RD Sharma Class 12 Ex 9.1 Solutions Chapter 9 Continuity

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter9
Exercise9.1
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 9.1 Solutions Chapter 9 Continuity

Question 1. Test the continuity of the following function at the origin:

  f(x)= \begin{cases}\frac{x}{|x|},& x \neq 0 \\1,& x=0\end{cases}

Solution:

Given that

f(x)= \begin{cases}\frac{x}{|x|},& x\neq0 \\1,& x=0\end{cases}

Now, let us consider LHL at x = 0

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}f(-h)
=\lim_{h\to0}\frac{-h}{|-h|}
=\lim_{h\to0}\frac{-h}{h} =-1

Now, let us consider RHL at x = 0

\lim_{x \to 0^+}f(x) =\lim_{h \to 0}f(0+h)
=\lim_{h \to 0}\frac{h}{|h|}=1

So, LHL ≠ RHL

Therefore, f(x) is discontinuous at origin and the discontinuity is of 1st kind.

Question 2. A function f(x) is defined as f(x)= \begin{cases}\frac{x^2-x-6}{x-3},& \text{if } x\neq3 \\5,& \text{if }x=3\end{cases}  . Show that f(x) is continuous at x = 3.

Solution:

Given that

f(x)= \begin{cases}\frac{x^2-x-6}{x-3},& \text{if } x\neq3 \\5,& \text{if }x=3\end{cases}

So, here we check the given f(x) is continuous at x = 3,

Now, let us consider LHL at x = 3

\lim_{x \to 3^-} f(x) =\lim_{h \to 0}f(3-h)
=\lim_{h \to 0}\frac{(3-h)^2-(3-h)-6}{(3-h)-3}
=\lim_{h \to 0}\frac{h^2-5h}{-h}
=\lim_{h\to0}-h+5=5

Now, let us consider RHL at x = 3

\lim_{x\to3^+} f(x) =\lim_{h\to0}f(3+h)
=\lim_{h\to0}\frac{(3+h)^2-(3+h)-6}{(3+h)-3}
=\lim_{h\to0}\frac{h^2+5h}{h}
=\lim_{h\to0}h+5=5

So, f(3) = 5

LHL= RHL = f(3) 

Therefore, f(x) is continuous at x = 3

Question 3. A function f(x) is defined as

 f(x)= \begin{cases}\frac{x^2-9}{x-3},& \text{if } x\neq3 \\6,& \text{if }x=3\end{cases}  

Show that f(x) is continuous at x = 3.

Solution:

Given that

f(x)= \begin{cases}\frac{x^2-9}{x-3},& \text{if } x\neq3 \\6,& \text{if }x=3\end{cases}

So, here we check the given f(x) is continuous at x = 3,

Now, let us consider LHL at x = 3

\lim_{x\to3^-}f(x)=\lim_{h\to0}f(3-h)
=\lim_{h\to0}\frac{(3-h)^2-9}{(3-h)-3}
=\lim_{h\to0}\frac{h^2-6h}{-h}
=\lim_{h\to0}-h+6=6

Now, let us consider RHL at x = 3

\lim_{x\to3^+}f(x) =\lim_{h\to0}f(3+h)
=\lim_{h\to0}\frac{(3+h)^2-9}{(3+h)-3}
=\lim_{h\to0}\frac{h^2+6h}{h}
=\lim_{h\to0}h+6=6

So, f(3) = 6

LHL= RHL= f(3)

Therefore, f(x) is continuous at x = 3

Question 4. f(x)= \begin{cases}\frac{x^2-1}{x-1},& \text{if } x\neq1 \\2,& \text{if }x=1\end{cases}  

Find whether f(x) is continuous at x = 1

Solution:

Given that

f(x)= \begin{cases}\frac{x^2-1}{x-1},& \text{if } x\neq1 \\2,& \text{if }x=1\end{cases}

So, here we check the given f(x) is continuous at x = 1,

Now, let us consider LHL at x = 1

\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)
=\lim_{h\to0}\frac{(1-h)^2-1}{(1-h)-1}
=\lim_{h\to0}\frac{1+h)^2-2h-1}{1-h-1}
=\lim_{h\to0}\frac{h^2-2h}{-h}
=\lim_{h\to0}\frac{h(h-2)}{-h}
=\lim_{h\to0}(2-h)=2

Now, let us consider RHL at x = 1

\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)
=\lim_{h\to0}\frac{(1+h)^2-1}{(1+h)-1}
=\lim_{h\to0}\frac{1+h^2+2h-1}{1+h-1}
=\lim_{h\to0}\frac{h^2+2h}{h}
=\lim_{h\to0}\frac{h(h+2)}{h}
=\lim_{h\to0}(2+h)=2

So, f(1) = 2

LHL= RHL = f(1)

Therefore, f(x) is continuous at x = 1

Question 5. If f(x)= \begin{cases}\frac{sin3x}{x},& \text{when } x\neq0 \\1,& \text{when }x=0\end{cases}  

 Find whether f(x) is continuous at x = 0.

Solution:

Given that

f(x)= \begin{cases}\frac{sin3x}{x},& \text{when } x\neq0 \\1,& \text{when }x=0\end{cases}

So, here we check the given f(x) is continuous at x = 0,

Now, let us consider LHL at x = 0

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{sin3(-h)}{-h}
=\lim_{h\to0}-\frac{sin3h}{-h}=3

Now, let us consider RHL at x = 0

\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)
=\lim_{h\to0}\frac{sin3h}{h}=3

So, f(0) = 1

LHL = RHL≠ f(0)

Therefore, f(x) is discontinuous at x = 0.

Question 6. If f(x)= \begin{cases}\\e^{\frac{1}{x}},& \text{if } x\neq0 \\1,& \text{if }x=0\end{cases}  

Find whether f is continuous at x = 0.

Solution:

Given that

f(x)= \begin{cases}\\e^{\frac{1}{x}},& \text{if } x\neq0 \\1,& \text{if }x=0\end{cases}

So, here we check the given f(x) is continuous at x = 0,

Now, let us consider LHL at x = 0

\lim_{x\to0^-}f(x) =\lim_{x\to0}f(x)
=\lim_{h\to0}f(0-h)
=\lim_{h\to0}e^\frac{1}{-h}=e^{-∞}=0

Now, let us consider RHL at x = 0

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}e^\frac{1}{h}=e^{-∞}=∞

So, LHL≠ RHL

Therefore, the f(x) is discontinuous at x = 0. 

Question 7. Let f(x)= \begin{cases}\frac{1-cosx}{x^2},& \text{when } x\neq0 \\1,& \text{when }x=0\end{cases}

Show that f(x) is discontinuous at x = 0.

Solution:

Given that

f(x)= \begin{cases}\frac{1-cosx}{x^2},& \text{when } x\neq0 \\1,& \text{when }x=0\end{cases}

So, here we check the given f(x) is discontinuous at x = 0,

Now, let us consider LHL at x = 0

\lim_{x\to0^-}f(x) =\lim_{x\to0}f(0-h)
=\lim_{h\to0}\frac{1-cos(-h)}{(-h)^2}
=\lim_{h\to0}\frac{1-cosh}{h^2}
=\lim_{h\to0}\frac{2sin^2(h/2)}{h^2}
=\lim_{h\to0}2(\frac{sin(h/2)}{h})^2

= 2 × 1/4 = 1/2                           

Now, let us consider RHL at x = 0

\lim_{x\to0^+}f(x) =\lim_{x\to0}f(0+h)
=\lim_{h\to0}\frac{1-cosh}{h^2}
=\lim_{h\to0}\frac{2sin^2(h/2)}{h^2}
=\lim_{h\to0}2(\frac{sin(h/2)}{h})^2

= 2 × 1/4 = 1/2                           

f(0) = 1

LHL= RHL ≠ f(0)

Therefore, the f(x) is discontinuous at x = 0. 

