Application of Derivatives
233 Questions
Start JEE Mains Test
2020
Q151
JEE Mains
MCQ
14 Mar 2026
The length of the perpendicular from the origin,
on the normal to the curve,
x2 + 2xy – 3y2 = 0 at the point (2,2) is
x2 + 2xy – 3y2 = 0 at the point (2,2) is
A.
$\sqrt 2 $
B.
$4\sqrt 2 $
C.
2
D.
$2\sqrt 2 $
2020
Q152
JEE Mains
MCQ
14 Mar 2026
Let ƒ(x) = xcos–1(–sin|x|), $x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$, then
which of the following is true?
A.
ƒ' is decreasing in $\left( { - {\pi \over 2},0} \right)$ and increasing
in $\left( {0,{\pi \over 2}} \right)$
B.
ƒ '(0) = ${ - {\pi \over 2}}$
C.
ƒ is not differentiable at x = 0
D.
ƒ' is increasing in $\left( { - {\pi \over 2},0} \right)$ and decreasing
in $\left( {0,{\pi \over 2}} \right)$
2020
Q153
JEE Mains
MCQ
14 Mar 2026
If c is a point at which Rolle's theorem holds
for the function,
f(x) = ${\log _e}\left( {{{{x^2} + \alpha } \over {7x}}} \right)$ in the interval [3, 4], where a $ \in $ R, then ƒ''(c) is equal to
f(x) = ${\log _e}\left( {{{{x^2} + \alpha } \over {7x}}} \right)$ in the interval [3, 4], where a $ \in $ R, then ƒ''(c) is equal to
A.
${1 \over {12}}$
B.
${{\sqrt 3 } \over 7}$
C.
$-{1 \over {12}}$
D.
$-{1 \over {24}}$
2020
Q154
JEE Mains
MCQ
14 Mar 2026
Let ƒ(x) be a polynomial of degree 5 such that x = ±1 are its critical points.
If $\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$, then which one of the following is not true?
If $\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$, then which one of the following is not true?
A.
ƒ(1) - 4ƒ(-1) = 4.
B.
x = 1 is a point of minima and x = -1 is a point of maxima of ƒ.
C.
x = 1 is a point of maxima and x = -1 is a point of minimum of ƒ.
D.
ƒ is an odd function.
2020
Q155
JEE Mains
MCQ
14 Mar 2026
The value of c in the Lagrange's mean value theorem for the function
ƒ(x) = x3 - 4x2 + 8x + 11, when x $ \in $ [0, 1] is:
ƒ(x) = x3 - 4x2 + 8x + 11, when x $ \in $ [0, 1] is:
A.
${2 \over 3}$
B.
${{\sqrt 7 - 2} \over 3}$
C.
${{4 - \sqrt 5 } \over 3}$
D.
${{4 - \sqrt 7 } \over 3}$
2020
Q156
JEE Mains
MCQ
14 Mar 2026
Let the function, ƒ:[-7, 0]$ \to $R be continuous on [-7,0] and differentiable on (-7, 0). If ƒ(-7) = -
3 and ƒ'(x) $ \le $ 2, for all x $ \in $ (-7,0), then for all such functions ƒ, ƒ(-1) + ƒ(0) lies in the interval:
A.
$\left[ { - 6,20} \right]$
B.
$\left( { - \infty ,\left. {20} \right]} \right.$
C.
$\left[ { - 3,11} \right]$
D.
$\left( { - \infty ,\left. {11} \right]} \right.$
2020
Q157
JEE Mains
Numerical
14 Mar 2026
If the lines x + y = a and x – y = b touch the
curve y = x2 – 3x + 2 at the points where the curve intersects the x-axis, then ${a \over b}$ is equal to _______.
curve y = x2 – 3x + 2 at the points where the curve intersects the x-axis, then ${a \over b}$ is equal to _______.
Correct Answer: 0.50
Explanation:
y = x2
– 3x + 2
$ \Rightarrow $ y = (x – 1)(x – 2)
At x-axis y = 0
$ \Rightarrow $ x = 1, 2
So this curve intersects the x-axis at A(1, 0) and B(2, 0).
