ()x
> 212
ln (tan )= c 2 sin ( )
> y
> 4
> x
> 2
x3dy + xy dx = x2dy + 2 ydx Integrating both sides with respect to , we get
Let
Putting
Putting
Putting
From equation , we get
Given i.e. at
Putting in equation , we get
Now putting , we get
Q9.
Put,
Alternate Solution :
This posses through the point (1, 1).
Thus, the equation of the curve is
y = x ( log |y| 1)
Q10.
dividing both sides by x 2
As it is of type
Put
from equations (i) and (ii), we get
integrating both sides,
.
Q11.
I.F.
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(x3 x2)dy =(2 x)ydx
= dx
> dy y
> ( 2 x)
> (x3x2)
x
= dx + k . . . . . . . . . . . .( i)
> dy y
> ( 2 x)
> x2(x1 )
= + +
> ( 2 x)
> x2(x1 )
> AxBx2
> C
> (x1 )
(2 x)= Ax (x 1)+ B(x 1)+ Cx 2
x = 0 2 = B B = 2
x = 1 2 1 = C C = 1
x = 2 2 2 = A(2)(1)+ B(1)+ C(22) 2 A + 2 = 0 A = 1 (i)
= ( + + )dx + k
> dy y
> 1
> x
> 2
> x2
> 1
> x1
ln y = ln x + + ln| x 1|+ k . . . . . . . . . .( ii )2
> x
y(2)= e x = 2, y = e
ln e = ln 2 + + ln|2 1|+ k22
1 = ln 2 + 1 + 0 + k
k = ln 2 (ii )ln y = ln x + + ln| x 1|+ ln 2 2
> x
x = 4 ln y = ln 4 + + ln|4 1|+ ln 2 24
ln y = 2 ln 2 + + ln 3 + ln 2 12
ln y = ln 2 + + ln 3 12
ln y = + ln 1232
ln =
> 2y
> 312
= e
> 2y
> 3
> 12
y = e32
y(4)= e32
=
> dy dx y2
> xyx 2
y = vx = v + x
> dy dx dv dx
v + x =dv dx v2
> v1
x = v =
> dv dx v2
> v1 vv1
( )dv =
> v1 vdx x
(1 )dv = 1vdx x
v n v = n x + c = n y + c
> yx
We have, (xy x2) = y2d y d x
y2 = xy x2d x d y
= 1
> x2
> d x d y
> 1
> x
> 1
> y
> 1
> y2
Put = v =1
> x
> 1
> x2
> d x d y dv
> d y
+ = , which is linear d v
> d y
> v
> y
> 1
> y2
IF = e d y = elog y = y
> 1
> y
The solution is vy = y d y + c1
> y2
= log |y| + c
> yx
y = x(log |y| + c)
1 = 1(log 1 + c)ie, c = 1
x2dy + xydy = x2dx + y2dx
dy + dy = dx + ( )2
dx
> yxyx
(1 + ) = 1 + ( )2
. . . .( i)
> yxdy dx yx
= f( )
> dy dx yx
= t = t + x . . . .( ii )
> yxdy dx dt dx
(1 + t)( t + x )= 1 + t2dt dx
x = ( )dt = . . . .( iii )dt dx
> 1 t
> 1+ t
> 1+ tt1
> dx x
( + )dt = 2
> t1
> t1
> t1
> dx x
2l n|t 1| t = l nx + C
2l n 1 = l nx + C
> yxyx
lnx = 2l ny x + + C
> yx
or, l nx = 2l nx y + + C
> yx
+ ycot x = , x (0, )
> dy dx
> 4xsinx
= e cotx = eln sinx = sinx
y sinx = . sinx dx 4xsinx
ysinx = 2 x2 + cQ12.
Given,
Dividing given equation by
Put
Now, integrating factor
, where, is the constant of integration.
.
Q13.
Given slope at (x, y) is
......
let
Differentiating this wrt
......
Now using equation and , we have
Integrating both sides.
Now Put value of in above equation.
.....
Given that curve passes through so put in above equation.
Now, put the value of in equation
Q14.
Given,
If
Put
If
Q15.
Given equation can be rewritten as
If
Required solution is
Put
Q16.
Given differential equation is
On substituting and , we get
Which is a ,
Now,
Required solution where is any constant.
Q17. The given equation is
or
On integrating, we get
Q18.
Differential Equations
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y( )= 0
> 2
0 1 = 2 ( )2
+ c
> 2
c = 2
> 2
y( )= =[( )( )] 2
> 62 ()2
> ()
> 6
> 2212
> 2
> 18
> 2
> 2
= 22
> 9
= 82
> 9
+ log z = (log z)2dz dx zxzx2
z(log z)2, + =1
> z( log z)2
> dz dx
> 1
> x
> 1log z
> 1
> x2
= t =1log z
> 1( log z)2
> 1
> zdz dx dt dx
+ =dt dx tx
> 1
> x2
= . . .( i)dt dx tx
> 1
> x2
I. F = e = e ln x = eln =
> dx x
> 1
> x1
> x
= dx = + ctx
> 1
> x3
> 12x2
c
= + c1
> xlog z
> 12x2
( )x =( )+cx 21log z
> 12
= + sec ( )
> dy dx yx
> yx
(i)= t y = xt
> yx
x
= t + x
> dy dx dt dx
(ii )(i) (ii )t + x = t + sec(t) dt dx
=dt
> sec ( t)
> dx x
cos( t)dt = dx 1x
sin( t)= ln( x)+c
t
sin ( )= ln( x)+c
> yx
(iii )
(1, )
> 6
sin ( )= ln(1)+c c =
> 612
c (iii )sin ( )= ln( x)+
> yx
> 12
+ = 1 x
> dy dx y
> ( 1 x) x
= e dx
> 1( 1 x) x
x = t
dx = dt 12 x
= e dt 21 t2
= e log = =
> 221+ t
> 1 t1+ t
> 1 t
> 1+ x
> 1 x
+ x =dx dy
> 1
> (1+ y2)
> etan 1 y
> (1+ y2)
= e dy
= etan 1 y
> 11+ y2
xe tan 1 y = dy etan 1 yetan 1 y
> 1+ y2
etan 1 y = t etan 1 y dy = dt 11+ y2
xe tan 1 y = t dt = + ct2
> 2
2 xe tan 1 y = e2 tan 1 y + k
2y + y2( )=
> dy dx
> 1
> xln x
> cos x
> ln x
+ =dz dx zx ln x
> cos x
> ln x
y2 = z 2y =
> dy dx dz dx
+ =dz dx zx ln x
> cos x
> ln x
LDE
y2 ln x = z ln x = sin x +Cy( )= 0 C = 1 3
> 2
y = 1+sin x
> ln x
y = 1+sin x
> ln x
( y > 0)
y = a 1+sin x
> ln x
a
(sin 2x)(2 ydy )+(2 sin x cos xdx )y2 = 2 xdx d(sin 2x y2)= 2 xdx
sin 2x y2 = x2 + CGiven that
We know that
As,
If
Q19.
