Exercise-6.2, Class 12th, Maths, Chapter 6, NCERT

NOTE – Definition -1 : Let $I$ be an interval contained in the domain of a real-valued function $f$. Then $f$ is said to be

(i) increasing on $I$

$$x_1<x_2\Rightarrow f(x_1)<f(x_2)\text{for all }{x}_1,x_2\in I$$

(ii) decreasing on $I$

$$x_1<x_2\Rightarrow f(x_1)>f(x_2)\text{for all }{x}_1,x_2\in I$$

(iii) constant on $I$

$$f(x)=c\text{ for all }x\in I$$

where $c$ is a constant.

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Question 1 

Show that the function given by
$f(x)=3x+17$ is increasing on $\mathbb{R}$

Solution

Let $x_1$ and $x_2$ be any two real numbers such that

$$x_1<x_2$$

Then,

$$3x_1<3x_2(\text{multiplying both sides by 3})$$

$$3x_1+17<3x_2+17$$

$$\Rightarrow f(x_1)<f(x_2)$$

Thus, by Definition 1 (Increasing Function), the function $f(x)=3x+17$ is strictly increasing on $\mathbb{R}$.

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Question 2

Show that the function given by

$$f(x)=e^{2x}$$

is increasing on $\mathbb{R}$.

Solution

We have

$$f(x)=e^{2x}$$

Differentiate with respect to $x$:

$$f^{\mathrm{\prime }}(x)=\frac{d}{dx}(e^{2x})=2e^{2x}$$Now, we know that:

$$e^{2x}>0\text{ for all }x\in \mathbb{R}$$

Therefore,

$$f^{\mathrm{\prime }}(x)=2e^{2x}>0\text{for all }x\in \mathbb{R}$$

Since $f^{\mathrm{\prime }}(x)>0$ for every real number $x$, by, the function $f$ is increasing on $\mathbb{R}$.

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Question 3 

Show that the function given by $f(x)=\sin x$ is
(a) increasing in $(0,\frac{\pi }{2})$
(b) decreasing in $(\frac{\pi }{2},\pi )$

Solution

We have:

$$f(x)=\sin x$$

Differentiate with respect to $x$:

$$f^{\mathrm{\prime }}(x)=\cos x$$

(a) Increasing in $(0,\frac{\pi }{2})$

In the interval $(0,\frac{\pi }{2})$,

$$\cos x>0$$

Therefore,

$$f^{\mathrm{\prime }}(x)=\cos x>0\text{for every }x\in (0,\frac{\pi }{2})$$

Since $f^{\mathrm{\prime }}(x)>0$ in this interval, by Theorem 1 (Increasing and Decreasing Test),

$$\boxed{{\displaystyle f(x)=\sin x\text{ is increasing in }(0,\frac{\pi }{2})}}$$

(b) Decreasing in $(\frac{\pi }{2},\pi )$

In the interval $(\frac{\pi }{2},\pi )$,

$$\cos x<0$$

Therefore,

$$f^{\mathrm{\prime }}(x)=\cos x<0\text{for every }x\in (\frac{\pi }{2},\pi )$$

Since $f^{\mathrm{\prime }}(x)<0$ in this interval, by Theorem 1,

$$\boxed{{\displaystyle f(x)=\sin x\text{ is decreasing in }(\frac{\pi }{2},\pi )}}$$

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Question 4 

Find the intervals in which the function

$$f(x)=2x^2-3x$$is
(a) increasing
(b) decreasing

Solution

We have:

$$f(x)=2x^2-3x$$

Differentiate with respect to $x$:

$$f^{\mathrm{\prime }}(x)=4x-3$$

Now set $f^{\mathrm{\prime }}(x)=0$ to find the critical point:

$$4x-3=0$$

$$x=\frac{3}{4}$$

The point $x=\frac{3}{4}$ divides the real line into two intervals:

$$(-\mathrm{\infty },\frac{3}{4})\text{and}(\frac{3}{4},\mathrm{\infty })$$

Check the sign of $f^{\mathrm{\prime }}(x)$ in these intervals

For $x<\frac{3}{4}$

Take any value, say $x=0$:

$$f^{\mathrm{\prime }}(0)=4(0)-3=-3<0$$

So $f^{\mathrm{\prime }}(x)<0$ in $(-\mathrm{\infty },\frac{3}{4})$

$$\Rightarrow f(x)\text{ is decreasing in }(-\mathrm{\infty },\frac{3}{4})$$

For $x>\frac{3}{4}$

Take $x=1$:$$f^{\mathrm{\prime }}(1)=4(1)-3=1>0$$

So $f^{\mathrm{\prime }}(x)>0$ in $(\frac{3}{4},\mathrm{\infty })$

$$\Rightarrow f(x)\text{ is increasing in }(\frac{3}{4},\mathrm{\infty })$$

Final Answer

$$\boxed{{\displaystyle \text{(a) Increasing in }(\frac{3}{4},\mathrm{\infty })}}$$
$$\boxed{{\displaystyle \text{(b) Decreasing in }(-\mathrm{\infty },\frac{3}{4})}}$$

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Question 5 

Find the intervals in which the function

$$f(x)=2x^3-3x^2-36x+7$$is
(a) increasing
(b) decreasing

Solution

We have:

$$f(x)=2x^3-3x^2-36x+7$$

Differentiate with respect to $x$:

$$f^{\mathrm{\prime }}(x)=6x^2-6x-36$$

Factorizing:

$$f^{\mathrm{\prime }}(x)=6(x^2-x-6)$$

$$f^{\mathrm{\prime }}(x)=6(x-3)(x+2)$$

Now set $f^{\mathrm{\prime }}(x)=0$:

$$(x-3)(x+2)=0\Rightarrow x=-2,3$$

So the real line is divided into intervals:

$$(-\mathrm{\infty },-2),(-2,3),(3,\mathrm{\infty })$$

Sign of $f^{\mathrm{\prime }}(x)$ in each interval

Final Answer

$$\boxed{{\displaystyle \text{(a) Increasing in }(-\mathrm{\infty },-2)\text{and}(3,\mathrm{\infty })}}$$
$$\boxed{{\displaystyle \text{(b) Decreasing in }(-2,3)}}$$

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Question 6

Find the intervals in which the following functions are strictly increasing or strictly decreasing:

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(a) $f(x)=x^2+2x-5$

$$f^{\mathrm{\prime }}(x)=2x+2$$

Set $f^{\mathrm{\prime }}(x)=0$:

$$2x+2=0\Rightarrow x=-1$$

Intervals: $(-\mathrm{\infty },-1)$ and $(-1,\mathrm{\infty })$

Pick a test point:

• For $x<-1$, say $x=-2$:

  $$f^{\mathrm{\prime }}(-2)=2(-2)+2=-2<0\Rightarrow \text{Decreasing}$$

• For $x>-1$, say $x=0$:

  $$f^{\mathrm{\prime }}(0)=2(0)+2=2>0\Rightarrow \text{Increasing}$$

Answer

$$\boxed{{\displaystyle \text{Decreasing on }(-\mathrm{\infty },-1)}}$$
$$\boxed{{\displaystyle \text{Increasing on }(-1,\mathrm{\infty })}}$$

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(b) $f(x)=10-6x-2x^2$

$$f^{\mathrm{\prime }}(x)=-6-4x=-(4x+6)$$

Set $f^{\mathrm{\prime }}(x)=0$:

$$4x+6=0\Rightarrow x=-\frac{3}{2}$$

• For $x<-\frac{3}{2}$, say $x=-2$:

  $$f^{\mathrm{\prime }}(-2)=-6-4(-2)=2>0\Rightarrow \text{Increasing}$$

• For $x>-\frac{3}{2}$, say $x=0$:

  $$f^{\mathrm{\prime }}(0)=-6<0\Rightarrow \text{Decreasing}$$

Answer

$$\boxed{{\displaystyle \text{Increasing on }(-\mathrm{\infty },-\frac{3}{2})}}$$
$$\boxed{{\displaystyle \text{Decreasing on }(-\frac{3}{2},\mathrm{\infty })}}$$

