RD Sharma Solutions Class 12 Mathematics Chapter 17 MCQ

# RD Sharma Solutions Class 12 Mathematics Chapter 17 MCQ

Edited By Satyajeet Kumar | Updated on Jan 21, 2022 02:57 PM IST

Class 12 RD Sharma chapter 17 exercise MCQ solution is a sought-after book that is used by almost all students in class 12. The RD Sharma class 12th exercise MCQ covers questions from the entire NCERT maths book and provides intensive knowledge and essential information for all concepts and chapters. RD Sharma Solutions It is designed to aid students to perform well in any school and entrance exam and expand their knowledge on the maths subject.

## Maxima and Minima Excercise: MCQ

Maxima and Minima exercise MCQ question 1

Answer: option (b) $e^{\frac{1}{e}}$
Hint: For local maxima or minima, we must have.$\Rightarrow \frac{dy}{dx}=0$
Given:$y = x^{\frac{1}{x}}$
Solution:
$y = x^{\frac{1}{x}}$
$\log y =\frac{1}{x} \log _e x$
$\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \cdot \frac{1}{x}+\log _{e} x\left(\frac{-1}{x^{2}}\right)$
For maximum or minimum $\Rightarrow \frac{dy}{dx}=0$
\begin{aligned} &\frac{d y}{d x}=y\left(\frac{1}{x^{2}}-\frac{1}{x^{2}}\left(\log _{e} x\right)\right)=0 \\ &\frac{1}{x^{2}}=\frac{1}{x^{2}} \log _{e} x \\ &\frac{1}{x^{2}}\left(\log _{e} x-1\right)=0 \end{aligned}
$\log _ex-1$
x = e is maximum value
Hence,
$x^{\frac{1}{x}}=e^{\frac{1}{e}}$
Maximum Value =$f(e)=e^{\frac{1}{e}}$.

Maxima and Minima exercise Multiple choice question, question 2

Answer: option (b) $ab\geq \frac{c^2}{4}$
Hint: For local maxima or minima, we must have f'(x) =0.
Given:$ax+\frac{b}{x}\geq c$
Solution:
We have,
$ax+\frac{b}{x}\geq c$
Minimum value of $ax+\frac{b}{x}= c$
Now,
$f(x)=ax+\frac{b}{x}$
$f'(x)=a-\frac{b}{x^2}$
$f'(x)=0$
$a-\frac{b}{x^2}=0$
$ax^2-b=0$
$x^2=\frac{b}{a}$
$x=\pm \sqrt{\frac{b}{a}}$
$f''(x)=\frac{2b}{x^3}$
Taking $x=\sqrt{\frac{x}{b}}$
\begin{aligned} &f^{\prime \prime}\left(\sqrt{\frac{b}{a}}\right)=\frac{2 b}{\left(\sqrt{\frac{b}{a}}\right)^{3}} \\ &f^{\prime \prime}\left(\sqrt{\frac{b}{a}}\right)=\frac{2 b(a)^{\frac{3}{2}}}{(b)^{\frac{3}{2}}}>0 \end{aligned}
So, $x=\sqrt{\frac{x}{b}}$ is a point of local minima.
\begin{aligned} &f\left(\frac{\sqrt{b}}{\sqrt{a}}\right)=a\left(\frac{\sqrt{b}}{\sqrt{a}}\right)+\frac{b}{\left(\frac{\sqrt{b}}{\sqrt{a}}\right)} \geq c \\ &f\left(\frac{\sqrt{b}}{\sqrt{a}}\right)=\sqrt{a b}+\sqrt{a b} \geq c \end{aligned}
$2\sqrt{ab}\geq c$
$\sqrt{ab}\geq \frac{c}{2}$
$ab\geq \frac{c^2}{4}$
Option (b) is correct.

Maxima and Minima exercise MCQ question 3

Hint: For local maxima or minima, we must have f'(x) =0.
Given:$f(x)=\frac{x}{\log _e x}$
Solution:
We have,
\begin{aligned} &f(x)=\frac{x}{\log _{e} x} \\ &f^{\prime}(x)=\frac{\log _{e} x-1}{\left(\log _{e} x\right)^{2}} \end{aligned}
For maxima and minima $f'(x)=0$
$\Rightarrow \frac{\log _e x-1}{(\log _ex)^2}=0$
$\Rightarrow \log _e x-1=0$
$\Rightarrow \log _e x=10$
$\Rightarrow x=e$
Now,
\begin{aligned} &f^{\prime \prime}(x)=\frac{-1}{x\left(\log _{e} x\right)^{2}}+\frac{2}{x\left(\log _{e} x\right)^{3}} \\ &f^{\prime \prime}(e)=\frac{-1}{e}+\frac{2}{e}=\frac{1}{e}>0 \end{aligned}
So, $x =e$ is a point of local minima.
Minimum Value of $f(e)=\frac{e}{\log _e e}=e$

