Question

What should be the value of k such that the function \(\rm f(x)=\left\{\begin{matrix} \frac{ksin(π -x)}{π -x} & if & x \neq π \\ 1 & if & x =π \end{matrix}\right.\) is continuous at x = π.

1. π
2. 1
3. -1
4. 0

Answer 1

Correct Answer - Option 2 : 1

Concept:

For a function say f, \(\mathop {\lim }\limits_{x \to a} f(x)\) exists

\( ⇒ \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = l = \;\mathop {\lim }\limits_{x \to a} f(x)\), where l is a finite value.

Any function say f is said to be continuous at point say a if and only if \(\mathop {\lim }\limits_{x \to a} f(x) = l = f\left( a \right)\), where l is a finite value.

Calculation:

Given: \(\rm f(x)=\left\{\begin{matrix} \frac{ksin(π -x)}{π -x} & if & x \neq π \\ 1 & if & x =π \end{matrix}\right.\) is continuous at x = π.

As we know that, if a function f is continuous at point say a then \(\mathop {\lim }\limits_{x \to a} f(x) = l = f\left( a \right)\)

\(\Rightarrow \rm \lim_{x\rightarrow π}\frac{ksin(π-x)}{π-x} = f(π) = 1\)

∵ \(\rm \lim_{x\rightarrow π}\frac{ksin(π-x)}{π-x} = \frac{0}{0}\), we can use L'Hospitals rule.

⇒ \(\rm \lim_{x\rightarrow π}\frac{-kcos(π-x)}{-1} = 1\)

⇒ \(\rm \frac{kcos(π-π)}{1} = 1\) 

⇒ k cos(0) = 1

⇒ k = 1.

Hence, option 2 is correct.