Exercise 5.1
Question 1
Examine the following functions for continuity at the indicated points:
(i) $f(x)=\left\{ \begin{array}{cl} x^3+1 &, x \neq 0 \\ 1 &, x=0 \end{array}\right.$
Sol :
Continuity at x=0
Here, f(x)=f(0)=1
LHL
$\lim_{x\rightarrow 0^-}~ f(x)=\lim_{h\rightarrow 0}~f(0-h)$
$=\lim_{h\rightarrow 0} ~(h)^3+3$=3
RHL
$\lim_{x\rightarrow 0^+}~ f(x)=\lim_{h\rightarrow 0}~f(0+h)$
$=\lim_{h\rightarrow 0}~ h^3+3$=3
$\lim_{x\rightarrow 0^-}~f(x)=\lim_{x\rightarrow 0^+}~f(x) \neq f(0)$
∴ f(x) is discontinuous at x =0
(ii) f(x)=x3+2x2-1 at x=1
Sol :
Continuity at x=0
Here, f(x)=f(1)=1+2-1=2
LHL
$\lim_{x\rightarrow 1^-}~ f(x)=\lim_{h\rightarrow 0}~f(1-h)$
$=\lim_{h\rightarrow 0} ~(1-h)^3+2(-h)^2-1$=1+2-1=2
RHL
$\lim_{x\rightarrow 1^+}~ f(x)=\lim_{h\rightarrow 0}~f(1+h)$
$=\lim_{h\rightarrow 0}~ (1+h)^3+2(1+h)^2-1$=1+2-1=2
LHL=RHL=f(1)
$\lim_{x\rightarrow 1^-}~f(x)=\lim_{x\rightarrow 1^+}~f(x) = f(1)$
∴ f(x) is continuous at x =0
Question 2
Prove that the function f(x)=5x-3 is continuous at x=0, at x=-3 and at x=5
Sol :
f(x)=5x-3
continuity at x=0
Here f(0)=-3
LHL
$\lim_{x\rightarrow 0^-}~ f(x)=\lim_{h\rightarrow 0}~f(0-h)$
$=\lim_{h\rightarrow 0}~5(0-h)-3$=-3
RHL
$\lim_{x\rightarrow 0^+}~ f(x)=\lim_{h\rightarrow 0}~f(0+h)$
$=\lim_{0^+}~5(0+h)-3$=-3
LHL=RHL=f(0)
∴ f(x) is continuous at x =0
continuity at x=-3
Here , f(-3)=-15-3=-18
LHL
$\lim_{x\rightarrow (-3)^-}~ f(x)=\lim_{h\rightarrow 0}~f(-3-h)$
$=\lim_{h \rightarrow 0}~5(-3-h)-3$
=-18
RHL
$\lim_{x\rightarrow (-3)^+}~ f(x)=\lim_{h\rightarrow 0}~f(-3+h)$
$=\lim_{h \rightarrow 0} ~5(-3+h)-3$=-18
LHL=RHL=f(-3)
∴ f(x) is continuous at x =-3
continuity at x=5
Here , f(5)=25-3=22
LHL
$\lim_{x\rightarrow 5^-}~ f(x)=\lim_{h\rightarrow 0}~f(5-h)$
$=\lim_{h\rightarrow 0}~5(5-h)-3$=22
RHL
$\lim_{x\rightarrow 1^+}~ f(x)=\lim_{h\rightarrow 0}~f(1+h)$
$=\lim_{h\rightarrow 0}~5(5+h)-3$=22
LHL=RHL=f(5)
∴ f(x) is continuous at x =5
Question 3
(i) If $f(x)=\frac{x^2-1}{x-1}$ for x≠1 and f(x)=2 when x=1, show that the function is continuous at x=1
Sol :
$f(x)=\left\{ \begin{array}{cl} \frac{x^2-1}{x-1} &, x\neq 1 \\ 2 & , x=1 \end{array} \right.$$\Rightarrow f(x)=\left\{ \begin{array}{cl} \frac{(x-1)(x+1)}{x-1} &, x\neq 1 \\ 2 & , x=1 \end{array} \right.$
$\Rightarrow f(x)=\left\{ \begin{array}{cl} x+1 &, x\neq 1 \\ 2 & , x=1 \end{array} \right.$
