RD Sharma Solutions Class 12 Mathematics Chapter 16 MCQ

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RD Sharma Solutions Class 12 Mathematics Chapter 16 MCQ

RD Sharma Solutions Class 12 Mathematics Chapter 16 MCQ

Kuldeep MauryaUpdated on 21 Jan 2022, 01:29 PM IST

The RD Sharma Solutions Class 12 Maths Chapter 16 Increasing and Decreasing Functions–Our experts plan Differentiation to help students understand the thoughts peddled in this chapter and systems to deal with issues in a more limited period. Points shrouded in class 12 RD Sharma chapter 16 exercise MCQ solution of common mathematical imbalances, rigorously expanding capacities, Strictly diminishing abilities, Monotonic powers, Finding the spans in which a degree is increasing or decreasing, Proving the monotonicity of a total on a given stretch, Finding the span in which a capacity is expanding or diminishing, and so forth At Career360, RD Sharma class 12th exercise MCQ helps students who try to get a fair, insightful score in the exam.

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RD Sharma Class 12 Solutions Chapter 16 MCQ Increasing and Decreasing Functions - Other Exercise
Increasing and Decreasing Functions Excercise:MCQ
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RD Sharma Class 12 Solutions Chapter 16 MCQ Increasing and Decreasing Functions - Other Exercise

Increasing and Decreasing Functions Excercise:MCQ

Increasing and Decreasing Functions exercise Multiple choice question , question 1

Answer:

Correct option (b)
Hint: If$f(x)$ is increasing function${f}'(x)>0$
Given:$f(x)=x-e^{x}+\tan \left(\frac{2 \pi}{7}\right)$
Explanation: It is given that
$f(x)=x-e^{x}+\tan \left(\frac{2 \pi}{7}\right)$
Differentiate w.r.t x
${f}'(x)=1-e^{x}$
Since$f(x)$ is increasing function${f}'(x)>0$
$\begin{aligned} &\Rightarrow 1-e^{x}>0 \\ &\Rightarrow 1>e^{x} \\ &\Rightarrow x<0 \\ &\Rightarrow x \in(-\infty, 0) \end{aligned}$
Thus, the required interval is $(-\infty, 0)$

Increasing and Decreasing Functions exercise Multiple choice question , question 2

Answer:

Correct option (c)
Hint: If$f(x)$ is increasing function${f}'(x)>0$
Given:$f(x)=\cot ^{-1} x+x$
Explanation: It is given that
$\begin{aligned} &f(x)=\cot ^{-1} x+x \\ &f^{\prime}(x)=\frac{-1}{1+x^{2}}+1 \end{aligned}$
$\because f(x)$ is increasing function${f}'(x)>0$
$\begin{aligned} &\Rightarrow \frac{-1}{1+x^{2}}+1>0 \\ &\Rightarrow \frac{x^{2}}{1+x^{2}}>0 \\ &\Rightarrow x \in(-\infty, \infty) \end{aligned}$
Thus, the required interval is $(-\infty, \infty)$

Increasing and Decreasing Functions exercise Multiple choice question , question 3

Answer:

Answer: Correct option (c)
Hint: If $f(x)$ is decreasing function ${f}'(x)<0$
Given:$f(x)=x^{x}$
Explanation: It is given that
$f(x)=x^{x}$
Taking log on both sides,
$\begin{aligned} &\log f(x)=\log x^{x} \\ &\log f(x)=x \log x \quad\left[\because \log a^{b}=b \log a\right] \end{aligned}$
Differentiate w.r.t x
$\begin{aligned} &\frac{1}{f(x)} \cdot f^{\prime}(x)=x \cdot \frac{1}{x}+\log x .1 \end{aligned}$ [$\because$ by using u.v. rule]
$\begin{aligned} &\frac{f^{\prime}(x)}{f(x)}=1+\log x\\ &\therefore f^{\prime}(x)=x^{x}( 1+\log x) \quad\left[\because f(x)=x^{x}\right] \end{aligned}$$\begin{aligned} &\frac{f^{\prime}(x)}{f(x)}=1+\log x\\ &\therefore f^{\prime}(x)=x^{x}( 1+\log x) \quad\left[\because f(x)=x^{x}\right] \end{aligned}$
$f(x)$ is decreasing function ${f}{}'(x)<0$
$\begin{aligned} &\Rightarrow x^{x}(1+\log x)<0 \\ &\Rightarrow 1+\log x<0 \\ &\Rightarrow \log x<-1 \\ &\Rightarrow x<e^{-1} \\ &\because e>1 \Rightarrow e^{-1}<1 \end{aligned}$
Thus the function is decreasing on $\left ( 0,\frac{1}{e} \right )$