The given differential equation is,
On integrating both sides, we get
If the above curve passes through the point , then,
The given curve is
Q20. The given equation is
On integration, we get
Q21. x2 y 2 = a 2
Product of slope of and slope of orthogonal trajectory of
is
Slope of orthogonal trajectory of is
logx = logy + logc
logxy = logc
xy = c
which is a rectangular hyperbola.
Q22.
Let,
Let, when (using )
Differential Equations
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x dy = y dx + y2dy
= dy
> ydx xdy y2
d( )=xyydx xdy y2
= y + cxy
y(1)= 1 1 = 1 + c
c = 2 = y + 2 xy
x = 3 3 = y2 + 2 y
y2 2 y 3 = 0 y = 3
y(1 + xy )dx xdy = 0 ( ydx xdy )+ xy 2dx = 0 + x dx = 0
> y dx x dy y2
d( )+x dx = 0 xy
d( )+ xdx = 0 xy
+ = Cxyx2
> 2
(1, 2) + = C C = 1 1212
+ = 1 xyx2
> 2
y = 2x
> 2 x2
f(x)= 2x
> 2 x2
=
> dy dx y'( x ) y 2
> ( x )
y ' (x)dx (x)dy = y 2dx = dx d [ ]= dx
> y'( x ) dx ( x ) dy y2
> ( x ) y
= x + c (x)= y(x + c)
> ( x ) y
2x 2y = 0
> dy dx
=
> dy dx xy
x2 y2 = a2
x2 y2 = a2 1
x2 y2 = a2
=
> dy dx yx
=
> dx xdy y
p = p(t)= 0. 5 p 450
> dp dt
= p900 2
> p
> 850
dp =
> t
> 0
dt 2
> p900
2|log| p 900|| p
> 850
= t
2|log| p 900| log|50||= t
2 log = t . . .(1)
> p900 50
= et/2 p900 50
p = 900 50 et/2
t = T , p = 0 (1) T = 2 ln 18 Q1. Let be two continuous differentiable functions
satisfying the relationships and
Let If , then
(1)
(2)
(3)
(4) None of these
Q2. If . Then
(1)
(2)
(3)
(4)
Q3. If is a parameter, then at
is equal to
(1)
(2)
(3)
(4)
Q4. If the derivative of w.r.t. at , where
and is then find ?
Q5.
(1)
(2)
(3)
(4)
Q6. If , then is equal to
(1)
(2)
(3)
(4)
Q7. Let Then the value of
is
(1)
(2)
(3)
(4)
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f(x), g(x)
f (x) = g(x) f (x) = f(x).
h(x) = [ f(x)] 2 + [ g(x)] 2. h(0) = 5 h(10) = 10 515 log( x + y) 2 xy = 0 y(0) = 11 20
x = et sin t, y = et cos t, t d2ydx 2
(1, 1) 12
14
0
> 12
f(tan x) g(sec x) x =
> 4
f (1) = 2 g(2) = 4 2
> k
k
(tan 1 ) =ddx
> cos x
> 1+sin x
1212
1 1
y = sin 1 ( )
> 5x+12 1 x2
> 13
> dy dx
> 1
> 1 x2
1
> 1 x2
> 3
> 1 x2
> 3
> 1 x2
f( x) = x + .12x+ 12x+ 12x+..
f(100). f (100) 100
> 1100
100 1100 Answer Key
Q1 (2) Q2 (1) Q3 (1) Q4 (2.00)
Q5 (1) Q6 (1) Q7 (1)
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Given are two continuous differentiable functions
satisfying and
Hence
Given
Differentiating the above equation,
, a constant
i.e.
for all .
Hence .
Q2.
Given,
Differentiate equation w.r.t
Hence,
Now put in equation
from equation
Q3.
Given that,
At point
On differentiating Equation w.r.t. , we get
and
Again differentiating w.r.t. , we get
At
Q4.
Let and
and
.
Q5.
Q6.
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f(x), g(x) f'(x)= g(x), f '' (x)= g '(x) f'' (x)= f(x) g'(x)= f(x)
h(x) = [ f(x)] 2 + [ g(x)] 2
h'(x)= 2f(x) f '(x) + 2g(x) g '(x)
h'( x) = 2[f(x) g(x) + g(x) [f(x)]]
h'( x) = 2[f(x) g(x) f(x) g(x)]= 0
h(x)= C
h(0)= C C = 5 h(x)= 5 xh(10)= 5 log( x + y)2 xy = 0 . . . . . . . . .(1) (1) x
(1 + )2 (x + y)= 0 1
> x+ydy dx dy dx
( 2 x) +( 2 y)= 0 1
> x+ydy dx
> 1
> x+y
=
> dy dx
> 12 xy 2 y2
> 12 xy 2 x2
y(0)= . . . . . . . .(2)
> 102 y( 0 ) 2
> 100
x = 0 (1) log(0 + y)= 0 y ( 0 ) = 1 (2)
y(0)= 1
x = et sin t, y = et cos t (1) (1, 1), 1 = et sin t, 1 = et cos t
tan t = 1 t =
> 4
(1) x
= et(cos t sin t)
> dy dt
= et(sin t + cos t)dx dt
= =
> dy dx
> dy dt dx dt
> cos tsin t
> cos t+sin t
x
= ( )
> d2ydx 2
> dx dt
> cos tsin t
> cos tsin tdt dx
=[ ]
> [ ( cos t+sin t) ( sin tcos t) ( cos tsin t) ( sin t+cos t) ] ( cos t+sin t)2
> dt dx
= .