${{dy} \over {dx}} = 2x - 3$
${\left( {{{dy} \over {dx}}} \right)_{x = 1}} = - 1$ and ${\left( {{{dy} \over {dx}}} \right)_{x = 2}} = 1$
Equation of tangent at A(1, 0) :
y = –1(x –1)
$ \Rightarrow $ x + y = 1
and equation of tangent at B(2, 0):
y = 1(x – 2)
$ \Rightarrow $ x – y = 2
So a = 1 and b = 2
$ \Rightarrow $ ${a \over b}$ = 0.5
$ \Rightarrow $ y = (x – 1)(x – 2)
At x-axis y = 0
$ \Rightarrow $ x = 1, 2
So this curve intersects the x-axis at A(1, 0) and B(2, 0).
${{dy} \over {dx}} = 2x - 3$
${\left( {{{dy} \over {dx}}} \right)_{x = 1}} = - 1$ and ${\left( {{{dy} \over {dx}}} \right)_{x = 2}} = 1$
Equation of tangent at A(1, 0) :
y = –1(x –1)
$ \Rightarrow $ x + y = 1
and equation of tangent at B(2, 0):
y = 1(x – 2)
$ \Rightarrow $ x – y = 2
So a = 1 and b = 2
$ \Rightarrow $ ${a \over b}$ = 0.5
2020
Q158
JEE Mains
Numerical
14 Mar 2026
Let ƒ(x) be a polynomial of degree 3 such that
ƒ(–1) = 10, ƒ(1) = –6, ƒ(x) has a critical point
at x = –1 and ƒ'(x) has a critical point at x = 1.
Then ƒ(x) has a local minima at x = _______.
Correct Answer: 3
Explanation:
Let f(x) = ax3
+ bx2
+ cx + d
Given f(-1) = 10, f(1) = -6
$ \therefore $ -a + b - c + d = 10 ....(i)
and a + b + c + d = -6 ......(ii)
adding (i) + (ii)
2(b + d) = 4
$ \Rightarrow $ b + d = 2 ....(iii)
f'(x) = 3ax2 + 2bx + c
Given f'(-1) = 0
$ \Rightarrow $ 3a - 2b + c = 0 .....(iv)
f"(x) = 6ax + 2b
Given f"(1) = 0
$ \therefore $ 6a + 2b = 0 ....(v)
$ \Rightarrow $ b = -3a
adding (iv) + (v), we get
9a + c = 0 ....(vi)
$ \Rightarrow $ $9\left( {{{ - b} \over 3}} \right)$ + c = 0
$ \Rightarrow $ c = 3b
f(x) = ${{{ - b} \over 3}{x^3}}$ + bx2 + 3bx + (2 - b)
$ \Rightarrow $ f'(x) = -bx2 + 2bx + 3b
= -b(x2 - 2x - 3)
At maxima and minima f'(x) = 0
$ \therefore $ (x2 - 2x - 3) = 0
$ \Rightarrow $ (x - 3) (x + 1) = 0
x = 3, -1
As a + b + c + d = -6
$ \Rightarrow $ ${{{ - b} \over 3}}$ + b + 3b + 2 - b = -6
$ \Rightarrow $ b = -3
$ \therefore $ f'(x) = 3(x2 - 2x - 3)
$ \Rightarrow $ f''(x) = 3(2x - 2)
At x = 3, f''(x) = 3(2.3 - 2) = 12 > 0
$ \therefore $ Minima at x = 3.
