3 Marks

Question 13 Marks

Write the set of values of a for which $\text{f}(\text{x})=\cos\text{x}+\text{a}^2\text{x}+\text{b}$ is strictly increasing on R.

Answer

$\text{f}(\text{x})=\cos\text{x}+\text{a}^2\text{x}+\text{b}$

$\text{f}'(\text{x})=\text{a}^2-\sin\text{x}$

Given: f(x) is strictly increasing on R.

$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in\text{R}$

$\Rightarrow\text{a}^2-\sin\text{x}>0,\forall\ \text{x}\in\text{R}$

$\Rightarrow\text{a}^2>\sin\text{x},\forall\ \text{x}\in\text{R}$

We know that the maximum value of $\sin\text{x}$ is 1.

Since, $\text{a}^2>\sin\text{x},\text{a}^2$ is always greater than 1.

$\Rightarrow\text{a}^2>1$

$\Rightarrow\text{a}^2-1>0$

$\Rightarrow(\text{a}+1)(\text{a}-1)>0$

$\Rightarrow\text{a}\in(-\infty,-1)\cup(1,\infty)$

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Question 23 Marks

Find the set of values of 'b' for which $\text{f}(\text{x})=\text{b}(\text{x}+\cos\text{x})+4$ is decreasing on R.

Answer

$\text{f}(\text{x})=\text{b}(\text{x}+\cos\text{x})+4$
$\text{f}'(\text{x})=\text{b}(1-\sin\text{x})$
Given: f(x) is decreasing on R.
$\Rightarrow\text{f}'(\text{x})<0$
$\Rightarrow\text{b}(1-\sin\text{x})<0\ ....(1)$
We know,
$\sin\text{x}\leq1$
$\Rightarrow1-\sin\text{x}\geq0$
$\Rightarrow\text{b}<0$ $[\text{Since }(1-\sin\text{x})\geq0,\text{b}(1-\sin\text{x})<0\Rightarrow\text{b}<0]$
$\Rightarrow\text{b}\in(-\infty,0)$

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Question 33 Marks

Find the intervals in which the following functions are increasing or decreasing.
f(x) = x⁸ + 6x²

Answer

We have

f(x) = x⁸ + 6x²

$\therefore$ f'(x) = 8x⁷ + 12x

Critical points

f'(x) = 0

⇒ 8x⁷ + 12x = 0

⇒ 4x(2x⁶ + 3) = 0

⇒ x = 0

Clearly, f'(x) > 0 if x < 0

f'(x) < 0 if x < 0

Thus, f(x) increases in $(0,\infty),$ decreases in $(-\infty,0).$

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Question 43 Marks

Find the set of values of 'a' for which $\text{f}(\text{x})=\text{x}+\cos\text{x}+\text{ax}+\text{b}$ is increasing on R.

Answer

$\text{f}(\text{x})=\text{x}+\cos\text{x}+\text{ax}+\text{b}$

$\text{f}'(\text{x})=1-\sin\text{x}+\text{a}$

For f(x) to be increasing we must have

$\text{f}'(\text{x})>0$

$\Rightarrow1-\sin\text{x}+\text{a}>0$

$\Rightarrow\sin\text{x}<1+\text{a}$

We know that the maximum value of $\sin\text{x}$ is 1.

$\Rightarrow1+\text{a}>1$

$\Rightarrow\text{a}>0$

$\Rightarrow\text{a}\in(0,\infty)$

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Question 53 Marks

Write the set of values of k for which $\text{f}(\text{x})=\text{k}\text{x}-\sin\text{x}$ is increasing on R.

Answer

$\text{f}(\text{x})=\text{k}\text{x}-\sin\text{x}$

$\text{f}'(\text{x})=\text{k}-\cos\text{x}$

For, f(x) to be increasing, we must have

$\text{f}'(\text{x})>0$

$\Rightarrow\text{k}-\cos\text{x}>0$

$\Rightarrow\cos\text{x}<\text{k}$

We know that the maximum value of $\cos\text{x}$ is 1.

Since $\cos\text{x}<\text{k},$ the minimum value of k is 1.