> 2 ( cos t+sin t)2
> 1
> et( sin t+cos t)
= .2 ( et cos t+et sin t )1( cos t+sin t ) 2
= . [from Equation(1)]
> 2
> x+y
> 1( cos t+sin t)2
t = , x = 1, y = 1
> 4
= .
> d2ydx 2
> 2 1+1 1
> (cos +sin )2
>
> 4
>
> 4
= =
> 1
> [+]
> 12 12
> 12
u = f(tan x) v = g(sec x) = f (tan x)sec 2 xdu dx
= g(sec x)sec x tan xdv dx
= / =du dv du dx dv dx f ( tan x ) sec 2 xg ( sec x ) sec x tan x
[ ]x= = =du dv
> 4
> f(tan )
> 4
> g(sec )sin
> 4
>
> 4
> f( 1 ) 2
> g(2 )
= =
> 22 412
= 12 2 2
= 2 2
= 2
> k
k = 2
[tan 1 ( )] ddx
> cos x
> 1+sin x
= [tan 1 ( )] ddx
> cos 2sin 2x
> 2
> x
> 2
> cos 2+sin 2+2 sin cos x
> 2
> x
> 2
> x
> 2
> x
> 2
= [tan 1 ( )] ddx
> cos sin
> x
> 2
> x
> 2
> cos +sin
> x
> 2
> x
> 2
= [tan 1 ( )] ddx
> 1tan ()x
> 2
> 1+tan ()x
> 2
= [tan 1 tan ( )] ddx
> 4
> x
> 2
= 12Given,
On putting and
we get
Differentiating both sides w.r.t. , we get
Q7. Let
Undefined y 2 = 1 + x 2
On differentiating
Differentiation
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y = sin 1 [ ]
> 5x+12 1 x2
> 13
x = sin , 5 = r cos
12 = r sin ,
y = sin 1 ( )r cos sin +r sin cos
> 13
= sin 1 ( ) ;[ r = 25 + 144 = 13 ]
> 13 sin ( +)13
= +
= sin 1 x + tan 1 ( ) ;[ tan = ]12 512 5
x
=
> dy dx
> 1
> 1 x2
y = f( x)= x + 12x+ 12x+ 12x+
y = x + =1
> x+yx2+xy +1
> x+y
xy + y2 = xy + x2 + 1 y2 x2 = 1 2 yy ' = 2 x
yy ' = x
f(100). f ' (100)= 100 Q1. If the domain of is , then the domain of
is
(1)
(2)
(3)
(4)
Q2. The domain of the function where
represents greatest integer function ), is:
(1)
(2)
(3)
(4)
Q3. The range of values of for which the line and the curve
enclose a region, is
(1)
(2)
(3) 1]
(4)
Q4. For and , if the number of natural numbers in the range of
is , then the value of is equal to
(1)
(2)
(3)
(4)
Q5. If the graph of the function is symmetrical about -axis,
then equals
(1)
(2)
(3)
(4)
Q6. The function is
(1) an odd function
(2) an even function
(3) neither an odd nor an even function
(4) a periodic function
Q7. The period of the function is
(1)
(2)
(3)
(4)
Q8. If , then is
(1)
(2)
(3)
(4) none of these
Q9. For , let and . If
, then is equal to:
(1)
(2)
(3)
(4)
Q10. The function is defined as . From the following
statements,
I. is one-one
II. is onto
III. is a decreasing function
the true statements are
(1) only I, II
(2) only II, III
(3) only I, III
(4) I, II, III
Q11. If defined as is surjective,
then is equal to
(1)
(2)
(3)
(4)
Q12. If and
, then the mapping is
(1) one-one but not onto
(2) onto but not one-one
(3) both one-one and onto
(4) neither one-one nor onto
Q13. Which of the following functions is inverse of itself?
(1)
(2)
(3)
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f(x) [1, 3] f (log 2(x2 + 3 x 2 ))
[5, 4] [1, 2] [13, 2] [ , 5 ]35
[4, 1] [2, 7] [3, 2]
f(x) = esin( x[ x]) + [ x] cos ( ), (
> [x+1]
[]
RR [1, 0]
R [0, 1]
R [1, 0)
m y = mx y = xx2+1
(1, 1) (0, 1) (1, )
p > 2 x Rf(x) = x2+2 x+px2+2 x+2 3 p
3456
f(x) = ax1
> xn(ax+1)
Yn
2
> 2314
13
f(x) = + + 1 xex1
> x
> 2
f() = sin + cos
> 3
>
> 2
3
6
9
12 f(x) = x
> 1+ x2
(fofof )( x)
> 3x
> 1+ x2
> x
> 1+3 x2
> 3x
> 1 x2
x (0, )
> 32
f(x) = x, g(x) = tan x h(x) = 1 x2
> 1+ x2
(x) = (hof) og) ( x) ( )
> 3
tan
> 12
tan 5
> 12
tan 7
> 12
tan 11
> 12
f : R R f(x) = 3 x
ffff : R A f(x) = tan 1 (4 ( x2 + x + 1) )
A
( , )
> 2
>
> 2
[0, )
> 2
[ , )
> 3
>
> 2
(0, ]
> 3
A = {x : x } , B = { y : 1 y 1} 25
> 2 5
f(x) = cos(5 x + 2) f : A Bf(t) = (1 t)(1+ t)
f(t) = (1 t2)(1+ t2)
f(t) = 4 log t(4)
Q14. The inverse of is
(1)
(2)
(3)
(4)
Q15. If and then is equal to
(1)
(2)
(3)
(4) None of these
Q16. If is a function such that then find value of
.