Increasing and Decreasing Functions exercise Multiple choice question , question 5

Answer:

Correct option (a)
Hint: If $f(x)$ is increasing function ${f}'(x)>0$
Given:$f(x)=2 x^{2}-k x+5$
Explanation: It is given that
$f(x)=2 x^{2}-k x+5$
Differentiate w.r.t x
$f^{\prime}(x)=2.2 x-k$
$\because$ If $f(x)$ is increasing function ${f}'(x)>0$
$\begin{aligned} &\therefore 4 x-k>0 \text { on }[1,2] \\ &k<4 x \end{aligned}$
So, the minimum value of k is 4x
$k<4\\ \therefore k\epsilon \left ( -\infty, 4 \right )$
Thus k lies in the interval $\left ( -\infty ,4 \right )$

Increasing and Decreasing Functions exercise Multiple choice question , question 6

Answer:

Correct option (c)
Hint: If $f(x)$ is increasing function ${f}'(x)>0$
Given: $f(x)=x^{3}+a x^{2}+b x+5 \sin ^{2} x$
Explanation: It is given that
$f(x)=x^{3}+a x^{2}+b x+5 \sin ^{2} x$
Differentiate w.r.t x
$\begin{aligned} f^{\prime}(x) &=3 x^{2}+2 a x+b+5 \cdot \sin x \cdot \cos x \\ &=3 x^{2}+2 a x+b+5 \cdot \sin 2 x \quad[\because \sin 2 x=2 \sin x \cos x] \end{aligned}$
$\because$ If $f(x)$ is increasing ${f}'(x)>0$
$\therefore 3 x^{2}+2 a x+b+5 \cdot \sin 2 x>0$
For the quadratic equation
Discriminant is
$\begin{aligned} &(2 a)^{2}-4 \times 3(b+5 \sin 2 x)<0 \\ &4 a^{2}-12 b-60 \sin 2 x<0 \\ &a^{2}-3 b-15 \sin 2 x<0 \end{aligned}$
$\because$ Minimum value of $\sin 2x=-1$
So, $a^{2}-3 b-15(-1)<0$
Thus, $a^{2}-3 b+15<0$
So, a & b satisfy equation $a^{2}-3 b+15<0$

Increasing and Decreasing Functions exercise Multiple choice question , question 7

Answer:

Correct option (b)
Hint: Use that, a function $f(x)$ is odd if $f(-x)=-f(x)$ & even if $f(-x)=f(x)$
& If $f(x)$ is increasing function ${f}'(x)>0$ , decreasing if ${f}'(x)<0$ .
Given: $f(x)=\log _{e}\left(x^{3}+\sqrt{x^{6}+1}\right)$
Explanation: It is given that
$f(x)=\log _{e}\left(x^{3}+\sqrt{x^{6}+1}\right) \ldots . .(\mathrm{i})$
So, $f(-x)=\log _{e}\left((-x)^{3}+\sqrt{(-x)^{6}+1}\right)$
$\begin{aligned} \therefore f(-x) &=\log _{e}\left((-x)^{3}+\sqrt{x^{6}+1}\right) \\ &=\log _{e}\left[\frac{\left(-x^{3}+\sqrt{x^{6}+1}\right)}{\left(x^{3}+\sqrt{x^{6}+1}\right)} \times\left(x^{3}+\sqrt{x^{6}+1}\right)\right] \end{aligned}$
$\begin{aligned} &=\log _{e}\left[\frac{1}{\left(x^{3}+\sqrt{x^{6}+1}\right)}\right] \\ &=-\log _{e}\left(x^{3}+\sqrt{x^{6}+1}\right) \\ &=-f(x) \\ \because f(-x) &=-f(x) \end{aligned}$
So, $f(x)$ is odd function
Now,Differentiate (i) w.r.t x
$f^{\prime}(x)=\frac{1}{\left(x^{3}+\sqrt{x^{6}+1}\right)} \times\left(3 x^{2}+\frac{1}{2 \sqrt{x^{6}+1}} \times 6 x^{5}\right)$
$\begin{aligned} &=\frac{1}{\left(x^{3}+\sqrt{x^{6}+1}\right)} \times 3 x^{2}\left(\frac{\sqrt{x^{6}+1}}{\sqrt{x^{6}+1}}+x^{3}\right) \\ &f^{\prime}(x)=\frac{3 x^{2}}{\sqrt{x^{6}+1}} \end{aligned}$
Here x² & x⁶ are even power of x
So, $f^{\prime}(x)=\frac{3 x^{2}}{\sqrt{x^{6}+1}}>0$
Thus,$f(x)$ is odd and increasing function.

