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The value of k for which the function f(x)={1−cos⁡8x16x2,if x≠0k,if x=0f(x) = \begin{cases} \frac{1-\cos 8x}{16x^2}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} is continuous at x=0x = 0 is:

Solution

✅ Correct Option: 2

For the function to be continuous at x=0x = 0:

lim⁡x→01−cos⁡8x16x2=k\lim_{x \to 0} \frac{1-\cos 8x}{16x^2} = k


Using the trigonometric identity 1−cos⁡θ=2sin⁡2(θ2)1 - \cos \theta = 2\sin^2\left(\frac{\theta}{2}\right) with θ=8x\theta = 8x:

1−cos⁡8x=2sin⁡2(4x)1 - \cos 8x = 2\sin^2(4x)


lim⁡x→01−cos⁡8x16x2\lim_{x \to 0} \frac{1-\cos 8x}{16x^2}

=lim⁡x→02sin⁡2(4x)16x2= \lim_{x \to 0} \frac{2\sin^2(4x)}{16x^2}

=lim⁡x→0sin⁡2(4x)8x2= \lim_{x \to 0} \frac{\sin^2(4x)}{8x^2}


Rewriting the expression:

sin⁡2(4x)8x2\frac{\sin^2(4x)}{8x^2}

=sin⁡2(4x)(4x)2×(4x)28x2= \frac{\sin^2(4x)}{(4x)^2} \times \frac{(4x)^2}{8x^2}

=[sin⁡(4x)4x]2×16x28x2= \left[\frac{\sin(4x)}{4x}\right]^2 \times \frac{16x^2}{8x^2}

=[sin⁡(4x)4x]2×2= \left[\frac{\sin(4x)}{4x}\right]^2 \times 2


Using the standard limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1:

lim⁡x→0sin⁡(4x)4x=1\lim_{x \to 0} \frac{\sin(4x)}{4x} = 1


lim⁡x→0[sin⁡(4x)4x]2×2\lim_{x \to 0} \left[\frac{\sin(4x)}{4x}\right]^2 \times 2

=(1)2×2= (1)^2 \times 2

=2= 2

Therefore, k=2k = 2

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