continuity at x=1
Here f(1)=2
LHL
$\lim_{x\rightarrow 1^-}=\lim_{h\rightarrow 0} f(1-h)$
$=\lim_{h\rightarrow 0}~(1-h)+1$=2
RHL
$\lim_{x\rightarrow 1^+}~f(x)=\lim_{h\rightarrow 0}$
$=\lim_{h\rightarrow 0}~(1+h)+1$=2
LHL=RHL=f(1)
∴ f(x) is continuous at x =1
(ii) A function f is defined as $f(x)= \left\{ \begin{array}{cl} \frac{x^2-x-6}{x-3} & ,x\neq 3 \\ 5 &, x=3 \end{array} \right.$ show that f is continuous at x =3
Sol :
$\Rightarrow f(x)= \left\{ \begin{array}{cl} \frac{x^2-3x+2x-6}{x-3} & ,x\neq 3 \\ 5 &, x=3 \end{array} \right.$
$\Rightarrow f(x)= \left\{ \begin{array}{cl} \frac{(x-3)(x+2)}{x-3} & ,x\neq 3 \\ 5 &, x=3 \end{array} \right.$
$\Rightarrow f(x)= \left\{ \begin{array}{cl} x+2 & ,x\neq 3 \\ 5 &, x=3 \end{array} \right.$
continuity at x=3
Here f(3)=5
LHL
$\lim_{x\rightarrow 3^-}~f(x)=\lim_{h\rightarrow 0}~f(3-h)$
$=\lim_{h \rightarrow 0}~3-h+2$=5
RHL
$\lim_{x\rightarrow 3^+}~f(x)=\lim_{h\rightarrow 0}~f(3+h)$
$\lim_{h\rightarrow 0}~(3+h)+2$=5
LHL=RHL=f(3)
∴f(x) is continuous at x =3
Question 4
Is the function f defined by $f(x)=\left\{ \begin{array}{cl} x & , if~ x \leq 1 \\ 5 & , x>1 \end{array} \right.$ continuous at
(i) x=0 (ii) x=1 (iii) x=2
Sol :
continuity at x=0
Here f(0)=0
LHL
$\lim_{x\rightarrow 0^-}~f(x)=\lim{h\rightarrow 0}~f(0-h)$
$=\lim_{h\rightarrow 0}~0-h$=0
RHL
$\lim_{x\rightarrow 0^+}~f(x)=\lim{h\rightarrow 0}~f(0+h)$
$=\lim_{h\rightarrow 0}~0+h$=0
LHL=RHL=f(0)
∴f(x) is continuous at x =0
(ii) $f(x)=\left\{ \begin{array}{cl} x & , if~ x\geq 1 \\ 5 & , if~ x>1 \end{array} \right.$
Sol :
continuity at x=1
Here f(1)=1
LHL
$\lim_{x\rightarrow 1^-}~f(x)=\lim_{h \rightarrow 0}~f(1-h)$
$=\lim_{h\rightarrow 0}~1-h$=1
RHL
$\lim_{x\rightarrow 1^+}~f(x)=\lim_{h \rightarrow 0}~f(1+h)$
$=\lim_{h\rightarrow 0}~5$=5
LHL≠RHL
∴f(x) is discontinuous at x =1
(iii) $f(x)=\left\{ \begin{array}{cl} x & , if~ x\geq 1 \\ 5 & , if~ x>1 \end{array} \right.$
Sol :
continuity at x=2
Here f(2)=5
LHL
$\lim_{x\rightarrow 2}~f(x)=\lim_{h \rightarrow 0}~f(2-h)$
$=\lim_{h\rightarrow 0}~5$=5
RHL
$\lim_{x\rightarrow 2}~f(x)=\lim_{h \rightarrow 0}~f(2+h)$
$=\lim_{h\rightarrow 0}~5$=5
LHL=RHL=f(2)
∴f(x) is continuous at x =2
Question 5
Is the function f defined by $f(x)=\left\{ \begin{array}{cl} 3x+5 &, if~x\geq 2 \\ x^2 &, if~x<2 \end{array}\right.$ continuous at x =2?
