2 Marks

Question 12 Marks

Show that $\text{f}(\text{x})=\text{x}+\cos\text{x}-\text{a}$ is an increasing function on R for all values of a.

Answer

We have,

$\text{f}(\text{x})=\text{x}+\cos\text{x}-\text{a}$

$\therefore\ \text{f}'(\text{x})=1-\sin\text{x}=\frac{2\cos^2\text{x}}{2}$

Now,

$\text{x}\in\text{R}$

$\Rightarrow\frac{\cos^2\text{x}}{2}>0$

$\Rightarrow\frac{2\cos^2\text{x}}{2}>0$

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

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

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Question 22 Marks

Find the values of b for which the function $\text{f}(\text{x})=\sin\text{x}-\text{b}\text{x}+\text{c}$ is a decreasing function on R.

Answer

$\text{f}(\text{x})=\sin\text{x}-\text{b}\text{x}+\text{c}$

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

Given: f(x) is decreasing on R.

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

$\Rightarrow\cos\text{x}-\text{b}<0\ \forall\ \text{x}\in\text{R}$

$\Rightarrow\cos\text{x}<\text{b},\forall\ \text{x}\in\text{R}$

$\Rightarrow\text{b}\geq1$ $[\because\ -1\leq\cos\text{x}\leq1]$

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Question 32 Marks

Write the set of values of a for which the function f(x) = ax + b is decreasing for all $\text{x}\in\text{R}.$

Answer

f(x) = ax + b
f'(x) = a
For f(x) to be decreasing, we must have
f'(x) < 0
⇒ a < 0
$\Rightarrow\text{a}\in(-\infty,0)$

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Question 42 Marks

Prove that the function $\text{f}(\text{x})=\log_{\text{e}}\text{x}$ is increasing on $(0,\infty).$

Answer

Let $\text{x}_1,\text{x}_2\in(0,\infty)$ such that x₁ < x₂. Then

x₁ < x₂

Implies that$\log_{\text{e}}\text{x}_1<\log_{\text{e}}\text{x}_2$

Implies that f(x₁) < f(x₂)

$\therefore$ x₁ < x₂ Implies that $\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,\infty)$

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

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Question 52 Marks

Prove that the function $\text{f}(\text{x})=\cos\text{x}$ is:
Strictly increasing in $(\pi,2\pi)$

Answer

$\text{f}(\text{x})=\cos\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}$
Here,
$\pi<\text{x}<2\pi$
$\Rightarrow\sin\text{x}<0$ $[\because$ sine function is negative in third and fourth quadrent$]$
$\Rightarrow-\sin\text{x}>0$
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in(\pi,2\pi)$
So, f(x) is strictly decreasing on $(\pi,2\pi).$

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Question 62 Marks

Find the value(s) of a for which f(x) = x³ - ax is an increasing function on R.

Answer

f(x) = x³ − ax
f'(x) = 3x² − a
Given: f(x) is increasing on R.
$\Rightarrow\text{f}'(\text{x})\geq0\ \forall\ \text{x}\in\text{R}$
$\Rightarrow3\text{x}^2-\text{a}\geq0\ \forall\ \text{x}\in\text{R}$
$\Rightarrow\text{a}\leq3\text{x}^2\ \forall\ \text{x}\in\text{R}$
The least value of 3x² is 0.
$\therefore\ \text{a}\leq0$

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Question 72 Marks

Prove that the function f(x) = x³ - 6x² + 12x - 18 is increasing on R.

Answer

f(x) = x³ - 6x² + 12x - 18
f'(x) = 3x² - 12x + 12
= 3(x² - 4x + 4)
$= 3(\text{x} - 2)^2\geq0,\forall\text{x}\in\text{R}$ $[3>0\ \&(\text{x}-2)^2\geq0]$
So, f(x) is increasing on R.

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Question 82 Marks

Show that $\text{f}(\text{x})=\tan\text{x}$ is an increasing function on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big).$

Answer

We have,
$\text{f}(\text{x})=\tan\text{x}$
$\therefore\ \text{f}'(\text{x})=\sec^2\text{x}$
Now,
$\text{x}\in\Big(\frac{-\pi}{2}\frac{\pi}{2}\Big)$
$\Rightarrow\sec^2\text{x}>0$
$\Rightarrow\text{f}'(\text{x})>0$
Hence, f(x) is increasing function on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big).$

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Question 92 Marks

Prove that the function $\text{f}(\text{x})=\cos\text{x}$ is:
Strictly decreasing in $(0,\pi).$

Answer

$\text{f}(\text{x})=\cos\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}$
Here,
$0<\text{x}<\pi$
$\Rightarrow\sin\text{x}>0$ $[\because$ sine function is positive in first and second quadrent$]$
$\Rightarrow-\sin\text{x}<0$
$\Rightarrow\text{f}'(\text{x})<0,\forall\ \text{x}\in(0,\pi)$
So, f(x) is strictly decreasing on $(0,\pi).$

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Question 102 Marks

If g (x) is a decreasing function on R and $\text{f}(\text{x})=\tan^{-1}[\text{g}(\text{x})].$ State whether f(x) is increasing or decreasing on R.