Increasing and Decreasing Functions exercise Multiple choice question , question 8

Answer:

Correct option (c)
Hint: If$f(x)$ is increasing function${f}'(x)>0$
Given: $f(x)=2 \tan x+(2 a+1) \log _{e}|\sec x|+(a-2)$
Explanation: It is given that
$f(x)=2 \tan x+(2 a+1) \log _{e}|\sec x|+(a-2)$ ………(i)
Case i:
If $\sec x>0$
Now,Differentiate (i) w.r.t x
$\begin{aligned} &f^{\prime}(x)=2 \sec ^{2} x+(2 a+1) \frac{1}{\sec x} \sec x \cdot \tan x+(a-2) \\ &f^{\prime}(x)=2 \sec ^{2} x+(2 a+1) \tan x+(a-2) \end{aligned}$
$\because f(x)$ is increasing, ${f}'(x)>0$
$\begin{aligned} &\Rightarrow 2 \sec ^{2} x+(2 a+1) \tan x+(a-2)>0 \\ &\Rightarrow 2\left(\tan ^{2} x+1\right)+(2 a+1) \tan x+(a-2)>0 \end{aligned}$
$\Rightarrow 2 \tan ^{2} x+(2 a+1) \tan x+(a-2)>0$
The above equation is quadratic in $\tan x$
Its discriminant is
$(2 a+1)^{2}-4 \times 2 a<0$
$\Rightarrow(2 a-1)^{2}<0$ , which is impossible.
Thus, if $\sec x <0$ then $\left | \sec x \right |=-\sec x$
$\therefore 2 \sec ^{2} x-(2 a+1) \tan x+(a-2) \geq 0$
$\Rightarrow(2 a-1)^{2} \leq 0$ , which is not possible
$\begin{aligned} &\therefore(2 a-1)^{2}=0 \\ &\therefore a=\frac{1}{2} \end{aligned}$

Increasing and decreasing functions exercise multiple choice quection, question 10

Answer:

Correct option (c)
Hint: Differentiate the given function w.r.t x then check the given conditions
Given:$f(x)=x^{3}-6 x^{2}+15 x+3$
Explanation: It is given that
$f(x)=x^{3}-6 x^{2}+15 x+3$
Now,Differentiate (i) w.r.t x
$\begin{aligned} &f^{\prime}(x)=3 x^{2}-12 x+15 \\ &f^{\prime}(x)=3\left(x^{2}-4 x+5\right) \\ &f^{\prime}(x)=3\left[(x-2)^{2}+1\right]>0 \end{aligned}$
Since $\begin{aligned} &f^{\prime}(x)>0 \end{aligned}$ , so $\begin{aligned} &f(x) \end{aligned}$ is increasing function.
As $\begin{aligned} &f(x) \end{aligned}$ is increasing, it is invertible
Thus, $\begin{aligned} &f(x) \end{aligned}$ is invertible function.

Increasing and decreasing functions exercise multiple choice quection, question 11

Answer:

Correct option (b)
Hint: If $f(x)$ is increasing function ${f}'(x)>0$
Given: $f(x)=x^{2} e^{-x}$
Explanation: It is given that
$f(x)=x^{2} e^{-x}$
Differentiate (i) w.r.t x
$\begin{aligned} &f^{\prime}(x)=-x^{2} e^{-x}+x e^{-x} \\ &f^{\prime}(x)=-e^{-x} x(x-2) \end{aligned}$
$f(x)$ is monotonically increasing,${f}'(x)>0$
$\begin{aligned} &-e^{-x} x(x-2)>0 \\ &x(x-2)>0 \end{aligned}$
So, $0<x<2$
Thus, $f(x)$ is monotonically increasing $0<x<2$

Increasing and decreasing functions exercise multiple choice quection, question 12

Answer:

Correct option (a)
Hint: If $f(x)$ is decreasing function ${f}'(x)<0$
Given:$f(x)=\cos x-2\lambda x$
Explanation: It is given that
$f(x)=\cos x-2\lambda x$ …….(i)
Differentiate (i) w.r.t x
${f}'(x)=-\sin x-2\lambda$
$\because f(x)$ is decreasing ${f}'(x)<0$
$\begin{aligned} &\therefore-\sin x-2 \lambda<0 \\ &-\sin x<2 \lambda \\ &\frac{-\sin x}{2}<\lambda \end{aligned}$
$\Rightarrow \frac{1}{2}<\lambda$
Thus, $f(x)$ is monotonically decreasing when $\frac{1}{2}>\lambda$

