If{[(1-sinx)(π-2x)^2.(log sinx)/(log(1+π^2-4πx+4x^2)),x ≠ π/2][k,x= π/2] is continuous at x = π/2, then k = A.-(1/16) B. -(1/32) ​

Question

If \(f(x) = \begin{cases} \frac{1-sinx}{(\pi-2x)^2}.\frac{log\,sin\,x}{(log(1+\pi^2-4\pi x+4x^2))},x ≠\frac{\pi}{2}\\k,x=\frac{\pi}{2} \end{cases} \) is continuous at x = π/2, then k =

A. \(-\frac{1}{16}\)

B. \(-\frac{1}{32}\)

C. \(-\frac{1}{64}\)

D. \(-\frac{1}{28}\)

Answer 1

Option : (C).

Formula : - 

(i) \(\lim\limits_{x \to 0}\frac{log(1-x)}{x}\) = 1 and \(\lim\limits_{x \to 0}\frac{sin\,x}{x}\) = 1

(ii) A function f(x) is said to be continuous at a point x=a of its domain, if \(\lim\limits_{x \to a}f(x)\) = f(a)

 \(\lim\limits_{x \to a^+}f(a+h)\) = \(\lim\limits_{x \to a^-}f(a-h)\) = f(a)

(iii) \(\lim\limits_{x \to a}{\{f(x).g(x)}\}\) = 1.m,

Where \(\lim\limits_{x \to a}f(x)\) = 1, \(\lim\limits_{x \to a}g(x)\) = m

Given :-

\(f(x) = \begin{cases} \frac{1-sinx}{(\pi-2x)^2}.\frac{log\,sin\,x}{(log(1+\pi^2-4\pi x+4x^2))},x ≠\frac{\pi}{2}\\k,x=\frac{\pi}{2} \end{cases} \) 

Function f(x) is continuous at x = \(\frac{\pi}{2}\)

\(\lim\limits_{x \to \frac{\pi}{2}}f(x)\) = \(f(\frac{\pi}{2})\) 

Using substitution method 

If \(\frac{\pi}{2}\)- x, then

Using standard limit formula (i)

k = \(-\frac{1}{64}\)