Sol :
continuity at x=2
Here , f(2)=3(2)+5=11
LHL
$\lim_{x \rightarrow 2^-}~f(x)=\lim_{h \rightarrow 0}~f(2-h)$
$=\lim_{h\rightarrow 0}~(2-h)^2$=4
RHL
$\lim_{x \rightarrow 2^+}~f(x)=\lim_{h \rightarrow 0}~f(2+h)$
$=\lim_{h\rightarrow 0}~[3(2+h)+5]$=11
LHL≠RHL
∴f(x) is discontinuous at x =2
Question 6
Is the function f defined by $f(x)=\left\{ \begin{array}{cl} \frac{2x^2-3x-2}{x-2} &, \text{if } x\neq 2 \\ 5 & , \text{if } x=2 \end{array} \right.$
Sol :
continuity at x =2
$\Rightarrow f(x)=\left\{ \begin{array}{cl} \frac{(2x+1)(x-2)}{x-2} &, \text{if } x\neq 2 \\ 5 & , \text{if } x=2 \end{array} \right.$
$\Rightarrow f(x)=\left\{ \begin{array}{cl} 2x+1 &, \text{if } x\neq 2 \\ 5 & , \text{if } x=2 \end{array} \right.$
Here , f(2)=2(2)+1=5
LHL
$\lim_{x \rightarrow 2^-}~f(x)=\lim_{h \rightarrow 0}~f(2-h)$
$=\lim_{h \rightarrow 0}~[2(2-h)+1]$=5
RHL
$\lim_{x \rightarrow 2^+}~f(x)=\lim_{h \rightarrow 0}~f(2+h)$
$=\lim_{h \rightarrow 0}~[2(2+h)+1]$=5
LHL=RHL=f(2)
∴f(x) is continuous at x=2
Question 7
Is the function f defined by $f(x)=\left\{ \begin{array}{cl} \frac{x}{\sin 2x} &, \text{when } x \neq 0 \\ 2 & , \text{when } x=0 \end{array}\right.$ continuous at x=0 ?
Sol :
continuity at x=0
Here , f(0)=2
LHL
$\lim_{x\rightarrow 0^-}~f(x)=\lim_{h \rightarrow 0}~f(0-h)$
$\lim_{h \rightarrow 0}~\frac{0-h}{\sin 2(0-h)}=\lim_{h \rightarrow 0}~\frac{-h}{-\sin 2h}$
$\lim_{h \rightarrow 0}~\frac{2h}{\sin 2h} \times \frac{1}{2}$
$=\frac{1}{2} \frac{1}{\lim_{2h \rightarrow 0}~ \frac{\sin 2h}{2h}}$ [as h→0 , then 2h→0]
$=\frac{1}{2} \times 1 \left[ \lim_{x\rightarrow 0} \frac{\sin x}{x}=1\right]$
since, LHL ≠ f(0)
∴f(x) is discontinuous at x=0
Question 8
If $f(x)=\left\{ \begin{array}{cl}kx^2 &,x\leq 2 \\ 3 & , x>2 \end{array} \right.$ , find k so that f may be continuous at x=2
Sol :
continuity at x=2
LHL=RHL=f(2)
Here f(2)=4k
RHL
$\lim_{x\rightarrow 2^+}~f(x)=\lim_{h\rightarrow 0}~f(2+h)$
$=\lim_{h \rightarrow 0}~3$=3
∴f(x) is continuous at x=2
⇒RHL=f(2)
⇒3=4k
⇒$k=\frac{3}{4}$
Question 9
If $f(x)=\left\{ \begin{array}{cl} 3x-8 & , x \leq 5 \\ 2k & , x> 5\end{array} \right.$ , find k so that f may be continuous at x =5
Sol :
$f(x)=\left\{ \begin{array}{cl} 3x-8 & , x \leq 5 \\ 2k & , x> 5\end{array} \right.$
continuous at x=5
Here , f(5)=3(5)-8=7
RHL
$\lim_{x\rightarrow 5^+}~f(x)=\lim_{h\rightarrow 0}~f(5+h)$
$=\lim_{h \rightarrow 0}~2k$=2k
∵f(x) is continuous at x=5
⇒RHL=f(5)
⇒2k=7
⇒$k=\frac{7}{2}$
Question 10
(i) If $f(x)=\left\{ \begin{array}{cl} \frac{x^2-2x-3}{x+1} & , x \neq -1 \\ k & , x=-1\end{array}\right.$ find k so that the function f may be continuous at x=-1