Maxima and Minima exercise MCQ question 4

Answer: (d)Maximum Value < Minimum Value
Hint: For local maxima or minima, we must havef'(x) =0.
Given:$f(x)=x+\frac{1}{x}$
Solution:
We have,
$f(x)=x+\frac{1}{x}$
$f'(x)=1-\frac{1}{x^2}$
For maxima and minima $f'(x)=0$
$\Rightarrow 1-\frac{1}{x^2}=0$
$\Rightarrow x^2-1=0$
$\Rightarrow x^2=1$
$\Rightarrow x=\pm 1$
Now,
$f''(x)=\frac{2}{x^3}$
$f''(1)=2>0$
So, x =1 is a point of local minima.
Also, $f''(-1)=-2<0$
So, x = -1 is a point of local maxima.
$\therefore$ Maximum Value < Minimum Value

Maxima and Minima exercise MCQ question 5
Answer: Option (b) a minimum at x = 1

Hint: For local maxima or minima, we must have f'(x) =0 .
Given: $f(x)=x^3+3x^2-9x+2$
Solution:
We have,
$f(x)=x^3+3x^2-9x+2$
$f'(x)=3x^2+6x-9$
For maxima and minima $f'(x)=0$
$\Rightarrow 3x^2+6x-9=0$
$\Rightarrow x^2+2x-3=0$
$\Rightarrow (x+3)(x-1)=0$
$\Rightarrow x=-3,1$
Now,
$f''(x)=6x+6$
$f''(1)=12>0$
So, x =1 is a point of local minima.
Also, $f''(-3)=-18 +6=-12<0$
So, x = -3 is a point of local maxima.

Maxima and Minima exercise MCQ question 6
Answer: option (b) 4

Hint: For local maxima or minima, we must have f'(x) =0.
Given:$f(x)=x^4-x^2-2x+6$
Solution:
We have,
$f(x)=x^4-x^2-2x+6$
$f'(x)=4x^3-2x-2$
$f'(x)=(x-1)(4x^2+4x+2)$
For maxima and minima $f'(x)=0$
$(x-1)(4x^2+4x+2)=0$
$\Rightarrow x-1=0$
$\Rightarrow x=1$
Now,
$f''(x)=12x^2-2$
$f''(1)=12-2=10>0$
So, x = 1 is a point of local minima.
The local minimum value is given by $f(1)=1-1-2+6 =4$

Maxima and Minima exercise MCQ question 7

Answer: option(a) $\frac{1}{2}$
Hint: For local maxima or minima, we must have f'(x) =0.
Given:$f(x)=x-x^2$
Solution:
We have,
$f(x)=x-x^2$
$f'(x)=1-2x$
For maxima and minima $f'(x)=0$
$\Rightarrow 1-2x=0$
$\Rightarrow x=\frac{1}{2}$
Now,
$f''(x)=-2<0$
So, $x=\frac{1}{2}$ is a point of local maxima.
The local maximum value is given by $f\left (\frac{1}{2} \right )=\frac{1}{2}-\frac{1}{4}=\frac{1}{2}$

Maxima and Minima exercise MCQ question 8
Answer: option(a) $\frac{a+b+c}{3}$

Hint: For local maxima or minima, we must have f'(x) =0.
Given:$f(x)=(x-a)^2+(x-b)^2+(x-c)^2$
Solution:
We have,
$f(x)=(x-a)^2+(x-b)^2+(x-c)^2$
$f'(x)=2(x-a)+2(x-b)+2(x-c)$
For maxima and minima $f'(x)=0$
$\Rightarrow 2(x-a)+2(x-b)+2(x-c)=0$
$\Rightarrow 2x-2a+2x-2b+2x-2c=0$
$\Rightarrow 6x-2a-2b-2c=0$
$\Rightarrow 6x=2(a+b+c)$
$\Rightarrow x=\frac{(a+b+c)}{3}$
Now,
$f''(x)=2+2+2=6>0$
So,$x=\frac{a+b+c}{3}$ is a point of local minima.

Maxima and Minima exercise MCQ question 9.

