Question

$\rm{Let \: f (x) = \begin{cases} \rm{x}^{2} , \: \: \: \: \: \: \: \: \: \: \: \: if < 2 \\ \rm mx + b, \: \: if \: x > 2\end {cases}} \\ \rm if \: f(x) \: is \: differentiable \: every \: where \: then$


$\rm (i) \: m = 4 \: and \: b = - 4 \\ \rm{ (ii) \: m = 4 \: and \: b = 4} \\ \rm (iii) \: m = - 4 \: and \: b = - 4 \\ \rm (iv) \: m = - 4 \:and \: b = 4$​

Answer 1

Appropriate Question :-

$\rm{Let \: f (x) = \begin{cases} \rm{x}^{2} , \: \: \: \: \: \: \: \: \: \: \: \: if \: x \: < 2 \\ \rm mx + b, \: \: if \: x \geqslant 2\end {cases}} \\ \rm if \: f(x) \: is \: differentiable \: every \: where \: then$

$\rm (i) \: m = 4 \: and \: b = - 4 \\ \rm{(ii) \: m = 4 \: and \: b = 4} \\ \rm (iii) \: m = - 4 \: and \: b = - 4 \\ \rm (iv) \: m = - 4 \:and \: b = 4$

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f (x) = \begin{cases} \rm{x}^{2} , \: \: \: \: \: \: \: \: \: \: \: \: if \: x < 2 \\ \\ \rm mx + b, \: \: if \: x \geqslant 2\end {cases}$

As it is given that f(x) is differentiable at x = 2.

We know, every differentiable function is always continuous.

So, f(x) is continuous at x = 2.

So, it means

$\rm \: \displaystyle\lim_{x \to 2^+}f(x) = \displaystyle\lim_{x \to 2^ - }f(x)$

$\rm \: \displaystyle\lim_{x \to 2^+} \: mx + b = \displaystyle\lim_{x \to 2^ - } \: {x}^{2}$

can be rewritten as

$\rm \: \displaystyle\lim_{h \to 0} \: m(2 + h) + b = \displaystyle\lim_{h \to 0 } \: {(2 - h)}^{2}$

$\rm \: 2m + b = 4 - - - (1)$

Now, Differentiability at x = 2

$\rm \: \displaystyle\lim_{x \to 2^+}\rm \: \frac{f(x) - f(2)}{x - 2} = \displaystyle\lim_{x \to 2^ - }\rm \: \frac{f(x) - f(2)}{x - 2}$

$\rm \: \displaystyle\lim_{x \to 2^+}\rm \: \frac{mx + b - (2m + b)}{x - 2} = \displaystyle\lim_{x \to 2^ - }\rm \: \frac{ {x}^{2} - (2m + b)}{x - 2}$

$\rm \: \displaystyle\lim_{x \to 2^+}\rm \: \frac{mx - 2m}{x - 2} = \displaystyle\lim_{x \to 2^ - }\rm \: \frac{ {x}^{2} - 4}{x - 2}$

$\rm \: \displaystyle\lim_{x \to 2^+}\rm \: \frac{m(x - 2)}{x - 2} = \displaystyle\lim_{x \to 2^ - }\rm \: \frac{(x - 2)(x + 2)}{x - 2}$

$\rm \: \displaystyle\lim_{x \to 2^+}\rm \: m \: = \: \displaystyle\lim_{x \to 2^ - }\rm \: (x + 2)$

$\rm\implies \:m \: = \: 4$

On substituting m = 4, in equation (1), we get

$\rm \: 8 + b = 4$

$\rm \: b = 4 - 8$

$\rm\implies \:b \: = \: - \: 4$

So, option (a) is correct.

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {x}^{n}\\ \\ \sf {nx}^{n - 1} & \sf {e}^{x} \end{array}} \\ \end{gathered}$