Application of Derivatives
584 Questions
Start JEE Mains Test
2019
Q401
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
Q402
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
Q403
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
Q404
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
Q405
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
Q406
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}$$\sqrt {{7 \over 3}} $
2019
Q407
JEE Mains
MCQ
14 Mar 2026
The shortest distance between the point $\left( {{3 \over 2},0} \right)$ and the curve y = $\sqrt x $, (x > 0), is -
A.
${{\sqrt 3 } \over 2}$
B.
${5 \over 4}$
C.
${3 \over 2}$
D.
${{\sqrt 5 } \over 2}$
2019
Q408
JEE Mains
MCQ
14 Mar 2026
The maximum volume (in cu.m) of the right circular cone having slant height 3 m is :
A.
2$\sqrt3$$\pi $
B.
3$\sqrt3$$\pi $
C.
6$\pi $
D.
${4 \over 3}\pi $
2019
Q409
JEE Advanced
MSQ
14 Mar 2026
Let f : R $ \to $ R be given by
$f(x) = (x - 1)(x - 2)(x - 5)$. Define
$F(x) = \int\limits_0^x {f(t)dt} $, x > 0
Then which of the following options is/are correct?
$f(x) = (x - 1)(x - 2)(x - 5)$. Define
$F(x) = \int\limits_0^x {f(t)dt} $, x > 0
Then which of the following options is/are correct?
A.
F(x) $ \ne $ 0 for all x $ \in $ (0, 5)
B.
F has a local maximum at x = 2
C.
F has two local maxima and one local minimum in (0, $\infty $)
D.
F has a local minimum at x = 1
2019
Q410
JEE Advanced
MSQ
14 Mar 2026
Let, $f(x) = {{\sin \pi x} \over {{x^2}}}$, x > 0
Let x1 < x2 < x3 < ... < xn < ... be all the points of local maximum of f and y1 < y2 < y3 < ... < yn < ... be all the points of local minimum of f.
Then which of the following options is/are correct?
Let x1 < x2 < x3 < ... < xn < ... be all the points of local maximum of f and y1 < y2 < y3 < ... < yn < ... be all the points of local minimum of f.
Then which of the following options is/are correct?
A.
$|{x_n} - {y_n}|\, > 1$ for every n
B.
${x_{n + 1}} - {x_n}\, > 2$ for every n
C.
x1 < y1
D.
${x_n} \in \left( {2n,\,2n + {1 \over 2}} \right)$ for every n
2018
Q411
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
Q412
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
Q413
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
Q414
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
Q415
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}$
2018
Q416
JEE Advanced
Numerical
14 Mar 2026
For each positive integer n, let
${y_n} = {1 \over n}(n + 1)(n + 2)...{(n + n)^{{1 \over n}}}$.
For x$ \in $R, let [x] be the greatest integer less than or equal to x. If $\mathop {\lim }\limits_{n \to \infty } {y_n} = L$, then the value of [L] is ..............
Correct Answer: 1
Explanation:
We have,
${y_n} = {1 \over n}(n + 1)(n + 2)...{(n + n)^{1/n}}$ and $\mathop {\lim }\limits_{n \to \infty } {y_n} = L$
$ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}{[(n + 1)(n + 2)(n + 3)...(n + n)]^{1/n}}$
$ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {\left[ {\left( {1 + {1 \over n}} \right)\left( {1 + {2 \over n}} \right)\left( {1 + {3 \over n}} \right)...\left( {1 + {n \over n}} \right)} \right]^{{1 \over n}}}$
$ \Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {\log \left( {1 + {1 \over n}} \right) + \log \left( {1 + {2 \over n}} \right)...\log \left( {1 + {n \over n}} \right)} \right]$
$ \Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {\log \left( {1 + {r \over n}} \right)} $
$ \Rightarrow \log L = \int_0^1 {_{II}^1 \times \mathop {\log }\limits_I } (1 + x)\,dx$
$ \Rightarrow \log L = (x.\log (1 + x)_0^1 - \int_0^1 {\left[ {{d \over {dx}}(\log (1 + x)\int {dx} } \right]} dx$
[by using integration by parts]
$ \Rightarrow \log L = [x\log (1 + x)]_0^1 - \int_0^1 {{x \over {1 + x}}} dx$