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(c) $f(x)=-2x^3-9x^2-12x+1$

$$f^{\mathrm{\prime }}(x)=-6x^2-18x-12=-6(x^2+3x+2)$$

$$=-6(x+1)(x+2)$$

Set $f^{\mathrm{\prime }}(x)=0\Rightarrow x=-1,-2$

Intervals:

$$(-\mathrm{\infty },-2),(-2,-1),(-1,\mathrm{\infty })$$

Sign test:

Answer

$$\boxed{{\displaystyle \text{Increasing on }(-2,-1)}}$$

$$\boxed{{\displaystyle \text{Decreasing on }(-\mathrm{\infty },-2)\text{ and }(-1,\mathrm{\infty })}}$$

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(d) $f(x)=6-9x-x^2$

$$f^{\mathrm{\prime }}(x)=-9-2x$$

Set $f^{\mathrm{\prime }}(x)=0$:

$$-9-2x=0\Rightarrow x=-\frac{9}{2}$$

Test sign:

• $x<-\frac{9}{2}$: $f^{\mathrm{\prime }}(x)>0$ → Increasing

• $x>-\frac{9}{2}$: $f^{\mathrm{\prime }}(x)<0$ → Decreasing

Answer

$$\boxed{{\displaystyle \text{Increasing on }(-\mathrm{\infty },-\frac{9}{2})}}$$

$$\boxed{{\displaystyle \text{Decreasing on }(-\frac{9}{2},\mathrm{\infty })}}$$

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(e) $f(x)=(x+1{)}^3(x-3{)}^3$

$$f(x)=[(x+1)(x-3){]}^3$$

Let $g(x)=(x+1)(x-3)=x^2-2x-3$

$$f^{\mathrm{\prime }}(x)=3[g(x){]}^2\cdot g^{\mathrm{\prime }}(x)$$

$$=3(x^2-2x-3{)}^2(2x-2)$$

$$=6(x^2-2x-3{)}^2(x-1)$$

Critical point at $x=1$

Since $(x^2-2x-3{)}^2\ge 0$ always and is zero only at $x=-1,3$, sign depends on $(x-1)$:

• $x<1$: $f^{\mathrm{\prime }}(x)<0$ → Decreasing

• $x>1$: $f^{\mathrm{\prime }}(x)>0$ → Increasing

Answer

$$\boxed{{\displaystyle \text{Decreasing on }(-\mathrm{\infty },1)}}$$

$$\boxed{{\displaystyle \text{Increasing on }(1,\mathrm{\infty })}}$$

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Question 7

Show that

$$y=\log (1+x)-\frac{2x}{2+x},x>-1$$

is an increasing function throughout its domain.

Solution

Given:$$y=\log (1+x)-\frac{2x}{2+x}$$

Differentiate with respect to $x$.

Step 1: Differentiate each term

$$\frac{d}{dx}(\log (1+x))=\frac{1}{1+x}$$

Now differentiate the second term using quotient rule:

$$\frac{d}{dx}\left(\frac{2x}{2+x}\right)=\frac{(2+x)\cdot 2-2x\cdot 1}{(2+x{)}^2}$$

$$=\frac{4+2x-2x}{(2+x{)}^2}=\frac{4}{(2+x{)}^2}$$

Step 2: Write derivative of $y$

$$y^{\mathrm{\prime }}=\frac{1}{1+x}-\frac{4}{(2+x{)}^2}$$

Step 3: Simplify

Take LCM $(1+x)(2+x{)}^2$:

$$y^{\mathrm{\prime }}=\frac{(2+x{)}^2-4(1+x)}{(1+x)(2+x{)}^2}$$

Expand:$$(2+x{)}^2=x^2+4x+4$$

So:$$y^{\mathrm{\prime }}=\frac{x^2+4x+4-4x-4}{(1+x)(2+x{)}^2}$$

Simplifying numerator:

$$x^2+4x+4-4x-4=x^2$$

Thus:$$y^{\mathrm{\prime }}=\frac{x^2}{(1+x)(2+x{)}^2}$$

Step 4: Check sign of $y^{\mathrm{\prime }}$

• $x^2\ge 0$ for all real $x$

• For domain $x>-1$, both denominators $(1+x)$ and $(2+x{)}^2$ are positive

Therefore:$$y^{\mathrm{\prime }}=\frac{x^2}{(1+x)(2+x{)}^2}\ge 0\text{for all }x>-1$$

And equality occurs only at $x=0$, otherwise positive.