Increasing and decreasing functions exercise multiple choice quection, question 13

Answer:

Correct option (b)
Hint: Use the condition for $f(x)$,
${f}'(x)>0$ , $f(x)$ is increasing
${f}'(x)<0$ ,$f(x)$ is decreasing
Given:$f(x)=2|x-1|+3|x-2|$
Explanation: We need to check$f(x)$ in interval (1,2)
Here, $f(x)=2|x-1|+3|x-2|$
$\because x\epsilon (1,2)$ So, $x>1$ & $x<2$
$\begin{aligned} &\Rightarrow x-1>0 \& x-2<0 \\ &f(x)=2|x-1|+3|x-2| \\ &f(x)=2(x-1)+3(x-2) \\ &f(x)=2 x-2-3 x+6 \\ &f(x)=-x+4 \\ &f^{\prime}(x)=-1<0 \end{aligned}$
Thus, $f(x)$ is monotonically decreasing in the interval (1,2)

Increasing and decreasing functions exercise multiple choice quection, question 14

Answer:

Correct option (d)
Hint: If $f(x)$ is monotonically increasing function${f}'(x)\geq 0$
Given: $f(x)=x^{3}-27x+5$
Explanation: It is given that
$f(x)=x^{3}-27x+5$ …..(i)
Differentiate (i) w.r.t x
$\begin{aligned} &f^{\prime}(x)=3 x^{2}-27 \\ &f^{\prime}(x)=3\left(x^{2}-9\right) \end{aligned}$
$\because f(x)$ is increasing ${f}'(x)\geq 0$
$\begin{aligned} &\Rightarrow 3\left(x^{2}-9\right) \geq 0 \\ &\Rightarrow x^{2}-9 \geq 0 \\ &\Rightarrow x^{2} \geq 9 \\ &\Rightarrow|x| \geq 3 \end{aligned}$
Thus,$f(x)$ is increasing when $\left | x \right |\geq 3$

Increasing and decreasing functions exercise multiple choice quection, question 15

Answer:

Correct option (d)
Hint: If $f(x)$ is decreasing function ${f}'(x)<0$
Given: $f(x)=2 x^{3}-9 x^{2}+12 x+29$
Explanation:It is given that
$f(x)=2 x^{3}-9 x^{2}+12 x+29$ …..(i)
Differentiate (i) w.r.t x
${f}'(x)=6 x^{2}-18x+12$
$f(x)$ is decreasing ${f}'(x)<0$
$\begin{aligned} &\Rightarrow 6 x^{2}-18 x+12<0 \\ &\Rightarrow 6\left(x^{2}-3 x+2\right)<0 \\ &\Rightarrow x^{2}-3 x+2<0 \\ &\Rightarrow(x-1)(x-2)<0 \\ &\Rightarrow 1<x<2 \end{aligned}$
Thus, the function is monotonically decreasing when $\begin{aligned} 1<x<2 \end{aligned}$

Increasing and decreasing functions exercise multiple choice quection, question 16

Answer:

Correct option (c)
Hint: If $f(x)$ is increasing function ${f}'(x)>0$
Given: $f(x)=k x^{3}-9 x^{2}+9 x+3$
Explanation:It is given that
$f(x)=k x^{3}-9 x^{2}+9 x+3$ …..(i)
Differentiate (i) w.r.t x
${f}'(x)=3 k x^{2}-18 x+9$
$\because f(x)$ is increasing ${f}'(x)>0$
$\begin{aligned} &3 k x^{2}-18 x+9>0 \\ &k x^{2}-6 x+3>0 \end{aligned}$
If $a>0\Rightarrow b^{2}-4ac<0$
So, $(-6)^{2}-4.k.3<0$
$\begin{aligned} &36-12 k<0 \\ &3-k<0 \\ &k>3 \end{aligned}$
Thus, the function is monotonic increasing if $k>3$

Increasing and decreasing functions exercise multiple choice quection, question 17

Answer:

