Question

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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{\dfrac{1 - coskx}{x \: sinx} \: \: when \: \: x \: \ne \: 0} \\ \\ &\sf{\dfrac{1}{2} \: \: when \: x \: = \: 0} \end{cases}\end{gathered}\end{gathered}$

Further given that,

↝  Function f(x) is continuous at x = 0

We know,

A function f(x) is said to be continuous at x = a iff

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to a}\rm f(x) = f(a) \: }}}$

So, here

$\red{\rm \implies\: \: \displaystyle\lim_{x \to 0}\rm f(x) = f(0) \: }$

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{1 - coskx}{x \: sinx} = \frac{1}{2}$

We know,

$\boxed{ \tt{ \: 1 - cos2x = {2sin}^{2}x \: }}$

So, using this identity,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{2 {sin}^{2}\bigg[\dfrac{kx}{2} \bigg]}{x \times \bigg[\dfrac{sinx}{x} \bigg] \times x} = \frac{1}{2}$

We know,

$\red{\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1 \: }}}$

So, using this identity, we get

$\rm :\longmapsto\:2\displaystyle\lim_{x \to 0}\rm \frac{ {sin}^{2} \bigg[\dfrac{kx}{2} \bigg]}{ {x}^{2} \times 1} = \frac{1}{2}$

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{ {sin}^{2} \bigg[\dfrac{kx}{2} \bigg]}{ {x}^{2}} = \frac{1}{4}$

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{ {sin}^{2} \bigg[\dfrac{kx}{2} \bigg]}{\bigg[\dfrac{ {k}^{2} }{4} \bigg]{x}^{2}} \times \bigg[\dfrac{ {k}^{2} }{4} \bigg] = \frac{1}{4}$

$\rm :\longmapsto\: {k}^{2} \displaystyle\lim_{x \to 0}\rm \frac{ {sin}^{2} \bigg[\dfrac{kx}{2} \bigg]}{\bigg[\dfrac{ {k}^{2} {x}^{2} }{4} \bigg]} = 1$

$\rm :\longmapsto\: {k}^{2} \times {1}^{2} = 1$

$\rm :\longmapsto\: {k}^{2} = 1$

$\bf\implies \:k \: = \: \pm \: 1$

• Hence, Option (a) is correct.

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Additional Information :-

$\red{\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1 \: }}}$

$\red{\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\rm \frac{sin {}^{n} x}{ {x}^{n} } = 1 \: }}}$

$\red{\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1 \: }}}$

$\red{\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\rm \frac{tan {}^{n} x}{ {x}^{n} } = 1 \: }}}$

$\red{\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1 \: }}}$

$\red{\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1 \: }}}$

$\red{\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga \: }}}$