Skip to content
Related Articles

Related Articles

Improve Article

Class 12 RD Sharma Solutions – Chapter 9 Continuity – Exercise 9.1 | Set 2

  • Last Updated : 26 May, 2021

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}   continuous at x = 2.

Solution:

Given that, 

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}   is continuous at x = 2.

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}

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}   continuous at x = 4, find a, b.

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

Attention reader! Don’t stop learning now. Join the First-Step-to-DSA Course for Class 9 to 12 students , specifically designed to introduce data structures and algorithms to the class 9 to 12 students




My Personal Notes arrow_drop_up
Recommended Articles
Page :