Question 8. Show that f(x)= \begin{cases}\frac{x-|x|}{2},& \text{when } x\neq0 \\2,& \text{when }x=0\end{cases}  is discontinuous at x = 0.

Solution:

Given that

f(x)= \begin{cases}\frac{x-|x|}{2},& \text{when } x\neq0 \\2,& \text{when }x=0\end{cases}

So, here we check the given f(x) is discontinuous at x = 0,

Now, let us consider LHL at x = 0

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{-h-|h|}{2}
=\lim_{h\to0}\frac{-h-h}{2}=0

Now, let us consider RHL at x = 0

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}\frac{h-|h|}{2}=0

f(0) = 2

Thus, LHL= RHL≠ f(0)

Therefore, f(x) is discontinuous at x = 0. 

Question 9. Show that f(x)= \begin{cases}\frac{|x-a|}{x-a},& \text{when } x\neq a \\1,& \text{when }x=a\end{cases}  is discontinuous at x = a.

Solution:

Given that

f(x)= \begin{cases}\frac{|x-a|}{x-a},& \text{when } x\neq a \\1,& \text{when }x=a\end{cases}

So, here we check the given f(x) is discontinuous at x = a,

Now, let us consider LHL at x = a

\lim_{x\to a^-}f(x)=\lim_{h\to0}f(a-h)
=\lim_{h\to0}\frac{|a-h-a|}{a-h-a}
=\lim_{h\to0}\frac{h}{-h}=-1

Now, let us consider RHL at x = a

\lim_{x\to a^+}f(x)=\lim_{h\to0}f(a+h)
=\lim_{h\to0}\frac{|a+h-a|}{a+h-a}
=\lim_{h\to0}\frac{h}{h}=1

Thus, LHS ≠ RHL

Therefore, the f(x) is discontinuous at x = a.

Discuss the continuity of the following functions at the indicated points(s):

Question 10 (i). f(x)= \begin{cases}|x|cos(\frac{1}{x}),& x\neq0 \\0,& x=0\end{cases}at\space x=0

Solution:

Given that

f(x)= \begin{cases}|x|cos(\frac{1}{x}),& x\neq0 \\0,& x=0\end{cases}at\space x=0

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0} (|-h|cos(-1/h)
=\lim_{h\to0}hcos(-1/h)=0

Now, let us consider RHL,

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}|h|cos(1/h)=0

f(0) = 0

Thus, LHL= RHL= f(0) = 0

Therefore, f(x) is continuous at x = 0.

Question 10 (ii). f(x)= \begin{cases}x^2sin(\frac{1}{x}),& \text{when } x\neq0 \\0,& \text{when }x=0\end{cases}  at x = 0

Solution:

Given that

f(x)= \begin{cases}x^2sin(\frac{1}{x}),& \text{when } x\neq0 \\0,& \text{when }x=0\end{cases}

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}(-h)^2sin(-1/h)=0

Now, let us consider RHL,

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}h^2sin(1/h)=0

f(0) = 0

Thus, LHL= RHL = f(0) = 0

Therefore, f(x) is continuous at x = 0.

Question 10 (iii). f(x)= \begin{cases}(x-a)sin(\frac{1}{x-a}),& \text{when } x\neq a \\0,& \text{when }x=a\end{cases}  at x = a

Solution:

Given that

f(x)= \begin{cases}(x-a)sin(\frac{1}{x-a}),& \text{when } x\neq a \\0,& \text{when }x=a\end{cases}

So, here we check the continuity of the given f(x) at x = a,

Let us consider LHL,

\lim_{x\to a^-}f(x)=\lim_{h\to0}f(a-h)
=\lim_{h\to0}(a-h-a)sin(\frac{1}{a-h-a})
=\lim_{h\to0}-hsin(1/h)=0

Now, let us consider RHL,

\lim_{x\to a^+}f(x)=\lim_{h\to0}f(a+h)
=\lim_{h\to0}(a+h-a)sin(\frac{1}{a+h-a})
=\lim_{h\to0}hsin(1/h)=0

f(a) = 0

Thus, LHL= RHL= f(a) = 0

Therefore, f(x) is continuous at x = 0.

Question 10 (iv). f(x)= \begin{cases}\frac{(e^x-1)}{log(1+2x)},& \text{when } x\neq0 \\7,& \text{when }x=0\end{cases}  at x = 0

Solution:

Given that

f(x)= \begin{cases}\frac{(e^x-1)}{log(1+2x)},& \text{when } x\neq0 \\7,& \text{when }x=0\end{cases}

So, here we check the continuity of the given f(x) at x = 0,

\lim_{x\to0}f(x)=\lim_{x\to0}\frac{e^x-1}{log(1+2x)}
=\lim_{x\to0}\frac{e^x-1}{\frac{2xlog(1+2x)}{2x}}
=(1/2)×\lim_{x\to0}\frac{(e^x-1)/x}{\frac{log(1+2x)}{2x}}
=(1/2)×\frac{\lim_{x\to0}(e^x-1)/x}{\lim_{x\to0}\frac{log(1+2x)}{2x}}

= 1/2 × 1/1 = 1/2                        

And, 

f(0) = 7

\lim_{x\to0}f(x)  ≠ f(0)

Therefore, f(x) is discontinuous at x = 0.

Question 10 (v). f(x)= \begin{cases}\frac{(1-x^n)}{(1-x)},& \text{when } x\neq1 \\n-1,& \text{when }x=1\end{cases}  n ∈ N at x = 1 

Solution:

Given that

f(x)= \begin{cases}\frac{(1-x^n)}{(1-x)},& \text{when } x\neq1 \\n-1,& \text{when }x=1\end{cases}

So, here we check the continuity of the given f(x) at x = 1,

Let us consider LHL,

\lim_{x\to1^-}f(x) =\lim_{h\to0}f(1-h)
=\lim_{h\to0}\frac{1-(1-h)^n}{1-(1-h)}
=\lim_{h\to0}\frac{1-[1-nh+\frac{n(n-1)}{2!}h^2+...]}{h}
=\lim_{h\to0}n-\frac{n(n-1)h}{2!}+...=n

Now, let us consider RHL,

\lim_{x\to1^+}f(x) =\lim_{h\to0}f(1+h)
=\lim_{h\to0}\frac{1-(1+h)^n}{1-(1+h)}
=\lim_{h\to0}\frac{1-[1+nh+\frac{n(n-1)}{2!}h^2+...]}{-h}
=\lim_{h\to0}n+\frac{n(n-1)h}{2!}+...=n

f(1) = n – 1

Thus, LHL = RHL ≠ f(1)

Therefore, f(x) is discontinuous at x = 1.

Question 10 (vi). f(x)= \begin{cases}\frac{|x^2-1|}{x-1},& \text{when } x\neq1 \\2,& \text{when }x=1\end{cases}  at x = 1

Solution:

Given that

f(x)= \begin{cases}\frac{|x^2-1|}{x-1},& \text{when } x\neq1 \\2,& \text{when }x=1\end{cases}

So, here we check the continuity of the given f(x) at x = 1,

Let us consider LHL,

\lim_{x\to1^-}f(x) =\lim_{h\to0}f(1-h)
=\lim_{h\to0}\frac{|(1-h)^2-1|}{(1-h)-1}
=\lim_{h\to0}\frac{|h^2-2h|}{-h}
=\lim_{h\to0}(h-2)=-2

Now, let us consider RHL,

\lim_{x\to1^+}f(x) =\lim_{h\to0}f(1+h)
=\lim_{h\to0}\frac{|(1+h)^2-1|}{(1+h)-1}
=\lim_{h\to0}\frac{h^2+2h}{h}=2

f(1) = 2

LHL= RHL = f(1) = 2

Therefore, f(x) is discontinuous at x = 1.