Answer: option (b) $\frac{1}{2}$
Hint: For local maxima or minima, we must have f'(x) =0.
Given:$f(x)=\frac{1}{x}+\frac{1}{y}$
Solution:
Let the two non-zero numbers be x and y.
x + y =8
y = 8 -x ...(i)
We have,
$f(x)=\frac{1}{x}+\frac{1}{y}$
$f(x)=\frac{1}{x}+\frac{1}{(8-x)}$ …(ii) From Equation (i)
On differentiating w.r.t x
$f'(x)=\frac{-1}{x^2}+\frac{1}{(8-x)^2}$
for maxima and minima$f'(x)=0$
$\Rightarrow \frac{-1}{x^2}+\frac{1}{(8-x)^2}=0$
$\Rightarrow \frac{-(8-x)^2+x^2}{x^2(8-x)^2}=0$
$\Rightarrow -64-x^2+16x+x^2=0$
$\Rightarrow 16x=64$
$\Rightarrow x=4$
Now, again differentiating (ii) w.r.t x
\begin{aligned} &f^{\prime \prime}(x)=\frac{2}{x^{3}}-\frac{2}{(8-x)^{3}} \\ &f^{\prime \prime}(4)=\frac{2}{4^{3}}-\frac{2}{(8-4)^{3}} \\ &f^{\prime \prime}(4)=\frac{2}{64}-\frac{2}{64}=0 \end{aligned}
Minimum Value =$\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$

Maxima and Minima exercise MCQ question 10.

Hint: For local maxima or minima, we must have f'(x) =0.
Given:$f(x)=\sum_{r-1}^{5}(x-r)^2$
Solution:
We have,
$f(x)=\sum_{r-1}^{5}(x-r)^2$
\begin{aligned} &f(x)=(x-1)^{2}+(x-2)^{2}+(x-3)^{2}+(x-4)^{2}+(x-5)^{2} \\ &f^{\prime}(x)=2(x-1+x-2+x-3+x-4+x-5) \\ &f^{\prime}(x)=2(5 x-15) \end{aligned}
For maxima and minima
$f'(x)=0$
$\Rightarrow 2(5x-15)=0$
$\Rightarrow 5x-15=0$
$\Rightarrow 5x=15$
$\Rightarrow x=3$
Now,
$f''(x)=10>0$
So, x =3 is a point of local minima.

Maxima and Minima exercise MCQ question 11

Answer: option (d) None of these.
Hint: For local maxima or minima, we must have f'(x) =0.
Given:$f(x)=2\sin 3x+3\cos 3x$
Solution:
We have,
$f(x)=2\sin 3x+3\cos 3x$
$f'(x)=6 \cos 3x-9 \sin 3x$
For maxima and minima $f'(x)=0$
$\Rightarrow 6 \cos 3x-9 \sin 3x =0$
$\Rightarrow 6 \cos 3x=9 \sin 3x$
$\Rightarrow \frac{\sin 3x}{\cos 3x}=\frac{2}{3}$
$\Rightarrow \tan 3x=\frac{2}{3}$
At $x=\frac{5\pi }{6}$, $\tan 3x=\tan \frac{5\pi }{2}=\tan \frac{\pi}{2}$ = not defined
So, $\tan 3x$ is not defined $\left [\tan 3x\neq \frac{2}{3} \right ]$
Thus, $x=\frac{5}{6}$ is not a critical point.

Maxima and Minima exercise MCQ question 12

Answer: option(c) x =1

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=x^2+x+1$
Solution:
We have,
$f(x)=x^2+x+1$
$f(x)=2x+1$
For maxima and minima $f'(x)=0$
$\Rightarrow 2x+1$
$\Rightarrow x=\frac{-1}{2}\euro [0,1]$
At extreme points,
$f(0)=1>0$
$f(1)=1+1+1=3>0$
So, x = 1 is a local minima.

Maxima and Minima exercise MCQ question 13.

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=x^3-18x^2+96x$
Solution:
We have,
$f(x)=x^3-18x^2+96x$
$\Rightarrow f'(x)=3x^2-36x+96$
For maxima and minima $f'(x)=0$
$\Rightarrow 3x^2-36x+96=0$
$\Rightarrow x^2-12x+32=0$
$\Rightarrow (x-4)(x-8)=0$
$\Rightarrow x=4,8$
At x = 4, $f(4)=(4)^3-18(4)^2+96(4)=64-288+384=160$
At x = 8, $f(4)=(8)^3-18(8)^2+96(8)=512-1152+768=128$
Now at the extreme points of [0, 9]
$f(0)=(0)^3-18(0)^2+96(0)=0$
$f(9)=(9)^3-18(9)^2+96(9)=729-1458=864=135$
Hence, 0 is the minimum value in the range [0, 9].

Maxima and Minima exercise MCQ question 14.