Given f(-1) = 10, f(1) = -6
$ \therefore $ -a + b - c + d = 10 ....(i)
and a + b + c + d = -6 ......(ii)
adding (i) + (ii)
2(b + d) = 4
$ \Rightarrow $ b + d = 2 ....(iii)
f'(x) = 3ax2 + 2bx + c
Given f'(-1) = 0
$ \Rightarrow $ 3a - 2b + c = 0 .....(iv)
f"(x) = 6ax + 2b
Given f"(1) = 0
$ \therefore $ 6a + 2b = 0 ....(v)
$ \Rightarrow $ b = -3a
adding (iv) + (v), we get
9a + c = 0 ....(vi)
$ \Rightarrow $ $9\left( {{{ - b} \over 3}} \right)$ + c = 0
$ \Rightarrow $ c = 3b
f(x) = ${{{ - b} \over 3}{x^3}}$ + bx2 + 3bx + (2 - b)
$ \Rightarrow $ f'(x) = -bx2 + 2bx + 3b
= -b(x2 - 2x - 3)
At maxima and minima f'(x) = 0
$ \therefore $ (x2 - 2x - 3) = 0
$ \Rightarrow $ (x - 3) (x + 1) = 0
x = 3, -1
As a + b + c + d = -6
$ \Rightarrow $ ${{{ - b} \over 3}}$ + b + 3b + 2 - b = -6
$ \Rightarrow $ b = -3
$ \therefore $ f'(x) = 3(x2 - 2x - 3)
$ \Rightarrow $ f''(x) = 3(2x - 2)
At x = 3, f''(x) = 3(2.3 - 2) = 12 > 0
$ \therefore $ Minima at x = 3.
2020
Q159
JEE Mains
Numerical
14 Mar 2026
Let the normal at a point P on the curve
y2 – 3x2 + y + 10 = 0 intersect the y-axis at $\left( {0,{3 \over 2}} \right)$ .
If m is the slope of the tangent at P to the curve, then |m| is equal to
y2 – 3x2 + y + 10 = 0 intersect the y-axis at $\left( {0,{3 \over 2}} \right)$ .
If m is the slope of the tangent at P to the curve, then |m| is equal to
Correct Answer: 4
Explanation:
Given curve : y2
– 3x2
+ y + 10 = 0
$ \Rightarrow $ 2y${{dy} \over {dx}}$ - 6x + ${{dy} \over {dx}}$ = 0
$ \Rightarrow $ ${{dy} \over {dx}}$ = ${{6x} \over {2y + 1}}$
Let P be (x1, y1)
Slope of tangent at P = ${{6{x_1}} \over {2{y_1} + 1}}$
$ \therefore $ Slope of normal at P = $ - {{2{y_1} + 1} \over {6{x_1}}}$
$ \Rightarrow $ Equation of normal (y – y1) = $ - \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$(x – x1)
This normal passes through point $\left( {0,{3 \over 2}} \right)$.
$ \therefore $ (${{3 \over 2}}$ – y1) = $ - \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$(0 – x1)
$ \Rightarrow $ y1 = 1
Put y1 = 1 in equation of curve , then we get x1 = $ \pm $2
$ \Rightarrow $ |m| = slope of tangent = $\left| {{{6{x_1}} \over {2{y_1} + 1}}} \right|$ = ${{12} \over 3}$ = 4
$ \Rightarrow $ 2y${{dy} \over {dx}}$ - 6x + ${{dy} \over {dx}}$ = 0
$ \Rightarrow $ ${{dy} \over {dx}}$ = ${{6x} \over {2y + 1}}$
Let P be (x1, y1)
Slope of tangent at P = ${{6{x_1}} \over {2{y_1} + 1}}$
$ \therefore $ Slope of normal at P = $ - {{2{y_1} + 1} \over {6{x_1}}}$
$ \Rightarrow $ Equation of normal (y – y1) = $ - \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$(x – x1)
This normal passes through point $\left( {0,{3 \over 2}} \right)$.
$ \therefore $ (${{3 \over 2}}$ – y1) = $ - \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$(0 – x1)
$ \Rightarrow $ y1 = 1
Put y1 = 1 in equation of curve , then we get x1 = $ \pm $2
$ \Rightarrow $ |m| = slope of tangent = $\left| {{{6{x_1}} \over {2{y_1} + 1}}} \right|$ = ${{12} \over 3}$ = 4
2019
Q160
JEE Mains
MCQ
14 Mar 2026
If m is the minimum value of k for which the function f(x) = x$\sqrt {kx - {x^2}} $ is increasing in the interval [0,3]
and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to :
A.
$\left( {5,3\sqrt 6 } \right)$
B.
$\left( {4,3\sqrt 3 } \right)$
C.
$\left( {4,3\sqrt 2 } \right)$
D.
