CBSE Class 12 Mathematics 2023: question 22(b)

Hints

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Step-by-step solution

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1. For $\lambda \ne 0$:

   ${\lim }_{x\to 0}\frac{{\sin }^2\lambda x}{x^2}={\lim }_{x\to 0}{\left(\frac{\sin \lambda x}{\lambda x}\right)}^2\cdot {\lambda }^2=1\cdot {\lambda }^2={\lambda }^2$ [using ${\lim }_{\theta \to 0}\frac{\sin \theta }{\theta }=1$]

   1 mark

2. Since $f$ is continuous at $x=0$, ${\lim }_{x\to 0}f(x)=f(0)$, so ${\lambda }^2=1$.

   ½ mark

3. Hence $\lambda =\pm 1$.

   ($\lambda =0$ is not possible: the limit would then be $0\ne 1$.)

   ½ mark

Answer: $\lambda =\pm 1$

Revise the chapter

• Chapter: Continuity and Differentiability (notes, formulas, where marks are lost and every past question)