f(x) = * 20 l k x 2 & if & x 2 3, & if & x > 2 . at x = 2.

We have, f(2) = 4k x f(x) = x 2 + h h 0 3 = 3 x f(x) = x 2 - h h 0 (2 - h) 2 k = 4k For continuity at x = 2, we have x f(x) = x f(x) = f(2) 4k = 3 k = 4

Continuity and Differentiability — Class 12 Maths Solution

ncert exercise SA NCERT Ex.5.1 ,Q.27,Page 161

Question

$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{k{x^2}}&{if}&{x \le 2}\\{3,}&{if}&{x > 2}\end{array}} \right.$ at $x = 2.$

Step-by-step Solution

We have,$f(2) = 4k$

$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 2 + h\atop\scriptstyle h \to 0} 3 = 3$

$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 2 - h\atop\scriptstyle h \to 0} {(2 - h)^2}k = 4k$

For continuity at x $=$ 2, we have

$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} f(x) = f(2)$

$\Rightarrow$ $4k = 3 \Rightarrow k = \cfrac{3}{4}$

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