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f(t) = 2 t
f(x) = 2310 x10 x
> 10 x+10 x
log 10 131+ x
> 1 x
log 10 122+3 x
> 23 x
log 10 132+3 x
> 23 x
log 10 1623 x
> 2+3 x
5f(x) + 3 f ( ) = x + 2 1
> x
y = xf (x) ( )x=1
> dy dx
> 1478
1
f 2f(x) + f(2 x) = x2
f(4) Answer Key
Q1 (1) Q2 (4) Q3 (2) Q4 (3)
Q5 (4) Q6 (2) Q7 (4) Q8 (2)
Q9 (4) Q10 (3) Q11 (3) Q12 (3)
Q13 (1) Q14 (2) Q15 (2) Q16 (9.33)
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Domain of is
It is the same as solving two inequalities.
Case
or,
or,
Case
or,
or,
Considering both Case and Case we have,
Q2.
We have,
For to be defined,
Hence, domain of is .
Q3.
Given:
and
Hence points of intersection are:
and
Q4. Here,
If there are natural numbers in the range
Q5.
since, is symmetrical about -axis.
( i.e., it is a even function.)
Hence, the value of which satisfy this relation, is
Q6.
Given,
for all .
is an even function.
Q7. Period of
And period of
Period of
Q8.
Given:
Then,
Q9.
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f(x) [1, 3] 1 log 2(x2 + 3 x 2 ) 3 2 x2 + 3 x 2 8 1 : x2 + 3 x 2 2
x2 + 3 x 4 0 (x + 4)( x 1) 0
x (, 4] [1, ) 2 : x2 + 3 x 2 8
x2 + 3 x 10 0 (x + 5)( x 2) 0
x [5, 2] 1 2
x [5, 4][1, 2]
f(x)= esin ( x [ x ] ) +[ x]cos ( )
> [x+1 ]
f(x)[x + 1] 0 [ x]+1 0 [ x] 1 x [ 1, 0)
f(x) x R [ 1, 0)
y = mx y = xx2+1
x = 0
x2 + 1 = 1
> m
x2 = 1 1
> m
1 > 0 1
> m
> 0 1 mm
< 0
> m1
> m
m (0, 1)
f(x)= 1 + (1, p 1]
> p2 (x+1 ) 2+1
3 p 1 = 4 p = 5
f(x)= ax1
> xn(ax+1 )
f(x) Y
f(x)= f( x) =
> ax1
> xn(ax+1 )
> ax1 ( x)n(ax+1 )
=
> ax1
> xn(ax+1 ) 1 ax
> ( x)n( 1+ ax)
xn = ( x)n
n 13
f(x)= + + 1 xex1
> x
> 2
= + 1 = + 1
> 2x+xe xx
> 2 ( ex1 )
> x+xe x
> 2 ( ex1 )
f( x)= + 1 = + 1
> xxe x
> 2 ( ex1 )
> x+xe x
> 2 ( ex1 )
f( x) = f(x) x
f(x)sin = 6
> 3
cos = 4
> 2
f(x)= LCM(6 , 4)= 12 f(x)= x
> 1+ x2
(fof )( x)= f[f(x)]= f( )= =x
> x2+1
> x
> 1+ x2
> 1+ x21+ x2
> x
> 2x2+1
(fofof )( x)= f[f{f(x)}]= f( )= =x
> 2x2+1
> x
> 2x2+1
> 1+ x22x2+1
> x
> 1+3 x2
f(x)= xg(x)= tan xh(x)= 1 x2
> 1+ x2
fog (x)= tan xhofog (x)= h(tan x)Q10.
Since, such that
Let and be two elements of such that
Since, if two images are equal, then their elements are equal, therefore it is
one-one function.
Since, is positive for every value of , therefore is into.
On differentiating w.r.t. , we get for every value of .
It is decreasing function.
Statement I and III are true.
Q11.
Range of is
Q12.
Let , then
A bijective function is both one-one and onto.
, which is bijective in since is
decreasing in and Range of is .
Hence, is bijective.
Q13. Let
Now,
i.e., or
Thus, this function is inverse of itself.
Q14.
Given,
Taking both sides, we get
.
Q15.
Replacing by
From
and from
Subtracting from
.
Q16.
Given,
Substitute by in equation we get,
Now, we get,
Then,
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= = tan ( x)
> 1tan x
> 1+tan x
> 4
(x)= tan ( x)
> 4
( )= tan ( )
> 3
>
> 4
>
> 3
= tan ( )= tan
> 12
>
> 12
= tan ( )= tan
> 12 11
> 12
f : R R f(x)= 3 x
y1 y2 f(x) y1 = y2
3x1 = 3 x2 x1 = x2
f(x) x f(x)
x = 3 x log 3 < 0
> dy dx
x
x2 + x + 1= (x + )2
+1234
4(x2 + x + 1 ) 3 4( x2 + x + 1) 3 tan 1 (4( x2 + x + 1) ) tan 1 (3 )
f(x)
> 3
f(x) [ , )
> 3
>
> 2
t = 5 x + 2 A =( t : 0 t )
f(t)= cos t [0, ] cos t
[0, ] cos t [1, 1]
f(x)
y = f(t) t = f 1 (y)
y = f(t)= y + ty = 1 t
> 1t
> 1+t
t + ty = 1 y t = 1 y
> 1+ y
f 1 (y)= 1 y
> 1+y
f 1 (t)= 1 t
> 1+ t
y = [ ]2310 x10 x
> 10 x+10 x
y = [ ]2310 2x1 10 2x+1
10 2x = 3y+2 23 y
log 2x log 10 (10)= log 10 ( )
> 3y+2 23 y
x = log 10 ( )122+3 y
> 23 y
f 1 (x)= log 10 ( )122+3 x
> 23 x
5(x)+3 ( )= x + 2 (1) 1
> x
x 1x
5( ) + 3 (x)= + 2 (2) 1x1x
(1) 25 (x)+15 ( )= 5 x + 10 (3) 1
> x
(2) 9(x)+15 ( )= + 6 (4) 1x3x
(4) (3)
16 (x)= 5 x + 43
> x
x(x)= = y
> 5x 23+4x 16
=
> dy dx
> 10 x+4 16
x=1 =
> dy dx
> 10+4 16
= 78
2f(x)+ f(2 x)= x2 . . .( i)
x (2 x) (i)2f(2 x)+ f(x)= (2 x)2 . . .( ii )(i)2 ( ii )3f(x)= 2 x2 (2 x)2
f(x)= [2x2 (2 x)2]13
f((4))= [2(4) 2 (2 4) 2]13
= [32 4]= = 9. 33 1328 3Q1. If is a differentiable function such that
, (where, is the constant of
integration) and , then equals
(1)
(2)
(3)
(4)
Q2.