$\Rightarrow\text{k}\in(1,\infty)$

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Question 63 Marks

Find the intervals in which f(x) is increasing or decreasing:
$\text{f}(\text{x})=\text{x}|\text{x}|,\text{x}\in\text{R}$

Answer

$\text{f}(\text{x})=\text{x}|\text{x}|,\text{x}\in\text{R}$

Case I:

When $\text{x}\geq0$

$\text{f}(\text{x})=\text{x}|\text{x}|=\text{x}(\text{x})=\text{x}^2$

$\Rightarrow\text{f}'(\text{x})=2\text{x}\geq0\ \forall\ \text{x}\geq0$

So, f(x) is increasing for $\text{x}\geq0.$

Case II:

When $\text{x}<0$

$\text{f}(\text{x})=\text{x}|\text{x}|=\text{x}(-\text{x})=-\text{x}^2$

$\Rightarrow\text{f}'(\text{x})=-2\text{x}\geq0\ \forall\ \text{x}<0$

So, f(x) is increasing for x < 0.

Hence f(x) is increasing for $\text{x}\in\text{R}.$

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Question 73 Marks

Show that $\text{f}(\text{x})=\log\sin\text{x}$ is increasing on $\Big(0,\frac{\pi}{2}\Big)$ and decreasing on $\Big(\frac{\pi}{2},\pi\Big).$

Answer

$\text{f}(\text{x})=\log\sin\text{x}$
$\text{f}'(\text{x})=\frac{1}{\sin\text{x}}\cos\text{x}=\cot\text{x}$
In interval $\Big(0,\frac{\pi}{2}\Big),\text{f}'(\text{x})=\cot\text{x}>0.$
$\therefore$ f is strictly increasing in $\Big(0,\frac{\pi}{2}\Big).$
In interval $\Big(\frac{\pi}{2},\pi\Big),\text{f}'(\text{x})=\cot\text{x}<0.$
$\therefore$ f is strictly decreasing in $\Big(\frac{\pi}{2},\pi\Big).$

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Question 83 Marks

Prove that the function $\text{f}(\text{x})=\log_{\text{e}}\text{x}$ is increasing on $(0,\infty)$ if a > 1 and decreasing on $(0,\infty)$ if 0 < a < 1.

Answer

Case I:

When $\text{a}>1$

Let $\text{x}_1,\text{x}_2\in(0,\infty)$

We have

$\text{x}_1<\text{x}_2$

$\Rightarrow\log_{\text{a}}\text{x}_1<\log_{\text{a}}\text{x}_2$

$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$

Thus, f(x) is increasing on $(0,\infty)$

Case II:

When $0<\text{a}<1$

$\text{f}(\text{x})=\log_{\text{a}}\text{x}=\frac{\log\text{x}}{\log\text{a}}$

When $\text{a}<1\Rightarrow\log_\text{a}<0$

Let $\text{x}_1<\text{x}_2$

_{$\Rightarrow\log\text{x}_1<\log\text{x}_2$}

_{$\Rightarrow\frac{\log\text{x}_1}{\log_\text{a}}>\frac{\log\text{x}_2}{\log_\text{a}}$ $[\because\ \log\text{a}<0]$}

_{$\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)$}

So, f(x) is increasing on $(0,\infty).$

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Question 93 Marks

Prove that the function f given by f(x) = x - [x] is increasing in (0, 1).

Answer

f(x) = x - [x]

Let $\text{x}_1,\text{x}_2\in(0,1)$ such that x₁ < x_2.Then

[x₁] = [x₂] = 0 ....(1)

Now,

x₁ < x₂

⇒ x₁ - [x₁] < x₂ - [x₂] [From eq. (1)]

⇒ f(x₁) < f(x₂)

$\therefore$ x₁ < x₂

$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,1)$

Hence, f(x) is increasing on (0, 1).

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Question 103 Marks

Show that f(x) = x⁹ + 4x⁷ + 11 is an increasing function for all $\text{x}\in\text{R}.$.

Answer

f(x) = x⁹ + 4x⁷ + 11
f'(x) = 9x⁸ + 28x⁶
= x⁶(9x² + 28)
Now,
$\text{x}\in\text{R}$
⇒ x⁶ > 0 and 9x² + 28 > 0
⇒ x⁶(9x² + 28) > 0
⇒ f'(x) > 0
So, f(x) is increasing on function for $\text{x}\in\text{R}.$

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Question 113 Marks

Show that f(x) = e^2x is increasing on R.

