# Find the value of k (if any), which makes the function f(x).x=1. f(x)=\begin{cases} { k^2 }...

## Question:

Find the value of k (if any), which makes the function {eq}f(x) {/eq} continuous at {eq}x=1 {/eq} {eq}\displaystyle f(x)=\begin{cases} { k^2 } x-\frac{5}{4}; & x\leq 1\\ \frac { \sqrt {x^2+3}-2}{x^{2}-1};& x >1 \end{cases} {/eq}

## Continuous Function:

A function {eq}f(x) {/eq} is said to be continuous at a given value {eq}x = a {/eq}, if the left-hand limit, the right-hand limit and the value of the function at {eq}x= a {/eq} are all same. i.e., the limit of the function at that value should exist and should be equal to the value of the function at the given value.

$$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a) $$

## Answer and Explanation:

The given function is:

$$f(x)=\left\{\begin{array}{ll}{k^{2} x-\frac{5}{4} ;} & {x \leq 1} \\ {\frac{\sqrt{x^{2}+3}-2}{x^{2}-1} ;} &...

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