$\left( {3,3\sqrt 3 } \right)$
2019
Q161
JEE Mains
MCQ
14 Mar 2026
A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate
25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the
horizontal ground when the top of the ladder is 1 m above the ground is :
A.
${{25} \over 3}$
B.
25
C.
25$\sqrt 3 $
D.
${{25} \over {\sqrt 3 }}$
2019
Q162
JEE Mains
MCQ
14 Mar 2026
A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of
50 cm3
/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice
decreases, is :
A.
${5 \over {6\pi }}$
B.
${1 \over {9\pi }}$
C.
${1 \over {36\pi }}$
D.
${1 \over {18\pi }}$
2019
Q163
JEE Mains
MCQ
14 Mar 2026
If the tangent to the curve $y = {x \over {{x^2} - 3}}$
, $x \in \rho ,\left( {x \ne \pm \sqrt 3 } \right)$, at a point ($\alpha $, $\beta $) $ \ne $ (0, 0) on it is parallel to the line
2x + 6y – 11 = 0, then :
A.
| 6$\alpha $ + 2$\beta $ | = 9
B.
| 2$\alpha $ + 6$\beta $ | = 11
C.
| 2$\alpha $ + 6$\beta $ | = 19
D.
| 6$\alpha $ + 2$\beta $ | = 19
2019
Q164
JEE Mains
MCQ
14 Mar 2026
A water tank has the shape of an inverted right
circular cone, whose semi-vertical angle is
${\tan ^{ - 1}}\left( {{1 \over 2}} \right)$. Water is poured into it at a constant
rate of 5 cubic meter per minute. The the rate
(in m/min.), at which the level of water is rising
at the instant when the depth of water in the tank
is 10m; is :-
A.
${1 \over {15\pi }}$
B.
${1 \over {5\pi }}$
C.
${1 \over {10\pi }}$
D.
${2 \over \pi }$
2019
Q165
JEE Mains
MCQ
14 Mar 2026
If ƒ(x) is a non-zero polynomial of degree four,
having local extreme points at x = –1, 0, 1; then
the set
S = {x $ \in $ R : ƒ(x) = ƒ(0)}
Contains exactly :
S = {x $ \in $ R : ƒ(x) = ƒ(0)}
Contains exactly :
A.
four rational numbers.
B.
four irrational numbers.
C.
two irrational and one rational number.
D.
two irrational and two rational numbes.
2019
Q166
JEE Mains
MCQ
14 Mar 2026
Let S be the set of all values of x for which the
tangent to the curve
y = ƒ(x) = x3 – x2 – 2x at (x, y) is parallel to the line segment joining the points (1, ƒ(1)) and (–1, ƒ(–1)), then S is equal to :
y = ƒ(x) = x3 – x2 – 2x at (x, y) is parallel to the line segment joining the points (1, ƒ(1)) and (–1, ƒ(–1)), then S is equal to :
A.
$\left\{ { {1 \over 3}, - 1} \right\}$
B.
$\left\{ { - {1 \over 3}, 1} \right\}$
C.
$\left\{ { - {1 \over 3}, - 1} \right\}$
D.
$\left\{ { {1 \over 3}, 1} \right\}$
2019
Q167
JEE Mains
MCQ
14 Mar 2026
If the tangent to the curve, y = x3 + ax – b at
the point (1, –5) is perpendicular to the line,
–x + y + 4 = 0, then which one of the following
points lies on the curve ?
A.
(2, –2)
B.
(2, –1)
C.
(–2, 2)
D.
(–2, 1)
2019
Q168
JEE Mains
MCQ
14 Mar 2026
The height of a right circular cylinder of maximum
volume inscribed in a sphere of radius 3 is
A.
$\sqrt 3 $
B.
$2\sqrt 3 $
C.
$\sqrt 6 $
D.
${2 \over 3} {\sqrt 3} $
2019
Q169
JEE Mains
MCQ
14 Mar 2026
Given that the slope of the tangent to a curve y
= y(x) at any point (x, y) is
$2y \over x^2$. If the curve passes through the centre of the circle x2 + y2 – 2x – 2y = 0, then its equation is :
A.
x loge|y| = 2(x – 1)
B.
x2 loge|y| = –2(x – 1)
C.
x loge|y| = x – 1
D.
x loge|y| = –2(x – 1)
2019
Q170
JEE Mains
MCQ
14 Mar 2026
Let ƒ : [0, 2] $ \to $ R be a twice differentiable
function such that ƒ''(x) > 0, for all x $ \in $ (0, 2).