Correct option (c)
Hint: If $f(x)$ is monotonic increasing function then ${f}'(x)>0$
Given: $f(x)=2 x-\tan ^{-1}-\log \left(x+\sqrt{x^{2}+1}\right)$
Explanation:It is given that
$f(x)=2 x-\tan ^{-1}-\log \left(x+\sqrt{x^{2}+1}\right)$ …..(i)
Differentiate (i) w.r.t x
$\begin{aligned} &f^{\prime}(x)=2-\frac{1}{1+x^{2}}-\frac{1}{x+\sqrt{x^{2}+1}}\left(1+\frac{2 x}{2 \sqrt{x^{2}+1}}\right) \\ &f^{\prime}(x)=2-\frac{1}{1+x^{2}}-\frac{1}{\sqrt{x^{2}+1}} \end{aligned}$
$\begin{aligned} &f^{\prime}(x)=\frac{1+2 x^{2}}{1+x^{2}}-\frac{1}{\sqrt{x^{2}+1}} \\ &f^{\prime}(x)=\frac{1+2 x^{2}-\sqrt{x^{2}+1}}{1+x^{2}} \end{aligned}$
$\because f(x)$ is monotonic increasing then ${f}'(x)>0$
$\begin{aligned} &\frac{1+2 x^{2}-\sqrt{x^{2}+1}}{1+x^{2}}>0 \\ &\Rightarrow 1+2 x^{2}-\sqrt{x^{2}+1}>0 \\ &\Rightarrow 1+2 x^{2}>\sqrt{x^{2}+1} \end{aligned}$
Squaring on both sides, we get
$\left(1+2 x^{2}\right)^{2}>x^{2}+1$
$\begin{aligned} &1+4 x^{2}+4 x^{4}>x^{2}+1 \\ &\therefore 3 x^{2}+4 x^{4}>0 \forall x \in R \end{aligned}$
Thus, the function is monotonically increasing when $\begin{aligned} x \in R \end{aligned}$

Increasing and decreasing functions exercise multiple choice quection, question 18

Answer:

Correct option (d)
Hint: Take 3 conditions $x<0, x>1 \& 0<x<1$ and check whether the function is increasing
Given:$f(x)=|x|-|x-1|$
Explanation:It is given that
$f(x)=|x|-|x-1|$
Case (i): If $x<0$
$\begin{aligned} &|x|=-x \&|x-1|=-(x-1) \\ &\Rightarrow f(x)=-x+x-1=-1 \\ &\Rightarrow f^{\prime}(x)=0 \end{aligned}$
So, the function is not monotonically increasing when $x<0$
Case (ii): If $x>1$
So,
$\begin{aligned} &\Rightarrow f(x)=1 \\ &\Rightarrow f^{\prime}(x)=0 \end{aligned}$
Thus,$f(x)$ is not increasing when $x>1$
Case (iii): If $0<x<1$
If $0<x<1$ then $|x|=x \&|x-1|=-(x-1)$
$\begin{aligned} &\Rightarrow f(x)=x+x-1=2 x-1 \\ &\Rightarrow f^{\prime}(x)=2 \end{aligned}$
So, the function is monotonically increasing when $0<x<1$

Increasing and Decreasing functions exercise Multiple choice question, question 19

Answer:

Correct option (a)
Hint: If $f(x)$ is invertible function ${f}'(x)>0$
Given: Every invertible function is
Explanation: We know,
A function is invertible in a given domain,
If it is continuous & one-one in the domain.
And if the function is one-one in the domain,
It has to be strictly monotonic .
Hence, every invertible function is monotonic.

Increasing and Decreasing functions exercise Multiple choice question, question 21

Answer:

Correct option (c)
Hint: If $f(x)$ is increasing function ${f}'(x)\geq 0$
Given: $f(x)=\cos |x|-2 a x+b$
Explanation:It is given that
$f(x)=\cos |x|-2 a x+b$ …..(i)
Differentiate (i) w.r.t x
$f^{\prime}(x)=-\sin x+2 a$
$\because f(x)$ is increasing function, ${f}'(x)\geq 0$
$\begin{aligned} &\Rightarrow-\sin x+2 a \geq 0 \\ &\Rightarrow \sin x \geq 2 a \end{aligned}$
$\because$ Minimum value $\sin x$ of is -1
$\therefore \frac{-1}{2} \geq a$

Increasing and Decreasing functions exercise Multiple choice question, question 22

Answer:

Correct option (a)
Hint: Take two conditions, $x<0,x>0$ to identify the type of function.
Given:$f(x)=\frac{x}{1+|x|}$
Explanation:It is given that
$f(x)=\frac{x}{1+|x|}$
Case (i):
If $x<0$
$\begin{aligned} &\Rightarrow|x|=-x \\ &\therefore f(x)=\left(\frac{x}{1-x}\right) \end{aligned}$
Differentiate (i) w.r.t x
$f^{\prime}(x)=\left(\frac{x}{1-x}\right)^{2}>0$
So, function is increasing.
Case (ii):
If $x>0$
$\begin{aligned} &\Rightarrow|x|=x \\ &\therefore f(x)=\left(\frac{x}{1+x}\right) \\ &f^{\prime}(x)=\left(\frac{x}{1+x}\right)^{2}>0 \end{aligned}$
So, function is increasing .
Hence, the given function is strictly increasing.