Conclusion

$$\boxed{{\displaystyle y^{\mathrm{\prime }}\ge 0\text{ for all }x>-1}}$$

$$\boxed{{\displaystyle \text{Hence, }y=\log (1+x)-\frac{2x}{2+x}\text{ is an increasing function throughout its domain}}}$$

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Question 8

Find the values of $x$ for which

is an increasing function.

Solution

Given:

Differentiate with respect to $x$:

Factorising:

Now factor further:

We want to find where $y^{\mathrm{\prime }}>0$.
So solve:

The critical points where the expression changes sign are:

These points divide the real line into intervals:

Sign Table

Final Answer

or

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Question 9

Prove that

$$y=\frac{4\sin \theta }{2+\cos \theta }-\theta$$

is an increasing function of $\theta$ in

$$[0,\frac{\pi }{2}]$$

Solution

Given:

$$y=\frac{4\sin \theta }{2+\cos \theta }-\theta$$

Differentiate with respect to $\theta$:

Step 1 – Differentiate the first term using quotient rule

Let:

$$u=4\sin \theta ,v=2+\cos \theta$$
$$u^{\mathrm{\prime }}=4\cos \theta ,v^{\mathrm{\prime }}=-\sin \theta$$

$$\frac{d}{d\theta }\left(\frac{u}{v}\right)=\frac{u^{\mathrm{\prime }}v-u{v}^{\mathrm{\prime }}}{v^2}$$

Substitute:

$$\frac{d}{d\theta }\left(\frac{4\sin \theta }{2+\cos \theta }\right)=\frac{(4\cos \theta )(2+\cos \theta )-(4\sin \theta )(-\sin \theta )}{(2+\cos \theta {)}^2}$$

Simplify the numerator:

$$=\frac{8\cos \theta +4{\cos }^2\theta +4{\sin }^2\theta }{(2+\cos \theta {)}^2}$$

$$=\frac{8\cos \theta +4({\cos }^2\theta +{\sin }^2\theta )}{(2+\cos \theta {)}^{\mathrm{2<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}}}$$

Using the identity ${\sin }^2\theta +{\cos }^2\theta =1$:

$$=\frac{8\cos \theta +4}{(2+\cos \theta {)}^2}$$

Step 2 – Differentiate second term

$$\frac{d}{d\theta }(-\theta )=-1$$

Therefore

$$y^{\mathrm{\prime }}=\frac{8\cos \theta +4}{(2+\cos \theta {)}^2}-1$$

Take LCM:

$$y^{\mathrm{\prime }}=\frac{8\cos \theta +4-(2+\cos \theta {)}^2}{(2+\cos \theta {)}^2}$$

Expand the square:

$$(2+\cos \theta {)}^2=4+4\cos \theta +{\cos }^2\theta$$

Substitute:

$$y^{\mathrm{\prime }}=\frac{8\cos \theta +4-4-4\cos \theta -{\cos }^2\theta }{(2+\cos \theta {)}^2}$$

Simplify numerator:

$$y^{\mathrm{\prime }}=\frac{4\cos \theta -{\cos }^2\theta }{(2+\cos \theta {)}^2}$$

Factor:

$$y^{\mathrm{\prime }}=\frac{\cos \theta (4-\cos \theta )}{(2+\cos \theta {)}^2}$$

Check sign of $y^{\mathrm{\prime }}$ on $[0,\frac{\pi }{2}]$

In this interval:

• $\cos \theta \ge 0$

• $4-\cos \theta >0$

• $(2+\cos \theta {)}^2>0$

Therefore,

$$y^{\mathrm{\prime }}=\frac{\cos \theta (4-\cos \theta )}{(2+\cos \theta {)}^2}\ge 0$$

Thus,

$$y^{\mathrm{\prime }}\ge 0\text{for all }\theta \in [0,\frac{\pi }{2}]$$

So $y$ is an increasing function on this interval.