(1)
(2)
(3)
(4)
Q3.
(1)
(2)
(3)
(4)
Q4.
(1)
(2)
(3)
(4)
Q5. The value of is (Where is
the constant of integration)
(1)
(2)
(3)
(4) None of these
Q6. is equal to
(1)
(2)
(3)
(4)
Q7. If a function is defined as
and , then which of the
following is correct?
(1) is an even function
(2) is an onto function
(3) is an odd function
(4) is many one function
Q8. The value of indefinite integral:
equals: (where is constant of integration)
(1)
(2)
(3)
(4)
Q9. If , then
(1)
(2)
(3)
(4) None of these
Q10.
(1)
(2)
(3)
(4)
Indefinite Integration
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f(x)
f(x)dx = 2[ f(x)] 2 + C Cf(1) = 1/4 f()
> 34
>
> 42376
dx =
> 1 x
> 1+ x
sin 1 x 1 x 2 + c 12
sin 1 x + 1 x 2 + c 12
sin 1 x 1 x2 + c
sin 1 x + 1 x 2 + c
ex tan 1 (ex)dx = f(x) log (1 + e2x) + c f(x) = 12
ex ex tan 1 (ex)
x2 + ex tan 1 (ex)ex tan 1 (ex)
x ex tan 1 (ex)
dx =
> cos xsin x
> 79 sin 2 x
log + c124 4+3(sin x+cos x)43(sin x+cos x)
log + c124 43(sin x+cos x)4+3(sin x+cos x)
log + c124 4(sin xcos x)4+(sin xcos x)
log + c124 4+(sin xcos x)4(sin xcos x)
dx
> f(x)(x)+ (x)f(x)(f(x) (x)+1) (x) f(x)1
C
cos 1 f(x)2 (x)2
tan 1 [f(x)(x)] sin 1 f(x)
> (x)
( cos 3/7 x) (sin 11/7 x) dx
log sin 4/7 x + c
tan 4/7 x + c47
tan 4/7 x + c74
log cos 3/7 x + cf : R Rf(x) = dx
> x8+4
> x42 x2+2
f(0) = 1
f(x)
f(x)
f(x)
f(x)
dx
> sin xsin 3x
> 1sin 3x
c
cos 1 (sin x) + c32
> 32
sin 1 (sin x) + c23
> 32
sin 1 (sin x) + c32
> 32
tan 1 (sin x) + c23
> 32
In = (ln x) ndx In + nI n1 =+ C
> (ln x) n
> x
x(ln x) n1 + C x(ln x) n + C
ex [ ] dx =
> 2+sin 2 x
> 1+cos 2 x
ex tan x + Cex + tan x + C
2ex tan x + Cex tan 2 x + CQ11. If , then the
value of is
(1)
(2)
(3)
(4)
Q12. Find the ordered triplet , If
(1)
(2)
(3)
(4)
Q13. Evaluate:
(1)
(2)
(3)
(4)
Q14. , equals (where
is constant of integration)
(1)
(2)
(3)
(4)
Indefinite Integration
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dx = dx + dx
> 3x+1 (x3)( x5) 5 (x3)
> B
> (x5)
B
3468(A, B, )
dx = A log e(| cos x + sin x 2|) + Bx + C2 cos xsin x+
> cos x+sin x2
( , , 1 )1232
( , , 1 )3212
( , 1, )1232
( , 1, )3212
{ }2
dx
> (log x1) 1+(log x)2
+ Cxx2+2
+ C
> log x
> (log x)2+1
+ Cx
> (log x)2+1
+ C
> xe x
> 1+ x2
ex sin x (x2 cos x + x sin x + 1 ) dx Cxe x2 sin x + Cx2ex sin x + Cxe x sin x + C
2x2ex sin x + CAnswer Key
Q1 (2) Q2 (4) Q3 (4) Q4 (1)
Q5 (4) Q6 (3) Q7 (2) Q8 (2)
Q9 (3) Q10 (1) Q11 (4) Q12 (2)
Q13 (3) Q14 (3)
Indefinite Integration
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If you want to solve these questions online, download the MARKS App from Google Play or visit https://web.getmarks.app Q1. Differentiating w.r.t. we get
or
Integrating, we have
As
Q2.
Multiply and divide by
Q3. Put
Q4.
Let
Q5.
Let
Let
Q6.
Put
Q7.