Question 10 (vii). f(x)= \begin{cases}\frac{2|x|+x^2}{x},& \text{when } x\neq0 \\0,& \text{when }x=0\end{cases}  at x = 0

Solution:

Given that

f(x)= \begin{cases}\frac{2|x|+x^2}{x},& \text{when } x\neq0 \\0,& \text{when }x=0\end{cases}

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{2(|-h|)+(-h)^2}{-h}
=\lim_{h\to0}\frac{2h+h^2}{-h}=-2

Let us consider RHL,

\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)
=\lim_{h\to0}\frac{2×|h|+h^2}{h}=2

Thus, LHL ≠ RHL

Therefore, f(x) is discontinuous at x = 0.

Question 10 (viii). f(x)= \begin{cases}|x-a|sin\frac{1}{(x-a)},& \text{when } x\neq a \\0,& \text{when }x=a\end{cases}  at x = a

Solution:

Given that, 

f(x)= \begin{cases}|x-a|sin\frac{1}{(x-a)},& \text{when } x\neq a \\0,& \text{when }x=a\end{cases}

f(x) = (x – a)sin{1/(x – a)}, x > 0

= (x – a)sin{1/(x – a)}, x < 0

= 0, x = a 

Let us consider LHL,

\lim_{x\to a^-}f(x)=-(a+a)sin(\frac{1}{-a+a})=0

Now, let us consider RHL,

\lim_{x\to a^+}f(x)=(a-a)sin(\frac{1}{a-a})=0

⇒ \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)

Therefore, f(x) is continuous at x = a.

Question 11. Show that f(x)=\begin{cases}1+x^2,& \text{if }0\leq x\leq1 \\2-x,& \text{if }x>1\end{cases} ” height=”79″ width=”346″>is discontinuous at x = 1.<span class=

Solution:

Given that, 

f(x)=\begin{cases}1+x^2,& \text{if }0\leq x\leq1 \\2-x,& \text{if }x>1\end{cases}

So, here we check the given f(x) is discontinuous at x = 1,

Let us consider LHL,

\lim_{x\to1^-}f(x) =\lim_{h\to0}f(1-h)
=\lim_{h\to0}1+(1-h)^2
=\lim_{h\to0}1+1-2h+h^2=2

Now, let us consider RHL,

\lim_{x\to1^+}f(x) =\lim_{h\to0}f(1+h)
=\lim_{h\to0}2-(1+h)=1

LHL ≠ RHL

Therefore, f(x) is discontinuous at x = 1.

Question 12. Show that  f(x)= \begin{cases}\frac{sin3x}{tan2x},& \text{if }x<0 \\(\frac{3}{2})&\text{if }x=0\\\frac{log(1+3x)}{e^{2x}-1},&\text{if }x>0\end{cases} ” height=”113″ width=”326″> is continuous at x = 0<span class=

Solution:

Given that,

f(x)= \begin{cases}\frac{sin3x}{tan2x},& \text{if }x<0 \\(\frac{3}{2})&\text{if }x=0\\\frac{log(1+3x)}{e^{2x}-1},&\text{if }x>0\end{cases}

So, here we check the given f(x) is continuous at x = 0,

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{sin(3(-h))}{tan(2(-h))}
=\lim_{h\to0}\frac{-sin3h}{-tan2h}
=\lim_{h\to0}\frac{\frac{sin3h}{3h}3h}{\frac{tan2h}{2h}2h}=3/2

Let us consider RHL,

\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)
=\lim_{h\to0}\frac{log(1+3h)}{e^{2h}-1}
=\lim_{h\to0}\frac{\frac{log(1+3h)}{3h}3h}{\frac{e^{2h}-1}{2h}2h}=3/2

f(0) = 3/2

Thus, LHL = RHL = f(0) = 3/2

Therefore, f(x) is continuous at x = 0.

Question 13. Find the value of ‘a’ for which the function f defined by 

f(x)=\begin{cases}asin\frac{π}{2}(x+1),& \text{if }x\leq0 \\\frac{tanx-sinx}{x^3},& \text{if }x>0\end{cases} ” height=”79″ width=”376″>  is continuous at x = 0.</h3><p><strong>Solution:</strong></p><p>Given that,</p><figure class=f(x)=\begin{cases}asin\frac{π}{2}(x+1),& \text{if }x\leq0 \\\frac{tanx-sinx}{x^3},& \text{if }x>0\end{cases}

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}f(-h)
=\lim_{h\to0}asin(\frac{π}{2})(-h+1)
=\lim_{h\to0}asin\frac{π}{2}=a

Now, let us consider RHL,

\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)=\lim_{h\to0}f(h)=\lim_{h\to0}\frac{tanh-sinh}{h^3}

⇒ \lim_{h\to0^+}f(x)=\lim_{h\to0}\frac{\frac{sinh}{cosh}-sinh}{h^3}

=\lim_{h\to0}\frac{\frac{sinh}{cosh}(1-cosh)}{h^3}
=\lim_{h\to0}\frac{(1-cosh)tanh}{h^3}
=\lim_{h\to0}\frac{2sin^2(\frac{h}{2})tanh}{4(\frac{h^2}{4})×h}
=(2/4)\lim_{h\to0}\frac{sin^2(\frac{h}{2})tanh}{(\frac{h^2}{4})×h}
=(1/2)\lim_{h\to0}(\frac{sin(\frac{h}{2})}{\frac{h}{2}})^2\lim_{h\to0}\frac{tanh}{h}

= (1/2) × 1 × 1

⇒ \lim_{x\to0^+}f(x)=1/2

If f(x) is continuous at x = 0, then

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

⇒ a = 1/2

Question 14. Examine the continuity of the function 

f(x)=\begin{cases}3x-2,& \text{if }x\leq 0 \\x+1,& \text{if }x>0\end{cases} ” height=”79″ width=”301″> at x = 0</h3><h3 class=Also sketch the graph of this function.

Solution:

Given that, 

f(x)=\begin{cases}3x-2,& \text{if }x\leq 0 \\x+1,& \text{if }x>0\end{cases}

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}3(-h)-2
=\lim_{h\to0}-3h-2=-2

Now, let us consider RHL,

\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)
=\lim_{h\to0}h+1=1

LhL ≠ RHL

So, the f(x) is discontinuous.

Question 15. Discuss the continuity of the function

 f(x)=\begin{cases}x,& \text{if }x>0 \\1,& \text{if }x=0 \\-x,& \text{if }x<0\end{cases}  at the point x = 0.

Solution:

Given that, 

f(x)=\begin{cases}x,& \text{if }x>0 \\1,& \text{if }x=0 \\-x,& \text{if }x<0\end{cases}

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

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

Now, let us consider RHL,

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

f(0) = 1

LHL = RHL ≠ f(0)

Hence, the f(x) is discontinuous at x = 0. 

Question 16. Discuss the continuity of the function 

f(x)=\begin{cases}x,&0\leq x<(\frac{1}{2}) \\(\frac{1}{2}),&x=(\frac{1}{2}) \\1-x,&(\frac{1}{2})<x\leq1\end{cases}  at the point x = 1/2.

Solution:

Given that, 

f(x)=\begin{cases}x,&0\leq x<(\frac{1}{2}) \\(\frac{1}{2}),&x=(\frac{1}{2}) \\1-x,&(\frac{1}{2})<x\leq1\end{cases}

So, here we check the continuity of the given f(x) at x = 1/2,

Let us consider LHL,

\lim_{x\to{\frac{1}{2}}^-}f(x) =\lim_{h\to0}f(\frac{1}{2}-h)
=\lim_{h\to0}\frac{1}{2}-h=\frac{1}{2}

Now, let us consider RHL,

\lim_{x\to{\frac{1}{2}}^+}f(x) =\lim_{h\to0}f(\frac{1}{2}+h)
=\lim_{h\to0}1-(\frac{1}{2}+h)=\frac{1}{2}

f(1/2) = 1/2

Thus, LHL= RHL = f(1/2) = 1/2

Hence, the f(x) is continuous at x = 1/2. 

Question 17. Discuss the continuity of f(x)=\begin{cases}2x-1,& \text{if }x<0 \\2x+1,& \text{if }x\geq0\end{cases}  at the point x = 0.

