Application of Derivatives — Page 13 | JEE Mathematics

Q121

The number of real solutions of

{x^7} + 5{x^3} + 3x + 1 = 0

is equal to ____________.

A 0

B 1

C 3

D 5

Correct Answer

Option B

Solution

$f^{\prime}(x)=7 x^{6}+15 x^{2}+3>0 \,\,\forall\,\, x \in R$ $f(x)$ is always increasing So clearly it intersects x-axis at only one point

Q122

The sum of the absolute minimum and the absolute maximum values of the function f(x) = |3x

-

x2 + 2|

-

x in the interval [

-

1, 2] is :

A

{{\sqrt {17} + 3} \over 2}

B

{{\sqrt {17} + 5} \over 2}

C 5

D

{{9 - \sqrt {17} } \over 2}

Correct Answer

Option A

Solution

$f(x)=\left|x^2-3 x-2\right|-x \forall x \in[-1,2]$ $\Rightarrow f(x)=\left\{\begin{array}{l}x^2-4 x-2 \text{ if }-1 \leq x<\dfrac{3-\sqrt{17}}{2} \\ -x^2+2 x+2 \text{ if } \dfrac{3-\sqrt{17}}{2} \leq x \leq 2\end{array}\right.$ $f(x)_{\max }=3$ $f(x)_{\min }=f\left(\dfrac{3-\sqrt{17}}{2}\right)$ $=\dfrac{\sqrt{17}-3}{2}$

Q123

Let S be the set of all the natural numbers, for which the line

{x \over a} + {y \over b} = 2

is a tangent to the curve

{\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2

at the point (a, b), ab

\ne

0. Then :

A S =

\phi

B n(S) = 1

C S = {2k : k

\in

N}

D S = N

Correct Answer

Option D

Solution

{\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2

Differentiating both sides with respect to x, we get

\Rightarrow n\,.\,{\left( {{x \over a}} \right)^{n - a}}\,.\,{1 \over a} + n\,.\,{\left( {{y \over b}} \right)^{n - 1}}\,.\,{1 \over b}\,.\,{{dy} \over {dx}} = 0

\Rightarrow {{dy} \over {dx}} = - {{{1 \over a}\,.\,{{\left( {{x \over a}} \right)}^{n - 1}}} \over {{1 \over b}{{\left( {{y \over b}} \right)}^{n - 1}}}}

= - {b \over a}{\left( {{{xb} \over {ya}}} \right)^{n - 1}}

Now,

{{dy} \over {dx}}

at (a, b)

= - {b \over a}{\left[ {{{ab} \over {ba}}} \right]^{n - 1}} = - {b \over a}

Equation of tangent at (a, b) is,

(y - b) = - {b \over a}(x - a)

\Rightarrow {{y - b} \over b} = - {{x - a} \over a}

\Rightarrow {y \over b} - 1 = - {x \over a} + 1

\Rightarrow {x \over a} + {y \over b} = 2

$\therefore$ It is tangent for all value of n.

Q124

Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :

A 2 : 5

B 19 : 45

C 3 : 8

D 19 : 15

Correct Answer

Option B

Solution

$\because$

{s_1} + {s_2} = k

76{x^2} + 3\pi {r^2} = k

$\therefore$

152x{{dx} \over {dr}} + 6\pi r = 0

$\therefore$

{{dx} \over {dr}} = {{ - 6\pi r} \over {152x}}

Now

V = 40{x^3} + {2 \over 3}\pi {r^3}

$\therefore$

{{dv} \over {dr}} = 120{x^2}\,.\,{{dx} \over {dr}} + 2\pi {r^2} = 0

\Rightarrow 120{x^2}\,.\,\left( {{{ - 6\pi } \over {152}}{r \over x}} \right) + 2\pi {r^2} = 0

\Rightarrow 120\left( {{x \over r}} \right) = 2\pi \left( {{{152} \over {6\pi }}} \right)

\Rightarrow \left( {{x \over r}} \right) = {{152} \over 3}{1 \over {120}} = {{19} \over {45}}