Answer: $\frac{1}{4}$
Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=\frac{x}{4-x+x^2}$
Solution:
We have,
\begin{aligned} &f(x)=\frac{x}{4-x+x^{2}} \\ &f^{\prime}(x)=\frac{4-x+x^{2}-x(-1+2 x)}{\left(4-x+x^{2}\right)^{2}} \end{aligned}
For maxima and minima $f'(x)=0$
$\Rightarrow \frac{4-x+x^{2}-x(-1+2 x)}{\left(4-x+x^{2}\right)^{2}}$
$\Rightarrow 4-x+x^2-x(-1+2x)=0$
$\Rightarrow 4-x^2=0$
$\Rightarrow x^2=4$
$\Rightarrow x=\pm 2\euro [-1,1]$
$f(-1)=\frac{-1}{4+1+1}=\frac{-1}{6}$
$f(1)=\frac{1}{4-1+1}=\frac{1}{4}$
Hence, the maximum value is $\frac{1}{4}$.
Note: option is not matching with the answer given in the book.

Maxima and Minima exercise MCQ question 15

Hint: Using Distance Formula, calculate the value of variables.
Given:$y^2=2x$
Solution:
Let the required point be (x,y) which is nearest to (2,1).
$y^2=2x$
$\Rightarrow x=\frac{y^2}{4}$ …(i)
Now,
$d=\sqrt{(x-2)^2+(y-1)^2}$
Squaring on both sides, we get
\begin{aligned} &d^{2}=(x-2)^{2}+(y-1)^{2} \\ &d^{2}=\left(\frac{y^{2}}{4}-2\right)^{2}+(y-1)^{2} \\ &d^{2}=\frac{y^{4}}{16}+4-y^{2}+y^{2}+1-2 y \end{aligned}
Now,
$Z=d^2=\frac{y^2}{16}+5-2y$
$\frac{dZ}{dy}=\frac{y^3}{4}-2$ …(ii)
For extrema$\frac{dZ}{dy}=0$
$\Rightarrow \frac{y^3}{4}-2=0$
$\Rightarrow \frac{y^3}{4}=2$
$\Rightarrow y^3=8$
$\Rightarrow y=2$
Substitute the value in Equation (i),
$x=\frac{2^2}{4}=1$
Now, differentiating (ii) w.r.t y
\begin{aligned} &\frac{d^{2} Z}{d y^{2}}=\frac{3 y^{2}}{4} \\ &\frac{d^{2} Z}{d y^{2}}=\frac{3(2)^{2}}{4}=3>0 \end{aligned}
The nearest point is (1, 2).

Maxima and Minima exercise MCQ question 16.

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given: x + y = 8
Solution:
We have,
$x+y =8\Rightarrow y=8-x$ …(i)
Let $f(x)=xy$
$\Rightarrow f(x)=x(8-x)$ (From Equation i)
$\Rightarrow f'(x)=8-2x$
For maxima and minima $f'(x)=0$
$\Rightarrow 8-2x=0$
$\Rightarrow 8=2x$
$\Rightarrow x=4$
$\Rightarrow y=8-x=8-4=4$
Now,
$f''(x)=-2$
$f''(4)=-2<0$
So, x = 4 is the local maxima.
Hence, $f(4)=4 \times 4=16$

Maxima and Minima exercise MCQ question 17.

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=x^3-6x^2+9x$
Solution:
We have,
$f(x)=x^3-6x^2+9x$
$\Rightarrow f'(x)=3x^2-12x+9$
For maxima and minima $f'(x)=0$
$\Rightarrow 3x^2-12x+9=0$
$\Rightarrow x^2-4x+3=0$
$\Rightarrow (x-1)(x-3)=0$
$\Rightarrow x=1,3$
Now,
$f(1)=1^3-6(1)^2+9(1)=1-6+9=4$
$f(3)=3^3-6(2)^2+9(6)=27-54+27=0$
And at the extreme point of [0, 6]
$f(0)=0^3-6(0)^2+9(0)=0$
$f(6)=6^3-6(6)^2+9(6)=216-216+54=54$
The least and greatest values of $f(x)=x^3-6x^2+9x$ in (0 ,6) are 0 and 54.
Note: Option is not matching with the answer, given in the book.

Maxima and Minima exercise MCQ question 18.

### Answer: option(c) $\frac{\pi}{6}$

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=\sin x+\sqrt{3}\cos x$
Solution:
We have,
$f(x)=\sin x+\sqrt{3}\cos x$
$f'(x)=\cos x-\sqrt{3}\sin x$
For maxima and minima $f'(x)=0$
$\Rightarrow \cos x-\sqrt{3}\sin x=0$
$\Rightarrow \cos x=\sqrt{3}\sin x$
$\Rightarrow \tan x=\frac{1}{\sqrt{3}}$
$\Rightarrow x =\frac{\pi}{6}$
Now,
\begin{aligned} &f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x \\ &f^{\prime \prime}\left(\frac{\pi}{6}\right)=-\sin \left(\frac{\pi}{6}\right)-\sqrt{3} \cos \left(\frac{\pi}{6}\right) \\ &f^{\prime \prime}\left(\frac{\pi}{6}\right)=-\frac{1}{2}-\frac{3}{2}=-2<0 \end{aligned}
So $x=\frac{\pi}{6}$ is local maxima.

