#### Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 41 maths textbook solution

The correct option is (d)

Hint:

A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a^{-}} f(x)=f(a) \end{aligned}

Given:

$f(x)= \begin{cases}5 x-4 &, 0

is continuous at every point of domain

Solution:

$f(x)= \begin{cases}5 x-4 &, 0

At x = 1  f(x)=1

So,

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0} s(1-h)-4 \quad[\because {f}(x)=5 x-4 \text { when } x<1] \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} 5-5 h-4\\ &=\lim _{h \rightarrow 0} 1-5 h\\ &=1-5 \times 0\\ &=1 \end{aligned}

\begin{aligned} \lim _{x \rightarrow 1^{+}} f(x) &=\lim _{h \rightarrow 0} f(1+h) \\ &=\lim _{h \rightarrow 0}\left\{4(1+h)^{2}+3 a(1+n)\right\} &\left[\because {f}({x})=4 {x}^{2}+3 {a} x \text { when } x>\mathbf{1}\right] \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0}\left\{4\left(1+h^{2}+2 b\right)+3 a+3 a h\right\} \quad\left[\because(\boldsymbol{a}+\boldsymbol{b})^{2}=\boldsymbol{a}^{2}+\boldsymbol{b}^{2}+2 \boldsymbol{a} \boldsymbol{b}\right] \\ &=\lim _{h \rightarrow 0}\left(4+4 h^{2}+8 h+3 a+3 a h\right) \end{aligned}

\begin{aligned} &=4+4 \times 0^{2}+8 x 0+3 a+3 a \times 0 \\ &=4+0+0+3 a+0=4+3 a \quad-(2) \\ &\text { Also, } f(1)=5 \cdot 1-4 \\ &\quad=5-4 \\ &\qquad=1 \quad-(3) \quad[\because {f}(x)=5 x-4 \text { when } x=1] \end{aligned}

It is given that the function f(x) is continuous at x=1

\begin{aligned} \Rightarrow \lim _{x \rightarrow 1^{-}} f(x) &=\lim _{x \rightarrow 1^{+}} f(x) \\ &=f(1) \\ \Rightarrow 1=4+3 a & \end{aligned}

\begin{aligned} &=1 \quad \text { [using equation (1), (2) and (3) }\\ &\text { So, } \Rightarrow 4+3 a=1 \\ \Rightarrow 3 a &=1-4 \\ &=-3 \\ \Rightarrow a &=\quad \frac{-3}{3}=-1 \\ & \therefore a=-1 \end{aligned}