Q125

Water is being filled at the rate of 1 cm3 / sec in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate (in cm2 / sec) at which the wet conical surface area of the vessel increases is

A 5

B

{{\sqrt {21} } \over 5}

C

{{\sqrt {26} } \over 5}

D

{{\sqrt {26} } \over {10}}

Correct Answer

Option C

Solution

$\because$

V = {1 \over 3}\pi {r^2}h

and

{r \over h} = {7 \over {35}} = {1 \over 5}

\Rightarrow V = {1 \over {75}}\pi {h^3}

{{dV} \over {dt}} = {1 \over {25}}\pi {h^2}{{dh} \over {dt}} = 1

\Rightarrow {{dh} \over {dt}} = {{25} \over {\pi {h^2}}}

Now,

S = \pi rl = \pi \left( {{h \over 5}} \right)\sqrt {{h^2} + {{{h^2}} \over {25}}} = {\pi \over {25}}\sqrt {26} {h^2}

\Rightarrow {{dS} \over {dt}} = {{2\sqrt {26} \pi h} \over {25}}\,.\,{{dh} \over {dt}} = {{2\sqrt {26} } \over h}

\Rightarrow {{dS} \over {d{t_{(h = 10)}}}} = {{\sqrt {26} } \over 5}

Q126

If the angle made by the tangent at the point (x0, y0) on the curve

x = 12(t + \sin t\cos t)

,

y = 12{(1 + \sin t)^2}

, $$0 0 is equal to:

A

6\left( {3 + 2\sqrt 2 } \right)

B

3\left( {7 + 4\sqrt 3 } \right)

C 27

D 48

Correct Answer

Option C

Solution

$\because$

{{dy} \over {dx}} = {{24(1 + \sin t)\cos t} \over {12(1 + \cos 2t)}} = {{1 + \sin t} \over {\cos t}} = \tan \left( {{\pi \over 4} + {t \over 2}} \right)

$\because$

{{dy} \over {d{x_{({x_0},{y_0})}}}} = \sqrt 3 = \tan \left( {{\pi \over 4} + {t \over 2}} \right)

\Rightarrow t = {\pi \over 6}

So,

{y_{0\left( {at\,t = {\pi \over 6}} \right)}} = 12{\left( {1 + \sin {\pi \over 6}} \right)^2} = 27

Q127

The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by

{{{x^2}} \over {xy - {x^2}{y^2} - 1}}

. If the curve passes through the point (1, 1), then e . y(e) is equal to

A

{{1 - \tan (1)} \over {1 + \tan (1)}}

B tan(1)

C 1

D

{{1 + \tan (1)} \over {1 - \tan (1)}}

Correct Answer

Option D

Solution

$\because$

- {{dx} \over {dy}} = {{{x^2}} \over {xy - {x^2}{y^2} - 1}}

$\therefore$

{{dy} \over {dx}} = {{{x^2}{y^2} - xy + 1} \over {{x^2}}}

Let

xy = v \Rightarrow y + x{{dy} \over {dx}} = {{dv} \over {dx}}

$\therefore$

{{dv} \over {dx}} - y = {{({v^2} - v + 1)y} \over v}

$\therefore$

{{dv} \over {dx}} = {{{v^2} + 1} \over x}

$\because$

y(1) = 1 \Rightarrow {\tan ^{ - 1}}(xy) = \ln x + {\tan ^{ - 1}}(1)

Put

x = e

and

y = y(e)

we get

{\tan ^{ - 1}}(e\,.\,y(e)) = 1 + {\tan ^{ - 1}}1

{\tan ^{ - 1}}(e\,.\,y(e)) - {\tan ^{ - 1}}1 = 1

$\therefore$

e(y(e)) = {{1 + \tan (1)} \over {1 - \tan (1)}}

Q128

Let

\lambda

^ *

be the largest value of

\lambda

for which the function

{f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36x + 48

is increasing for all x

\in

R. Then

{f_{{\lambda ^ * }}}(1) + {f_{{\lambda ^ * }}}( - 1)

is equal to :

