RD Sharma Class 12 Ex 9.1 Solutions Chapter 9 Continuity

Here we provide RD Sharma Class 12 Ex 9.1 Solutions Chapter 9 Continuity for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 9.1 Solutions Chapter 9 Continuity book pdf download. Now you will get step-by-step solutions to each question.

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><span 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><span 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

⇒ 2k× 1 = 8

⇒ k= 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) = (x– 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.

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.