$ \Rightarrow \log L = \log 2 - \int_0^1 {\left( {{{x + 1} \over {x + 1}} - {1 \over {x + 1}}} \right)} dx$
$ \Rightarrow \log L = \log 2 - [x]_0^1 + [\log (x + 1)]_0^1$
$ \Rightarrow \log L = \log 2 - 1 + \log 2 - 0$
$ \Rightarrow \log L = \log 4 - \log e = \log {4 \over e}$
$ \Rightarrow L = {4 \over e}$
$ \Rightarrow [L] = \left[ {{4 \over e}} \right] = 1$
${y_n} = {1 \over n}(n + 1)(n + 2)...{(n + n)^{1/n}}$ and $\mathop {\lim }\limits_{n \to \infty } {y_n} = L$
$ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}{[(n + 1)(n + 2)(n + 3)...(n + n)]^{1/n}}$
$ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {\left[ {\left( {1 + {1 \over n}} \right)\left( {1 + {2 \over n}} \right)\left( {1 + {3 \over n}} \right)...\left( {1 + {n \over n}} \right)} \right]^{{1 \over n}}}$
$ \Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {\log \left( {1 + {1 \over n}} \right) + \log \left( {1 + {2 \over n}} \right)...\log \left( {1 + {n \over n}} \right)} \right]$
$ \Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {\log \left( {1 + {r \over n}} \right)} $
$ \Rightarrow \log L = \int_0^1 {_{II}^1 \times \mathop {\log }\limits_I } (1 + x)\,dx$
$ \Rightarrow \log L = (x.\log (1 + x)_0^1 - \int_0^1 {\left[ {{d \over {dx}}(\log (1 + x)\int {dx} } \right]} dx$
[by using integration by parts]
$ \Rightarrow \log L = [x\log (1 + x)]_0^1 - \int_0^1 {{x \over {1 + x}}} dx$
$ \Rightarrow \log L = \log 2 - \int_0^1 {\left( {{{x + 1} \over {x + 1}} - {1 \over {x + 1}}} \right)} dx$
$ \Rightarrow \log L = \log 2 - [x]_0^1 + [\log (x + 1)]_0^1$
$ \Rightarrow \log L = \log 2 - 1 + \log 2 - 0$
$ \Rightarrow \log L = \log 4 - \log e = \log {4 \over e}$
$ \Rightarrow L = {4 \over e}$
$ \Rightarrow [L] = \left[ {{4 \over e}} \right] = 1$
2017
Q417
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
Q418
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
Q419
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
Q420
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
Q421
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
2017
Q422
JEE Advanced
MCQ
14 Mar 2026
Which of the following options is the only INCORRECT combination?
A.
(I) (iii) (P)
B.
(II) (iv) (Q)
C.
(II) (ii) (P)
D.
(III) (i) (R)
2017
Q423
JEE Advanced
MCQ
14 Mar 2026
Which of the following options is the only CORRECT combination?
A.
(I) (ii) (R)
B.
(III) (iv) (P)
C.
(II) (iii) (S)
D.
(IV) (i) (S)
2017
Q424
JEE Advanced
MCQ
14 Mar 2026
Which of the following options is the only CORRECT combination?
A.
(III) (iii) (R)
B.
(IV) (iv) (S)
C.
(II) (ii) (Q)
D.
(I0 (i) (P)
2017
Q425
JEE Advanced
MSQ
14 Mar 2026
f : R $ \to $ R is a differentiable function such that f'(x) > 2f(x) for all x$ \in $R, and f(0) = 1 then
A.
f(x) > e2x in (0, $\infty $)
B.
f'(x) < e2x in (0, $\infty $)
C.
f(x) is increasing in (0, $\infty $)
D.
f(x) is decreasing in (0, $\infty $)
2017
Q426
JEE Advanced
MSQ
14 Mar 2026
If $f(x) = \left| {\matrix{
{\cos 2x} & {\cos 2x} & {\sin 2x} \cr
{ - \cos x} & {\cos x} & { - \sin x} \cr
{\sin x} & {\sin x} & {\cos x} \cr
} } \right|$,
then
then
A.
f(x) attains its minimum at x = 0
B.
f(x) attains its maximum at x = 0
C.
f'(x) = 0 at more than three points in ($-$$\pi $, $\pi $)
D.
f'(x) = 0 at exactly three points in ($-$$\pi $, $\pi $)
2016
Q427
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
Q428
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
Q429
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
Q430
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
Q431
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
Q432
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)$
2016
Q433
JEE Advanced
MCQ
14 Mar 2026
The least value of a $ \in R$ for which $4a{x^2} + {1 \over x} \ge 1,$, for all $x>0$. is
A.