Answer

Given: g(x) is decreasing on R.
$\Rightarrow\text{x}_1<\text{x}_2$
$\Rightarrow\text{g}(\text{x}_1)>\text{g}(\text{x}_2)$
Applying $\tan^{-1}$ on both sides we get,
$\Rightarrow\tan^{-1}\{\text{g}(\text{x}_1)\}>\tan^{-1}\{\text{g}(\text{x}_2)\}$
$\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)$

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Question 112 Marks

State whether $\text{f}(\text{x})=\tan\text{x}-\text{x}$ is increasing or decreasing its domain.

Answer

$\text{f}(\text{x})=\tan\text{x}-\text{x}$

$\text{f}'(\text{x})=\sec^2\text{x}-1$

$\tan^{2}\text{x}\geq0,\forall\ \text{x}\in[0,2\pi]$

So, f(x) is increasing in its domain.

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Question 122 Marks

Show that f(x) = x³ - 15x² + 75x - 50 is an increasing function for all $\text{x}\in\text{R}.$

Answer

f(x) = x³ - 15x² + 75x - 50
f'(x) = 3x² - 30x + 75
= 3(x²- 10x + 25)
$=3(\text{x}-5)^2>0,\forall\ \text{x}\in\text{R}$ $[\because$ Square of any function is always greater than zero$]$
So, f(x) is an increasing function for all $\text{x}\in\text{R}.$

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Question 132 Marks

Let f defined on [0, 1] be twice differentiable such that $|\text{f}'(\text{x})|\leq1$ for all $\text{x}\in[0,1].$ If f(0) = f(1), then show that $|\text{f}'(\text{x})|<1$ for all $\text{x}\in[0,1].$

Answer

If a function is continuous and differentiable and f(0) = f(1) in given domain $\text{x}\in[0,1],$
Then by Rolle's theorem:
f'(x) = 0 for some $\text{x}\in[0,1]$
Given: $|\text{f}'(\text{x})|\leq1$
On integrating boty sides we get,
$|\text{f}'(\text{x})|\leq\text{x}$
Now, within interval $\text{x}\in[0,1]$
We get, $|\text{f}'(\text{x})|<1.$

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Question 142 Marks

Prove that the following function are increasing on R.
f(x) = 3x⁵ + 40x³ + 240x

Answer

f(x) = 3x⁵ + 40x³ + 240x
f'(x) = 15x⁴ + 120x² + 240
= 15(x⁴ + 8x² +16)
$=15(\text{x}^2+4)^2>0,\forall\ \text{x}\in\text{R}$ $[\because\ 15>0\text{ and }(\text{x}^2+4)^2>0]$
So, f(x) increasing on R.

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Question 152 Marks

Prove that the function $\text{f}(\text{x})=\cos\text{x}$ is:
Neither increasing nor decreasing in $(0,2\pi)$

Answer

$\text{f}(\text{x})=\cos\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}$
$\Rightarrow\text{f}'(\text{x})<0,\forall\ \text{x}\in(0,\pi)\ ....(1)$
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in(\pi,2\pi)\ ....(2)$
From eqs. (1) and (2), we get
f(x) is strictly decreasing on $(0,\pi)$ and is strictly increasing on $(\pi,2\pi).$
So, f(x) Neither increasing nor decreasing on $(0,2\pi).$

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Question 162 Marks

Show that $\text{f}(\text{x})=\log_{\text{a}}\text{x},0<\text{a}<1$ is a decreasing function for all x > 0.

Answer

$\text{f}(\text{x})=\log_{\text{a}}\text{x}$
$=\frac{\log\text{x}}{\log\text{a}}$
$\text{f}'(\text{x})=\frac{1}{\text{x}\log\text{a}}$
Since, $0<\text{a}<1$ and $\text{f}'(\text{x})=\frac{1}{\text{x}\log\text{a}}<0.$
Hence, f(x) is decreasing function for all x > 0.

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Question 172 Marks

Show that $\text{f}(\text{x})=\sin\text{x}$ is an increasing function on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big).$

Answer

$\text{f}(\text{x})=\sin\text{x}$
$\text{f}'(\text{x})=\cos\text{x}>0\ \forall\ \text{x}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$
$[\because$ Cos function is positive in first and fourth quadrant$]$
So, f(x) is increasing on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big).$

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Question 182 Marks

Prove that the following function are increasing on R.
f(x) = 4x³ + 18x² + 27x - 27

Answer

f(x) = 4x³ + 18x² + 27x - 27
⇒ f'(x) = 12x² + 36x + 27
⇒ f'(x) = 3(4x² - 12x +9)
$\Rightarrow\text{f}'(\text{x})=3(2\text{x}-3)^2>0,\forall\ \text{x}\in\text{R}$
So, f(x) increasing on R.

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Question 192 Marks

Show that $\text{f}(\text{x})=\text{x}^2-\text{x}\sin\text{x}$ is an increasing function on $\Big(0,\frac{\pi}{2}\Big).$

Answer

We have,
$\text{f}(\text{x})=\text{x}^2-\text{x}\sin\text{x}$
$\therefore\ \text{f}'(\text{x})=2\text{x}-\sin\text{x}-\text{x}\cos\text{x}$
Now,
$\text{x}\in\Big(0,\frac{\pi}{2}\Big)$
$\Rightarrow0\leq\sin\text{x}\leq1,0\leq\cos\text{x}\leq1$
$\Rightarrow2\text{x}-\sin\text{x}-\text{x}\cos\text{x}>0$
$\Rightarrow\text{f}'(\text{x})\geq0$
So, f(x) is strictly increasing function on $\Big(0,\frac{\pi}{2}\Big).$

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