#### Explain Solution R.D. Sharma Class 12 Chapter 9 Differentiability  Exercise 9 .1 Question 11 Maths Textbook Solution.

Answer: $a=3,b=5$

Hint: For differentiability, LHD = RHD where LHD is left hand derivative and RHD is right hand derivative. Also, LHD and RHD should exist in given limit.

Given: $f\left ( x \right )=$$\left\{\begin{array}{c} x^{2}+3 x+a, \text { if } x \leq 1 \\ b x+2, \text { if } x>1 \end{array}\right.$

Solution:

\begin{aligned} &\text { LHD at } \mathrm{x}=1: \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{1-h-1} \\ &=\lim _{h \rightarrow 0} \frac{\left[(1-h)^{2}+3(1-h)+a\right]-[1+3+a]}{-h} \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left[1+h^{2}-2 h+3-3 h+a\right]-[4+a]}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h^{2}-5 h}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h(h-5)}{-h} \\ &=\lim _{h \rightarrow 0}-(h-5) \\ &=5 \end{aligned}

$RHD\; at\: x=0:\lim_{x\rightarrow 1^{+}}\frac{f\left ( x \right )-f\left ( 1 \right )}{x-1}$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{1+h-1} \\ &=\lim _{h \rightarrow 0} \frac{[b(1+h)+2]-(b+2)}{h} \\ &=\lim _{h \rightarrow 0} \frac{[b(1+h)+2]-(b+2)}{h} \\ &=\lim _{h \rightarrow 0} \frac{b h}{h} \\ &=b \end{aligned}

Since f(x) is differentiable, so (LHD at x = 1) = (RHD at x = 1)

$\Rightarrow b=5$

$\Rightarrow f\left ( 1 \right )=1+3+a$

$=4+a$

\begin{aligned} &\mathrm{LHL} \Rightarrow \lim _{x \rightarrow 1^{-}} f(x) \\ &\Rightarrow \lim _{h \rightarrow 0} f(1-h) \\ &\Rightarrow \lim _{h \rightarrow 0}(1-h)^{2}+3(1-h)+a \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \\ &=\lim _{h \rightarrow 0} 1+h^{2}-2 h+3-3 h+a \\ &=4+a \end{aligned}

\begin{aligned} &\mathrm{RHL} \Rightarrow \lim _{x \rightarrow 1^{+}} f(x) \\ &\Rightarrow \lim _{h \rightarrow 0} f(1+h) \\ &\Rightarrow \lim _{h \rightarrow 0} b(1+h)+2 \\ &=\lim _{h \rightarrow 0} b+b h+2 \end{aligned}

$=b+2$

Since f(x) is differentiable, f(x) is continuous.

LHL = RHL

$b+2=4+a$

we know that $b=5$

$5+2=4+a$

$7=4+a$

$a=3$

Hence, $a=3,$ and $b=5$