Answer

f(x) = e^2x
f'(x) = 2e^2x
Now,
$\text{x}\in\text{R}$
Since the value of e^2x is always positive for any real value of x, e^2x > 0.
⇒ 2e^2x > 0
⇒ f'(x) > 0
So, f(x) is increasing on R.

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Question 123 Marks

Find the intervals in which the following functions are increasing or decreasing.
f(x) = 2x³ - 9x² - 12x + 1

Answer

f(x) = 2x³ - 9x² - 12x + 1
f'(x) = -6x² - 18x - 12
Critical points
f'(x) = 0
⇒ -6x² - 18x - 12 = 0
⇒ x² + 3x + 2 = 0
⇒ (x + 2)(x + 1) = 0
⇒ x = -2, -1
Clearly, f'(x) > 0 if x < -1 and x > -2
f'(x) < 0 if -2 < x < -1
Thus, f(x) increasing in (-2, -1) decreasing in $(-\infty,-2)\cup(-1,\infty).$

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Question 133 Marks

Show that $\text{f}(\text{x})=\frac{1}{\text{x}}$ is decreasing function on $(0,\infty).$

Answer

We have,

$\text{f}(\text{x})=\frac{1}{\text{x}}$

Let, $\text{x}_1,\text{x}_2\in(0,\infty)$ and $\text{x}_1>\text{x}_2$

$\Rightarrow\frac{1}{\text{x}_1}<\frac{1}{\text{x}_2}$

$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$

Thus, $\text{x}_1>\text{x}_2\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$

So, f(x) is decreasing function.

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Question 143 Marks

Find the intervals in which f(x) is increasing or decreasing:
$\text{f}(\text{x})=\sin\text{x}+|\sin\text{x}|,0<\text{x}\leq2\pi$

Answer

$\text{f}(\text{x})=\sin\text{x}+|\sin\text{x}|,0<\text{x}\leq2\pi$

Case I:

When $\text{x}\in(0,\pi)$

$\text{f}(\text{x})=\sin\text{x}+\sin\text{x}=2\sin\text{x}$

$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}$

As, $\cos\text{x}>0$ for $\text{x}\in\Big(0,\frac{\pi}{2}\Big)$ and $\cos\text{x}<0$ for $\text{x}\in\Big(\frac{\pi}{2},\pi\Big)$

So, $\text{f}'(\text{x})>0$ for $\text{x}\in\Big(0,\frac{\pi}{2}\Big)$ and $\text{f}'(\text{x})<0$ for $\text{x}\in\Big(\frac{\pi}{2},\pi\Big)$

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

Case II:

When $\text{x}\in(\pi,2\pi)$

$\text{f}(\text{x})=\sin\text{x}-\sin\text{x}=0$

$\Rightarrow\text{f}'(\text{x})=0$

$\therefore$ f(x) is neither increasing nor decreasing on $(\pi,2\pi).$

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Question 153 Marks

Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.

Answer

Here,
f(x) = ax + b
Let $\text{x}_1,\text{x}_2\in\text{R}$ such that x₁< x₂. Then,
x₁< x₂
⇒ ax₁ < ax₂$[\because\ \text{a}>0]$
⇒ ax₁ + b < ax₂ + b
⇒ f(x₁) < f(x₂)
$\therefore$ x₁ < x₂
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\text{x}_1,\text{x}_2\in\text{R}$
So, f(x) is increasing on R.

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Question 163 Marks

Find 'a' for which $\text{f}(\text{x})=\text{a}(\text{x}+\sin\text{x})+\text{a}$ is increasing on R.

Answer

$\text{f}(\text{x})=\text{a}(\text{x}+\sin\text{x})+\text{a}$
$\text{f}'(\text{x})=\text{a}(1+\cos\text{x})$
For f(x) to be increasing, we must have
$\text{f}'(\text{x})>0$
$\Rightarrow\text{a}(1+\cos\text{x})>0\ ....(1)$
We know,
$-1\leq\cos\text{x}\leq1,\forall\ \text{x}\in\text{R}$
$\Rightarrow0\leq(1+\cos\text{x})\leq2,\forall\ \text{x}\in\text{R}$
$\therefore\ \text{a}>0$ [From eq. (1)]
$\Rightarrow\text{a}\in(0,\infty)$

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Question 173 Marks

Find the intervals in which the following functions are increasing or decreasing.
f(x) = 2x³- 9x² + 12x - 5

Answer

We have,
f(x) = 2x³- 9x² + 12x - 5
$\therefore$ f'(x) = 6x² - 18x + 12
Critical points
f'(x) = 0
⇒ 6(x² - 3x + 2) = 0
⇒ (x - 2)(x - 1) = 0
⇒ x = 2, 1
Clearly, f'(x) > 0 if x < 1 and x > 6
f'(x) < 0 if 1 < x < 2
Thus, f(x) increases in $(-\infty,1)\cup(2,\infty),$ decreases in (1, 2).

