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If \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {Kx^2}&{if}&{x \le 2}\\ 3&{if}&{x > 2} \end{array}} \right.\) is continuous at x = 2, then the value of K is
1. 3 / 4
2. 4
3. 4 / 3
4. 3
5. None of these

1 Answer

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Best answer
Correct Answer - Option 1 : 3 / 4

Concept:

A function f(x) is said to be continuous at a point x = a, in its domain if  exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a ⇔ \(\rm \lim_{x\rightarrow a^{+}}f(x)=\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a}f(x)\)

Calculation:

\(\rm f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {Kx^2}&{if}&{x \le 2}\\ 3&{if}&{x > 2} \end{array}} \right.\) 

    

\(\rm \lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2}K x^{2}\)

⇒ \(\rm \lim_{x\rightarrow 2^{-}}f(x)=4 K\)                               

Similarly, 

\(\rm \lim_{x\rightarrow 2^{+}}f(x)=3 \)

Function is continuos at x = 2,

So, \(\rm \lim_{x\rightarrow 2^{+}}f(x)=\lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2}f(x)\)

⇒ 4K = 3 

K = 3/4

The correct option is 1.

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