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

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

Answers (1)

Answer: -4
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]

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