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Question 183 Marks

Find the intervals in which the following functions are increasing or decreasing.
f(x) = x³ - 6x² - 36x + 2

Answer

f(x) = x³ - 6x² - 36x + 2
$\therefore$ f'(x) = 3x² - 12x - 36
Critical point
f'(x) = 0
⇒ 3(x² - 4x - 12) = 0
⇒ (x - 6)(x + 2) = 0
⇒ x = 6, -2
Clearly, f'(x) > 0 if x < -2 and x > 6
f'(x) < 0 if -2x < x < 6
Thus, f(x) increases in $(-\infty,-2)\cup(6,\infty),$ decreases in (-2, 6).

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Question 193 Marks

State when a function f(x) is said to be increasing on an interval [a, b]. Test whether the function f(x) = x² - 6x + 3 is increasing on the interval [4, 6].

Answer

A function f(x) is said to be increasing on an interval [a, b] if it is increasing at x = a and x = b.
Here,
f(x) = x² - 6x + 3
f'(x) = 2x - 6
f'(x) = 2(x - 3)
Now, f'(4) = 2(4 - 3)
= 2
$\therefore$ f'(4) > 0
So, f(x) is increasing on x = 4
f'(6) = 2(6 - 3)
= 6
$\therefore$ f'(6) > 0
So, f(x) is increasing on x = 6
Hence, f(x) is increasing on [4, 6].

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Question 203 Marks

Without using the derivative show that the function f(x) = 7x - 3 is strictly increasing function on R.

Answer

Here,
$\text{f}(\text{x})=7\text{x}-3$
Let $\text{x}_1,\text{x}_2\in\text{R}$ such that $\text{x}_1<\text{x}_2.$ Then,
$\text{x}_1<\text{x}_2$
$\Rightarrow7\text{x}_1<7\text{x}_2$ $[\because\ 7>0]$
$\Rightarrow7\text{x}_1-3<7\text{x}_2-3$
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$
$\therefore\ \text{x}_1<\text{x}_2\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in\text{R}$
So, f(x) is strictly increasing on R.

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Question 213 Marks

Determine whether $\text{f}(\text{x})=\frac{-\pi}{2}+\sin\text{x}$ is a increasing or decreasing on $\Big(\frac{-\pi}{3},\frac{\pi}{3}\Big).$

Answer

$\text{f}(\text{x})=\frac{-\pi}{2}+\sin\text{x}$
$\text{f}'(\text{x})=\frac{-1}{2}+\cos\text{x}$
Here,
$\frac{-\pi}{3}<\text{x}<\frac{\pi}{3}$
$\Rightarrow\cos\text{x}>\frac{1}{2}$
$\Rightarrow\frac{-1}{2}+\cos\text{x}>0$
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in\Big(\frac{-\pi}{3},\frac{\pi}{3}\Big)$
So, f(x) is increasing on $\Big(\frac{-\pi}{3},\frac{\pi}{3}\Big).$

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Question 223 Marks

Without using the derivative, show that the function f(x) = |x| is

Strictly increasing in $(0,\infty)$
Strictly decreasing in $(-\infty,0)$

Answer

We have,

$\text{f}(\text{x})=|\text{x}|=\begin{cases}\text{x},&\text{x}>0\\\text{-x},&\text{x}<0\end{cases}$

Let $\text{x}_1,\text{x}_2\in(0,\infty)$ and $\text{x}_1>\text{x}_2$

$\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)$

So, f(x) is increasing in $(0,\infty).$

Let $\text{x}_1,\text{x}_2\in (-\infty,0)$ and $\text{x}_1>\text{x}_2$

$\Rightarrow-\text{x}_1<-\text{x}_2$

$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$

So, f(x) is decreasing on $(-\infty,0).$

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Question 233 Marks

Write the interval in which $\text{f}(\text{x})=\sin\text{x}+\cos\text{x},\text{x}\in\Big[0,\frac{\pi}{2}\Big]$ is increasing.