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Question 10

Prove that the logarithmic function is increasing on $(0,\mathrm{\infty })$.

Solution

Let

$$f(x)=\log x,x>0$$

Differentiate with respect to $x$:

$$f^{\mathrm{\prime }}(x)=\frac{1}{x}$$

Now examine the sign of $f^{\mathrm{\prime }}(x)$ over its domain $(0,\mathrm{\infty })$:

• For all $x>0$, $\frac{1}{x}>0$

Therefore:$$f^{\mathrm{\prime }}(x)>0\text{for all }x\in (0,\mathrm{\infty })$$

According to Theorem 1 on Increasing and Decreasing Functions, if

$$f^{\mathrm{\prime }}(x)>0$$

for every $x$ in an interval, then $f(x)$ is increasing in that interval.

So:

$$\boxed{{\displaystyle \text{The function }\log x\text{ is increasing on }(0,\mathrm{\infty })\mathrm{.<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}}}$$

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Question 11

Prove that the function $f(x)=x^2-x+1$ is neither strictly increasing nor strictly decreasing on $(-1,1)$.

Solution

Given:

$$f(x)=x^2-x+1$$

Differentiate with respect to $x$:

$$f^{\mathrm{\prime }}(x)=2x-1$$

Now set $f^{\mathrm{\prime }}(x)=0$ to locate the critical point:

$$2x-1=0$$
$$x=\frac{1}{2}$$

This point lies inside the interval $(-1,1)$.
So the interval $(-1,1)$ is divided into two parts:

$$(-1,\frac{1}{2})\text{and}(\frac{1}{2},1)$$

Test the sign of $f^{\mathrm{\prime }}(x)$ in these intervals

For $x\in (-1,\frac{1}{2})$

Choose $x=0$:

$$f^{\mathrm{\prime }}(0)=2(0)-1=-1<0$$

$$\Rightarrow f(x)\text{ is decreasing on }(-1,\frac{1}{2})$$

For $x\in (\frac{1}{2},1)$

Choose $x=\frac{3}{4}$:

$$f^{\mathrm{\prime }}(\frac{3}{4})=2(\frac{3}{4})-1=\frac{3}{2}-1=\frac{1}{2}>0$$
$$\Rightarrow f(x)\text{ is increasing on }(\frac{1}{2},1)$$

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Question 12

Which of the following functions are decreasing on

$$(0,\frac{\pi }{2})?$$

(A) $\cos x$
(B) $\cos 2x$
(C) $\cos 3x$
(D) $\tan x$

Solution

A function is decreasing if its derivative is negative in the interval.

Option (A) $f(x)=\cos x$

$$f^{\mathrm{\prime }}(x)=-\sin x$$

In $(0,\frac{\pi }{2})$, $\sin x>0$, so

$$f^{\mathrm{\prime }}(x)=-\sin x<0\Rightarrow \cos x\text{ is decreasing}$$

Option (B) $f(x)=\cos 2x$

$$f^{\mathrm{\prime }}(x)=-2\sin 2x$$

In $(0,\frac{\pi }{2})$, $2x\in (0,\pi )$, so $\sin 2x>0$, hence

$$f^{\mathrm{\prime }}(x)=-2\sin 2x<0\Rightarrow \cos 2x\text{ is decreasing}$$

Option (C) $f(x)=\cos 3x$

$$f^{\mathrm{\prime }}(x)=-3\sin 3x$$

In $(0,\frac{\pi }{2})$, $3x\in (0,\frac{3\pi }{2})$, and in this range $\sin 3x>0$, so

$$f^{\mathrm{\prime }}(x)=-3\sin 3x<0\Rightarrow \cos 3x\text{ is decreasing}$$

Option (D) $f(x)=\tan x$

$$f^{\mathrm{\prime }}(x)={\sec }^{2}x$$

In $(0,\frac{\pi }{2})$, ${\sec }^{2}x>0$

So

$$f^{\mathrm{\prime }}(x)>0\Rightarrow \tan x\text{ is increasing, not decreasing}$$