${1 \over {64}}$
B.
${1 \over {32}}$
C.
${1 \over {27}}$
D.
${1 \over {25}}$
2016
Q434
JEE Advanced
MSQ
14 Mar 2026
Let f: R $ \to \left( {0,\infty } \right)$ and g : R $ \to $ R be twice differentiable functions such that f'' and g'' are continuous functions on R. Suppose f'$(2)$ $=$ g$(2)=0$, f''$(2)$$ \ne 0$ and g'$(2)$ $ \ne 0$. If
$\mathop {\lim }\limits_{x \to 2} {{f\left( x \right)g\left( x \right)} \over {f'\left( x \right)g'\left( x \right)}} = 1,$ then
$\mathop {\lim }\limits_{x \to 2} {{f\left( x \right)g\left( x \right)} \over {f'\left( x \right)g'\left( x \right)}} = 1,$ then
A.
$f$ has a local minimum at $x=2$
B.
$f$ has a local maximum at $x=2$
C.
$f''(2)>f(2)$
D.
$f(x)-f''(x)=0$ for at least one $x \in R$
2015
Q435
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
Q436
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.
2015
Q437
JEE Advanced
MSQ
14 Mar 2026
Let $f, g :$ $\left[ { - 1,2} \right] \to R$ be continuous functions which are twice differentiable on the interval $(-1, 2)$. Let the values of f and g at the points $-1, 0$ and $2$ be as given in the following table:
| X = -1 | X = 0 | X = 2 | |
|---|---|---|---|
| f(x) | 3 | 6 | 0 |
| g(x) | 0 | 1 | -1 |
In each of the intervals $(-1, 0)$ and $(0, 2)$ the function $(f-3g)''$ never vanishes. Then the correct statement(s) is (are)
A.
$f'\left( x \right) - 3g'\left( x \right) = 0$ has exactly three solutions in $\left( { - 1,0} \right) \cup \left( {0,2} \right)$
B.
$f'\left( x \right) - 3g'\left( x \right) = 0$ has exactly one solution in $(-1, 0)$
C.
$f'\left( x \right) - 3g'\left( x \right) = 0$ has exactly one solution in $(0, 2)$
D.
$f'\left( x \right) - 3g'\left( x \right) = 0$ has exactly two solutions in $(-1, 0)$ and exactly two solutions in $(0, 2)$
2015
Q438
JEE Advanced
Numerical
14 Mar 2026
A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of $V$ $m{m^3}$, has a $2$ mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness $2$ mm and is of radius equal to the outer radius of the container.
If the volume of the material used to make the container is minimum when the inner radius of the container is $10 $ mm,
then the value of ${V \over {250\pi }}$ is
Correct Answer: 4
Explanation:
Given: The inner volume of cylinder $=\mathrm{Vmm}^3$
Thickness of wall $=2 \mathrm{~mm}$
Thickness of bottom circular disc $=2 \mathrm{~mm}$
Let the inner radius of cylinder $=r \mathrm{~mm}$ and height of the inner cylinder $=h \mathrm{~mm}$.
$ \Rightarrow \mathrm{V}=\pi r^2 h $
Now, volume of the material used $=$ volume of outer cylinder - volume of the inner cylinder + Volume of the circular disc
$ \begin{gathered} \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi(r+2)^2 h-\pi r^2 h+\pi(r+2)^2 2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h\left\{(r+2)^2-r^2\right\}+2 \pi(r+2)^2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h(4 r+4)+2 \pi(r+2)^2 \\\\ \left\{\because(a+b)^2=a^2+b^2+2 a b\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{2 h(r+1)+(r+2)^2\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi r^2}(r+1)+(r+2)^2\right\} \\\\ \left\{\because \mathrm{V}=\pi r^2 h\right\} \end{gathered} $
For $\mathrm{V}_{\mathrm{m}}$ to be minimum, $\frac{d v_m}{d r}=0$
Differentiating the above equation w.r.t.r,
$ \begin{aligned} \quad & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \frac{d}{d r}\left\{\frac{r+1}{r^2}\right\}+\frac{d}{d r}(r+2)^2\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2(1)-(r+1)(2 r)}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2-2 r^2-2 r}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi\left\{\frac{\mathrm{V}}{\pi r^3}(-r-2)+(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi(r+2)\left(-\frac{\mathrm{V}}{\pi r^3}+1\right) \\\\ \because & \frac{d \mathrm{~V}_m}{d r}=0 \\\\ \Rightarrow & 4 \pi(r+2)\left(\frac{-\mathrm{V}}{\pi r^3}+1\right)=0 \end{aligned} $
Also, given that $\mathrm{V}_{\mathrm{m}}$ is minimum at $r=10 \mathrm{~mm}$
$ \begin{array}{ll} \Rightarrow & 4 \pi(10+2)\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & 48 \pi\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & \frac{\mathrm{V}}{10^3 \pi}=1 \\\\ \Rightarrow & \frac{\mathrm{V}}{250 \pi}=4 \end{array} $
Thickness of wall $=2 \mathrm{~mm}$
Thickness of bottom circular disc $=2 \mathrm{~mm}$
Let the inner radius of cylinder $=r \mathrm{~mm}$ and height of the inner cylinder $=h \mathrm{~mm}$.