Answer

$\text{f}(\text{x})=\sin\text{x}+\cos\text{x},\text{x}\in\Big[0,\frac{\pi}{2}\Big]$

$\text{f}'(\text{x})=\cos\text{x}-\sin\text{x}$

For f(x) to be increasing, we must have

$\text{f}'(\text{x}) >0$

$\Rightarrow\cos\text{x}-\sin\text{x}>0$

$\Rightarrow\sin\text{x}<\cos\text{x}$

$\Rightarrow\frac{\sin\text{x}}{\cos\text{x}}<1$

$\Rightarrow\tan\text{x}<1$

$\Rightarrow\text{x}\in\Big[0,\frac{\pi}{4}\Big)$

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Question 243 Marks

Show that $\text{f}(\text{x})=\cos^2\text{x}$ is a decreasing function on $\Big(0,\frac{\pi}{2}\Big).$

Answer

We have,
$\text{f}(\text{x})=\cos^2\text{x}$
$\therefore\ \text{f}'(\text{x})=2\cos\text{x}(-\sin\text{x})$
$\Rightarrow\text{f}'(\text{x})=-2\sin\text{x}\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=-\sin2\text{x}$
Now,
$\text{x}\in\Big(0,\frac{\pi}{2}\Big)$
$\Rightarrow2\text{x}\in(0,\pi)$
$\Rightarrow\sin2\text{x}>0$ when $2\text{x}\in(0,\pi)$
$\Rightarrow-\sin2\text{x}<0$
$\Rightarrow\text{f}'(\text{x})<0$
So, f(x) is decreasing function on $\Big(0,\frac{\pi}{2}\Big).$

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Question 253 Marks

Find the values of 'a' for which hte function $\text{f}(\text{x})=\sin\text{x}-\text{a}\text{x}+4$ is increasing function on R.

Answer

$\text{f}(\text{x})=\sin\text{x}-\text{a}\text{x}+4$
$\text{f}'(\text{x})=\cos\text{x}-\text{a}$
Given: f(x) is increasing on R.
$\Rightarrow\text{f}'>\cos\text{x}-\text{a}$
$\Rightarrow\cos\text{x}>\text{a}$
We know,
$\cos\text{x}\geq-1,\forall\ \text{x}\in\text{R}$
$\Rightarrow\text{a}<-1$
$\Rightarrow\text{a}\in(-\infty,-1)$

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Question 263 Marks

Find the intervals in which the following functions are increasing or decreasing.
f(x) = 6+ 12x + 3x² - 2x³

Answer

f(x) = 6+ 12x + 3x² - 2x³
f'(x) = 12 + 6x - 6x²
= -6(x² - x - 2)
= -6(x - 2)(x + 1)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ -6(x - 2)(x + 1) > 0
⇒ (x - 2)(x + 1) < 0
[Since, -6 < 0, -6(x - 2)(x + 1) > 0 ⇒ (x - 2)(x + 1) < 0]
⇒ -1 < x < 2
$\Rightarrow\text{x}\in(-1,2)$
So, f(x) is increasing on (-1, 2).
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ -6(x - 2)(x + 1) < 0
⇒ (x - 2)(x + 1) < 0
[Since, -6 < 0, -6(x - 2)(x + 1) > 0 ⇒ (x - 2)(x + 1) > 0]
⇒ x < -1 or x > 2
$\Rightarrow\text{x}\in(-\infty,-1)\cup(2,\infty)$
So, f(x) is decreasing on $(-\infty,-1)\cup(2,\infty).$

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Question 273 Marks

Show that f(x) = (x - 1)e^x + 1 is an increasing function for all x > 0.

Answer

f(x) = (x - 1)e^x + 1

f'(x) = (x - 1)e^x + e^x

= xe^x - e^x + e^x

= xe^x

Given: x > 0

We know,

e^x> 0

⇒ xe^x > 0

⇒ f'(x) > 0, for all x > 0

So, f(x) is increasing on for all x > 0.