Maxima and Minima exercise MCQ question 19

Answer: option(d) $\frac{2}{3}$
Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$h=R+\sqrt{R^2-r^2}$
Solution:
We have,
Solution:

$AB=2r$
$OC=r$
$CD =R+h$
We have,
$h=R+\sqrt{R^2-r^2}$
$h-R=\sqrt{R^2-r^2}$
Squaring on both sides,
$(h-R)^=(\sqrt{R^2-r^2})^2$
$h^2+R^2-2hR=R^2-r^2$
$r^2=2hR-h^2$ …(i)
Now,
$V=\frac{1}{3}\pi r ^2h$
$V=\frac{\pi}{3}(2hR-h^2)h$ …using equation(i)
\begin{aligned} &V=\frac{\pi}{3}\left(2 h^{2} R-h^{3}\right) \\ &\frac{d V}{d h}=\frac{\pi}{3}\left(4 h R-3 h^{2}\right) \end{aligned}
For maxima and minima $\frac{dV}{dh}=0$
\begin{aligned} &\Rightarrow \frac{\pi}{3}\left(4 h R-3 h^{2}\right)=0 \\ &\Rightarrow 4 h R-3 h^{2}=0 \\ &\Rightarrow 3 h^{2}=4 h R \\ &\Rightarrow h=\frac{4 R}{3} \end{aligned}
Now,
$\frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}(4 R-6 h)=\frac{\pi}{3}\left(4 R-6 \times \frac{4 R}{3}\right)=\frac{\pi}{3} 4 R\left(1-\frac{6}{3}\right)=\frac{\pi}{3} 4 R\left(\frac{-3}{3}\right)=-\frac{4 \pi R}{3}<0$
Volume is maximum, when $h=\frac{4R}{3}$
$h=\frac{2(2R)}{3}$
$\frac{h}{2R}=\frac{2}{3}$

Maxima and Minima exercise MCQ question 20

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=x^2+\frac{250}{x}$
Solution:
We have,
$f(x)=x^2+\frac{250}{x}$
$f'(x)=2x-\frac{250}{x^2}$
For maxima and minima $f'(x)=0$
$\Rightarrow 2x-\frac{250}{x^2}=0$
$\Rightarrow 2x^3=250$
$\Rightarrow x^3=125$
$\Rightarrow x=5$
Now,
$f^{\prime \prime}(x)=2+\frac{500}{x^{3}}=2+\frac{500}{5^{3}}=2+\frac{500}{125}=\frac{750}{125}=6>0$
So, x = 5 is a local minima.
$f(x)_{min}=(5)^2+\frac{250}{5}=25+50=75$

Maxima and Minima exercise MCQ question 21.

Answer: option(d) None of these
Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=x+\frac{1}{x}$
Solution:
We have,
$f(x)=x+\frac{1}{x}$
$f'(x)=1-\frac{1}{x^2}$
For maxima and minima $f'(x)=0$
$\Rightarrow 1-\frac{1}{x^2}=0$
$\Rightarrow x^2-1=0$
$\Rightarrow x^2=1$
$\Rightarrow x=\pm 1$
$\Rightarrow x= 1(x>0)$
Now,
$f''(x)=\frac{2}{x^3}$
$f''(1)=\frac{2}{1^3}=2>0$
So, x = 1 is a local minima.

Maxima and Minima exercise MCQ question 22

Answer: option(a) $\frac{4}{3}$
Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=\frac{1}{4x^2+2x+1}$
Solution:
We have,
Maximum Value of $\frac{1}{4x^2+2x+1}$ = Minimum Value of $4x^2+2x+1$
$f(x)=4x^2+2x+1$
$\Rightarrow f'(x) =8x+2$
For maxima and minima $f'(x)=0$
$\Rightarrow 8x+2=0$
$\Rightarrow x=\frac{-2}{8}$
$\Rightarrow x=\frac{-1}{4}$
Now,
$f''(x)=8$
$f''(1)=8>0$
So,$x=\frac{-1}{4}$ is a local minima.
Thus, $\frac{1}{4x^2+2x+1}$ is maximum at $x=\frac{-1}{4}$
Maximum Value of
$\frac{1}{4 x^{2}+2 x+1}=\frac{1}{4\left(\frac{-1}{4}\right)^{2}+2\left(\frac{-1}{4}\right)+1}=\frac{1}{4\left(\frac{1}{16}\right)+2\left(\frac{-1}{4}\right)+1}=\frac{1}{\frac{1}{4}+\frac{-1}{2}+1}=\frac{1}{\frac{-1}{4}+1}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

