Rd sharma solution class 12 chapter continuity

Exercise 9.2

Question 1

Prove that the function f(x)=\begin{cases}\dfrac{sin~x}{x},x<0\\~x+1,x\geq0\end{cases} is everywhere continuous .

Sol :

When x<0 , we have \dfrac{sin~x}{x}

We know that sin x as well as the identity function are everywhere continuous .

So , the quotient function \dfrac{sin~x}{x} is continuous at each point x<0

When x>0 , we have f(x)=x+1 , which is a polynomial function .

Therefore, f(x) is continuous at each point x>0

Now , Let us consider the point x=0

Given ,f(x)=\begin{cases}\dfrac{sin~x}{x},x<0\\~x+1,x\geq0\end{cases}

We have , (LHL at x = 0)

=\lim_{x\to0^{-}}~f(x)

=\lim_{h\to0}~f(0-h)

=\lim_{h\to0}~f(-h)

=\lim_{h\to0}~\dfrac{sin~(-h)}{h}

=\lim_{h\to0}~\dfrac{-sin~(h)}{h}

=\lim_{h\to0}~\dfrac{sin~h}{h}

= 1

 

(RHL at x = 0)

=\lim_{x\to0^{+}}~f(x)

=\lim_{h\to0}~f(0+h)

=\lim_{h\to0}~f(h)

=\lim_{h\to0}~(h+1)

= 1

Also , f(0)=0+1=1

\lim_{x\to0^{-}}~f(x)=\lim_{x\to0^{+}}=f(0)

Thus , f(x) is continuous at 0

Hence, f(x) is everywhere continuous 

 

 

Question 2

Discuss the continuity of the function f(x)=\begin{cases}\dfrac{x}{|x|},x\not=0\\~~0~~,x=0\end{cases}

Sol :

|x|=\begin{cases}\phantom{-}x~,~x\geq0\\-x~,~x<0\end{cases} 

f(x)=\begin{cases}\dfrac{x}{x}~~~,x>0\\\dfrac{\phantom{-}x}{-x},x<0\\~~0~~~,x=0\end{cases}

f(x)=\begin{cases}\phantom{-}1~,~x>0\\-1~,~x<0\\\phantom{-}0~,~x=0\end{cases}

We have (LHL at x = 0)

=\lim_{x\to0^{-}}~f(x)

=\lim_{h\to0}~(-1)

= -1

 

(RHL at x = 0)

=\lim_{x\to0^{+}}~f(x)

=\lim_{h\to0}~(1)

= 1

\lim_{x\to0^{-}}\not=\lim_{x\to0^{+}}

Thus , f(x) is discontinuous at x=0

 

 

Question 3

Find the points of discontinuity, if any, of the following functions:

(i) f(x)=\begin{cases}x^3-x^2+2x-2~,~x\not=1\\~~~~~~~~~~~~~~~4~~~~~~~~~~~~,~x=1\end{cases}

Sol :

When x\not=1 then f(x)=x^3-x^2+2x-2 

We know that a polynomial function is everywhere continuous 

So , f(x)=x^3-x^2+2x-2 is continuous at each point at x\not=1

Now , at x=1 we have

(LHL at x = 1)

=\lim_{x\to1^{-}}~f(x)

=\lim_{h\to0}~f(1-h)

=\lim_{h\to0}~\left[(1-h)^3-(1-h)^2+2(1-h)-2\right]

=1-1+2-2

=0

( RHL at x = 1 )

=\lim_{x\to1^{+}}~f(x)

=\lim_{h\to0}~f(1+h)

=\lim_{h\to0}~\left[(1+h)^3-(1+h)^2+2(1+h)-2\right]

=1-1+2-2

=0

Also , f(1)=4

\lim_{x\to1^{-}}~f(x)=\lim_{x\to1^{+}}~f(x)\not=f(1)

Thus , f(x) is discontinuous at x=1

Hence , the only point of discontinuity for f(x)\text{ is } x=1

 

(ii) (ii) f(x)=\begin{cases}\dfrac{x^4-16}{x-2},x\not=2\\\\~~~~16~~~,x=2\end{cases}

Sol :

When x\not=2 , then f(x)=\dfrac{x^4-16}{x-2} =\dfrac{(x^2)^2-(2^2)^2}{x-2} =\dfrac{[(x^2)+(2^2)][(x^2)-(2^2)]}{x-2} =\dfrac{[(x^2)+(2^2)][(x-2)(x+2)]}{x-2} =(x^2+4)(x+2)

We know that a polynomial function is everywhere continuous . Therefore , the functions (x^2+4)\text{ and }(x+2) are everywhere continuous .

So , the product function (x^2+4)(x+2) is continuous at every x\not=2

Now at x = 2 , we have

(LHL at x=2)

=\lim_{x\to2^{-}}~f(x)

=\lim_{h\to0}~f(2-h)

=\lim_{h\to0}~[(2-h)^2+4](2-h+2)

= 8\times 4

= 32

 

(RHL at x=2)

=\lim_{x\to2^{+}}~f(x)

=\lim_{h\to0}~f(2+h)

=\lim_{h\to0}~[(2+h)^2+4](2+h+2)

= 8\times 4

= 32

Also , f(2)=16

\lim_{x\to2^{-}}=\lim_{x\to2^{+}}\not=f(2)

Thus , f(x) is discontinuous for f(x)\text{ is }x=2

 

(iii)

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