ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.2

Students can track their progress and improvement through regular use of ISC Maths Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.2

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.2

Question 1.
Is the function f defined by f(x) = |x| continuous at
(i) x = 0
(ii) x = 1
(iii) x = \(\frac {1}{2}\) ?
Solution:
(i) Here, \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) [x]
= 0 – 1
= – 1
and \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) [x] = 0
∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\)
Thus, f is discontinuous at x = 0.

(ii) Given, f(x) = [x]
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) [x]
= 1 – 1 = 0
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)
Thus, f is discontinuous at x = 1.

(iii) \(\underset{x \rightarrow \frac{1}{2}^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow \frac{1}{2}^{-}}{\mathrm{Lt}}\) [x] = 0
[as x → \(\frac{1}{2}\)^–
⇒ x < \(\frac{1}{2}\)
⇒ [x] = 0]
\(\underset{x \rightarrow \frac{1}{2}^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow \frac{1}{2}^{+}}{\mathrm{Lt}}\) [x] = 0
[as x → \(\frac{1}{2}\)⁺
⇒ x > latex]\frac{1}{2}[/latex]
∴ [x] = 0]

Aliter:
\(\mathrm{Lt}_{x \rightarrow \frac{1}{2}}\) f(x) = \(\mathrm{Lt}_{x \rightarrow \frac{1}{2}}\)[x]
= [latex]\frac{1}{2}[/latex] = 0
∴ \(\underset{x \rightarrow \frac{1}{2}^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow \frac{1}{2}^{-}}{\mathrm{Lt}}\) f(x)
= 0
= f(\(\frac{1}{2}\))
= [latex]\frac{1}{2}[/latex]
= 0
Thus f is continuous at x = \(\frac{1}{2}\).

Examine the following (2 to 7) functions for continuity :

Question 2.
(i) f(x) = \(\left\{\begin{array}{cl}
x & , \text { if } x \geq 0 \\
x^2 & \text {, if } x<0 \end{array}\right.\) (NCERT) (ii) f(x) = \(\left\{\begin{array}{lll} x+5 & \text { if } & x \leq 1 \\ x-5 & \text { if } & x>1
\end{array}\right.\) (NCERT)
Solution:
(i) Clearly D_f = R, so we examine the continuity of f at all x ∈ R.
Let c ∈ R be arbitrary real number. Then three cases arises.

Case – I:
When c > 0, f(c) = c
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x = c = f(c)
∴ f is continuous for all c > 0

Case – II:
When c < 0, f(c) = c²
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x²
= c²
= f(c)
∴ f is continuous for all c < 0

Case – III:
When c = 0 then f(c) = f(0) = 0
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x² = 0 = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x = 0
∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = 0
Thus f is continuous at x = 0
Thus on combining all three cases, f is continuous ∀ c ∈ R i.e. at all x ∈ R.
Hence, f is a continuous function.

(ii) Clearly D_f = R, So we examine f for continuity at all x ∈ R. Let c ∈ R be any arbitrary real number. Three cases arises.

Case – I:
When c < 1, f(c) = c + 5
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x + 5
= c + 5
= f(c)
∴ f is continuous for all c < 1. Case – II: When c > 1, f(c) = c – 5
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x – 5
= c – 5
= f(c)
∴ f is continuous for all c > 1

Case – III:
When c = 1, then f(c) = f(1)
= 1 + 5 = 6
L.H.L. = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) x + 5
= 1 + 5
= 6
R.H.L. = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) x – 5
= 1 – 5
= – 4
∴ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x)
∴ f is discontinuous at x = 1.
Hence on combining all threee cases, f is continuous for all c ∈ R – {1} i.e. ∀ x ∈ R – {1}

Question 3.
(i) f(x) = \(\left\{\begin{array}{cc}
2 x-1, & \text { if } x<2 \\
\frac{3}{2} x & \text {, if } x \geq 2
\end{array}\right.\) (NCERT)
(ii) f(x) = \(\left\{\begin{array}{cc}
\sin x-\cos x & , x \neq 0 \\
-1 & , x=0
\end{array}\right.\) (NCERT)
Solution:
(i) When x < 2,
f(x) = 2x – 1, which is polynomial in x and hence continuous everywhere.
∴ f(x) is continuous for each x < 2. Where x > 2 ;
f(x) = \(\frac{3 x}{2}\) which is polynomial in x and hence continuous for each x > 2.

at x = 2 :
L.H.L = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 2}{\mathrm{Lt}}\) 2x – 1
= 4 – 1
= 3
and R.H.L = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 2}{\mathrm{Lt}}\) \(\frac{3 x}{2}\)
= \(\frac{3 \times 2}{2}\) = 3
also f(2) = 3
∴ L.H.L = R.H.L = f(2)
Hence f(x) is continuous at x = 2.
Thus f(x) is continuous everywhere.