If $\phi $(x) = ƒ(x) + ƒ(2 – x), then $\phi $ is :
A.
decreasing on (0, 2)
B.
decreasing on (0, 1) and increasing on (1, 2)
C.
increasing on (0, 2)
D.
increasing on (0, 1) and decreasing on (1, 2)
2019
Q171
JEE Mains
MCQ
14 Mar 2026
If S1 and S2 are respectively the sets of local
minimum and local maximum points of the function,
ƒ(x) = 9x4 + 12x3 – 36x2 + 25, x $ \in $ R, then :
ƒ(x) = 9x4 + 12x3 – 36x2 + 25, x $ \in $ R, then :
A.
S1 = {–1}; S2 = {0, 2}
B.
S1 = {–2}; S2 = {0, 1}
C.
S1 = {–2, 0}; S2 = {1}
D.
S1 = {–2, 1}; S2 = {0}
2019
Q172
JEE Mains
MCQ
14 Mar 2026
The tangent to the curve y = x2 – 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point :
A.
$\left\{ {{1 \over 4},{7 \over 2}} \right\}$
B.
$\left( { - {1 \over 8},7} \right)$
C.
$\left( {{7 \over 2},{1 \over 4}} \right)$
D.
$\left( {{1 \over 8}, - 7} \right)$
2019
Q173
JEE Mains
MCQ
14 Mar 2026
If the function f given by f(x) = x3 – 3(a – 2)x2 + 3ax + 7, for some a$ \in $R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, ${{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)$ is :
A.
$-$ 7
B.
5
C.
7
D.
6
2019
Q174
JEE Mains
MCQ
14 Mar 2026
Let f(x) = ${x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\,$ x $\, \in $ R, where a, b and d are non-zero real constants. Then :
A.
f is an increasing function of x
B.
f is neither increasing nor decreasing function of x
C.
f ' is not a continuous function of x
D.
f is a decreasing function of x
2019
Q175
JEE Mains
MCQ
14 Mar 2026
The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = $\left\{ {x\, \in R:{x^2} + 30 \le 11x} \right\}$ is :
A.
$-$ 222
B.
$-$ 122
C.
$122$
D.
222
2019
Q176
JEE Mains
MCQ
14 Mar 2026
The tangent to the curve, y = xex2 passing through the point (1, e) also passes through the point
A.
$\left( {{4 \over 3},2e} \right)$
B.
(3, 6e)
C.
(2, 3e)
D.
$\left( {{5 \over 3},2e} \right)$
2019
Q177
JEE Mains
MCQ
14 Mar 2026
A helicopter is flying along the curve given by y – x3/2 = 7, (x $ \ge $ 0). A soldier positioned at the point $\left( {{1 \over 2},7} \right)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is -
A.
${1 \over 6}\sqrt {{7 \over 3}} $
B.
${{\sqrt 5 } \over 6}$
C.
${1 \over 2}$
D.
${1 \over 3}$$\sqrt {{7 \over 3}} $
2019
Q178
JEE Mains
MCQ
14 Mar 2026
The shortest distance between the point $\left( {{3 \over 2},0} \right)$ and the curve y = $\sqrt x $, (x > 0), is -
A.
${{\sqrt 3 } \over 2}$
B.
${5 \over 4}$
C.
${3 \over 2}$
D.
${{\sqrt 5 } \over 2}$
2019
Q179
JEE Mains
MCQ
14 Mar 2026
The maximum volume (in cu.m) of the right circular cone having slant height 3 m is :
A.
2$\sqrt3$$\pi $
B.
3$\sqrt3$$\pi $
C.
6$\pi $
D.
${4 \over 3}\pi $
2018
Q180
JEE Mains
MCQ
14 Mar 2026
Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x3 $-$ 9x2 + 12x + 5 in the interval [0, 3]. Then M $-$m is equal to :
A.
5
B.
