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

Question 16. Discuss the continuity of the function 

 at the point x = 1/2.

Solution:

Given that, 

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

Let us consider LHL,

Now, let us consider RHL,

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  at the point x = 0.

Solution:

Given that, 

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

Let us consider LHL,

Now, let us consider RHL,

Thus, LHL ≠ RHL

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

Question 18. For what value of k is the function  continuous at x = 1 ?

Solution:

Given that, 

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

So, 

LHL = RHL = f(1)        ……(i)

Let us consider LHL,

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

  continuous at x = 1.

Solution:

Given that, 

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

So, LHL = RHL = f(1)        …..(i)

Let us consider LHL,

f(1) = k

From eq(i), we get

LHL = F(1)

Therefore, k = -1

Question 20. For what value of k is the function  continuous at x = 0 ?

Solution:

Given that, 

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

So, LHL = RHL = f(0)        …..(i)

Let us consider LHL,

 

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

2\end{cases} " title="Rendered by QuickLaTeX.com" height="79" width="268" style="vertical-align: -33px;"> continuous at x = 2.

Solution:

Given that, 

2\end{cases} " title="Rendered by QuickLaTeX.com" height="79" width="268" style="vertical-align: -33px;">

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

Then, f(2) = k(2)² = 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

 is continuous at x = 0.

Solution: 

Given that, 

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

So, LHL = RHL = f(0)        ….(i)

Let us consider LHL,

f(0) = k

From eq(i), we get

k = 2/5

Question 23. Find the values of a so that the function 2\end{cases} " title="Rendered by QuickLaTeX.com" height="79" width="301" style="vertical-align: -33px;"> is continuous at x = 2.

Solution: 

Given that, 

2\end{cases} " title="Rendered by QuickLaTeX.com" height="79" width="301" style="vertical-align: -33px;">

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

So, LHL = RHL = f(2)        …….(i)

Let us consider LHL,

= 2a + 5

Now, let us consider RHL,

From eq(i), we get

2a + 5 = 1 

⇒ a = -2

Question 24. Prove that the function

 remains discontinuous at x = 0, regardless the choice of k.

Solution: 

Given that, 

We have, at x = 0

Let us consider LHL,

f(0) = k

Now, let us consider RHL,

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 

Solution:

Given that, 

Also, f(x) is continuous at x = π/2

LHL = RHL

⇒ 

⇒ 

⇒ 

⇒ 

⇒ k/2 = 3

⇒ k = 6

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

0\end{cases}" title="Rendered by QuickLaTeX.com" height="118" width="402" style="vertical-align: -52px;">

is continuous at x = 0.

Solution: 

Given that, 

0\end{cases}" title="Rendered by QuickLaTeX.com" height="118" width="402" style="vertical-align: -52px;">

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

So, LHL = RHL = f(0)        …..(i)

f(0) = 0

Let us consider LHL,

= 

=       

= 

= a + 1 + 1 = a + 2

Now, let us consider RHL,

= 

= 

= 

= 

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  is continuous at x = 0, find k.

Solution: 

Given that, 

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

So, LHL = RHL = f(0)       …….(i)

f(0) = 1/2

Let us consider LHL,

= k²/2

Using eq(i) we get,

k²/2 = 1/2 ⇒ k = ±1

Question 28. If 4\end{cases} " title="Rendered by QuickLaTeX.com" height="127" width="321" style="vertical-align: -56px;"> continuous at x = 4, find a, b.

Solution: 

Given that, 

4\end{cases} " title="Rendered by QuickLaTeX.com" height="127" width="321" style="vertical-align: -56px;">

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

So, LHL = RHL = f(4)        ……(i)

f(4) = a + b     …..(ii)

Let us consider LHL,

= a – 1        ……(iii)

Now, let us consider RHL,

= 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

 continuous at x = 0 ?

Solution: 

Given that, 

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

So, LHL = RHL = f(0)        …..(i)

f(0) = k

Let us consider LHL,

Using eq(i), we get 

k = 2

Question 30. Let f(x) = , x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

Solution: 

Given that,

f(x) = 

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

So, LHL=RHL=f(0)        ….(i)

Let us consider LHL,

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

From eq(i), we get

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