Solution:

Given that, 

f(x)=\begin{cases}2x-1,& \text{if }x<0 \\2x+1,& \text{if }x\geq0\end{cases}

So, here we check the continuity of the given f(x) at x = 10,

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}2(-h)-1=-1

Now, let us consider RHL,

\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)
=\lim_{h\to0}2h+1=1

Thus, LHL ≠ RHL

Hence, the f(x) is discontinuous at x = 0. 

Question 18. For what value of k is the function f(x)=\begin{cases}\frac{x^2-1}{x-1},& x\neq1 \\k,& x=1\end{cases}  continuous at x = 1 ?

Solution:

Given that, 

f(x)=\begin{cases}\frac{x^2-1}{x-1},& x\neq1 \\k,& x=1\end{cases}

Also, f(x) is continuous at x = 1

So, 

LHL = RHL = f(1)        ……(i)

Let us consider LHL,

\lim_{x\to1^-}f(x) =\lim_{h\to0}f(1-h)
=\lim_{h\to0}\frac{(1-h)^2-1}{(1-h)-1}
=\lim_{h\to0}\frac{h^2-2h}{-h}=2

f(1) = k

From eq(i), we get

LHL = F(1)

Therefore, k = 2

Question 19. Determine the value of the constant k so that the function

 f(x)=\begin{cases}\frac{x^2-3x+2}{x-1},& \text{if }x\neq1 \\k,& \text{if }x=1\end{cases}  continuous at x = 1.

Solution:

Given that, 

f(x)=\begin{cases}\frac{x^2-3x+2}{x-1},& \text{if }x\neq1 \\k,& \text{if }x=1\end{cases}

Also, f(x) is continuous at x = 1

So, LHL = RHL = f(1)        …..(i)

Let us consider LHL,

\lim_{x\to1^-}f(x) =\lim_{h\to0}f(1-h)
=\lim_{h\to0}\frac{(1-h)^2-3(1-h)+2}{(1-h)-1}
=\lim_{h\to0}\frac{h^2+h}{-h}
=\lim_{h\to0}-h-1=-1

f(1) = k

From eq(i), we get

LHL = F(1)

Therefore, k = -1

Question 20. For what value of k is the function f(x)=\begin{cases}\frac{sin5x}{3x},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}  continuous at x = 0 ?

Solution:

Given that, 

f(x)=\begin{cases}\frac{sin5x}{3x},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)        …..(i)

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{sin5(-h)}{3(-h)}
=\lim_{h\to0}\frac{-sin5h}{-3h}
=\lim_{h\to0}\frac{sin5h}{3h}\frac{5h}{3h}=\frac{5}{3}

f(0) = k

Thus, from eq(i), we get

k = 5/3

Therefore, k = 5/3

Question 21. Determine the value of the constant k so that the function

f(x)=\begin{cases}kx^2,& \text{if }x\leq2 \\3,& \text{if }x>2\end{cases} ” height=”79″ width=”268″> continuous at x = 2.</h3><p><strong>Solution:</strong></p><p>Given that, </p><figure class=f(x)=\begin{cases}kx^2,& \text{if }x\leq2 \\3,& \text{if }x>2\end{cases}

Also, f(x) is continuous at x = 2

Then, f(2) = k(2)2 = 4k

\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=f(2)

⇒ \lim_{x\to2^-}(kx^2)=\lim_{x\to2^+}(3)=4k

⇒ k × 22 = 3 = 4k

⇒ 4k = 3 = 4k

⇒ 4k = 3

⇒ k = 3/4

Hence, the value of k is 3/4

Question 22. Determine the value of the constant k so that the function

f(x)=\begin{cases}\frac{sin2x}{5x},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}  is continuous at x = 0.

Solution: 

Given that, 

f(x)=\begin{cases}\frac{sin2x}{5x},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)        ….(i)

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{sin2(-h)}{5(-h)}
=\lim_{h\to0}\frac{-sin2h}{-5h}
=\lim_{h\to0}\frac{sin2h}{5h}×\frac{2h}{5h}=2/5

f(0) = k

From eq(i), we get

k = 2/5

Question 23. Find the values of a so that the function f(x)=\begin{cases}ax+5,& \text{if }x\leq2 \\x-1,& \text{if }x>2\end{cases} ” height=”79″ width=”301″> is continuous at x = 2.<span class=

Solution: 

Given that, 

f(x)=\begin{cases}ax+5,& \text{if }x\leq2 \\x-1,& \text{if }x>2\end{cases}

Also, f(x) is continuous at x = 2

So, LHL = RHL = f(2)        …….(i)

Let us consider LHL,

\lim_{x\to2^-}f(x) =\lim_{h\to0}f(2-h)
=\lim_{h\to0}a(2-h)+5

= 2a + 5

Now, let us consider RHL,

\lim_{x\to2^+}f(x) =\lim_{h\to0}f(2+h)
=\lim_{h\to0}2+h-1=1

From eq(i), we get

2a + 5 = 1 

⇒ a = -2

Question 24. Prove that the function

f(x)=\begin{cases}\frac{x}{|x|+2x^2},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}  remains discontinuous at x = 0, regardless the choice of k.

Solution: 

Given that, 

f(x)=\begin{cases}\frac{x}{|x|+2x^2},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}

We have, at x = 0

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{-h}{|-h|+2(-h)^2}
=\lim_{h\to0}\frac{-h}{h+2h^2}
=\lim_{h\to0}\frac{-1}{1+2h}=-1

f(0) = k

Now, let us consider RHL,

\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)
=\lim_{h\to0}\frac{h}{|h|+2h^2}
=\lim_{h\to0}\frac{1}{1+2h}=1

Since, LHL ≠ RHL, 

Therefore, f(x) will remain discontinuous at x = 0, regardless the value of k.

Question 25. Find the value of k if f(x) is continuous at x = π/2, where 

f(x)=\begin{cases}\frac{kcosx}{π-2x},& \text{if }x\neq(\frac{π}{2}) \\3,& \text{if }x=(\frac{π}{2})\end{cases}

Solution:

Given that, 

f(x)=\begin{cases}\frac{kcosx}{π-2x},& \text{if }x\neq(\frac{π}{2}) \\3,& \text{if }x=(\frac{π}{2})\end{cases}

Also, f(x) is continuous at x = π/2

LHL = RHL

⇒ \lim_{x\to{\frac{π}{2}}^-}f(x)=\lim_{x\to{\frac{π}{2}}^+}f(x)=\lim_{h\to{\frac{π}{2}}}f(x)=f(\frac{π}{2})

⇒ \lim_{x\to{\frac{π}{2}}^-}\frac{kcosx}{π-2x}=3

⇒ k\lim_{x\to\frac{π}{2}}\frac{sin(\frac{π}{2}-x)}{2(\frac{π}{2}-x)}=3

⇒ \frac{k}{2}\lim_{x\to\frac{π}{2}}\frac{sin(\frac{π}{2}-x)}{(\frac{π}{2}-x)}=3

⇒ k/2 = 3

⇒ k = 6

Question 26. Determine the values of a, b, c for which the function 

f(x)=\begin{cases}\frac{sin(a+1)x+sinx}{x},& \text{if }x<0 \\c,& \text{for }x=0\\\frac{\sqrt{x+bx^2}-√x}{bx^{3/2}},&\text{for }x>0\end{cases}” height=”118″ width=”402″></h3><h3 class=is continuous at x = 0.