Increasing and Decreasing functions exercise Multiple choice question, question 23

Answer:

Correct option (d)
Hint: If $f(x)$ is increasing function ${f}'(x)>0$
Given: $f(x)=\frac{\lambda \sin x+2 \cos x}{\sin x+\cos x}$

Explanation:It is given that
$f(x)=\frac{\lambda \sin x+2 \cos x}{\sin x+\cos x}$ …..(i)
Differentiate (i) w.r.t x
$\begin{aligned} &f^{\prime}(x)=(\lambda-2) \sin ^{2} x+(\lambda-2) \cos ^{2} x>0 \\ &f^{\prime}(x)=(\lambda-2)\left(\sin ^{2} x+\cos ^{2} x\right) \\ &f^{\prime}(x)=(\lambda-2)>0 \\ &\therefore \lambda>2 \end{aligned}$
So, the function is increasing if$\begin{aligned} & \lambda>2 \end{aligned}$

Increasing and Decreasing functions exercise Multiple choice question, question 25

Answer:

Correct option (b)
Hint: If $f(x)$ is increasing function${f}'(x)>0$
Given: $f(x)=\log _{a} x$
Explanation:It is given that
$f(x)=\log _{a} x$
$\Rightarrow a^{f(x)}=x$ …..(i)
Differentiate (i) w.r.t x
$\begin{aligned} &\Rightarrow a^{f(x)} \cdot \log a \cdot f^{\prime}(x)=1 \\ &\Rightarrow f^{\prime}(x)=\frac{1}{a^{f(x)} \cdot \log a} \\ &\Rightarrow f^{\prime}(x)=\frac{1}{x \log a} \end{aligned}$

Since the function is increasing on R

$\begin{aligned} &\frac{1}{x \log a}>0 \\ &\Rightarrow a>1 \end{aligned}$
Thus the function is increasing if $\begin{aligned} & a>1 \end{aligned}$

Increasing and Decreasing functions exercise Multiple choice question, question 26

Correct option (b)
Hint: First differentiate $Q(x)$ w.r.to then using the relation between ${f}'{x}$ & $f(2a-x)$identify type of $Q(x)$
Given:$Q(x)=f(x)+f(2 a-x)$
Explanation:It is given that
$Q(x)=f(x)+f(2 a-x)$ …..(i)
Differentiate (i) w.r.t x
$Q^{\prime}(x)=f^{\prime}(x)-f^{\prime}(2 a-x)$
Since $f^{\prime}(x)>0 \Rightarrow f^{\prime \prime}(x)>0$
Here $x\epsilon \left [ 0,a \right ]$
$\begin{aligned} &x \leq 2 a-x \\ &f^{\prime}(x) \leq f^{\prime}(2 a-x) \end{aligned}$
Also, $Q^{\prime}(x)=f^{\prime}(x)-f^{\prime}(2 a-x)$
$Q(x)$ is decreasing on [0,a]

Increasing and Decreasing functions exercise Multiple choice question, question 27

Answer:

Correct option (b)
Hint: If$f(x)$ is increasing function${f}'(x)>0$
Given:$f(x)=x^{2}-k x+5$
Explanation:It is given that
$f(x)=x^{2}-k x+5$ …..(i)
Differentiate (i) w.r.t x
${f}'(x)=2x-k$
$\because f(x)$ is increasing,${f}'(x)>0$
$\begin{aligned} &\Rightarrow 2 x-k>0 \\ &\Rightarrow 2 x>k \\ &\because x \in[2,4] \end{aligned}$
So, Maximum value of k is 4
$k \in(-\infty, 4)$
Thus if the function is increasing then $k \in(-\infty, 4)$

Increasing and decreasing function exercise multiple choice question, question 28

Answer:

Correct option (a)
Hint: If$f(x)$ is increasing function${f}'(x)>0$
And if $f(x)$ is increasing function then ${f}'(x)<0$
Given: $f(x)=\frac{-x}{2}+\sin x$
Explanation:It is given that
$f(x)=\frac{-x}{2}+\sin x$ …..(i)
Differentiate (i) w.r.t x
$f^{\prime}(x)=\frac{-1}{2}+\cos x$
Since $\because x \in\left[\frac{-\pi}{3}, \frac{\pi}{3}\right]$
$\begin{aligned} &\Rightarrow f^{\prime}(x)>0 \\ &\Rightarrow \frac{-1}{2}+\cos x>0 \end{aligned}$
Thus function is increasing.