Sol :
continuous at x =-1
$f(x)=\left\{ \begin{array}{cl} \frac{(x+1)(x-3)}{x+1} &, x \neq -1 \\ k & , x=-1 \end{array} \right.$
Here , f(-1)=k
LHL
$\lim_{x\rightarrow -1^-}~f(x)=\lim_{h \rightarrow 0}~f(-1-h)$
$=\lim_{(-1-h)-3}$=-4
∵f(x) is continuous at x=-1
⇒LHL = f(-1)
⇒-4=k
⇒k=-4
(ii) $f(x)=\left\{ \begin{array}{cl} \frac{(x-3)^2-36}{x-3} &, x\neq 3 \\ k & , x=3 \end{array}\right.$
Sol :
$f(x)=\left\{\begin{array}{cl} \frac{(x+3)^2-6^2}{x-3} &, x \neq 3 \\ k & , x=3 \end{array}\right.$
continuous at x=3
$f(x)=\left\{ \begin{array}{cl} \frac{[(x+3)+6] [(x+3)-6]}{x-3} & ,x\neq 3 \\ k & , x=3 \end{array}\right.$
$f(x)=\left\{ \begin{array}{cl} \frac{(x+9)(x-3)}{x-3} &, x \neq 3 \\ k &, x=3 \end{array} \right.$
$f(x)=\left\{ \begin{array}{cl} x+9 & , x\neq 3 \\ k & , x=3 \end{array} \right.$
Here , f(3)=k
LHL
$\lim_{x\rightarrow 3^-}~f(x)=\lim_{h\rightarrow 0}~f(3-h)$
$=\lim_{h\rightarrow 0}~[(3-h)+9]$=12
∵f(x) is continuous at x=3
⇒LHL=f(3)
⇒12=k
⇒k=12
(iii) If $f(x)=\left\{ \begin{array}{cl} \frac{x^2-x-6}{x^2-2x-3} &, x\neq 3 \\ k & , x=3 \end{array}\right.$ find k so that the function f may be continuous at x=3
Sol :
$f(x)=\left\{ \begin{array}{cl} \frac{x^2-3x+2x-6}{x^2-3x+x-3} &, x\neq 3 \\ k & , x=3 \end{array}\right.$
$f(x)=\left\{ \begin{array}{cl} \frac{x(x-3)+2(x-3)}{x(x-3)+1(x-3)} &, x\neq 3 \\ k & , x=3 \end{array}\right.$
$f(x)=\left\{ \begin{array}{cl} \frac{(x+3)(x-3)}{(x+1)(x-3)} &, x\neq 3 \\ k & , x=3 \end{array}\right.$
$f(x)=\left\{ \begin{array}{cl} \frac{x+2}{x+1} &, x\neq 3 \\ k & , x=3 \end{array}\right.$
Here, f(3)=k
RHL
$\lim_{x \rightarrow 3^+}~f(x)=\lim_{h \rightarrow 0}~f(3+h)$
$=\lim_{h \rightarrow 0}~\frac{(3+h)+2}{(3+h)+1}=\frac{5}{4}$
∵f(x) is continuous at x=3
⇒RHL=f(3)
⇒$\frac{5}{4}=k$
⇒$k=\frac{5}{4}$
Question 11
If $f(x)=\left\{ \begin{array}{cl} \frac{\tan 3x}{kx} &, x \neq 0 \\ 1 & , x=0 \end{array} \right.$ , find k so that the function f may be continuous at x=0
Sol :
$f(x)=\left\{ \begin{array}{cl} \frac{\tan 3x}{kx} &, x \neq 0 \\ 1 & , x=0 \end{array} \right.$
continuous at x=0
Here f(0)=1
RHL
$\lim_{x \rightarrow 0^+}~f(x)=\lim_{h \rightarrow 0}~f(0+h)$
$=\lim_{h \rightarrow 0}~\frac{\tan 3h}{kh}$
$=\frac{3}{h} \lim_{h\rightarrow 0}~ \frac{\tan 3h}{3h}$
$=\frac{3}{h} \lim_{3h \rightarrow 0}~\frac{\tan 3h}{3h}$ [as h→0, then 3h→0]
$=\frac{3}{k} \times 1$ [As $\lim_{x\rightarrow 0}~\frac{\tan x}{x}=1$]
Since, f(x) is continuous at x=0
⇒RHL=f(0)
⇒$\frac{3}{k}=1$
⇒k=3
Question 12
Is the function f defined by f(x)=tan x continuous at $x=\frac{\pi}{2}$?