Solution: 

Given that, 

f(x)=\begin{cases}\frac{sin(a+1)x+sinx}{x},& \text{if }x<0 \\c,& \text{for }x=0\\\frac{\sqrt{x+bx^2}-√x}{bx^{3/2}},&\text{for }x>0\end{cases}

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)        …..(i)

f(0) = 0

Let us consider LHL,

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

= \lim_{h\to0}\frac{sin(a+1)(-h)+sin(-h)}{-h}

= \lim_{h\to0}\frac{-sin(ah+h)-sinh}{-h}       

= \lim_{h\to0}\frac{sin(a+1)h}{h}+\lim_{h\to0}\frac{sinh}{h}

= a + 1 + 1 = a + 2

Now, let us consider RHL,

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

= \lim_{h\to0}\frac{\sqrt{h+bh^2}-√h}{bh^{3/2}}

= \lim_{h\to0}\frac{\sqrt{h+bh^2}-√h}{bh^{3/2}}\frac{\sqrt{h+bh^2}+√h}{\sqrt{h+bh^2}+√h}

= \lim_{h\to0}\frac{h+bh^2-h}{bh^{3/2}(\sqrt{h+bh^2}+√h)}

= \lim_{h\to0}\frac{bh^2}{bh^2(\sqrt{1+bh}+1)}=1/2

From eq(i), we get

a + 2 = 1/2 ⇒ a = -3/2

c = 1/2 and b ∈ R -{0}

Hence, a = -3/2, b ∈ R -{0}, c =1/2

Question 27. If f(x)=\begin{cases}\frac{1-coskx}{xsinx},& \text{if }x\neq0 \\(\frac{1}{2}),& \text{if }x=0\end{cases}  is continuous at x = 0, find k.

Solution: 

Given that, 

f(x)=\begin{cases}\frac{1-coskx}{xsinx},& \text{if }x\neq0 \\(\frac{1}{2}),& \text{if }x=0\end{cases}

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)       …….(i)

f(0) = 1/2

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{1-cosk(-h)}{-hsin(-h)}
=\lim_{h\to0}\frac{1-coskh}{hsinh}
=\lim_{h\to0}\frac{2sin^2(\frac{kh}{2})}{h.2sin(\frac{h}{2}).cos(\frac{h}{2})}
=\lim_{h\to0}(\frac{sin(\frac{kh}{2})}{\frac{kh}{2}})^2\frac{\frac{k^2h^2}{4}}{\frac{sin(\frac{h}{2})}{\frac{h}{2}}(\frac{h}{2})}.(\frac{1}{h})
=\lim_{h\to0}(\frac{sin(\frac{kh}{2})}{\frac{kh}{2}})^2\frac{\frac{k^2}{4}}{\frac{sin(\frac{h}{2})}{\frac{h}{2}}(\frac{1}{2})}

= k2/2

Using eq(i) we get,

k2/2 = 1/2 ⇒ k = ±1

Question 28. If f(x)=\begin{cases}\frac{x-4}{|x-4|},& \text{if }x<4 \\a+b,& \text{if }x=4\\\frac{x-4}{|x-4|}+b& \text{if }x>4\end{cases} ” height=”127″ width=”321″> continuous at x = 4, find a, b.<span class=

Solution: 

Given that, 

f(x)=\begin{cases}\frac{x-4}{|x-4|},& \text{if }x<4 \\a+b,& \text{if }x=4\\\frac{x-4}{|x-4|}+b& \text{if }x>4\end{cases}

Also, f(x) is continuous at x = 4

So, LHL = RHL = f(4)        ……(i)

f(4) = a + b     …..(ii)

Let us consider LHL,

\lim_{x\to4^-}f(x) =\lim_{h\to0}f(4-h)
=\lim_{h\to0}\frac{(4-h)-4}{|(4-h)-4|}+a
=\lim_{h\to0}\frac{-h}{h}+a

= a – 1        ……(iii)

Now, let us consider RHL,

\lim_{x\to4^+}f(x) =\lim_{h\to0}f(4+h)
=\lim_{h\to0}\frac{(4+h)-4}{|(4+h)-4|}+b
=\lim_{h\to0}\frac{h}{h}+b

= b + 1           ……(iv)

From eq(i), we get

a – 1 = b + 1 ⇒ a – b = 2         …..(v)

From eq(ii) and eq(iii), we get

a + b = a – 1 ⇒ a – b = -1

From eq(ii) and (iv), we get

a + b = b + 1 ⇒ a = 1

Thus, a = 1 and b = -1

Question 29. For what value of k is the function

f(x)=\begin{cases}\frac{sin2x}{x},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}  continuous at x = 0 ?

Solution: 

Given that, 

f(x)=\begin{cases}\frac{sin2x}{x},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)        …..(i)

f(0) = k

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{sin2(0-h)}{-h}
=\lim_{h\to0}\frac{-sin2h}{-h}=2

Using eq(i), we get 

k = 2

Question 30. Let f(x) = \frac{log(1+\frac{x}{a})-log(1-\frac{x}{b})}{x} , x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

Solution: 

Given that,

f(x) = \frac{log(1+\frac{x}{a})-log(1-\frac{x}{b})}{x}

Also, f(x) is continuous at x = 0

So, LHL=RHL=f(0)        ….(i)

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{log(1-\frac{h}{a})-log(1+\frac{h}{b})}{-h}
=\lim_{h\to0}\frac{log(1+(\frac{-h}{a}))}{(\frac{-h}{a})a}+\frac{log(1+\frac{h}{b})}{h}

= 1/a + 1/b = (a + b)/ab

From eq(i), we get

f(0) = (a + b)/ab

Question 31. If f(x)=\begin{cases}\frac{2^{x+2}-16}{4^x-16},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases}  is continuous at x = 2, find k.

Solution: 

Given that,

f(x)=\begin{cases}\frac{2^{x+2}-16}{4^x-16},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases}

Also, f(x) is continuous at x = 2

So, LHL = RHL = f(2)        …..(i)

Now,

f(2) = k  ……(ii)

Let us consider LHL,

\lim_{x\to2^-}f(x) =\lim_{h\to0}f(2-h)
=\lim_{h\to0}\frac{2^{(2-h)+2}-16}{4^{(2-h)}-16}
=\lim_{h\to0}\frac{2^{4-h}-16}{4^{(2-h)}-16}
=\lim_{h\to0}\frac{2^4.2^{-h}-16}{4^2.4^{-h}-16}
=\lim_{h\to0}\frac{16.2^{-h}-16}{16.4^{-h}-16}
=\lim_{h\to0}\frac{16(2^{-h}-1)}{16(4^{-h}-1)}
=\lim_{h\to0}\frac{2^{-h}-1}{(2^{-h})^2-1^2}

=\lim_{h\to0}\frac{2^{-h}-1}{(2^{-h}-1)(2^{-h}+1)}=1/2     ……(iii)

Using eq(i), (ii) and (iii), we get

k = 1/2

Question 32. If f(x)=\begin{cases}\frac{cos^2x-sin^2x-1}{\sqrt{x^2+1}-1},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}  is continuous at x = 0, find k.

Solution:

Given that, 

f(x)=\begin{cases}\frac{cos^2x-sin^2x-1}{\sqrt{x^2+1}-1},& \text{if }x\neq0 \\k,& \text{if }x=0\end{cases}

Also, f(x) is continuous at x = 2

So, LHL = RHL

Now,

\lim_{x\to0}f(x)=f(0)

⇒ \lim_{x\to0}\frac{cos^2x-sin^2x-1}{\sqrt{x^2+1}-1}=k

⇒ \lim_{x\to0}\frac{1-sin^2x-sin^2x-1}{\sqrt{x^2+1}-1}=k

⇒ \lim_{x\to0}\frac{-2sin^2x}{\sqrt{x^2+1}-1}=k

⇒ \lim_{x\to0}\frac{-2(sin^2x)(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)}=k

⇒ \lim_{x\to0}\frac{-2(sin^2x)(\sqrt{x^2+1}+1)}{x^2}=k

⇒ -2\lim_{x\to0}\frac{(sin^2x)(\sqrt{x^2+1}+1)}{x^2}=k

⇒ -2\lim_{x\to0}(\frac{sinx}{x})^2\lim_{x\to0}(\sqrt{x^2+1}+1)=k

⇒ -2 × 1 × (1 + 1) = k

⇒ k = -4

Question 33. Extend the definition of the following by continuity f(x) = \frac{1-cos7(x-π)}{5(x-π)^2}  at the point x = π.