9
C.
4
D.
1
2018
Q181
JEE Mains
MCQ
14 Mar 2026
If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is :
A.
${9 \over 2}$
B.
6
C.
${7 \over 2}$
D.
4
2018
Q182
JEE Mains
MCQ
14 Mar 2026
Let $f\left( x \right) = {x^2} + {1 \over {{x^2}}}$ and $g\left( x \right) = x - {1 \over x}$,
$x \in R - \left\{ { - 1,0,1} \right\}$.
If $h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$, then the local minimum value of h(x) is
$x \in R - \left\{ { - 1,0,1} \right\}$.
If $h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$, then the local minimum value of h(x) is
A.
$2\sqrt 2 $
B.
3
C.
-3
D.
$-2\sqrt 2 $
2018
Q183
JEE Mains
MCQ
14 Mar 2026
If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is :
A.
$6\sqrt 2 \pi $
B.
$6\sqrt 3 \pi $
C.
$8\sqrt 2 \pi $
D.
$8\sqrt 3 \pi $
2018
Q184
JEE Mains
MCQ
14 Mar 2026
If $\beta $ is one of the angles between the normals to the ellipse, x2 + 3y2 = 9 at the points (3 cos $\theta $, $\sqrt 3 \sin \theta $) and ($-$ 3 sin $\theta $, $\sqrt 3 \,\cos \theta $); $\theta \in \left( {0,{\pi \over 2}} \right);$ then ${{2\,\cot \beta } \over {\sin 2\theta }}$ is equal to :
A.
${2 \over {\sqrt 3 }}$
B.
${1 \over {\sqrt 3 }}$
C.
$\sqrt 2 $
D.
${{\sqrt 3 } \over 4}$
2017
Q185
JEE Mains
MCQ
14 Mar 2026
The function f defined by
f(x) = x3 $-$ 3x2 + 5x + 7 , is :
f(x) = x3 $-$ 3x2 + 5x + 7 , is :
A.
increasing in R.
B.
decreasing in R.
C.
decreasing in (0, $\infty $) and increasing in ($-$ $\infty $, 0)
D.
increasing in (0, $\infty $) and decreasing in ($-$ $\infty $, 0)
2017
Q186
JEE Mains
MCQ
14 Mar 2026
A tangent to the curve, y = f(x) at P(x, y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 and f(1) = 1, then the curve also passes through the point :
A.
$\left( {{1 \over 3},24} \right)$
B.
$\left( {{1 \over 2},4} \right)$
C.
$\left( {2,{1 \over 8}} \right)$
D.
$\left( {3,{1 \over 28}} \right)$
2017
Q187
JEE Mains
MCQ
14 Mar 2026
The tangent at the point (2, $-$2) to the curve, x2y2 $-$ 2x = 4(1 $-$ y) does not pass through the point :
A.
$\left( {4,{1 \over 3}} \right)$
B.
(8, 5)
C.
($-$4, $-$9)
D.
($-$2, $-$7)
2017
Q188
JEE Mains
MCQ
14 Mar 2026
The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes
through the point :
A.
$\left( {{1 \over 2},{1 \over 2}} \right)$
B.
$\left( {{1 \over 2}, - {1 \over 3}} \right)$
C.
$\left( {{1 \over 2},{1 \over 3}} \right)$
D.
$\left( { - {1 \over 2}, - {1 \over 3}} \right)$
2017
Q189
JEE Mains
MCQ
14 Mar 2026
Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the
maximum area (in sq. m) of the flower-bed, is :
A.
10
B.
25
C.
30
D.
12.5
We have
2016
Q190
JEE Mains
MCQ
14 Mar 2026
Let C be a curve given by y(x) = 1 + $\sqrt {4x - 3} ,x > {3 \over 4}.$ If P is a point
on C, such that the tangent at P has slope ${2 \over 3}$, then a point through which the normal at P passes, is :
A.
(2, 3)
B.
(4, $-$3)
C.
(1, 7)
D.
(3, $-$ 4),
2016
Q191
JEE Mains
MCQ
14 Mar 2026
Let f(x) = sin4x + cos4 x. Then f is an increasing function in the interval :
A.