We have,
Now,
Therefore,
Range of is ,
So, is an onto function
,
So, is not even
,
So, is not odd
,
Indefinite Integration
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x f(x)= 4 f(x) f (x)
f (x)= 0 = f(x)14
dx = f (x)dx 14
= f(x)+ Cx
4
f(1)= C = 0 14
f(x)= x
4
f()=
4
dx
1x 1+x
1 x = dx = dx
1x
1x 2
dx
1x 2
x
1x 2
= sin 1 x + dx 12( 2x )
1x 2
= sin 1 x + 1 x 2 + c
ex = t, exdx = dt , dx = dt 1
t
= dt
tan 1 tt2 = tan 1 t( ) ( ) dt
1
t
1
t
11+ t2
= + dt
tan 1 tt2
122tt2 ( 1+ t2 ) t2 = z, 2tdt = dz
= + dz
tan 1 tt
12( z+1 ) zz ( 1+ z ) = + [ dz ]
tan 1 tt
121
z+1
= + log +c
tan 1 tt
12
t2
1+ t2 = + log +c
tan 1 tt
12
e2x
1+ e2x
= extan 1 (ex) log 1 + e2x+ 2x + c1212
= x extan 1 (ex) log 1 + e2x+ c12
I = dx
cos xsin x
79 sin 2 x
I = dx cos xsin x
79 [ ( 1+sin 2 x ) 1 ]
I = cos xsin x
79 [ ( sin 2 x+cos 2 x+2 sin x cos x ) 1 ]
I = dx
cos xsin x
79[ ( sin x+cos x ) 21 ]
sin x + cos x = t (cos x sin x)dx = dt
I = dt
79 ( t21 )
I = dt
42 ( 3 t ) 2
I = log +c124 134+3 t
43 t
I = log +c124 4+3 ( sin x+ )cos x
43 ( sin x+ )cos x
I = dx
f(x)(x)+ (x)f (x)( f(x) (x)+1 ) (x) f(x)1
(x) f(x) 1 = t2
( (x) f (x) + f(x)(x)) dx = 2 tdt
I = = 2tdt
(t2+2 )t2
2dt t2+(2 )2
= 2 tan 1 + C12
t
2
= 2 tan 1 ( )+C
(x) f(x)1 2
( cos 3/7 x)(sin 11/7 x)dx = . sec 2 x dx
sin 11/7 x
cos 11/7 x
= tan 11/7 x sec 2 x dx
tan x = t sec 2 x dx = dt
I = t11/7 dt = tan 4/7 x + c
74
f(x)= dx
x8+4
x42 x2+2
f(x)= dx
( x8+4+4 x4 ) ( 4x4 )
x42 x2+2
= ( )dx
( x4+2 ) 2 ( 2x2 ) 2
x42 x2+2
= dx
( x4+2 x2+2 ) ( x42 x2+2 )( x42 x2+2 )
f(x)= + + 2 x + C
x5
52x3
3
f(0)= 0 + 0 + 0 + C = 1 C = 1 f(x)= + + 2 x + 1
x5
52x3
3
f(x) Rf(x)
f( x) f(x)
f(x)
f( x) f(x)
f(x)
f '(x)> 0 x RSo, is one-one
Q8.
Let
Put
Q9. Integrate by parts by taking as
Q10.
Let,
We know that
Using the above formulas we can write
We know that
Using the above formula we can write
Q11.
Given,
So, let
Putting in the above equation, we get
or .
Q12.
We have,
Differentiating w.r.t. , we get
So,
Q13.
Method1: by cross checking the options
Consider
Indefinite Integration
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f(x)
I = dx
> sin xsin 3x
> 1sin 3x
I = { } dx
> (sin x)1sin 2x
> 1sin 3x
I = { } dx
> (sin x)cos x
> 1sin 3x
sin x = t sin x cos xdx = dt
> 3232
I = 23
> dt
> 1 t2
I = sin 1 t + c23
I = sin 1 (sin x)+c23
> 32
(ln x) ndx In = (ln x )n.1. dx In = x( ln x) n dx
> x ( n ) ( ln x ) n1
> x
= x( ln x) n n I ( n1 )
I n + n I n1 = x( ln x) n
I = ex[ ] dx
> 2+ si n 2x
> 1+ co s 2x
sin 2 x = 2 sin x cos x & cos 2 x = 2 cos 2 x 1
I = ex( )dx
> 2+2 sin xcos x
> 2 cos 2x
I = ex( )dx
> 2 ( 1+sin xcos x)2 cos 2x
I = ex( + )dx 1cos 2 x
> sin xcos x
> cos 2x
I = ex (tan x + sec 2x) dx
ex(f(x)+f (x) )dx = e x f + C
I = ex tan x + C
dx = dx + dx 3x+1 ( x3 ) ( x5 ) 5 ( x3 )
> B
> (x5 )
= +
> 3x+1 (x3 ) ( x5 ) 5
> x3
> Bx5
3 x + 1 = 5( x 5)+ B(x 3)
x = 5 3(5)+1 = 0 + B(5 3) 16 = 2 B
B = 8
dx = A log e(|cos x + sin x 2|)+ Bx + C2 cos xsin x+
> cos x+sin x2
R. H. S. x
[A log e(|cos x + sin x 2|)+ Bx + C]ddx
= A + B
> cos xsin x
> cos x+sin x2
= A cos xA sin x+B cos x+B sin x2 B
> cos x+sin x2
=2 cos xsin x+
> cos x+sin x2
> Acos xAsin x+Bcos x+Bsin x2 B
> cos x+sin x2
A + B = 2, B A = 1, = 2 B
A = , B = , = 1 3212
f(x)= x
> ( log x)2+1
f (x)= 1+ ( log x ) 2 2 x log xx
> (1+ ( log x)2)2
f (x)= = ( )
> 21+ ( log x)22 log x
> (1+log 2x)2
> ( log x1 ) ( 1+log x)2
( )dx = f (x)dx = f(x)+ C
> ( log x1 ) 2
> 1+ ( log x)2
( )
> 2
dx = + C
> log x1 1+ ( log x)2
> x
> 1+ ( log x)2
Hence option 3 is the correct answer and we can check the
other choices by the similar argument.
Alternate solution
Put
Q14.
Let
Let
Using integration by parts,
, where is
the constant of integration.