A 36

B 48

C 64

D 72

Correct Answer

Option D

Solution

$\because$

{f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36\lambda + 48

$\therefore$

f{'_\lambda }(x) = 12(\lambda {x^2} - 6\lambda x + 3)

For

{f_\lambda }(x)

increasing :

{(6\lambda )^2} - 12\lambda \le 0

$\therefore$

\lambda \in \left[ {0,\,{1 \over 3}} \right]

$\therefore$

\lambda^ * = {1 \over 3}

Now,

f_\lambda ^*(x) = {4 \over 3}{x^3} - 12{x^2} + 36x + 48

$\therefore$

f_\lambda ^*(1) + f_\lambda ^*( - 1) = 73{1 \over 2} - 1{1 \over 2} = 72

Q129

The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :

A 9

B 10

C 11

D 12

Correct Answer

Option A

Solution

We know, Surface area of balloon (s) = 4 $\pi$ r2 $\therefore$

{{ds} \over {dt}} = {d \over {dt}}(4\pi {r^2})

\Rightarrow {{ds} \over {dt}} = 4\pi (2r) \times {{dr} \over {dt}}

\Rightarrow {{ds} \over {dt}} = 8\pi r \times {{dr} \over {dt}}

Given that, surface area of balloon is increasing in constant rate. $\therefore$

{{ds} \over {dt}}

= constant = k (Assume)

\Rightarrow k = 8\pi r\,.\,{{dr} \over {dt}}

\Rightarrow \int {k\,dt = \int {8\pi r\,dr} }

\Rightarrow kt = 8\pi \times {{{r^2}} \over 2} + c

\Rightarrow kt = 4\pi {r^2} + c

..... (1) Given at t = 0, radius r = 3 So,

0 = 4\pi ({3^2}) + c

\Rightarrow c = - 36\pi

$\therefore$ Equation (1) becomes

kt = 4\pi {r^2} - 36\pi

Also given, at t = 5, radius r = 7 $\therefore$

k(5) = 4\pi {(7)^2} - 36

\Rightarrow k = 32\pi

$\therefore$ Equation (1) is

32\pi t = 4\pi {r^2} - 36\pi

Now at

t = 9

\Rightarrow 32\pi (9) = 4\pi {r^2} - 36\pi

\Rightarrow 8 \times 9 = {r^2} - 9

\Rightarrow {r^2} = 81 \Rightarrow r = 9

Q130

For the function

f(x) = 4{\log _e}(x - 1) - 2{x^2} + 4x + 5,\,x > 1

, which one of the following is NOT correct?

A f is increasing in (1, 2) and decreasing in (2,

\infty

)

B f(x) =

-

1 has exactly two solutions

C

f'(e) - f''(2) < 0

D f(x) = 0 has a root in the interval (e, e + 1)

Correct Answer

Option C

Solution

Lets draw the curve $y=f(x)=4 \log _e(x-1)-2 x^2$ $+4 x+5, x>1$

\begin{aligned} &f(x)=4 \log _e(x-1)-2 x^2+4 x+5, x > 1 \\\\ &f^{\prime}(x)=\frac{4}{x-1}-4(x-1) \\\\ &f^{\prime \prime}(x)=\frac{-4}{(x-1)^2}-4\\\\ &\text{For} 1 < x< 2 \Rightarrow f^{\prime}(x) > 0 \\\\ &\text{For}~ x > 2 \Rightarrow f^{\prime}(x)<0 \text{ (option } A \text{ is correct) } \\\\ &f(x)=-1 \text{ has two solution (option } B \text{ is correct) } \\\\ &f(e) > 0 \\\\ &f(e+1) < 0 \\\\ &f(e) \cdot f(e+1)<0(\text{ option } D \text{ is correct) } \\\\ &f^{\prime}(e)-f^{\prime \prime}(2)=\frac{4}{e-1}-4(e-1)+8 > 0 \end{aligned}

(option C is incorrect)