$ \Rightarrow \mathrm{V}=\pi r^2 h $
Now, volume of the material used $=$ volume of outer cylinder - volume of the inner cylinder + Volume of the circular disc
$ \begin{gathered} \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi(r+2)^2 h-\pi r^2 h+\pi(r+2)^2 2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h\left\{(r+2)^2-r^2\right\}+2 \pi(r+2)^2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h(4 r+4)+2 \pi(r+2)^2 \\\\ \left\{\because(a+b)^2=a^2+b^2+2 a b\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{2 h(r+1)+(r+2)^2\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi r^2}(r+1)+(r+2)^2\right\} \\\\ \left\{\because \mathrm{V}=\pi r^2 h\right\} \end{gathered} $
For $\mathrm{V}_{\mathrm{m}}$ to be minimum, $\frac{d v_m}{d r}=0$
Differentiating the above equation w.r.t.r,
$ \begin{aligned} \quad & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \frac{d}{d r}\left\{\frac{r+1}{r^2}\right\}+\frac{d}{d r}(r+2)^2\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2(1)-(r+1)(2 r)}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2-2 r^2-2 r}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi\left\{\frac{\mathrm{V}}{\pi r^3}(-r-2)+(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi(r+2)\left(-\frac{\mathrm{V}}{\pi r^3}+1\right) \\\\ \because & \frac{d \mathrm{~V}_m}{d r}=0 \\\\ \Rightarrow & 4 \pi(r+2)\left(\frac{-\mathrm{V}}{\pi r^3}+1\right)=0 \end{aligned} $
Also, given that $\mathrm{V}_{\mathrm{m}}$ is minimum at $r=10 \mathrm{~mm}$
$ \begin{array}{ll} \Rightarrow & 4 \pi(10+2)\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & 48 \pi\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & \frac{\mathrm{V}}{10^3 \pi}=1 \\\\ \Rightarrow & \frac{\mathrm{V}}{250 \pi}=4 \end{array} $
2014
Q439
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
Q440
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.
$2f'\left( c \right) = 3g'\left( c \right)$
2014
Q441
JEE Advanced
Numerical
14 Mar 2026
The slope of the tangent to the curve ${\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2}$ at the point $(1, 3)$ is
Correct Answer: 8
Explanation:
Given curve
$ \begin{aligned} & \left(y-x^5\right)^2=x\left(1+x^2\right)^2 \\\\ & \Rightarrow 2\left(y-x^5\right)\left(\frac{d y}{d x}-5 x^4\right)=\left(1+x^2\right)^2+2 x\left(1+x^2\right) \cdot 2 x \end{aligned} $
Now putting $(1,3)$ in it, we get
$ 2(3-1)\left(\frac{d y}{d x}-5\right)=1\{2(2) 2\}+(1+1)^2 $
$\Rightarrow 4\left(\frac{d y}{d x}-5\right)=8+4 \Rightarrow \frac{d y}{d x}=8$
Thus, the slope at $(1,3)$ is 8 .
$ \begin{aligned} & \left(y-x^5\right)^2=x\left(1+x^2\right)^2 \\\\ & \Rightarrow 2\left(y-x^5\right)\left(\frac{d y}{d x}-5 x^4\right)=\left(1+x^2\right)^2+2 x\left(1+x^2\right) \cdot 2 x \end{aligned} $
Now putting $(1,3)$ in it, we get
$ 2(3-1)\left(\frac{d y}{d x}-5\right)=1\{2(2) 2\}+(1+1)^2 $
$\Rightarrow 4\left(\frac{d y}{d x}-5\right)=8+4 \Rightarrow \frac{d y}{d x}=8$
Thus, the slope at $(1,3)$ is 8 .
