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  • Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 39

Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 39

Answers (1)

Answer:

 The correct option is (b)

Hint:

(i) 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(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}

(ii) Standard limits

\lim _{x \rightarrow 0} \frac{\sin x}{x}=1

Given:

 f(x)= \begin{cases}\frac{1-\sin ^{2} x}{3 \cos ^{2} x} &, x<\frac{\pi}{2} \\ a & , x=\frac{\pi}{2} \\ \frac{b(1-\sin x)}{(\pi-2 x)^{2}} &, x>\frac{\pi}{2}\end{cases}

Solution:

f(x)= \begin{cases}\frac{1-\sin ^{2} x}{3 \cos ^{2} x} &, x<\frac{\pi}{2} \\ a & , x=\frac{\pi}{2} \\ \frac{b(1-\sin x)}{(\pi-2 x)^{2}} &, x>\frac{\pi}{2}\end{cases}

So it is given that the function f(x) is continuous at

x=\frac{\pi}{2}
So, we have

\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}}+f(x)=\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)-(1) \\ &\text { L.H.L at } x=\frac{\pi}{2} \\ &{\text { i.e }} \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{h \rightarrow 0} f(\frac{\pi}{2}-h) \end{aligned}

=\lim _{h \rightarrow 0} \frac{1-\sin ^{2}(\pi / 2-h)}{3 \cos ^{2}(\pi / 2-h)} \quad\left[\therefore \boldsymbol{f}(x)=\frac{1-\sin ^{2} x}{3 \cos ^{2} x}, x<\pi / 2\right]

=\lim _{h \rightarrow 0} \frac{1-\{\sin (\pi / 2-h)\}^{2}}{3\{\cos (\pi / 2-h)\}^{2}}

=\lim _{h \rightarrow 0} \frac{1-(\cos h)^{2}}{3(\sin h)^{2}} \quad\left[\begin{array}{c} \because \sin (\pi / 2-\theta)=\cos \theta, \\ \cos (\pi / 2-\theta)=\sin \theta \end{array}\right]

=\lim _{h \rightarrow 0} \frac{1-\cos ^{2} h}{3 \sin ^{2} h}

\begin{aligned} &=\lim _{n \rightarrow 0} \frac{\sin ^{2} h}{3 \sin ^{2} h} \quad\left[\because \cos ^{2} \theta+\sin ^{2} \theta=\perp \Rightarrow 1-\cos ^{2} \theta=\sin ^{2} \theta\right] \\ &=\frac{1}{3} \lim _{h \rightarrow 0} 1=\frac{1}{3} . \quad {-(2)} \end{aligned}

\begin{aligned} &\text { R.H.I at } x=\pi / 2 \\ &{\text { i.e }} \lim _{x \rightarrow \pi / 2^{+}} f(x)=\lim _{h \rightarrow 0} f(\pi / 2+h) \end{aligned}

=\lim _{h \rightarrow 0} \frac{b\{1-\sin (\pi / 2+h)\}}{\{\pi-2(\pi / 2+h)\}^{2}} \quad\left[\because {f}(x)=\frac{{b}(\mathbf{1}-\sin x)}{(\pi-2 x)^{2}}, x>\pi / 2\right]

=\lim _{h \rightarrow 0} \frac{b\{1-\cosh \}}{\left\{\pi-2\left(\frac{\pi}{2}\right)-2 h\right\}^{2}} \quad[\because \sin (\pi / 2+\theta)=\cos \theta]

=\lim _{h \rightarrow 0} \frac{b\left\{2 \sin ^{2} h / 2\right\}}{(\pi-\pi-2 h)^{2}} \quad\left[\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right]

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\frac{2 b \sin ^{2} h}{2}}{(-2 h)^{2}} \\ &=\lim _{h \rightarrow 0} \frac{\frac{2 b \sin ^{2} h}{2}}{4 h^{2}} \\ &=\frac{26}{4} \lim _{h \rightarrow 0} \frac{\frac{\sin ^{2} n}{2}}{h^{2}} \end{aligned}

\begin{aligned} &=\frac{b}{2} \lim _{h \rightarrow 0} \frac{\frac{\sin ^{2} h}{2}}{\frac{h^{2}}{4} x^{4}} \\ &=\frac{b}{8} \lim _{h \rightarrow 0}\left(\frac{\frac{\sin h}{2}}{\frac{h}{2}}\right)^{2} \\ &=\frac{6}{8}\left(\lim _{h \rightarrow 0} \frac{\sin n / 2}{h / 2}\right)^{2} \end{aligned}

\begin{aligned} &=\frac{b}{8} \times 1^{2} \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ &=\frac{b}{8} \times 1 \\ &=\frac{b}{8}-(3) \end{aligned}

\begin{aligned} &\text { also, } f(x) a t-x=\pi / 2\\ &{\text { i.e }} f(\pi / 2)=a \quad -(4) &[\because {f}({x})={a}, {x}=\pi / 2] \end{aligned}

Putting the value of equation 2 ,3 , 4 in (1)

we get ,

\begin{aligned} &-\frac{1}{3}=\frac{b}{8}=a \\ &\text { Sor } a=\frac{1}{3} \text { or } \frac{b}{8}=\frac{1}{3} \Rightarrow b=\frac{8}{3} \\ &\therefore a=\frac{1}{3}, b=\frac{8}{2} \end{aligned}

Posted by

Gurleen Kaur

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