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Let f(x)={1−cos⁡4xx2,if x<02λ,if x=0x16+x−4,if x>0f(x)=\begin{cases}\frac{1-\cos4x}{x^2}, & \text{if } x<0\\2\lambda, & \text{if } x=0\\\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{if } x>0\end{cases} such that ff continuous at x=0x=0. Then, the value of λ\lambda is equal to:

Solution

✅ Correct Option: 1

Left limit: 1−cos⁡4xx2=2sin⁡22xx2→8\frac{1-\cos4x}{x^2}=\frac{2\sin^2 2x}{x^2}\to8. Right limit: rationalising the denominator gives 16+x+4→8\sqrt{16+\sqrt x}+4\to8. Continuity needs 2λ=82\lambda=8, so λ=4\lambda=4.

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