Final Answer

$$\boxed{{\displaystyle \text{(A) }\cos x,\text{(B) }\cos 2x,\text{(C) }\cos 3x}}$$
$$\boxed{{\displaystyle \tan x\text{ is not decreasing}}}$$

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Question 13

On which of the following intervals is the function

$$f(x)=x^{100}+\sin x-1$$

decreasing?

(A) $(0,1)$
(B) $(\frac{\pi }{2},\pi )$
(C) $(0,\frac{\pi }{2})$
(D) None of these

Solution

Differentiate $f(x)$:

$$f^{\mathrm{\prime }}(x)=100x^{99}+\cos x$$

For $f(x)$ to be decreasing, we need:

$$f^{\mathrm{\prime }}(x)<0$$

Analyze sign of each term

• $100x^{99}>0$ for all $x>0$, because any positive number raised to any power remains positive.

• $\cos x$ in different intervals:

  • In $(0,\frac{\pi }{2})$, $\cos x>0$

  • In $(\frac{\pi }{2},\pi )$, $\cos x<0$

So the only interval where $f^{\mathrm{\prime }}(x)$ might be negative is:

$$(\frac{\pi }{2},\pi )$$

Check sign in this interval:

• $100x^{99}$ is a very large positive number

• $\cos x$ is negative, but lies between $-1$ and $0$

Thus:

$$f^{\mathrm{\prime }}(x)=100x^{99}+\cos x>100x^{99}-1>0$$

Therefore, $f^{\mathrm{\prime }}(x)>0$ everywhere on the given intervals.

So there is no interval from the options where the function is decreasing.

Final Answer

$$\boxed{{\displaystyle \text{(D) None of these}}}$$

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Question 14

For what values of $a$ the function

$$f(x)=x^2+ax+1$$

is increasing on $[1,2]$?

Solution

Differentiate $f(x)$ with respect to $x$:

$$f^{\mathrm{\prime }}(x)=2x+a$$

For $f(x)$ to be increasing on $[1,2]$, we require:

$$f^{\mathrm{\prime }}(x)\ge 0\text{for all }x\in [1,2]$$

That is:

$$2x+a\ge 0$$

Now check the minimum value of $2x+a$ in $[1,2]$.

Since $2x+a$ is a linear increasing function of $x$, the minimum occurs at the left endpoint $x=1$:

$$2(1)+a\ge 0$$

$$2+a\ge 0$$

$$a\ge -2$$

Final Answer

$$\boxed{{\displaystyle f(x)=x^2+ax+1\text{ is increasing on }[1,2]\text{ when }a\ge -2}}$$

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Question 15

Let $I$ be any interval disjoint from $[-1,1]$. Prove that the function

$$f(x)=x+\frac{1}{x}$$is increasing on $I$.

Solution

Given:$$f(x)=x+\frac{1}{x}$$Differentiate with respect to $x$:

$$f^{\mathrm{\prime }}(x)=1-\frac{1}{x^2}$$Rewrite:$$f^{\mathrm{\prime }}(x)=\frac{x^2-1}{x^2}$$
$$f^{\mathrm{\prime }}(x)=\frac{(x-1)(x+1)}{x^2}$$

We need to determine where $f^{\mathrm{\prime }}(x)>0$.

Sign Analysis

• $x^2>0$ for all $x\mathrm{\ne }0$

• So the sign of $f^{\mathrm{\prime }}(x)$ depends on the sign of $(x-1)(x+1)$

$$(x-1)(x+1)>0$$

This product is positive when both factors are positive or both are negative.

Thus:

$$x>1\text{or}x<-1$$

Intervals to Check

Given: The interval $I$ is disjoint from $[-1,1]$.