2013
Q442
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$
2013
Q443
JEE Mains
MCQ
14 Mar 2026
The real number $k$ for which the equation, $2{x^3} + 3x + k = 0$ has two distinct real roots in $\left[ {0,\,1} \right]$
A.
lies between 1 and 2
B.
lies between 2 and 3
C.
lies between $ - 1$ and 0
D.
does not exist.
2013
Q444
JEE Advanced
MCQ
14 Mar 2026
Let $f:\left[ {0,1} \right] \to R$ (the set of all real numbers) be a function. Suppose the function $f$ is twice differentiable,
$f(0) = f(1)=0$ and satisfies $f''\left( x \right) - 2f'\left( x \right) + f\left( x \right) \ge .{e^x},x \in \left[ {0,1} \right]$.
$f(0) = f(1)=0$ and satisfies $f''\left( x \right) - 2f'\left( x \right) + f\left( x \right) \ge .{e^x},x \in \left[ {0,1} \right]$.
Which of the following is true for $0 < x < 1?$
A.
$0 < f\left( x \right) < \infty $
B.
$ - {1 \over 2} < f\left( x \right) < {1 \over 2}$
C.
$ - {1 \over 4} < f\left( x \right) < 1$
D.
$ - \infty < f\left( x \right) < 0$
2013
Q445
JEE Advanced
MCQ
14 Mar 2026
Let $f:\left[ {0,1} \right] \to R$ (the set of all real numbers) be a function. Suppose the function $f$ is twice differentiable,
$f(0) = f(1)=0$ and satisfies $f''\left( x \right) - 2f'\left( x \right) + f\left( x \right) \ge .{e^x},x \in \left[ {0,1} \right]$.
$f(0) = f(1)=0$ and satisfies $f''\left( x \right) - 2f'\left( x \right) + f\left( x \right) \ge .{e^x},x \in \left[ {0,1} \right]$.
If the function ${e^{ - x}}f\left( x \right)$ assumes its minimum in the interval $\left[ {0,1} \right]$ at $x = {1 \over 4}$, which of the following is true?
A.
$f'\left( x \right) < f\left( x \right),{1 \over 4} < x < {3 \over 4}$
B.
$f'\left( x \right) > f\left( x \right),0 < x < {1 \over 4}$
C.
$f'\left( x \right) < f\left( x \right),0 < x < {1 \over 4}$
D.
$f'\left( x \right) < f\left( x \right),{3 \over 4} < x < 1$
2013
Q446
JEE Advanced
MSQ
14 Mar 2026
The function $f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|$ has a local minimum or a local maximum at x =
A.
$-$2
B.
${{ - 2} \over 3}$
C.
2
D.
${{ 2} \over 3}$
2013
Q447
JEE Advanced
MSQ
14 Mar 2026
A rectangular sheet of fixed perimeter with sides having their lengths in the ratio $8:15$ is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is $100$, the resulting box has maximum volume. Then the lengths of the vsides of the rectangular sheet are
A.
$24$
B.
$32$
C.
$45$
D.
$60$
2012
Q448
JEE Mains
MCQ
14 Mar 2026
A line is drawn through the point $(1, 2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a triangle $OPQ,$ where $O$ is the origin. If the area of the triangle $OPQ$ is least, then the slope of the line $PQ$ is :
A.
$-{1 \over 4}$
B.
$-4$
C.
$-2$
D.
$-{1 \over 2}$
2012
Q449
JEE Mains
MCQ
14 Mar 2026
Let $a,b \in R$ be such that the function $f$ given by $f\left( x \right) = In\left| x \right| + b{x^2} + ax,\,x \ne 0$ has extreme values at $x=-1$ and $x=2$
Statement-1 : $f$ has local maximum at $x=-1$ and at $x=2$.
Statement-2 : $a = {1 \over 2}$ and $b = {-1 \over 4}$
A.
Statement - 1 is false, Statement - 2 is true.
B.
Statement - 1 is true , Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.
C.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
D.
Statement - 1 is true, Statement - 2 is false.
2012
Q450
JEE Mains
MCQ
14 Mar 2026
A spherical balloon is filled with $4500\pi $ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72\pi $ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $49$ minutes after the leakage began is :
A.
${{9 \over 7}}$
B.
${{7 \over 9}}$
C.
${{2 \over 9}}$
D.
${{9 \over 2}}$