Indefinite Integration
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{ }
> 2
dx
> log ( x) 1 1+ ( log x)2
log( x)= t x = et dx = etdt
= et{ }dt
> (t1 ) 2
> (t2+1 )2
et{ }dt
> t2+12 t
> (t2+1 )2
= et{ +( )} dt 1
> t2+1 2 t
> (t2+1 )2
= + Cet
> t2+1
[ ex(f(x)+ f (x)) dx = exf(x)+ C]
= x
> ( log x)2+1
I = ex sin x(x2 cos x + x sin x + 1 )dx
I = ( x.( x cos x + sin x)ex sin x + 1. ex sin x)dx ex sin x = t ex sin x(sin x + x cos x)dx = dt I = (x + t)dx = x t + C = x e x sin x + Cdt dx CQ1. At , the function has
(1) A minimum
(2) A discontinuity
(3) A point of inflexion
(4) A maximum
Q2. The value of is equal to
Q3. If , then
(1)
(2)
(3)
(4)
Q4. equals
(1)
(2)
(3)
(4) Does not exist
Q5. The value of , (where
represents greatest integral function less than or equal to )
is Then the value of is
Q6. is equal to
(1)
(2)
(3)
(4)
Q7. If , then (where, [.]
is the greatest integer function)
(1) is equal to
(2) is equal to
(3) does not exist
(4) None of these
Q8. is equal to
Q9. The value of is equal to
(Use )
Q10. The value of is equal to (take
Q11. (where denotes the fractional
part of ) is equal to
(1)
(2)
(3)
(4) None of these
Q12. The value of is
(1)
(2)
(3)
(4)
Q13. Let be a positive increasing function
with . Then
(1)
(2)
(3)
(4)
Limits
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t = 0 f(t) = sin tt
lim x1
> 5
> x22 5x+1 4( x1) 2
lim x1 = 2
> ax 2+bx +c
> (x1) 2
(a, b, c)(2, 4, 2) (2, 4, 2) (2, 4, 2) (2, 4, 2) Lim x
> x(log x)3
> 1+ x+x2
01 1lim x0 ([ ] + [ ]) 100x sin x 99 sin x x [x] x99 .
lim x0 sin 1 ( )1
> x
> 2x
> 1+ x2
2 02
f(x) = [tan x]2, x (0, )
> 3
f ( )
> 4
10lim x2 3x+3 3 x12 3 3 1 xx
> 2
lim x
> ex+1 log (x3ex+1 )
> 10 x3
e = 2.7 lim x0 (cos x + sin x) 1
> x
e = 2.71) lim x0 {(1 + x) }
> 2
> x
{. }
xe2 7
e2 8
e2 6 lim x0 dt
> x20sin tx3
02/9 1/3 2/3
f : R R
lim x = 1
> f(3 x)
> f(x)
lim x =
> f(2 x)
> f(x)
1
> 2332
3Answer Key
Q1 (4) Q2 (0.01) Q3 (1) Q4 (1)
Q5 (2) Q6 (3) Q7 (3) Q8 (36.00)
Q9 (0.27) Q10 (2.71) Q11 (1) Q12 (4)
Q13 (1)
Limits
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At t = 0, first we will check continuity of the function.
Now,
And
Since,
So, the function is continuous at t = 0
Now, we check the function is maximum or minimum
and
For maximum or minimum value of , put
Now,
[using L' Hospital rule]
So, function f(t) is maximum at t = 0
Q2.
(Putting )
Q3. Given,
This limit will exist, if
Q4.
(By D.L. Hospital rule)
(By D.L. Hospital rule)
(By D.L. Hospital rule)
(By D.L. Hospital rule)
(By D.L. Hospital rule)
(By D.L. Hospital rule)
Q5.
We know that,
then,
Hence
Hence,
Limits
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f(t)= sin tt
LHL = f(0 h)= lim
> h0 sin(0 h)( 0 h)
= lim
> h0 sin h
> h
= 1
RHL = f(0 + h)= lim
> h0 sin(0+ h)( 0+ h)
= lim
> h0
= 1 sin hh
f(0)= 1
LHL = RHL = f(0)
f ' (t)= cos t sin t1
> t
> 1
> t2
f '' (t)= sin t cos t cos t + sin t
> 1
> t
> 1
> t2
> 1
> t2
> 2
> t3
= +
> sin tt
> 2 cos tt2
> 2 sin tt2
f(x)
f ' (x)= 0 = 0 cos tt
> sin tt2
= 1 tan tt
lim
> t0
f " ( t)
= lim
> t0
( )2 lim
> t0
( ) [ from ]sin ttt cos tsin tt3
> 00
= 1 2 lim
> t0
( )
> cos tsin tcos t
> 3t2
= 1 + lim
> t0 23sin tt
= 1 + 1 = < 0 231 3
lim
> x1
= lim
> y1
> 5x22 5x+1 4 ( x1 ) 2
> y22 y+1 4(y51 )2
> 5
x = y; as x 1, y 1 lim
> y1 (y1 ) 2
> 4 ( y1 ) 2(y4+y3+y2+y+1 )2
= = 0. 01 1254
lim
> x1
= 2
> ax 2+bx +c
> (x1 ) 2
ax 2 + bx + c = 2( x 1) 2
ax 2 + bx + c = 2 x2 4 x + 2 a = 2, b = 4, c = 2 Lim
> x ( log x ) 3+x.3 ( log x ) 21x
> 1+2x
Lim
> x 3 ( log x ) 2+6 ( log x ) 1x1x
> 2
Lim
> x 3 ( log x ) 2+6 log x 2x
Lim
> x 6 log x +1x6x
> 2
Lim
> x 6 log x+6 2x
Lim
> x 6()+0
> 1x
> 2
= = 0
> 6
> 2
lim
> x0
1 sin x x
lim
> x0
([ ]) lim
> x0
([ ]) = 100 100x sin x 100
> sin x
> x
lim
> x0
([ ]) lim
> x0
([ 99 ]) = 98 99 sin x xsin x
> x
lim
> x0
[ ] + [ ] = 100 + 98 = 198 100 x
> sin x
> 99 sin xx
99 = 198
= 2 Q6. Put
As
Q7.
does not exist.
Q8.
Let,
Q9.
Given,
Q10.
So,
Q11.
We know,
Also,
Q12.
Let
(by Leibnitz rule)
Hence, .