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Question 283 Marks

Show that $\text{f}(\text{x})=\text{x}-\sin\text{x}$ is increasing for all $\text{x}\in\text{R}.$

Answer

$\text{f}(\text{x})=\text{x}-\sin\text{x}$
$\text{f}'(\text{x})=1-\cos\text{x}$
For f(x) to be increasing, we must have
$\text{f}'(\text{x})>0$
$\Rightarrow1-\cos\text{x}>0$
$\Rightarrow\text{f}'(\text{x})\geq0$ for all $\text{x}\in\text{R}$ $[\because\ \cos\text{x}\leq1]$
So, f(x) is increasing for all $\text{x}\in\text{R}.$

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Question 293 Marks

Show that $\text{f}(\text{x})=\tan^{-1}\text{x}-\text{x}$ is a decreasing function on R.

Answer

$\text{f}(\text{x})=\tan^{-1}\text{x}-\text{x}$
$\text{f}'(\text{x})=\frac{1}{1+\text{x}^2}-1$
$=\frac{1-1-\text{x}^2}{1+\text{x}^2}$
$=\frac{-\text{x}^2}{1+\text{x}^2}$
We know,
$\text{x}^2\geq0,1+\text{x}^2>0,\forall\ \text{x}\in\text{R}$
$\therefore\ \frac{-\text{x}^2}{1+\text{x}^2}<0,\forall\ \text{x}\in\text{R}$
$\Rightarrow\text{f}'(\text{x})<0,\forall\ \text{x}\in\text{R}$
Hence, f(x) is decreasing on R.

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Question 303 Marks

Show that the function f given by $\text{f}(\text{x})=10^\text{x}$ is increasing for all x.

Answer

We have,
$\text{f}(\text{x})=10^\text{x}$
$\therefore\ \text{f}'(\text{x})=10^{\text{x}}\times\log10$
Now,
$\text{x}\in\text{R}$
$\Rightarrow10^\text{x}>0$
$\Rightarrow10^\text{x}\log10>0$
$\Rightarrow\text{f}'(\text{x})>0$
Hence, f(x) in an increasing function for all x.

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Question 313 Marks

Prove that the function f given by f(x) = x³ - 3x² + 4x is strictly increasing on R.

Answer

f(x) = x³ - 3x² + 4x
f'(x) = 3x² - 6x + 4
= 3(x² - 2x) + 4
= 3(x² - 2x + 1) - 3 + 4
$=2(\text{x}-1)^2+1>0,\forall\ \text{x}\in\text{R}$
Hence, f(x) is strictly increasing on R.

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Question 323 Marks

Show that $\text{f}(\text{x})=\text{e}^{\frac{1}{\text{x}}},\text{x}\neq0$ is a decreasing function for all $\text{x}\neq0.$

Answer

We have,
$\text{f}(\text{x})=\text{e}^{\frac{1}{\text{x}}},\text{x}\neq0$
$\text{f}'(\text{x})=\text{e}^{\frac{1}{\text{x}}}\times\Big(\frac{-1}{\text{x}^2}\Big)$
$\therefore\ \text{f}'(\text{x})=-\frac{\text{e}^{\frac{1}{\text{x}}}}{\text{x}^2}$
Now,
$\text{x}\in\text{R},\text{x}\neq0$
$\Rightarrow\frac{1}{\text{x}^2}>0\text{ and }\text{e}^{\frac{1}{\text{x}}}>0$
$\Rightarrow\frac{\text{e}^{\frac{1}{\text{x}}}}{\text{x}^2}>0$
$\Rightarrow-\frac{\text{e}^{\frac{1}{\text{x}}}}{\text{x}^2}<0$
$\Rightarrow\text{f}'(\text{x})<0$
Hence, f(x) is decreasing function for all $\text{x}\neq0.$

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Question 333 Marks

Prove that f(x) = ax + b, where a, b are constants and a < 0 is an decreasing function on R.

Answer

Here,

f(x) = ax + b

Let $\text{x}_1,\text{x}_2\in\text{R}$ such that x₁ < x₂.

Then,

x₁ < x₂

⇒ ax₁ > ax₂$(\because\ \text{a}<0)$

⇒ ax₁ + b > ax₂ + b

⇒ f(x₁) > f(x₂)

Thus, x₁ < x₂

$\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in\text{R}$

So, f(x) is decreasing on R.

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