Sol :
f(x)=tan x
continuity at $x=\frac{\pi}{2}$
$f\left( \frac{\pi}{2}\right)=\tan \frac{\pi}{2}=\infty$
$f\left( \frac{\pi}{2}\right) does not exist
∴f(x) is discontinuous at $x=\frac{\pi}{2}$
LHL=RHL=f(x)
Question 13
Is the function f defined by f(x)=|x| continuous at x=0 ?
Sol
f(x)=|x|
continuity at x=0
$f(x)=\left\{ \begin{array}{cl} x & , x\geq 0 \\ -x & , x<0 \end{array} \right.$
f(0)=0
LHL
$\lim_{x\rightarrow 0^-}~f(x)=\lim_{h \rightarrow 0}~(0-h)$
$=\lim_{h\rightarrow 0}~-(0-h)=\lim_{h \rightarrow 0}h=0$
RHL
$\lim_{x\rightarrow 0^+}~f(x)=\lim_{h \rightarrow 0}~(0+h)$
$=\lim_{h \rightarrow 0}~(0+h)=0$
⇒LHL=RHL=f(0)
⇒f(x) is continuous at x=0
Question 14
Is the function f defined by f(x)=|x-1| continuous at x=1 ?
Sol :
f(x)=|x-1|
continuity at x=1
$f(x)=\left\{ \begin{array}{cl| l} x-1 & , x-1 \geq 0 & x \geq 1 \\ -(x-1) & , x-1<0 & x<1 \end{array} \right.$
Here f(1)=1-1=0
LHL
$\lim_{x\rightarrow 1^-}~f(x)=\lim_{h \rightarrow 0}~f(1-h)$
$=\lim_{h \rightarrow 0}~-(1-h-1)=\lim_{h \rightarrow 0}h=0$
RHL
$\lim_{x\rightarrow 1^+}~f(x)=\lim_{h \rightarrow 0}~f(1+h)$
$=\lim_{h \rightarrow 0}~(1+h-1)=\lim_{h \rightarrow 0}h=0$
⇒LHL=RHL=f(1)
⇒f(x) is continuous at x=1
Question 15
Is the function f defined by f(x)=x-|x| continuous at x=0
Sol :
f(x)=x-|x|
continuity at x=0
$|x|=\left\{ \begin{array}{cl} x &, x \geq 0 \\ -x & , x<0 \end{array} \right.$
$f(x)=\left\{ \begin{array}{cl}x-x , & x \geq 0 \\ x-(-x) & , x<0 \end{array} \right.$
$f(x)=\left\{ \begin{array}{cl}0 , & x \geq 0 \\ 2x & , x<0 \end{array} \right.$
Here f(0)=0
LHL
$=\lim_{x\rightarrow 0^-}f(x)=\lim_{h \rightarrow 0}~f(0-h)$
$=\lim_{h \rightarrow 0}~2(0-h)$=0
RHL
$=\lim_{x\rightarrow 0^+}f(x)=\lim_{h \rightarrow 0}~f(0+h)$
$=\lim_{h \rightarrow 0}~0$=0
LHL=RHL=f(0)
⇒f(x) is continuous at x=0
working......
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