Solution: 

Given that, 

\frac{1-cos7(x-π)}{5(x-π)^2}

As we know that a f(x) is continuous at x = π if,

LHL = RHL = f(π)  ……(i)

Let us consider LHL,

\lim_{x\toπ^-}f(x) =\lim_{h\to0}f(π-h)
=\lim_{h\to0}\frac{1-cos7(π-h-π)}{5((π-h)-π)^2}
=\lim_{h\to0}\frac{2sin^2(7/2)h}{5h^2}
=\lim_{h\to0}(2/5)(\frac{sin(7/2)h}{(7/2)h})^2×(7/2)^2

= (2/5) × (49/4) = 49/10

 Thus, from eq(i) we get,

f(π) = 49/10

Hence, f(x) is continuous at x = π

Question 34. If f(x) = \frac{2x+3sinx}{3x+2sinx} , x ≠ 0 is continuous at x = 0, then find f(0).

Solution:

Given that, 

f(x) = \frac{2x+3sinx}{3x+2sinx}

Also, f(x) is continuous at x = 0

So, LHL = RHL = f(0)     ……(i)

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{2(-h)+3sin(-h)}{3(-h)+2sin(-h)}
=\lim_{h\to0}\frac{-2h-3sinh}{-3h-2sinh}
=\lim_{h\to0}\frac{\frac{2h+3sinh}{h}}{\frac{3h+2sinh}{h}}
=\lim_{h\to0}\frac{2+3\frac{sinh}{h}}{3+2\frac{sinh}{h}}=\frac{2+3}{3+2}=1

From eq(i) we get,

f(0) = 1

Question 35. Find the value of k for which f(x)=\begin{cases}\frac{1-cos4x}{8x^2} ,& \text{when }x\neq0 \\k,& \text{when }x=0\end{cases}  is continuous at x = 0

Solution:

Given that, 

f(x)=\begin{cases}\frac{1-cos4x}{8x^2} ,& \text{when }x\neq0 \\k,& \text{when }x=0\end{cases}

Also, f(x) is continuous at x = 0

LHL = RHL = f(0)     …..(i)

f(0) = k

Let us consider LHL,

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}\frac{1-cos4(-h)}{8(-h)^2}
=\lim_{h\to0}\frac{1-cos4h}{8h^2}
=\lim_{h\to0}\frac{2sin^22h}{8h^2}
=\lim_{h\to0}(\frac{sin2h}{2h})^2=1

Thus, from eq(i) we get,

k = 1

Question 36. In each of the following, find the value of the constant k so that the given function is continuous at the indicated point:

(i) f(x)=\begin{cases}\frac{1-cos2kx}{x^2},& \text{if }x\neq0 \\8,& \text{if }x=0\end{cases}  at x = 0

Solution:

Given that, 

f(x)=\begin{cases}\frac{1-cos2kx}{x^2},& \text{if }x\neq0 \\8,& \text{if }x=0\end{cases}

Also, f(x) is continuous at x = 0

\lim_{x\to0}f(x) =f(0)

⇒ \lim_{x\to0}\frac{1-cos2kx}{x^2}=8

⇒ \lim_{x\to0}\frac{2k^2sin^2kx}{k^2x^2}=8

⇒ 2k^2\lim_{x\to0}(\frac{sinkx}{kx})^2=8

⇒ 2k2 × 1 = 8

⇒ k2 = 4

⇒ k = ±2

(ii) f(x)=\begin{cases}(x-1)\frac{tanπx}{2},& \text{if }x\neq1 \\k,& \text{if }x=1\end{cases}  at x = 1

Solution:

Given that, 

f(x)=\begin{cases}(x-1)\frac{tanπx}{2},& \text{if }x\neq1 \\k,& \text{if }x=1\end{cases}

Also, f(x) is continuous at x = 1

\lim_{x\to1}f(x) =f(1)

⇒ \lim_{x\to1}(x-1)tan(πx/2)=k

Now, on putting x – 1 = y, we get

\lim_{y\to0}ytan\frac{π(y+1)}{2}=k

⇒ \lim_{y\to0}ytan(\frac{πy}{2}+π/2)=k

⇒ \lim_{y\to0}ytan(\frac{π}{2}+\frac{πy}{2})=k

⇒ -\lim_{y\to0}ycot(\frac{π}{2})=k

⇒ \frac{-2}{π}\lim_{y\to0}\frac{\frac{πy}{2}cos(\frac{πy}{2})}{sin\frac{πy}{2}}=k

⇒ \frac{-2}{π}\frac{\lim_{y\to0}cos(\frac{πy}{2})}{\lim_{y\to0}(\frac{sin(\frac{πy}{2})}{\frac{πy}{2}})}=k

⇒ (-2/π) × (1/1) = k

⇒ k = (-2/π)

(iii) f(x)=\begin{cases}k(x^2-2x),& \text{if }x<0 \\cosx,& \text{if }x\geq0\end{cases}  at x = 0

Solution:

Given that, 

f(x)=\begin{cases}k(x^2-2x),& \text{if }x<0 \\cosx,& \text{if }x\geq0\end{cases}

Also, f(x) is continuous at x = 0

Let us consider LHL, at x = 0

\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)
=\lim_{h\to0}f(-h)
=\lim_{h\to0}k(h^2+2h)=0

Let us consider RHL at x = 0

\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)
=\lim_{h\to0}f(h)
=\lim_{h\to0}cosh=1
\lim_{x\to0^-}f(x)≠\lim_{x\to0^+}f(x)

Hence, no value of k exists for which function is continuous at x = 0.

(iv) f(x)=\begin{cases}kx+1,& \text{if }x\leqπ \\cosx,& \text{if }x>π\end{cases} ” height=”79″ width=”281″>at x = π<span class=

Solution:

Given that, 

f(x)=\begin{cases}kx+1,& \text{if }x\leqπ \\cosx,& \text{if }x>π\end{cases}

Also, f(x) is continuous at x = π

Let us consider LHL 

\lim_{x\toπ^-}f(x) =\lim_{h\to0}f(π-h)
=\lim_{h\to0}k(π-h)+1=kπ+1

Let us consider RHL 

\lim_{x\toπ^+}f(x) =\lim_{h\to0}f(π+h)
=\lim_{h\to0}cos(π+h)

cosπ = -1

As we know that f(x) is continuous at x = π, so

\lim_{x\toπ^-}f(x)=\lim_{x\toπ^+}f(x)

⇒ kπ + 1 = -1

⇒ k = (-2/π)

(v) f(x)=\begin{cases}kx+1,& \text{if }x\leq5 \\3x-5,& \text{if }x>5\end{cases} ” height=”79″ width=”302″>at x = 5<span class=

Solution:

Given that, 

f(x)=\begin{cases}kx+1,& \text{if }x\leq5 \\3x-5,& \text{if }x>5\end{cases}

Also, f(x) is continuous at x = 5

Let us consider LHL 

\lim_{x\to5^-}f(x)=\lim_{h\to0}f(5-h)
=\lim_{h\to0}k(5-h)+1

= 5k + 1

Let us consider RHL 

\lim_{x\to5^+}f(x)=\lim_{h\to0}f(5+h)
=\lim_{h\to0}3(5+h)-5

= 10

As we know that f(x) is continuous at x = 5, so

\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)

⇒ 5k + 1 = 10

⇒ k = 9/5

(vi) f(x)=\begin{cases}\frac{x^2-25}{x-5},& \text{if }x\neq5 \\k,& \text{if }x=5\end{cases}  at x = 5 

Solution:

Given that, 

f(x)=\begin{cases}\frac{x^2-25}{x-5},& \text{if }x\neq5 \\k,& \text{if }x=5\end{cases}

