Find the value of a for which the function f defined by f(x) = asinπx+12, x≤0 and tanx-sinxx3,x>0 is continuous at x = 0.

LHL = limx→0-asinπx+12=aRHL = limx→0+tanx-sinxx3=12⇒a=12

Find the value of a for which the function f defined by f(x) = asinπx+12, x≤0 and tanx-sinxx3,x>0 is continuous at x = 0.

LHL = limx→0-asinπx+12=aRHL = limx→0+tanx-sinxx3=12⇒a=12

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Find the value of a for which the function f defined by f(x) = $\{asin (\frac{π (x + 1)}{2}), x \leq 0 and \frac{tanx - sinx}{x^{3}}, x > 0\}$ is continuous at x = 0.

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\frac{1}{2}

- \frac{1}{2}

-2

answer is A.

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Detailed Solution

LHL =

\lim_{x \to 0^{-}} asin (\frac{π (x + 1)}{2}) = a

RHL =

\lim_{x \to 0^{+}} \frac{tanx - sinx}{x^{3}} = \frac{1}{2}

\Rightarrow a = \frac{1}{2}

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Find the value of a for which the function f defined by f(x) = asinπx+12, x≤0 and tanx-sinxx3,x>0 is continuous at x = 0.

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Find the value of a for which the function f defined by f(x) = $\{asin (\frac{π (x + 1)}{2}), x \leq 0 and \frac{tanx - sinx}{x^{3}}, x > 0\}$ is continuous at x = 0.