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  • Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 42

Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 42

Answers (1)

Answer:

 The correct option is (a)

Hint:

(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}

(ii) Standard limits

\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ &(\text { iii }) \lim _{x \rightarrow 0}\{f(x) \pm g(x)\}=1 \pm m, \\ &{\text { Where, }} \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}

Given:

 f(x)= \begin{cases}\frac{(\sin (\cos x)-\cos x)}{(\pi-2 x)^{2}} &, x \neq \frac{\pi}{2} \\ k & , x<\frac{\pi}{2}\end{cases}

Solution:

f(x)= \begin{cases}\frac{(\sin (\cos x)-\cos x)}{(\pi-2 x)^{2}} &, x \neq \frac{\pi}{2} \\ k & , x<\frac{\pi}{2} \end{cases} \qquad -(1)

\begin{aligned} &\text { At } x=\frac{\pi}{2}\\ &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)\\ &\Rightarrow \lim _{x \rightarrow \pi / 2} \frac{\sin (\cos x)-\cos x)}{(\pi-2 x)^{2}}=k &\text { [using 1] } \end{aligned}

Let

\begin{aligned} &\pi-2 x=2 y\\ &\Rightarrow y=\frac{\pi-2 x}{2}\\ &\Rightarrow y=\frac{\pi}{2}-x\\ &\text { Since, when } x \rightarrow \frac{\pi}{2} \text { then } y \rightarrow 0 \text {. } \end{aligned}

\Rightarrow \lim _{y \rightarrow 0} \frac{\sin (\cos x)-\cos x}{(2 y)^{2}}=k \quad[\because \pi-2 x=y]

\Rightarrow \lim _{y \rightarrow 0} \frac{\sin \{\cos (\pi / 2-y)\}-\cos (\pi / 2-y)}{4 y^{2}}=k \quad\left[\because y=\frac{\pi}{2}-x \Rightarrow x=\frac{\pi}{2}-y\right]

\Rightarrow \lim _{y \rightarrow 0} \frac{\sin (\sin y)-\sin y}{4 y^{2}}=k \quad\left[\because \cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta\right]

\Rightarrow \lim _{y \rightarrow 0} \frac{2 \sin \left(\frac{\sin y-y}{2}\right) \cdot \cos \left(\frac{\sin y+y}{2}\right)}{4 y^{2}}=k \quad\left[\because \sin c-\sin D=2 \sin \left(\frac{C-D}{2}\right) \cdot \cos \left(\frac{c+D}{2}\right)\right]

\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0} \frac{\left(\frac{\sin y-y}{2}\right) \cdot \sin \left(\frac{\sin y-y}{2}\right)}{y^{2} \cdot\left(\frac{\sin y-y}{2}\right)} \cdot \cos \left(\frac{\sin y+y}{2}\right)

\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0}\left(\frac{\frac{\sin y-y}{2}}{y}\right) \cdot\left\{\frac{\sin \left(\frac{\sin y-y}{2}\right)}{\left(\frac{\sin y-y}{2}\right)} \mid \cdot\left\{\frac{\cos \left(\frac{\sin y+y}{2}\right)}{y}\right\}=k\right.

\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0} \frac{\sin y-y}{2 y} \times \lim _{y \rightarrow 0} \frac{\sin \left(\frac{\sin y-y}{2}\right)}{\left(\frac{\sin y-y}{2}\right)} \times \lim _{y \rightarrow 0} \frac{\cos \left(\frac{\sin y+y}{2}\right)}{y}=k

\left[\because \lim _{x \rightarrow a} f(x) \cdot g(x) \cdot h(x)=\lim _{x \rightarrow a} f(x) \lim _{x \rightarrow a} g(x) \lim _{x \rightarrow a} h(x)\right]

\Rightarrow \frac{1}{4}(1-1) \times 1 \times \lim _{y \rightarrow 0} \frac{\cos \left(\frac{\sin y+y}{2}\right)}{y}=k \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

\begin{aligned} &\Rightarrow \frac{1}{4} \times 0 \times 1 \times \lim _{y \rightarrow 0} \frac{\cos \left(\frac{\sin y+y}{2}\right)}{y^{\circ}}=R \\ &\Rightarrow 0=k \end{aligned}

\begin{aligned} \therefore k=0 \end{aligned}

So, the correct option is (a)

Posted by

Gurleen Kaur

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