$] 0, \frac{\pi}{4}[$
B.
$] \frac{\pi}{4}, \frac{\pi}{2}[$
C.
$] \frac{\pi}{2}, \frac{5 \pi}{8}[$
D.
$] \frac{5 \pi}{8}, \frac{3 \pi}{4}[$
2016
Q192
JEE Mains
MCQ
14 Mar 2026
The minimum distance of a point on the curve y = x2−4 from the origin is :
A.
${{\sqrt {19} } \over 2}$
B.
$\sqrt {{{15} \over 2}} $
C.
${{\sqrt {15} } \over 2}$
D.
$\sqrt {{{19} \over 2}} $
2016
Q193
JEE Mains
MCQ
14 Mar 2026
If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3−1, t $ \in $ R, meets the curve again at a point Q, then the coordinates of Q are :
A.
(t2 + 3, − t3 −1)
B.
(4t2 + 3, − 8t3 −1)
C.
(t2 + 3, t3 −1)
D.
(16t2 + 3, − 64t3 −1)
2016
Q194
JEE Mains
MCQ
14 Mar 2026
A wire of length $2$ units is cut into two parts which are bent respectively to form a square of side $=x$ units and a circle of radius $=r$ units. If the sum of the areas of the square and the circle so formed is minimum, then:
A.
$x=2r$
B.
$2x=r$
C.
$2x = \left( {\pi + 4} \right)r$
D.
$\left( {4 - \pi } \right)x = \pi \,\, r$
2016
Q195
JEE Mains
MCQ
14 Mar 2026
Consider :
f $\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).$
f $\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).$
A normal to $y = $ f$\left( x \right)$ at $x = {\pi \over 6}$ also passes through the point:
A.
$\left( {{\pi \over 6},0} \right)$
B.
$\left( {{\pi \over 4},0} \right)$
C.
$(0,0)$
D.
$\left( {0,{{2\pi } \over 3}} \right)$
2015
Q196
JEE Mains
MCQ
14 Mar 2026
Let $f(x)$ be a polynomial of degree four having extreme values
at $x=1$ and $x=2$. If $\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$, then f$(2)$ is equal to :
at $x=1$ and $x=2$. If $\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$, then f$(2)$ is equal to :
A.
$0$
B.
$4$
C.
$-8$
D.
$-4$
2015
Q197
JEE Mains
MCQ
14 Mar 2026
The normal to the curve, ${x^2} + 2xy - 3{y^2} = 0$, at $(1,1)$
A.
meets the curve again in the third quadrant.
B.
meets the curve again in the fourth quadrant.
C.
does not meet the curve again.
D.
meets the curve again in the second quadrant.
2014
Q198
JEE Mains
MCQ
14 Mar 2026
If $x=-1$ and $x=2$ are extreme points of $f\left( x \right) = \alpha \,\log \left| x \right|+\beta {x^2} + x$ then
A.
$\alpha = 2,\beta = - {1 \over 2}$
B.
$\alpha = 2,\beta = {1 \over 2}$
C.
$\alpha = - 6,\beta = {1 \over 2}$
D.
$\alpha = - 6,\beta = -{1 \over 2}$
2014
Q199
JEE Mains
MCQ
14 Mar 2026
If $f$ and $g$ are differentiable functions in $\left[ {0,1} \right]$ satisfying
$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$ and $f\left( 1 \right) = 6,$ then for some $c \in \left] {0,1} \right[$
$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$ and $f\left( 1 \right) = 6,$ then for some $c \in \left] {0,1} \right[$
A.
$f'\left( c \right) = g'\left( c \right)$
B.
$f'\left( c \right) = 2g'\left( c \right)$
C.
$2f'\left( c \right) = g'\left( c \right)$
D.
$2f'\left( c \right) = 3g'\left( c \right)$
2013
Q200
JEE Mains
MCQ
14 Mar 2026
The intercepts on $x$-axis made by tangents to the curve,
$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $ which are parallel to the line $y=2x$, are equal to :
$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $ which are parallel to the line $y=2x$, are equal to :
A.
$ \pm 1$
B.
$ \pm 2$
C.
$ \pm 3$
D.
$ \pm 4$