#### provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise  Fill in the blanks question 6

Hint: Use the formula of $\cos 3 x=4 \cos ^{3} x-3 \cos x$
Given:

$f(x)=\left\{\begin{array}{cl} \frac{\cos 3 x-\cos x}{x^{2}}, & x \neq 0 \\ \lambda & , x=0 \end{array}\right.$    is continuous at $x=0$
Solution:

We know that

$\lim _{x \rightarrow 0} f(x)=f(0),$ when function is continuous

$\therefore \lim _{x \rightarrow 0} f(x)=f(0)$

\begin{aligned} &\lim _{x \rightarrow 0} \frac{\cos 3 x-\cos x}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{\left(4 \cos ^{3} x-3 \cos x\right)-\cos x}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{4 \cos ^{3} x-4 \cos x}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{4 \cos x\left(\cos ^{2} x-1\right)}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{4 \cos x\left(-\sin ^{2} x\right)}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0}-4 \cos x=\lambda \\ &-4(1)=\lambda \end{aligned}

$\lambda=-4 \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \cos 3 x=4 \cos ^{3} x-3 \cos x \\ \because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \because \cos 0^{\circ}=1 \end{array}\right]$