Maxima and Minima exercise MCQ question 23

Answer: option (b) 2
Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=x+\frac{1}{x}$
Solution:
We have,
xy =1
$\Rightarrow y=\frac{1}{x}$
$f(x)=x+\frac{1}{x}$
$f'(x)=1-\frac{1}{x^2}$
For maxima and minima $f'(x)=0$
$\Rightarrow 1-\frac{1}{x^2}=0$
$\Rightarrow x^2-1=0$
$\Rightarrow x^2=1$
$\Rightarrow x=1$
$\Rightarrow \frac{1}{y}=1$
$\Rightarrow y=1$
Now,
$f''(x)=\frac{2}{x^3}$
$f''(1)=\frac{2}{1^3}=2>0$
So, x = 1 is a local minima.
Minimum Value of $f(1)=1+\frac{1}{1}=1+1=2$

Maxima and Minima exercise MCQ question 24

Answer: option (a) Minimum at $x=\frac{\pi}{2}$

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given: $f(x)=1+2\sin x+3\cos^2x$
Solution:
We have,
$f(x)=1+2\sin x+3\cos^2x$
$f'(x)=2\cos x-6\cos x\sin x$
$f'(x)=2\cos x(1-3 \sin x)$ (i)
For maxima and minima $f'(x)=0$
\begin{aligned} &\Rightarrow 2 \cos x(1-3 \sin x)=0 \\ &\Rightarrow 2 \cos x=0 \underset{\text { or }} \Rightarrow(1-3 \sin x)=0 \end{aligned}
$\Rightarrow \cos x=0\Rightarrow \sin x =\frac{1}{3}$
$\Rightarrow x=\frac{\pi}{2}$ $\Rightarrow x=\sin ^{-1}\left ( \frac{1}{3} \right )$
Now,
\begin{aligned} &f^{\prime}(x)=2 \cos x(-3 \cos x)+(1-3 \sin x)(-2 \sin x) \\ &=-6 \cos ^{2} x-2 \sin x+6 \sin ^{2} x \\ &=-6\left(\cos ^{2} x-\sin ^{2} x\right)-2 \sin x \\ &-6 \cos 2 x-2 \sin x \quad\left[\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right] \end{aligned}
\begin{aligned} &\text { at } x=\frac{\pi}{2}, f^{\prime \prime}\left(\frac{\pi}{2}\right) \\ &=-6 \cos \pi-2 \sin \frac{\pi}{2} \\ &=-6(-1)-2 \times 1 \quad\left[\because \cos \pi=-1, \sin \frac{\pi}{2}=1\right] \\ &-2+6=4>0 \end{aligned}
So, $x=\frac{\pi}{2}$ is local minima
\begin{aligned} &\text { at } x=\sin ^{-1}\left(\frac{1}{3}\right) \\ &f^{\prime \prime}(x)=f^{\prime \prime}\left(\sin ^{-1}\left(\frac{1}{3}\right)\right) \\ &=-2 \sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)-6\left(1-2 \sin ^{2} x\right) \quad\left[\because \cos 2 x=1-2 \sin ^{2} x\right] \end{aligned}
\begin{aligned} &=-2 \times \frac{1}{3}-6+12(\sin x)^{2} \\ &=\frac{-2}{3}-6+12\left(\sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)\right)^{2} \\ &=\frac{-2}{3}-6+12 \times\left(\frac{1}{3}\right)^{2}=\frac{-2}{3}-6+12 \times \frac{1}{9} \end{aligned}
\begin{aligned} &=\frac{-2}{3}-6-\frac{4}{3}=\frac{-2-18-4}{3} \\ &=\frac{-24}{3}=-8 \\ &f^{\prime \prime}(x)<0 \end{aligned}
So, $x=\sin ^{-1} \left (\frac{1}{3} \right )$ is local maxima

Maxima and Minima exercise MCQ question 25

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given: $f(x)=2x^3-15x^2+36x+4$
Solution:
We have,
$f(x)=2x^3-15x^2+36x+4$
$\Rightarrow f'(x)=6x^2-30x+36$
For maxima and minima $f'(x)=0$
$\Rightarrow 6x^2-30x+36=0$
$\Rightarrow x^2-5x+6=0$
$\Rightarrow (x-2)(x-3)=0$
$\Rightarrow x=2,3$
Now,
$f''(x)=12x-30$
$f''(2)=24-30=-6<0$
$f''(3)=36-30=6>0$
So, x= 2 is the local maxima.