Also, f(x) is continuous at x = 5

So, 

f(x) = (x2 – 25)/(x – 5), if x ≠ 5 & f(x) = k, if x = 5

⇒ f(x)= {(x – 5)(x+5)/(x-5)}, if x ≠ 5 & f(x) = k, if x = 5

⇒ f(x)= (x + 5), if x ≠ 5 & f(x) = k, if x = 5

As we know that f(x) is continuous at x = 5, so

\lim_{x\to5}f(x)=f(5)

⇒ \lim_{x\to5}(x+5)=k

⇒ k = 5 + 5 = 10

(vii) f(x)=\begin{cases}kx^2,& \text{if }x\geq1 \\4,& \text{if }x<1\end{cases}  at x = 1

Solution:

Given that, 

f(x)=\begin{cases}kx^2,& \text{if }x\geq1 \\4,& \text{if }x<1\end{cases}

Also, f(x) is continuous at x = 1

Let us consider LHL 

\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)
=\lim_{h\to0}4=4

Let us consider RHL 

\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)
=\lim_{h\to0}k(1+h)^2

= k

As we know that f(x) is continuous at x = 1, so

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

⇒ k = 4

(viii) f(x)=\begin{cases}k(x^2+2),& \text{if }x\leq0 \\3x+1,& \text{if }x>0\end{cases} ” height=”79″ width=”335″> at x = 0<span class=

Solution:

Given that, 

f(x)=\begin{cases}k(x^2+2),& \text{if }x\leq0 \\3x+1,& \text{if }x>0\end{cases}

Also, f(x) is continuous at x = 0

Let us consider LHL 

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}k((-h)^2+2)

= 2k

Let us consider RHL 

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}3h+1

= 1

As we know that f(x) is continuous at x = 0, so

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

⇒ 2k = 1

⇒ k = 1/2

(ix) f(x)=\begin{cases}\frac{x^3+x^2-16x+20}{(x-2)^2},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases}  at x = 2

Solution:

Given that, 

f(x)=\begin{cases}\frac{x^3+x^2-16x+20}{(x-2)^2},& \text{if }x\neq2 \\k,& \text{if }x=2\end{cases}

Also, f(x) is continuous at x = 2

f(x)= \frac{x^3+x^2-16x+20}{(x-2)^2} , if x ≠ 2 & f(x) = k, if x = 2

⇒ f(x)= \frac{x^3+x^2-16x+20}{x^2-4x+4}  , if x ≠ 2 & f(x) = k, if x = 2

⇒ f(x)= \frac{(x+5)(x^2-4x+4)}{x^2-4x+4}  , if x ≠ 2 & f(x) = k, if x = 2

⇒ f(x)= (x + 5), if x ≠ 2 & f(x) = k, if x = 2

As we know that f(x) is continuous at x = 2, so

\lim_{x\to2}f(x)=f(2)

⇒ \lim_{x\to2}(x+5)=f(2)

⇒ k = 2 + 5 = 7

Question 37. Find the values of a and b so that the function f given by 

f(x)=\begin{cases}1,& \text{if }x\leq3 \\ax+b,& \text{if }3<x<5\\7, &\text{if }x\geq5\end{cases}  is continuous at x = 3 and x = 5.

Solution:

Given that, 

f(x)=\begin{cases}1,& \text{if }x\leq3 \\ax+b,& \text{if }3<x<5\\7, &\text{if }x\geq5\end{cases}

Let us consider LHL at x = 3,

\lim_{x\to3^-}f(x)=\lim_{h\to0}f(3-h)
=\lim_{h\to0}(1)

= 1

Let us consider RHL at x = 3,

\lim_{x\to3^+}f(x)=\lim_{h\to0}f(3+h)
=\lim_{h\to0}a(3+h)+b

= 3a + b

Let us consider LHL at x = 5,

\lim_{x\to5^-}f(x)=\lim_{h\to0}f(5-h)
=\lim_{h\to0}(a(5-h)+b)

= 5a + b

Let us consider RHL at x = 5,

\lim_{x\to5^+}f(x)=\lim_{h\to0}f(5+h)
=\lim_{h\to0}7

= 7

It is given that f(x) is continuous at x = 3 and x = 5, then

\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)  and \lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)

⇒ 1 = 3a + b …..(i) 

and 5a + b = 7 …….(ii)

On solving eq(i) and (ii), we get

a = 3 and b = -8

Question 38. If f(x)=\begin{cases}\frac{x^2}{2},& \text{if }0\leq x \leq 1 \\2x^2-3x+(\frac{3}{2}),& \text{if }1<x\leq2\end{cases} . Show that f is continuous at x = 1.

Solution:

Given that, 

f(x)=\begin{cases}\frac{x^2}{2},& \text{if }0\leq x \leq 1 \\2x^2-3x+(\frac{3}{2}),& \text{if }1<x\leq2\end{cases}

So, 

Let us consider LHL at x = 1,

\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)
=\lim_{h\to0}\frac{(1-h)^2}{2}

= 1/2

Let us consider RHL at x = 1,

\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)
=\lim_{h\to0}[2(1+h)^2-3(1+h)+3/2]

= 2 – 3 + 3/2 = 1/2

Also,

f(1) = (1)2/2 = 1/2

\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)

LHL = RHL = f(1)

Hence, the f(x) is continuous at x = 1

Question 39. Discuss the continuity of the f(x) at the indicated points:

(i) f(x) = |x| + |x – 1| at x = 0, 1.

Solution:

Given that, 

f(x) = |x| + |x – 1|

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL at x = 0,

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}[|0-h|+|0-h-1|]=1

Let us consider RHL at x = 0,

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}[|0+h|+|0+h-1|]=1

Also,

f(0) = |0| + |0 – 1| = 0 + 1 = 1

LHL = RHL = f(0)

Now, we check the continuity of the given f(x) at x = 1,

Let us consider LHL at x = 1,

\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)
=\lim_{h\to0}f(|1-h|+|1-h-1|)=1+0

= 1

Let us consider RHL at x = 1

\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)
=\lim_{h\to0}f(|1+h|+|1+h-1|)=1+0

= 1

Also,

f(1) = |1| + |1 – 1| = 1 + 0 = 1

LHL = RHL = f(1)

Hence, f(x) is continuous at x = 0, 1.

(ii) f(x) = |x – 1| + |x + 1| at x = -1, 1.

Solution:

Given that, 

f(x) = |x – 1| + |x + 1| at x = -1, 1.

So, here we check the continuity of the given f(x) at x = -1,

Let us consider LHL at x = -1,

\lim_{x\to-1^-}f(x)=\lim_{h\to0}f(-1-h)
=\lim_{h\to0}[|-1-h-1|+|-1-h+1|]=2+0=2

Let us consider RHL at x = -1,

\lim_{x\to-1^+}f(x)=\lim_{h\to0}f(-1+h)
=\lim_{h\to0}[|-1+h-1|+|-1+h+1|]=2+0=2

Also,

f(-1) = |-1 – 1| + |-1 + 1| = |-2| = 2

LHL = RHL = f(-1)

Now, we check the continuity of the given f(x) at x = 1,

Let us consider LHL at x = 1,

\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)
=\lim_{h\to0}f(|1-h-1|+|1-h+1|)=0+2

= 2

\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)
=\lim_{h\to0}f(|1+h-1|+|1+h+1|)=0+2

= 2

Also,

f(1) = |1 + 1| + |1 – 1| = 2

LHL = RHL = f(1)

Hence, f(x) is continuous at x = -1, 1.

Question 40. Prove that f(x)=\begin{cases}\frac{x-|x|}{x},& \text{if }x\neq0 \\2,& \text{if }x=0\end{cases}  is discontinuous at x = 0.

Solution:

Prove that f(x)=\begin{cases}\frac{x-|x|}{x},& \text{if }x\neq0 \\2,& \text{if }x=0\end{cases}  is discontinuous at x = 0.

