Question 13 Marks
Show that $f(x) = x^9 + 4x^7 + 11$ is an increasing function for all $\text{x}\in\text{R}.$.
Answer$f(x) = x^9 + 4x^7 + 11$
$f'(x) = 9x^8 + 28x^6$
$= x^6(9x^2 + 28)$
Now,
$\text{x}\in\text{R}$
$\Rightarrow x^6 > 0$ and $9x^2 + 28 > 0$
$\Rightarrow x^6(9x^2 + 28) > 0$
$\Rightarrow f'(x) > 0$
So, $f(x)$ is increasing on function for $\text{x}\in\text{R}.$
View full question & answer→Question 23 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 $x, e^{2x} > 0.$
$\Rightarrow 2e^{2x} > 0$
$\Rightarrow f'(x) > 0$
So, $f(x)$ is increasing on $R.$
View full question & answer→Question 33 Marks
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = 2x^3 - 9x^2 - 12x + 1$
Answer$f(x) = 2x^3 - 9x^2 - 12x + 1$
$f'(x) = -6x^2 - 18x - 12$
Critical points
$f'(x) = 0$
$\Rightarrow -6x^2 - 18x - 12 = 0$
$\Rightarrow x^2 + 3x + 2 = 0$
$\Rightarrow (x + 2)(x + 1) = 0$
$\Rightarrow x = -2, -1$
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 $(-\infty,-2)\cup(-1,\infty).$
View full question & answer→Question 43 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,
$\sin\text{x}\leq1$
$\Rightarrow1-\sin\text{x}\geq0$
$\Rightarrow\text{b}<0$ $[\text{Since }(1-\sin\text{x})\geq0,\text{b}(1-\sin\text{x})<0\Rightarrow\text{b}<0]$
$\Rightarrow\text{b}\in(-\infty,0)$
View full question & answer→Question 53 Marks
Show that $\text{f}(\text{x})=\frac{1}{\text{x}}$ is decreasing function on $(0,\infty).$
AnswerWe 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.
View full question & answer→Question 63 Marks
State when a function $f(x)$ is said to be increasing on an interval $[a, b].$ Test whether the function $f(x) = x^2 - 6x + 3$ is increasing on the interval $[4, 6].$
AnswerA 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) = x^2 - 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].$
View full question & answer→Question 73 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).$
View full question & answer→Question 83 Marks
Prove that $f(x) = ax + b,$ where a, b are constants and $a > 0$ is an increasing function on $R$.
AnswerHere,
$f(x) = ax + b$
Let $\text{x}_1,\text{x}_2\in\text{R}$ such that $x_1< x_2.$ Then,
$x_1< x_2$
$\Rightarrow ax_1 < ax_2 [\because\ \text{a}>0]$
$\Rightarrow ax_1 + b < ax_2 + b$
$\Rightarrow f(x_1) < f(x_2)$
$\therefore x_1 < 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 increasing on $R.$
View full question & answer→Question 93 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)$
View full question & answer→Question 103 Marks
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = 2x^3- 9x^2 + 12x - 5$
AnswerWe have,
$ f(x)=2 x^3-9 x^2+12 x-5 $
$ \therefore f^{\prime}(x)=6 x^2-18 x+12$
Critical points
$ f^{\prime}(x)=0 $
$ \Rightarrow 6\left(x^2-3 x+2\right)=0 $
$ \Rightarrow(x-2)(x-1)=0 $
$ \Rightarrow x=2,1$
Clearly, $f^{\prime}(x)>0$ if $x<1$ and $x>6$
$f^{\prime}(x)<0 \text { if } 1<x<2$
Thus, $f(x)$ increases in $(-\infty, 1) \cup(2, \infty)$, decreases in $(1,2)$.
View full question & answer→Question 113 Marks
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = x^8 + 6x^2$
AnswerWe have$ f(x) = x^8 + 6x^2$
$\therefore f'(x) = 8x^7 + 12x$
Critical points $f'(x) = 0$
$\Rightarrow 8x^7 + 12x = 0 $
$\Rightarrow 4x(2x^6 + 3) = 0 $
$\Rightarrow 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).$
View full question & answer→Question 123 Marks
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = x^3 - 6x^2 - 36x + 2$
Answer$f(x)=x^3-6 x^2-36 x+2 $
$ \therefore f^{\prime}(x)=3 x^2-12 x-36$
Critical point
$ f^{\prime}(x)=0 $
$ \Rightarrow 3\left(x^2-4 x-12\right)=0 $
$ \Rightarrow(x-6)(x+2)=0 $
$ \Rightarrow x=6,-2$
Clearly, $\mathrm{f}^{\prime}(\mathrm{x})>0$ if $\mathrm{x}<-2$ and $x>6$
$f^{\prime}(x)<0 \text { if }-2 x<x<6$
Thus, $f(x)$ increases in $(-\infty,-2) \cup(6, \infty)$, decreases in $(-2,6)$.
View full question & answer→Question 133 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}.$
View full question & answer→Question 143 Marks
Prove that $f(x) = ax + b,$ where $a, b$ are constants and $a < 0$ is an decreasing function on $R.$
AnswerHere, $f(x) = ax + b$ Let $\text{x}_1,\text{x}_2\in\text{R}$
such that $x_1 < x_2.$
Then, $x_1 < x_2$_
$\Rightarrow ax_1 > ax_2(\because\ \text{a}<0)$
$\Rightarrow ax_1 + b > ax_2 + b $
$\Rightarrow f(x_1) > f(x_2)$
Thus, $x_1 < 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 decreasing on $R.$
View full question & answer→Question 153 Marks
Without using the derivative show that the function f(x) = 7x - 3 is strictly increasing function on R.