So $I$ must lie entirely in one of the following:

$$(-\mathrm{\infty },-1)\text{or}(1,\mathrm{\infty })$$

In these intervals:

So,

$$f^{\mathrm{\prime }}(x)>0\text{on both intervals}$$

Conclusion

$$\boxed{{\displaystyle f^{\mathrm{\prime }}(x)>0\text{ everywhere on any interval disjoint from }[-1,1]}}$$

Therefore,

$$\boxed{{\displaystyle f(x)=x+\frac{1}{x}\text{ is increasing on any such interval }I\mathrm{.}}}$$

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Question 16

Prove that the function

$$f(x)=\log (\sin x)$$

is increasing on $(0,\frac{\pi }{2})$ and decreasing on $(\frac{\pi }{2},\pi )$.

Solution

Given:$$f(x)=\log (\sin x)$$

Differentiate using the chain rule:

$$f^{\mathrm{\prime }}(x)=\frac{1}{\sin x}\cdot \cos x=\cot x$$

So:$$f^{\mathrm{\prime }}(x)=\cot x$$

We analyze the sign of $f^{\mathrm{\prime }}(x)$ in the given intervals.

1. In the interval $(0,\frac{\pi }{2})$

In this interval,

• $\sin x>0$

• $\cos x>0$

• Therefore, $\cot x=\frac{\cos x}{\sin x}>0$

So:

$$f^{\mathrm{\prime }}(x)>0\text{for all }x\in (0,\frac{\pi }{2})$$

Hence,

$$\boxed{{\displaystyle f(x)=\log (\sin x)\text{ is increasing on }(0,\frac{\pi }{2})}}$$

2. In the interval $(\frac{\pi }{2},\pi )$

In this interval,

• $\sin x>0$

• $\cos x<0$

• Therefore, $\cot x=\frac{\cos x}{\sin x}<0$

So:$$f^{\mathrm{\prime }}(x)<0\text{for all }x\in (\frac{\pi }{2},\pi )$$

Hence,

$$\boxed{{\displaystyle f(x)=\log (\sin x)\text{ is decreasing on }(\frac{\pi }{2},\pi )}}$$

Final Answer

$$\boxed{{\displaystyle \log (\sin x)\text{ is increasing on }(0,\frac{\pi }{2})\text{ and decreasing on }(\frac{\pi }{2},\pi )}}$$

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Question 17

Prove that the function

$$f(x)=\log \mathrm{\mid }\cos x\mathrm{\mid }$$

is decreasing on $(0,\frac{\pi }{2})$ and increasing on $(\frac{3\pi }{2},2\pi )$.

Solution

Given:

$$f(x)=\log \mathrm{\mid }\cos x\mathrm{\mid }$$

Differentiate using the chain rule:

$$f^{\mathrm{\prime }}(x)=\frac{1}{\mathrm{\mid }\cos x\mathrm{\mid }}\cdot \frac{d}{dx}(\mathrm{\mid }\cos x\mathrm{\mid })$$

Derivative of $\mathrm{\mid }\cos x\mathrm{\mid }$

$$\frac{d}{dx}(\mathrm{\mid }\cos x\mathrm{\mid })=\frac{\cos x}{\mathrm{\mid }\cos x\mathrm{\mid }}\cdot (-\sin x)$$

Therefore:

$$f^{\mathrm{\prime }}(x)=\frac{1}{\mathrm{\mid }\cos x\mathrm{\mid }}\cdot \left(\frac{\cos x}{\mathrm{\mid }\cos x\mathrm{\mid }}\right)(-\sin x)$$

Simplifying:

$$f^{\mathrm{\prime }}(x)=\frac{-\sin x}{\mathrm{\mid }\cos x\mathrm{\mid }}$$

So:

$$\boxed{{\displaystyle f^{\mathrm{\prime }}(x)=\frac{-\sin x}{\mathrm{\mid }\cos x\mathrm{\mid }}}}$$

