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  • Explain solution RD Sharma class 12 chapter Continuity exercise 8.2 question 5 maths

Explain solution RD Sharma class 12 chapter Continuity exercise 8.2 question 5 maths

Answers (1)

Answer:

a=-1, b=1 \text { or } a=1, b=1 \pm \sqrt{2} 

Hint:

 p at LHL = RHL at

x =1 ,\: \sqrt{2}

Given:

 f(x)= \begin{cases}\frac{x^{2}}{9} & 0 \leq x<1 \\ 9 & 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{2}} & \sqrt{2} \leq x<\infty\end{cases}

Explanation:

\begin{aligned} &\text { At } x=1, L . H . L=R . H . L=f(1) \quad ...(A)\\ &f(1)=A \quad ...(1) \end{aligned}

L . H . L=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0}(1-h)=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{a}=\frac{1}{a}

Using (A)

\begin{aligned} &a=\frac{1}{a} \Rightarrow a^{2}=1 \Rightarrow a=\pm 1\\ &\text { At } x=\sqrt{2}, \quad \text { L.H.L }=\text { R.H.L }=f(\sqrt{2}) \quad ....(B) \end{aligned}

f(\sqrt{2})=\frac{2 b^{2}-4 b}{(\sqrt{2})^{2}}=\frac{2 b^{2}-4 b}{2}=b^{2}-2 b \quad ....(2)

\text { L.H.L }=\lim _{x \rightarrow \sqrt{2}^{2}} f(x)=\lim _{h \rightarrow 0}(\sqrt{2}-h)=\lim _{h \rightarrow 0} a=a

So using (B) we get

\begin{aligned} &b^{2}-2 b=\pm 1 \\ &\qquad b^{2}-2 b=1 \text { or } \quad b^{2}-2 b=-1 \\ &\qquad b^{2}-2 b-1=0 \quad \text { or } \quad b^{2}-2 b+1=0 \end{aligned}

\begin{aligned} &b=\frac{1 \pm \sqrt{2}}{1} \quad \text { Or } \quad(b-1)^{2}=0\\ &b=1 \pm \sqrt{2} \quad \text { Or } \quad b=1 \end{aligned}

Hence,

a=-1, b=1 \quad \text { or } \quad a=1, b=1 \pm \sqrt{2}

Posted by

Gurleen Kaur

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