Maxima and Minima exercise MCQ question 26

Answer: option(c) $\frac{1}{6}$
Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=\frac{x}{4+x+x^2}$
Solution:
We have,
$f(x)=\frac{x}{4+x+x^2}$
$f'(x)=\frac{4+x+x^2-x(1-2x)}{(4+x+x^2)^2}$
For maxima and minima $f'(x)=0$
$\Rightarrow \frac{4+x+x^2-x(1-2x)}{(4+x+x^2)^2}=0$
\begin{aligned} &\Rightarrow 4+x+x^{2}-x-2 x^{2}=0 \\ &\Rightarrow 4-x^{2}=0 \\ &\Rightarrow x^{2}=4 \\ &\Rightarrow x=\pm 2 \notin[-1,1] \\ &f(1)=\frac{1}{4+1+1}=\frac{1}{6} \\ &f(-1)=\frac{-1}{4-1+1}=\frac{-1}{4} \end{aligned}
So, $\frac{1}{6}$ is the maximum value.

Maxima and Minima exercise MCQ question 27

Answer: option (b) -1
Hint: For maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=2x^3-3x^2-12x+5$
Solution:
We have,
$f(x)=2x^3-3x^2-12x+5$
$f'(x)=6x^2-6x-12$
For maxima and minima $f'(x)=0$
\begin{aligned} &\Rightarrow 6 x^{2}-6 x-12=0 \\ &\Rightarrow x^{2}-x-2=0 \\ &\Rightarrow(x-2)(x+1)=0 \\ &\Rightarrow x=2,-1 \end{aligned}
Now,
$f''(x)=12x-6$
$f''(-1)=12(-1)-6=-12-6=-18<0$
So, x = -1 is the local maxima.
$f''(2)=12(2)-6=24-6=18>0$
So, x = 2 is the local minima.

Maxima and Minima exercise MCQ question 28

Answer: option(c) $\frac{-1}{e}$
Hint: For local maxima or minima, we must have$f'(x)=0$.
Given: $f(x)=x \log _ex$
Solution:
We have,
$f(x)=x \log _ex$
$\Rightarrow f'(x)= \log _ex+1$
For maxima and minima $f'(x)=0$
$\Rightarrow \log _ex+1=0$
$\Rightarrow \log _ex=-1$
$\Rightarrow x=e^{-1}$
Now,
$f''(x)=\frac{1}{x}$
$f^{\prime \prime}\left(e^{-1}\right)=\frac{1}{e^{-1}}=e>0$
So,$x=e^{-1}$is the local minima.
Minimum value of $f(x)=f\left(e^{-1}\right)=e^{-1} \log _{e}\left(e^{-1}\right)=-e^{-1}=\frac{-1}{e}$

Maxima and Minima exercise MCQ question 29

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given: $f(x)=2x^3-21x^2+36x-20$
Solution:
We have,
$f(x)=2x^3-21x^2+36x-20$
$f'(x)=6x^2-42x+36$
For maxima and minima $f'(x)=0$
$\Rightarrow 6x^2-42x+36=0$
$\Rightarrow x^2-7x+6=0$
$\Rightarrow (x-1)(x-6)=0$
$\Rightarrow x=1,6$
Now,
\begin{aligned} &f^{\prime \prime}(x)=12 x-42 \\ &f^{\prime \prime}(1)=12-42=-30<0 \end{aligned}
So, x = 1 is the local maxima.
$f''(6)=72-42=30>0$
So, x = 6 is the local minima.
$f(6)=2(6)^{3}-21(6)^{2}+36(6)-20=432-756+216-20=-128$

Maxima and Minima exercise MCQ question 30

### Answer: option(d) f (x) is an increasing function.

Hint: For local maxima or minima, we must have $f'(x)=0$ .
Given: $f(x)=2x+\cos x$
Solution:
We have,
$f(x)=2x+\cos x$
$\Rightarrow f'x=2-\sin x>0$
Hence, f (x) is an increasing function.