Q13. As f is a positive increasing function, we have
Dividing by leads to
As , we have by squeeze theorem
or sandwich theorem,
Limits
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x = tan = tan 1 xx 0 0
lim
0
sin 1 ( )1tan
2 tan
1+tan 2
= lim
0
sin 1 (sin 2 )1tan
= lim
0
= 2 2
tan
LHD : f ( )= lim
h0 +
4
f ( h ) f ( )
> 4
>
> 4
h
= lim
h0 +
[ tan ( h ) ] 2
[ tan ]
> 2
>
> 4
>
> 4
h
= lim
h0 +
01 h
f ( )
4
3 = t, x 2 t 3
> x
> 2
lim
t3
= lim
t3
t2 12
> 27
> t2
> 1
> t
> 3
> t2
t4+2712 t2
t3
lim
t3
= 6 6 = 36
( t23 ) ( t+3 ) ( t3 ) ( t3 )
lim
x
eexlog ( +1 )
> x3
> ex
10 ( x ) 3
= lim
x
e log ( +1 )
> x3
> ex
( )
> x3
> ex
110
= e 1 = 0. 27 110 ( 0 if x )x3
ex
lim xa [f(x)] ( x ) = elim
> xa
( x ) [ f ( x ) 1 ]
as f(x) 1 & (x) as x a
lim
x0
(cos x + sin x) = elim
> x0
( cos x+sin x1 ) 1
> x
> 1
> x
= elim
> x0
= e
> ( sin x+cos x)1
{(1 + x)2/ x}= (1 + x)2/ x [(1 + x)2/ x]
lim
x0
(1 + x)2/ x = e2
lim
x0
{(1 + x)2/ x}= lim
x0
((1 + x)2/ x) lim
x0
((1 + x)2/ x)
= e2 [e2]= e2 7 L = lim
x0
dt
x20 sin t x3
= lim
x0 ( sin x ) 2x 3x 2
= lim
x0 23sin x x
= (1) 23
= 23
L = 23
f(x)< f(2x)< f(3x)
f(x) 1 < <
f ( 2x )
f ( x )
f ( 3x )
f ( x )
lim
x = 1
f ( 3x )
f ( x )
lim
x = 1
f ( 2x )
f ( x ) Q1. If be a relation defined as iff , then the relation is
(1) Reflexive
(2) Symmetric
(3) Transitive
(4) Symmetric and transitive
Q2. A relation is defined as for , where
is the set of all integers. Then the relation is:
(1) reflexive but not symmetric
(2) svmmetric but not reflexive
(3) reflexive and symmetric both
(4) equivalence relation
Q3. Given
and , then the value of
(1)
(2)
(3)
(4)
Q4. Out of students, the number of students taking Mathematics is and the
number of students taking both Mathematics and Biology is . Then, the number
of students taking the only Biology is
(1)
(2)
(3)
(4)
Q5. In a class of students, students play cricket and students play tennis
and students play both the games, then the number of students who play neither
is
(1)
(2)
(3)
(4)
Q6. Let and
be two sets. Then
(1) and
(2)
(3)
(4)
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R aRb |a b| > 0
R (x, y) R xy = yx x, y I {0} IRn(A) = 11, n(B) = 13, n(C) = 16, n(A B) = 3, n(B C) = 6, n(A C) = 5
n(A B C) = 2 n [Ac ( BC)] = 4713 23 64 45 10 18 19 20 17 60 25 20 10 45 025 35
P = { : sin cos = 2 cos }
Q = { : sin + cos = 2 sin }
P Q P
Q PP QP = QAnswer Key
Q1 (2) Q2 (3) Q3 (3) Q4 (2)
Q5 (3) Q6 (4)
Sets and Relations
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Reflexive : iff
Which is not true. So, is not reflexive.
Symmetric : iff
Now, iff
Thus, is symmetric.
Transitive : iff
iff
& which is not correct, hence relation is not transitive.
Q2.
as
is reflexive
Now
is symmetric
Now and
is not transitive.
Q3.
Given
Solving all equations, we get
To find:
Q4.
Let denote the sets of students taking Mathematics and
Biology respectively.
Given
Drawing the Venn diagram :
Therefore, the number of students taking only Biology .
Q5. Let student play cricket
Student play tennis
and total number of students
and
Now,
The number of students who play neither game
Q6. In set P,
In set
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R aRb |a b|> 0
aRa |a a|> 0
RaRb |a b|> 0
bRa |b a|> 0 |a b|> 0 aRb RaRb |a b|> 0
bRc |b c|> 0
aRb bRa aRa
(x, y) R xy = yx
(x, x) R xx = xx x I {0}
R
(x, y) R xy = yx yx = xy
( y, x) R R
(x, y) R xy = yx (y, z) R yz = zy
xy = yx y = x yx
yz = zx (x )z
= zx x = zx
> yxyz x
xyz = zx2
(x, z) RRn(A) = 11 a + b + d + e = 11
n(B) = 13 b + c + e + f = 13
n(C) = 16 d + e + f + g = 16
n(A B) = 3 b + e = 3
n(B C) = 6 e + f = 6
n(A C) = 5 d + e = 5
n(A B C) = 2 e = 2
e = 2, d = 3, f = 4, b = 1, a = 5, c = 6, g = 7
n[Ac ( BC )]
BC = b + c + d + gAc ( BC ) = c + gAc ( BC ) = c + g = 6 + 7 = 13
M & Bn(M)= 45, n(M B)= 10, n(M B)= 64
n(B)= n(M B) n(M)+ n(M B) n(B)= 64 45 + 10 = 29
n(only B)= n(B A)= n(B) n(M B)= 29 10 = 19 = 19 = C
= T
= S
n(S)= 60, n(C)= 25, n(T )= 20
n(C T )= 10
n(C T )= n(C)+ n(T ) n(C T )= 25 + 20 10 = 35
= n(C T )' = n(S) n(C T )= 60 35 = 25 sin =(2 + 1 )cos tan = 2 + 1
Q, (2 1 )sin = cos
tan = = 2 + 1 P = Q121