Increasing and decreasing function exercise multiple choice question, question 29

Answer:

Correct option (a)
Hint: If$f(x)$ is increasing function ${f}'(x)>0$
Given:$f(x)=x^{3}-9 k x^{2}+27 x+30$
Explanation:It is given that
$f(x)=x^{3}-9 k x^{2}+27 x+30$ …..(i)
Differentiate (i) w.r.t x
$\begin{aligned} &f^{\prime}(x)=3 x^{2}-9 k \cdot 2 x+27 \\ &f^{\prime}(x)=3\left(x^{2}-6 k x+9\right) \end{aligned}$
$\because f(x)$ is increasing
$\begin{aligned} &3\left(x^{2}-6 k x+9\right)>0 \\ &x^{2}-6 k x+9>0 \end{aligned}$
In $a x^{2}+b x+c=0$ , if $a>0$ then$b^{2}-4 a c<0$
$\begin{aligned} &\Rightarrow(-6 k)^{2}-4 \times 1 \times 9<0 \\ &36 k^{2}-36<0 \\ &k^{2}-1<0 \\ &(k+1)(k-1)<0 \\ &\Rightarrow-1<k<1 \end{aligned}$
Thus , if $f(x)$ is increasing function then $-1\leq k< 1$
Note:- option (a) has to be $-1\leq k< 1$

Increasing and decreasing function exercise multiple choice question, question 30

Answer:

Correct option (a)
Hint: If$f(x)$ is increasing function${f}'(x)>0$
Given:$f(x)=x^{9}+3 x^{7}+64$
Explanation:It is given that
$f(x)=x^{9}+3 x^{7}+64$ …..(i)
Differentiate (i) w.r.t x
$\begin{aligned} &f^{\prime}(x)=9 x^{8}+21 x^{6} \\ &f^{\prime}(x)=3 x^{6}\left(3 x^{2}+7\right) \end{aligned}$
$\because f(x)$ is increasing
$\Rightarrow 3 x^{6}\left(3 x^{2}+7\right)>0$
Thus, the given function is increasing on R

Increasing and decreasing function exercise multiple choice question, question 31

Answer:

$[-2,-1]$ , Correct option (b)
Hints: Take a derivative of given equation
Given: The interval on which $f(x)=2 x^{3}+9 x^{2}+12 x-1$ is decreasing, is
Solution:
We have,
$f(x)=2 x^{3}+9 x^{2}+12 x-1$
$\begin{aligned} f^{\prime}(x) &=6 x^{2}+18 x+12 \\ &=6\left(x^{2}+3 x+2\right) \\ &=6(x+2)(x+1) \end{aligned}$
f(x) decreases when $\begin{aligned} f^{\prime}(x) \leq 0 \end{aligned}$

∴ f’(x) =

From the sign scheme $\begin{aligned} f^{\prime}(x) \leq 0 \end{aligned}$
When $\begin{aligned} x\epsilon \left [ -2,-1 \right ] \end{aligned}$

Increasing and decreasing function exercise multiple choice question, question 32

Answer:

1<x<3, Option (a)
Hint: Take a derivative of given equation
Given: $y=x(x-3)^{2}$ decreases for the values
Solution: We have, $y=x(x-3)^{2}$
$\begin{aligned} &y=x\left(x^{2}-6 x+9\right) \\ &y=x^{3}-6 x^{2}+9 x \end{aligned}$
$\begin{aligned} \frac{\mathrm{dy}}{\mathrm{dx}} &=3 \mathrm{x}^{2}-12 \mathrm{x}+9 \\ \frac{\mathrm{dy}}{\mathrm{dx}} &=3\left(\mathrm{x}^{2}-4 \mathrm{x}+3\right) \\ &=3(\mathrm{x}-3)(\mathrm{x}-1) \end{aligned}$
Y = f(x) decreases when $\begin{aligned} \frac{\mathrm{dy}}{\mathrm{dx}}<0 \end{aligned}$
The sign scheme of $\begin{aligned} \frac{\mathrm{dy}}{\mathrm{dx}} \end{aligned}$ is shown,