AnswerHere,
$\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.
View full question & answer→Question 163 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).$
View full question & answer→Question 173 Marks
Without using the derivative, show that the function f(x) = |x| is
- Strictly increasing in $(0,\infty)$
- Strictly decreasing in $(-\infty,0)$
AnswerWe 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).$ View full question & answer→Question 183 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. $\Rightarrow\text{a}^2>1$ $\Rightarrow\text{a}^2-1>0$ $\Rightarrow(\text{a}+1)(\text{a}-1)>0$ $\Rightarrow\text{a}\in(-\infty,-1)\cup(1,\infty)$
View full question & answer→Question 193 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 $x_1 < x_{2.}$ Then
$[x_1] = [x_2] = 0 ....(1)$
Now,
$x_1 < x_2$
$\Rightarrow x_1 - [x_1] < x_2 - [x_2] [$From eq. $(1)]$
$\Rightarrow f(x_1) < f(x_2)$
$\therefore x_1 < x_2$
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,1)$
Hence, $f(x)$ is increasing on $(0, 1).$
View full question & answer→Question 203 Marks
Prove that the function f given by $f(x) = x^3 - 3x^2 + 4x$ is strictly increasing on $R.$
Answer$f(x) = x^3 - 3x^2 + 4x$
$f'(x) = 3x^2 - 6x + 4$
$= 3(x^2 - 2x) + 4$
$= 3(x^2 - 2x + 1) - 3 + 4$
$=2(\text{x}-1)^2+1>0,\forall\ \text{x}\in\text{R}$
Hence, f(x) is strictly increasing on $R$.
View full question & answer→Question 213 Marks
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = 6+ 12x + 3x^2 - 2x^3$
Answer$f(x) = 6+ 12x + 3x^2 - 2x^3$
$f'(x) = 12 + 6x - 6x^2$
$= -6(x^2 - x - 2)$
$= -6(x - 2)(x + 1)$
For $f(x)$ to be increasing, we must have
$f'(x) > 0$
$\Rightarrow -6(x - 2)(x + 1) > 0$
$\Rightarrow (x - 2)(x + 1) < 0$
$[$Since,$ -6 < 0, -6(x - 2)(x + 1) > 0 \Rightarrow (x - 2)(x + 1) < 0]$
$\Rightarrow -1 < x < 2$
$\Rightarrow\text{x}\in(-1,2)$
So, $f(x)$ is increasing on $(-1, 2).$
For $f(x)$ to be decreasing, we must have,
$f'(x) < 0$
$\Rightarrow -6(x - 2)(x + 1) < 0$
$\Rightarrow (x - 2)(x + 1) < 0$
$[$Since, $ -6 < 0, -6(x - 2)(x + 1) > 0 \Rightarrow (x - 2)(x + 1) > 0]$
$\Rightarrow x < -1$ or $x > 2$
$\Rightarrow\text{x}\in(-\infty,-1)\cup(2,\infty)$
So, $f(x)$ is decreasing on $(-\infty,-1)\cup(2,\infty).$
View full question & answer→Question 223 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 $\Big(\frac{\pi}{2},\pi\Big).$
View full question & answer→Question 233 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 $\text{x}\in\text{R}.$
View full question & answer→Question 243 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.
$\Rightarrow\text{k}\in(1,\infty)$
View full question & answer→Question 253 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$\text{f}'(\text{x}) >0$
$\Rightarrow\cos\text{x}-\sin\text{x}>0$ $\Rightarrow\sin\text{x}<\cos\text{x}$ $\Rightarrow\frac{\sin\text{x}}{\cos\text{x}}<1$ $\Rightarrow\tan\text{x}<1$ $\Rightarrow\text{x}\in\Big[0,\frac{\pi}{4}\Big)$
View full question & answer→Question 263 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 $0 < a < 1$.
AnswerCase 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 $(0,\infty).$
View full question & answer→Question 273 Marks
Show that $\text{f}(\text{x})=\cos^2\text{x}$ is a decreasing function on $\Big(0,\frac{\pi}{2}\Big).$
AnswerWe 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 $\Big(0,\frac{\pi}{2}\Big).$
View full question & answer→Question 283 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,
$\cos\text{x}\geq-1,\forall\ \text{x}\in\text{R}$
$\Rightarrow\text{a}<-1$
$\Rightarrow\text{a}\in(-\infty,-1)$
View full question & answer→Question 293 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,
$e^x> 0$
$\Rightarrow xe^x > 0$
$\Rightarrow f'(x) > 0,$ for all $x > 0$
So, $f(x)$ is increasing on for all $x > 0.$
View full question & answer→Question 303 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.
View full question & answer→Question 313 Marks
Show that the function f given by $\text{f}(\text{x})=10^\text{x}$ is increasing for all x.
AnswerWe 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.
View full question & answer→Question 323 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. $\Rightarrow1+\text{a}>1$ $\Rightarrow\text{a}>0$ $\Rightarrow\text{a}\in(0,\infty)$
View full question & answer→Question 333 Marks
Show that $\text{f}(\text{x})=\text{e}^{\frac{1}{\text{x}}},\text{x}\neq0$ is a decreasing function for all $\text{x}\neq0.$
AnswerWe 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 $\text{x}\neq0.$
View full question & answer→