Question

State $T$ for true and $F$ for false.
(i) $\underset{x\to 1/2}{\lim }$ $\left(\frac{8x-3}{2x-1}-\frac{4x^2+1}{4x^2-1}\right)=\frac{7}{2}$
(ii) $\underset{x\to 0}{\lim }$ $\frac{sinx-2sin3x+sin5x}{x}=1$
(iii) The derivative of $\frac{x^{2}cos\left(\frac{\pi }{4}\right)}{sinx}$ is
$xcos\frac{\pi }{4}\left[\frac{2sinx-xcosx}{si{n}^{2}x}\right]$.
(iv)The derivative of $5sinx-6cosx+7$ is $5cosx-7sinx$.

	(i)	(ii)	(iii)	(iv)
(a)	$F$	$T$	$F$	$T$
(b)	$T$	$F$	$T$	$F$
(c)	$T$	$T$	$F$	$F$
(d)	$F$	$F$	$T$	$T$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

(i) True
We have, $\underset{x\to 1/2}{\lim }$ $\left(\frac{8x-3}{2x-1}-\frac{4x^2+1}{4x^2-1}\right)$
$=\underset{x\to 1/2}{\lim }$ $\left[\frac{\left(8x—3\right)\left(2x+1\right)—\left(4x^2+1\right)}{\left(4x^2-1\right)}\right]$
$=\underset{x\to 1/2}{\lim }$$\left[\frac{16x^2+8x-6x-3-4x^2-1}{4x^2-1}\right]$
$=\underset{x\to 1/2}{\lim }$$\frac{2\left(6x^2+x-2\right)}{4x^2-1}$
$=\underset{x\to 1/2}{\lim }$ $\frac{2\left(3x+2\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}$
$=\underset{x\to 1/2}{\lim }$ $\frac{2\left(3x+2\right)}{2x+1}=\frac{2\left(3\times \frac{1}{2}+2\right)}{2\times \frac{1}{2}+1}$
$=\frac{3}{2}+2=\frac{7}{2}$

(ii) False
We have, $\underset{x\to 0}{\lim }$ $\frac{sinx-2sin3x+sin5x}{x}$
$=\underset{x\to 0}{\lim }$ $\frac{sin5x+sinx-2sin3x}{x}$
$=\underset{x\to 0}{\lim }$ $\frac{2sin3xcos2x-2sin3x}{x}$
$=\underset{x\to 0}{\lim }$ $\frac{2sin3x\left(cos2x-1\right)}{x}$
$=\underset{x\to 0}{\lim }$ $\frac{2sin3x}{\frac{1}{3}\times 3x}\left(cos2x-1\right)$
$=6\underset{x\to 0}{\lim }$ $\frac{sin3x}{3x}\cdot$ $\underset{x\to 0}{\lim }(cos2x-1)$
$=6\times 1\times 0=0$

(iii) True
Let $f\left(x\right)=\frac{x^{2}cos\left(\frac{\pi }{4}\right)}{sinx}\dots \left(1\right)$
Differentiating $\left(1\right)$ with respect to $x$, we get
$\frac{d}{dx}\left\{f\left(x\right)\right\}=cos\left(\frac{\pi }{4}\right)\left\{\frac{sinx\cdot 2x-x^{2}cosx}{si{n}^{2}x}\right\}$
$=xcos\left(\frac{\pi }{4}\right)\left[\frac{2sinx-xcosx}{si{n}^{2}x}\right]$

(iv) False
Let $f(x)=5sinx-6cosx+7\dots \left(1\right)$
Differentiating $(1)$ with respect to $x$, we get
$f^{\prime }(x)=5cosx-6(-sinx)+0$
$\therefore f^{\prime }(x)=5cosx+6sinx$.