Maxima and Minima exercise MCQ question 31

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$y=x^2-8x+17$
Solution:
We have,
$y=x^2-8x+17$
$\Rightarrow f'(y)=2x-8$
For maxima and minima $f'(x)=0$
$\Rightarrow 2x-8=0$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
$f(4)=4^{2}-8(4)+17=16-32+17=33-32=1>0$
Hence, minimum value f(x) of is 1 at x = 4

Maxima and Minima exercise MCQ question 32
Answer: option(b) $\frac{1}{2}$

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=\sin x \cos x$
Solution:
We have,
$f(x)=\sin x \cos x$
We know that,
\begin{aligned} &\sin 2 x=2 \sin x \cos x \\ &f(x)=\frac{1}{2} \sin 2 x \\ &\Rightarrow f^{\prime}(x)=\frac{1}{2} \frac{d}{d x}(\sin 2 x) \end{aligned}
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{1}{2}(\cos 2 x) \times 2 \\ &\Rightarrow f^{\prime}(x)=\cos 2 x \end{aligned}
For maxima and minima $f'(x)=0$
$\Rightarrow \cos 2x=0$
$\Rightarrow \cos 2x=\cos \frac{\pi}{2}$
$\Rightarrow 2x= \frac{\pi}{2}$
$\Rightarrow x= \frac{\pi}{4}$
\begin{aligned} &f^{\prime \prime}(x)=-2 \sin 2 x \\ &f^{\prime \prime}\left(\frac{\pi}{4}\right)=-2 \sin 2\left(\frac{\pi}{4}\right)=-2 \sin \left(\frac{\pi}{2}\right)=-2<0 \end{aligned}
$=\therefore x= \frac{\pi}{4}$ is local maxima.
$f\left(\frac{\pi}{4}\right)=\sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}$
Hence, maximum value of f (x) is $\frac{1}{2}$

Maxima and Minima exercise MCQ question 33

Answer: ( c) $e^{\frac{1}{e}}$

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=\left (\frac{1}{x} \right )^x$
Solution:
We have,
$f(x)=\left (\frac{1}{x} \right )^x$
$f^{\prime}(x)=\left(\frac{1}{x}\right)^{x}\left(\log \left(\frac{1}{x}\right)-1\right)$
For maxima and minima $f'(x)=0$
$\Rightarrow \left(\frac{1}{x}\right)^{x}\left(\log \left(\frac{1}{x}\right)-1\right)=0$
$\log 1- \log x=1$
$\frac{1}{x}=e$
$x=\frac{1}{e}$

### Maxima and Minima exercise MCQ question 34

Answer: (b) $\frac{1}{e}$
Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$y=x^x$
Solution:
Let us consider
$y=x^x$
Applying log on both sides,
$\log y=x \log x, (x>0)$
On differentiating, we get
\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=1 . \log x+x \cdot \frac{1}{x} \\ &\frac{d y}{d x}=(x)^{x}(1+\log x) \\ &\frac{d y}{d x}=0 \\ &(x)^{x}(1+\log x)=0 \end{aligned}
$1+\log x=0$
$\log x=-1$
$x=e^{-1}$
$x=\frac{1}{e}$
Therefore, the stationary point is $x=\frac{1}{e}$

Maxima and Minima exercise MCQ question 35

Hint: For local maxima or minima, we must have $f'(x)=0$.
Given: Maximum slope of the curve $y=-x^3+3x^2+9x-27$
Solution:
Slope of the curve is given by
$m=\frac{dy}{dx}=-3x^2+6x+9$
$\frac{dm}{dx}=-6x+6$
$\frac{d^2m}{d^2x}=-6<0$ ∀x
So, m is maximum, when x = 1
Putting x = 1 in the slope of the curve, we get, m = -3+6+9=12
The slope is maximum at the point (1, 12).
Maximum value of slope is m = 12

Maxima and Minima exercise MCQ question 36

Answer: (c) One Maximum and One Minimum
Hint: For local maxima or minima, we must have $f'(x)=0$.
Given:$f(x)=2x^3-3x^2-12x+4$
Solution:
Slope of the curve is given by,
$f(x)=2x^3-3x^2-12x+4$
$f'(x)=6x^2-6x-12$
For maxima and minima $f'(x)=0$
\begin{aligned} &\Rightarrow 6 x^{2}-6 x-12=0 \\ &\Rightarrow x^{2}-x-2=0 \\ &\Rightarrow x=\frac{1 \pm \sqrt{1-4 \times 1 \times-2}}{2} \\ &x=\frac{1 \pm 3}{2} \\ &x=2,-1 \\ &f^{\prime \prime}(x)=12 x-6 \\ &f^{\prime \prime}(2)=12(2)-6=24-6=18>0 \end{aligned}
So, x = 2 is local minima.
$f''(-1)=12(-1)-6=-12-6=-18<0$
So, x = -1 is local maxima.
Therefore, One Maxima and One Minima is the answer.\

RD Sharma class 12th exercise MCQ comprises 36 MCQs that incorporates the entire syllabus of chapter 17. The solutions are elaborate yet easy to understand. These solutions will work to solve all your queries and doubts. It include the following concepts:

• Maximum and minimum values

• Local maxima

• Local minima

• First derivative test for local maxima and minima

• Higher-order derivative test

• Theorem and algorithm based on higher derivative test

• Application based problems on maxima and minima

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