Check sign of $f^{\mathrm{\prime }}(x)$in the given intervals

1. On $(0,\frac{\pi }{2})$

• $\sin x>0$

• $\cos x>0\Rightarrow \mathrm{\mid }\cos x\mathrm{\mid }=\cos x>0$

So:$$f^{\mathrm{\prime }}(x)=\frac{-\sin x}{\cos x}=-\tan x<0$$

Therefore:

$$\boxed{{\displaystyle f(x)\text{ is decreasing on }(0,\frac{\pi }{2})}}$$

2. On $(\frac{3\pi }{2},2\pi )$

• $\sin x<0$

• $\cos x>0\Rightarrow \mathrm{\mid }\cos x\mathrm{\mid }=\cos x>0$

So:$$f^{\mathrm{\prime }}(x)=\frac{-\sin x}{\cos x}=-\tan x>0$$

(negative of a negative becomes positive)

Therefore:

$$\boxed{{\displaystyle f(x)\text{ is increasing on }(\frac{3\pi }{2},2\pi )}}$$

Final Answer

$$\boxed{{\displaystyle \log \mathrm{\mid }\cos x\mathrm{\mid }\text{ is decreasing on }(0,\frac{\pi }{2})\text{ and increasing on }(\frac{3\pi }{2},2\pi )}}$$

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Question 18

Prove that the function

$$f(x)=x^3-3x^2+3x-100$$

is increasing in $\mathbb{R}$ (the set of all real numbers).

Solution

Differentiate $f(x)$ with respect to $x$:

$$f^{\mathrm{\prime }}(x)=3x^2-6x+3$$

Factorize:

$$f^{\mathrm{\prime }}(x)=3(x^2-2x+1)$$

$$f^{\mathrm{\prime }}(x)=3(x-1{)}^2$$

Analyze the sign of $f^{\mathrm{\prime }}(x)$

• $(x-1{)}^2\ge 0$ for all $x\in \mathbb{R}$, since the square of any real number is non-negative.

• Therefore:

$$3(x-1{)}^2\ge 0\text{ for all }x$$

Thus:

$$f^{\mathrm{\prime }}(x)\ge 0\text{ for all }x\in \mathbb{R}$$

Since the derivative is never negative and is zero only at a single point $x=1$, the function does not decrease anywhere.

Conclusion

$$\boxed{{\displaystyle f^{\mathrm{\prime }}(x)\ge 0\text{ for all real numbers }x}}$$
$$\boxed{{\displaystyle \text{Hence, }f(x)=x^3-3x^2+3x-100\text{ is increasing on }\mathbb{R}\mathrm{.}}}$$

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Question 19

The interval in which

$$y=x^2{e}^{-x}$$

is increasing is:

(A) $(-\mathrm{\infty },\mathrm{\infty })$
(B) $(-2,0)$
(C) $(2,\mathrm{\infty })$
(D) $(0,2)$

Solution

Given:

$$y=x^2{e}^{-x}$$

Differentiate using product rule:

$$y^{\mathrm{\prime }}=(x^2{)}^{\mathrm{\prime }}{e}^{-x}+x^2(e^{-x}{)}^{\mathrm{\prime }}$$

$$y^{\mathrm{\prime }}=2x{e}^{-x}+x^2(-e^{-x})$$

$$y^{\mathrm{\prime }}=e^{-x}(2x-x^2)$$

Factor further:

$$y^{\mathrm{\prime }}=e^{-x}x(2-x)$$

Analyze the sign of $y^{\mathrm{\prime }}$

$$y^{\mathrm{\prime }}=e^{-x}x(2-x)$$

• $e^{-x}>0$ for all real $x$

• So the sign of $y^{\mathrm{\prime }}$ depends only on $x(2-x)$

$$x(2-x)>0$$

Solve inequality:

• Product is positive when both factors have the same sign.

Case 1:

$$x>0\text{and}2-x>0\Rightarrow x<2$$

So:

$$0<x<2$$

Case 2 would be:

$$x<0\text{and}2-x<0\Rightarrow x>2$$

Impossible.

Therefore, the function is increasing only in:

$$(0,2)$$

Final Answer

$$\boxed{{\displaystyle \text{The function is increasing on }(0,2)}}$$

$$\boxed{{\displaystyle \text{Correct option: (D)}}}$$