Proof:

Let us consider LHL at x = 0,

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}f(-h)
=\lim_{h\to0}2=2

Let us consider RHL at x = 0,

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

LHL ≠ RHL

Hence, f(x) is discontinuous at x = 0.

Question 41. If f(x)=\begin{cases}2x^2+k,& \text{if }x\geq0 \\-2x^2+k,& \text{if }x<0\end{cases}  then what should be the value of k so that f(x) is continuous at x = 0.

Solution:

Given that,

f(x)=\begin{cases}2x^2+k,& \text{if }x\geq0 \\-2x^2+k,& \text{if }x<0\end{cases}

Let us consider LHL at x = 0,

\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}f(-h)
=\lim_{h\to0}-2(-h)^2+k

= k

Let us consider RHL at x = 0,

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}f(h)
=\lim_{h\to0}f(2h^2+k)

= k

It is given that f(x) is continuous at x = 0.

LHL = RHL = f(0)

⇒ \lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=k

k can be any real number.

Question 42. For what value λ of is the function

f(x)=\begin{cases}λ(x^2-2x),& \text{if }x\leq0 \\4x+1,& \text{if }x>0\end{cases} ” height=”79″ width=”335″> continuous at x = 0 ? What about continuity at x = ±1?</h3><p><strong>Solution:</strong></p><p>Given that,</p><figure class=f(x)=\begin{cases}λ(x^2-2x),& \text{if }x\leq0 \\4x+1,& \text{if }x>0\end{cases}

Check for x = 0, 

Hence, there is no value of λ for which f(x) is continuous at x = 0.

Now for x = 1,

f(1) = 4x + 1 = 4 × 1 + 1 = 5

Hence, for any values of λ, f is continuous at x = 1.

Now for x = -1, 

f(-1) = λ(1 + 2)= 3λ

=\lim_{x\to-1}λ(1+2)=3λ
=\lim_{x\to-1}f(x)=f(-1)

Hence, for any values of λ, f is continuous at x=-1.

Question 43. For what values of k is the following function continuous at x = 2? 

 f(x)=\begin{cases}2x+1,& \text{if }x<2 \\k,& \text{if }x=2\\3x-1,& \text{if }x>2\end{cases}” height=”111″ width=”303″><span class=

Solution:

Given that,

f(x)=\begin{cases}2x+1,& \text{if }x<2 \\k,& \text{if }x=2\\3x-1,& \text{if }x>2\end{cases}

We have,

Let us consider LHL at x = 2,

=\lim_{x\to2^-}f(x)=\lim_{h\to0}f(2-h)
=\lim_{h\to0}(2(2-h)+1)

= 5

Let us consider RHL at x = 2,

\lim_{x\to2^+}f(x)=\lim_{h\to0}f(2+h)
=\lim_{h\to0}3(2+h)-1

= 5

Also,

f(2) = k

It is given that f(x) is continuous at x = 2.

LHL = RHL = f(2)

⇒ 5 = 5 = k

Hence, for k = 5, f(x) is continuous at x = 2.

Question 44. Let f(x)=\begin{cases}1-sin^3x3cos^2x,& \text{if }x<(\frac{π}{2}) \\a,& \text{if }x=(\frac{π}{2})\\\frac{b(1-sinx)}{(π-2x)^2},& \text{if }x>(\frac{π}{2})\end{cases} ” height=”117″ width=”437″> If f(x) is continuous at x = (π/2), find a and b.<span class=

Solution:

Given that,

f(x)=\begin{cases}1-sin^3x3cos^2x,& \text{if }x<(\frac{π}{2}) \\a,& \text{if }x=(\frac{π}{2})\\\frac{b(1-sinx)}{(π-2x)^2},& \text{if }x>(\frac{π}{2})\end{cases}

Let us consider LHL at x = π/2

=\lim_{x\to(\frac{π}{2})^-}f(x)=\lim_{h\to0}f(\frac{π}{2}-h)
=\lim_{h\to0}\frac{1-sin^3(\frac{π}{2}-h)}{3cos^2(\frac{π}{2}-h)}
=\lim_{h\to0}\frac{1-cos^3h}{3sin^2h}
=\frac{1}{3}\lim_{h\to0}(\frac{(1-cosh)(1+cos^2h+cosh)}{(1-cosh)(1+cosh)})
=\frac{1}{3}\lim_{h\to0}(\frac{(1+cos^2h+cosh)}{(1+cosh)})
=\frac{1}{3}(\frac{1+1+1}{1+1})

= 1/2

Let us consider RHL at x = π/2

=\lim_{x\to(\frac{π}{2})^+}f(x)=\lim_{h\to0}f(\frac{π}{2}+h)
=\lim_{h\to0}(\frac{b[1-sin(\frac{π}{2}+h)]}{[π-2(\frac{π}{2}+h)]^2})
=\lim_{h\to0}(\frac{b(1-cosh)}{[-2h]^2})
=\lim_{h\to0}(\frac{2bsin^2(\frac{h}{2})}{4h^2})
=\lim_{h\to0}(\frac{2bsin^2(\frac{h}{2})}{\frac{16h^2}{4}})
=(\frac{b}{8})\lim_{h\to0}(\frac{sin(\frac{h}{2})}{\frac{h}{2}})^2

= b/8 × 1

= b/8

Also,

f(π/2) = a

It is given that f(x) is continuous at x = π/2.

LHL = RHL = f(π/2)

So, 

⇒ 1/2 = b/8 = a

⇒ a = 1/2 and b = 4

Question 45. If the functions f(x), defined below is continuous at x = 0, find the value of k,

f(x)=\begin{cases}\frac{1-cos2x}{2x^2},& \text{if }x<0 \\k,& \text{if }x=0\\\frac{x}{|x|},& \text{if }x>0\end{cases}” height=”117″ width=”312″></h3><p><strong>Solution:</strong></p><p>Given that,</p><figure class=f(x)=\begin{cases}\frac{1-cos2x}{2x^2},& \text{if }x<0 \\k,& \text{if }x=0\\\frac{x}{|x|},& \text{if }x>0\end{cases}

Let us consider LHL at x = 0,

=\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)
=\lim_{h\to0}(\frac{1-cos2(-h)}{2(-h)^2})
=\lim_{h\to0}(\frac{1-cos2h}{2h^2})
=\frac{1}{2}\lim_{h\to0}(\frac{2sin^2h}{h^2})
=\frac{2}{2}\lim_{h\to0}(\frac{sin^2h}{h^2})
=\frac{2}{2}\lim_{h\to0}(\frac{sinh}{h})^2

= 1 × 1

Let us consider RHL at x = 0,

\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)
=\lim_{h\to0}f(h)
=\lim_{h\to0}1=1

Also,

f(0) = k

It is given that f(x) is continuous at x = 0,

LHL = RHL = f(0)

So, 

⇒ 1 = 1 = k

Hence, the required value of k is 1.

Question 46. Find the relationship between ‘a’ and ‘b’ so that function ‘f’ defined by

f(x)=\begin{cases}ax+1,& \text{if }x\leq3 \\bx+3,& \text{if }x>3\\\end{cases} ” height=”79″ width=”301″> is continuous at x = 3.</h3><p><strong>Solution:</strong></p><p>Given that,</p><figure class=f(x)=\begin{cases}ax+1,& \text{if }x\leq3 \\bx+3,& \text{if }x>3\\\end{cases}

Let us consider LHL at x = 3,

\lim_{x\to3^-}f(x)=\lim_{h\to0}f(3-h)
=\lim_{h\to0}a(3-h)+1

= 3a + 1

Let us consider RHL at x = 3,

\lim_{x\to3^+}f(x)=\lim_{h\to0}f(3+h)
=\lim_{h\to0}b(3+h)+3

= 3b + 3

It is given that f(x) is continuous at x = 3,

LHL = RHL = f(3)

So, 

⇒ 3a + 1 = 3b + 3

⇒ 3a – 3b = 2

Hence, the required relationship between a and b is 3a – 3b = 2.

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