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Questions  >>  CBSE XII  >>  Math  >>  Relations and Functions

For what value of $a$ and $b$ is the function $f(x) = \begin{cases} x^2 ,& x \leq c \\ ax+b , & x> c \end{cases} $ is differentiable at $x=c$

$(A)\; a = \frac{c}{2} , b = c^2$
$(B)\; a = 2c , b = c^2$
$(C)\; a= 2c, b = \frac{c}{2}$
$(D)\; a = 2c, b = -c^2$

1 Answer

Given : $f(x)$ is differentiable at c. Every differentiable function is continuous.
Hence $f(x)$ is continuous at $x=c.$
\[\lim_{x \to c^{-}} f(x) = \lim_{x \to c^+} f(x) = f(c)\]
\[ \lim_{x \to c} x^2 = \lim_{x \to c} ax+b = c^2 \]
$\implies c^2 =ac+b -----(1)$
$f(x)$ is differentiable at $x=c$
LHD at $x=c $ = RHD at $x=c$
\[ \lim_{x \to c^{-}} \frac{f(x)-f(c)}{x-c} = \lim_{x \to c^{+}} \frac{f(x)-f(c)}{x-c}\]
\[ \lim_{ x \to c} \frac{x^2 - c^2}{x-c} = \lim_{x \to c} \frac{(ax+b) - c^2}{x-c} \]
\[ \lim_{x \to c} x+c = \lim_{x \to c } \frac{a(x-c)}{x-c} \]
\[ \lim_{x \to c} (x+c) = \lim_{x \to c} a \]
$\implies 2c = a -------(2)$
from (1) and (2) we get
$c^2 = 2c^2 +b$
$\implies b = -c^2$
$\therefore a = 2c$ and $b = -c^2$
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