Question

State $T$ for true and $F$ for false.
(i) $\displaystyle \lim_{x \to 1/2}$ $\left(\frac{8x-3}{2x-1}-\frac{4x^{2}+1}{4x^{2}-1}\right)=\frac{7}{2}$
(ii) $\displaystyle \lim_{x \to 0}$ $\frac{sin \,x -2 \,sin \,3x +sin\, 5x}{x}=1$
(iii) The derivative of $\frac{x^{2}\,cos\left(\frac{\pi}{4}\right)}{sin\,x}$ is
$x\,cos \frac{\pi}{4}\left[\frac{2\,sin\,x-x\,cos\,x}{sin^{2}\,x}\right]$.
(iv)The derivative of $5 \,sin\, x - 6 \,cos\, x + 7$ is $5 \,cos\,x - 7\, sin\,x$.

	(i)$\quad$	(ii)$\quad$	(iii)$\quad$	(iv)
(a)$\quad$	$F$	$T$	$F$	$T$
(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, $\displaystyle \lim_{x \to 1/2}$ $\left(\frac{8x-3}{2x-1}-\frac{4x^{2}+1}{4x^{2}-1}\right)$
$=\displaystyle \lim_{x \to 1/2}$ $\left[\frac{\left(8x — 3\right)\left(2x +1\right) — \left(4x^{2} +1\right)}{\left(4x^{2}-1\right)}\right]$
$=\displaystyle \lim_{x \to 1/2}$$\left[\frac{16x^{2} +8x-6x- 3-4x^{2} -1}{4x^{2}-1}\right]$
$=\displaystyle \lim_{x \to 1/2}$$\frac{2\left(6x^{2}+x-2\right)}{4x^{2}-1}$
$=\displaystyle \lim_{x \to 1/2}$ $\frac{2\left(3x+2\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}$
$=\displaystyle \lim_{x \to 1/2}$ $\frac{2\left(3x+2\right)}{2x+1}=\frac{2\left(3\times\frac{1}{2}+2\right)}{2\times\frac{1}{2}+1}$
$=\frac{3}{2}+2=\frac{7}{2}$

(ii) False
We have, $\displaystyle \lim_{x \to 0}$ $\frac{sin\,x-2\,sin\,3x+sin\,5x}{x}$
$=\displaystyle \lim_{x \to 0}$ $\frac{sin\,5x+sin\,x-2\,sin\,3x}{x}$
$=\displaystyle \lim_{x \to 0}$ $\frac{2\,sin\,3x\,cos\,2x-2\,sin\,3x}{x}$
$=\displaystyle \lim_{x \to 0}$ $\frac{2\,sin\,3x\left(cos\,2x-1\right)}{x}$
$=\displaystyle \lim_{x \to 0}$ $\frac{2\,sin\,3x}{\frac{1}{3}\times3x}\left(cos\,2x-1\right)$
$=6\,\displaystyle \lim_{x \to 0}$ $\frac{sin\,3x}{3x}\cdot$ $\displaystyle \lim_{x \to 0} (cos2x-1) $
$= 6\times 1 \times 0 = 0$

(iii) True
Let $f\left(x\right)=\frac{x^{2}\,cos \left(\frac{\pi}{4}\right)}{sin\,x}\quad\ldots\left(1\right)$
Differentiating $\left(1\right)$ with respect to $x$, we get
$\frac{d}{dx}\left\{f\left(x\right)\right\}=cos\left(\frac{\pi}{4}\right)\left\{\frac{sin\,x\cdot2x-x^{2}\,cos\,x}{sin^{2}\,x}\right\}$
$=x\,cos\left(\frac{\pi}{4}\right)\left[\frac{2\,sin\,x-x\,cos\,x}{sin^{2}\,x}\right]$

(iv) False
Let $f(x) = 5 \,sin\, x - 6\, cos \,x + 7\quad\ldots\left(1\right)$
Differentiating $(1)$ with respect to $x$, we get
$f'(x) = 5\, cos \,x - 6 (- sin\, x)+ 0$
$\therefore f'(x)=5\,cos\,x+6\,sin\,x$.