∴ f’(x) =

From the sign scheme $\begin{aligned} \frac{\mathrm{dy}}{\mathrm{dx}}<0 \end{aligned}$
For $x\epsilon (1,3)$
$y=x(x-3)^{2}$ decreases when $x\epsilon 1<x<3$

Increasing and decreasing function exercise multiple choice question, question 33

Answer:

decreasing in $\left ( \frac{\pi}{2},\pi \right )$
Hints: Find the derivative of given equation.
Given: $f(x)=4 \sin ^{3} x-6 \sin ^{2} x+12 \sin x+100$
Solution: We have,
$f(x)=4 \sin ^{3} x-6 \sin ^{2} x+12 \sin x+100$
$\begin{aligned} f^{\prime}(x) &=12 \sin ^{2} x \cos x-12 \sin x \cos x+12 \cos x \\ &=12 \cos x\left[\sin ^{2} x-\sin x+1\right] \\ &=12 \cos x\left[\sin ^{2} x+(1-\sin x)\right] \end{aligned}$

Now $1-\sin x \geq 0$ and $\sin^{2} x \geq 0$
$\sin ^{2} x+1-\sin x \geq 0$
Hence, $f'{x}>0$ when $\cos x>0$ , i.e., $x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
So, $f(x)$ is increasing when $x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
$f'{x}<0$ , when $\cos x>0$ , i.e., $x\epsilon \left ( \frac{\pi}{2}, \frac{3\pi}{2} \right )$
Hence, $f(x)$ is decreasing when $x\epsilon \left ( \frac{\pi}{2}, \frac{3\pi}{2} \right )$
f(x) is decreasing in $x\epsilon \left ( \frac{\pi}{2},\pi \right )$

Increasing and decreasing function exercise multiple choice question, question 34

Answer:

$\cos x$ , Option (c)
Hints: Check all the options and choose is satisfies
Given: Function is decreasing in $\left ( 0,\frac{\pi}{2} \right )$
Solution:
Option (A)
$\begin{aligned} &f(x)=\sin 2 x \\ &f^{\prime}(x)=2 \cos 2 x \end{aligned}$
$f(x)$ increases from ‘0’ to ‘1’ in $\left ( 0,\frac{\pi}{2} \right )$
Option (B)
$\begin{aligned} &f(x)=\tan x \\ &f^{\prime}(x)=\sec ^{2} x \end{aligned}$
In interval $\left ( 0,\frac{\pi}{2} \right )$ , $f^{\prime}(x)=-\sin x<0$
$f(x)=\cos x$ is strictly increasing in interval $\left ( 0,\frac{\pi}{2} \right )$
Option (C)
$\begin{aligned} &f(x)=\cos x \\ &f^{\prime}(x)=-\sin x \end{aligned}$
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x)=-\sin x<0$
$f(x)=\cos x$ is strictly decreasing in $\left ( 0,\frac{\pi}{2} \right )$
Option (D)
$\begin{aligned} &f(x)=\cos 3 x \\ &f^{\prime}(x)=-3 \sin 3 x \end{aligned}$
Now,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\sin 3 x=0 \\ &3 x=\pi \end{aligned}$
As $x\epsilon \left ( 0,\frac{\pi}{2} \right )$
$x=\frac{\pi}{3}$
$f(x)=\cos 3 x$ is decreases only when $3 x \in\left(0, \frac{\pi}{2}\right)$
And $x \in\left(0, \frac{\pi}{6}\right)$
Therefore, Option (C) =cos x satisfies because $f(x)=\cos x$ is strictly decreasing in $\left ( 0,\frac{\pi}{2} \right )$

Increasing and decreasing function exercise multiple choice question, question 35

Answer:

Always increases (Option a)
Hints: Find the derivative of functions
Given: $f(x)=\tan x-x$
Solution:
We have, $\begin{aligned} &f(x)=\tan x-x \\ &f^{\prime}(x)=\sec ^{2} x-1>0 \end{aligned}$
Or
$\begin{aligned} &\sec ^{2} x>1 \end{aligned}$
Now
$\begin{aligned} &\cos x \in[-1,1] \\ &\sec x \in(-\infty,-1] \cup[1, \infty) \end{aligned}$
Thus,
$\sec ^{2} x \in[1, \infty)$
Hence,
$f^{\prime}(x)>0 \forall x$
Function $f(x)=\tan x-x$ is always increasing.

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