Let f(x)=sin(πcos2x)x2,x≠0. The value of f(0) so that f is continuous function is—

∵f is continuous at x = 0⇒f(0)=limx→0 sin⁡πcos2⁡xx2=limx→0 sin⁡π−πsin2⁡xx2=limx→0 sin⁡πsin2⁡xx2=π

Let f(x)=sin(πcos2x)x2,x≠0. The value of f(0) so that f is continuous function is—

∵f is continuous at x = 0⇒f(0)=limx→0 sin⁡πcos2⁡xx2=limx→0 sin⁡π−πsin2⁡xx2=limx→0 sin⁡πsin2⁡xx2=π

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Let $f (x) = \frac{\sin (π \cos^{2} x)}{x^{2}}, x \neq 0$ . The value of f(0) so that f is continuous function is—

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$π$

$- \frac{π}{2}$

$\frac{π}{2}$

$- π$

answer is B.

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Detailed Solution

$∵$ f is continuous at x = 0
$\begin{array}{l} \Rightarrow f (0) = lim_{x \to 0} \{\frac{\sin (π \cos^{2} x)}{x^{2}}\} \\ = lim_{x \to 0} \{\frac{\sin (π - π \sin^{2} x)}{x^{2}}\} \\ = lim_{x \to 0} \{\frac{\sin (π \sin^{2} x)}{x^{2}}\} = π \end{array}$