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  • Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 21 maths textbook solution

Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 21 maths textbook solution

Answers (1)

Answer:

 The correct option is (c)

Hint:

 If a function is continuous at x=2, then

\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)

Given:

 The function f(x)

=\left\{\begin{array}{l} x^{2} a, 0 \leq x<1 \\ a, 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{2}}, \sqrt{2 \leq x<\infty} \end{array}\right. \text { is continuous for } 0 \leq x<\infty

Solution:

Step 1: Calculate f(x) is continuous for

x = 1, \: \sqrt{2}

Calculate for f(x) is continuous at x = 1

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1+} f(x)=f(1) \cdots(i) \\ &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{a} \end{aligned}

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

\begin{aligned} &=\frac{1}{a}+0-2 \frac{1}{a} \times 0 \\ &=\frac{3}{a} \end{aligned}

f(a) = a

\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+n)=\lim _{n \rightarrow 0} a=a \\ &\text { from (i) } \Rightarrow \frac{1}{a}=a=a \end{aligned}

\begin{aligned} &\Rightarrow \frac{1}{a}=a \\ &\Rightarrow a^{2}=1 \\ &\Rightarrow a=\pm 1 \end{aligned}

Step 2: Calculate for f(x) is continuous at

\begin{aligned} x=\sqrt{2} \end{aligned}

Therefore,

\begin{aligned} &\lim _{x \rightarrow \sqrt{2}}-f(x)=\lim _{x \rightarrow \sqrt{2}} f(x)=f(\sqrt{2})-\text { (ii) } \\ &\Rightarrow \lim _{x \rightarrow \sqrt{2}}-f(x)=\lim _{h \rightarrow 0} f(\sqrt{2}-h)=\lim _{h \rightarrow 0} a=a \end{aligned}

\begin{aligned} &\text { and } \lim _{x \rightarrow \sqrt{2}+} f(x)=\lim _{h \rightarrow 0} f(\sqrt{2}+h) \\ &=\lim _{h \rightarrow 0} \frac{2 b^{2}-4 b}{(\sqrt{2}+h)^{2}} \\ &=\left(2 b^{2}-4 b\right) \lim _{h \rightarrow 0} \frac{1}{(\sqrt{2}+4)^{2}} \end{aligned}

\begin{aligned} &=\left(2 b^{2}-4 b\right) \cdot \frac{1}{(\sqrt{2}+0)^{2}} \\ &=\left(2 b^{2}-4 b\right) \cdot \frac{1}{(\sqrt{2})^{2}} \\ &=\frac{2 b^{2}-4 b}{2} \\ &\text { also, } f(\sqrt{2})=\frac{2 b^{2}-4 b}{(\sqrt{2})^{2}} \end{aligned}

\begin{aligned} &=\frac{2 b^{2}-4 b}{2} \end{aligned}

from(ii), we have

\begin{aligned} &\Rightarrow a=\frac{2 b^{2}-4 b}{2} \\ &=\frac{2 b^{2}-4 b}{2} \\ &\Rightarrow a=\frac{2 b^{2}-4 b}{2} \\ &\Rightarrow 2 a-2 b^{2}+4 b=0 \\ &\Rightarrow b^{2}-a-2 b=0 \end{aligned}

\begin{aligned} &\text { When }-a=1 \text { then } \\ &\Rightarrow b^{2}-2 b-1=0 \\ &\Rightarrow b=\frac{2 \pm \sqrt{4-(-4)}}{2} \\ &=\frac{2 \pm \sqrt{4+4}}{2} \\ &=\frac{2 \pm \sqrt{8}}{2} \\ &=1 \pm \sqrt{2} \end{aligned}

\begin{aligned} &\text { Also, when } a=-1 \text { then }\\ &\Rightarrow b^{2}-2 b+1=0\\ &\Rightarrow b=\frac{2 \pm \sqrt{4-4}}{2}\\ &=\frac{2}{2}\\ &=1\\ &\text { Thus } \mathrm{a}=-1 \text { and } \mathrm{b}=1 \end{aligned}

The correct option is (c)

Posted by

Gurleen Kaur

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