(iii) When x ≠ 0 i.e. x < 0 and x > 0
f (x) = sin x – cos x,
since sin x and cos x.
both are continuous everywhere
∴ f(x) = sin x – cos x is also continuous for each x ≠ 0.
since difference of two continuous functions is continuous.

at x = 0 :
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) f(- h)
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [sin (- h) – cos (- h)]
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [- sin h – cos h]
= 0 – 1
= – 1

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) f(h)
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [sin h – cos h]
= 0 – 1
= – 1
and f(0) = – 1
∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = f(0) = – 1
Thus f is continuous at x = 0.
Hence f(x) is continuous everywhere and nowhere discontinuous.

Question 4.
(i) f(x) = |x| (NCERT)
(ii) f(x) = |x – 5| (NCERT)
Solution:
(i) Modulus function is given by
|x| = \(\left[\begin{array}{rl}
-x & ; \quad x<0 \\
x & ; \quad x \geq 0
\end{array}\right.\)
Here D_f = R

\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) (0 + h) = 0
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – x
= – \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) (0 – h) = 0
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
Thus, modulus function is continuous at x = 0
f(x) = |x| is continuous at x = 0.

(ii) f(x) = \(\left\{\begin{array}{cc}
\frac{x}{x}=1 ; & x>0 \\
\frac{x}{x}=-1 ; & x<0 \\
0 & ; x=0
\end{array}\right.\)

When x < 0 :
f(x) = – 1, which is a constant function and hence continuous everywhere.
∴ f(x) is continuous for all x < 0.
When x > 0 ;
f(x) = 1, which is a constant function and hence continuous for all each x > 0.

at x = 0;
L.H.L = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) – 1
= – 1
and R.H.L = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) 1 = 1
∴ L.H.L. ≠ R.H.L.
Thus, f(x) is not continuous at x = 0.
Hencef(x) is continuous at each point except x = 0.

Question 5.
(i) f(x) = \(\frac{x^2-25}{x+5}\) (NCERT)
(ii) f(x) = \(\frac{1}{x-5}\) (NCERT)
Solution:
(i) Given f(x) = \(\frac{x^2-25}{x+5}\)
∴ D_f = R – {- 5}
Let c be any arbitrary point of its domain.
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow c} \frac{x^2-25}{x+5}\)
= \(\frac{c^2-25}{c+5}\)
= c – 5
= f(c)
Thus f is continuous at x = c, but c is arbitrary.
Hence f is continuous at every point of its domain.

(ii) Given, f(x) = \(\frac{1}{x-5}\),
Now f is not defined at x = 5.
∴ D_f = R – {- 5}.
Let c be any arbitrary point of its domain.
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow c} \frac{1}{x-5}\)
= \(\frac{1}{c-5}\)
= f(c)
Therefore f is continuous at x = c, but c is arbitrary.
Hence f is continuous at every point of its domain.

Question 6.
(i) f(x) = \(\left\{\begin{array}{cc}
\frac{\sin 2 x}{x}, & \text { if } x>0 \\
2 & \text {, if } x \leq 2
\end{array}\right.\) (NCERT)
(ii) f(x) = \(\left\{\begin{array}{cc}
\frac{x}{|x|}, & x \neq 0 \\
0, & x=0
\end{array}\right.\) (NCERT)
Solution:
(i) f(x) = \(\left\{\begin{array}{cc}
\frac{\sin 2 x}{x}, & \text { if } x>0 \\
2 & \text {, if } x \leq 2
\end{array}\right.\)
Here D_f = R, so we examine f for continuity at all x ∈ R.
Let c ∈ Rbe any arbitrary real number. Three cases arises.

Case – I :
When c < 0 ; f(c) = 2
and \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2
= 2 = f(c)
∴ f is continuous for all c < 0. Case – II : When c > 0 ;
f(c) = \(\frac{\sin 2 c}{c}\)
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) \(\frac{\sin 2 x}{x}\)
= \(\frac{\sin 2 c}{c}\)
= f(c)
∴ f is continuous for all c > 0.

Case – III :
When c = 0
Then f(c) = f(0) = 2
\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\sin 2 x}{x}\)
= \(\ {Lt}_{x \rightarrow 0} 2 \times \frac{\sin 2 x}{2 x}\)
= 2 \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\sin 2 x}{x}\)
= 2 × 1 = 2
= f(0)
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\sin \theta}{\theta}\) = 1]
∴ f is continuous for all c = 0.
So on combining all three cases, f is continous for all c ∈ R i.e. for all x ∈ R.

(ii) f(x) = \(\left\{\begin{array}{cc}
\frac{x}{x}=1 ; & x>0 \\
\frac{x}{x}=-1 & ; x<0 \\
0 & ; x=0
\end{array}\right.\)

When x < 0
f(x) = – 1, which is a constant function and hence continuous everywhere.
∴ f(x) is continuous for all x < 0. When x > 0 ;
f(x) = 1. which is a constant function and hence continuous for all each x > 0.
at x = 0;
L.H.L = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) – 1 = – 1
and R.H.L. = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) 1 = 1
∴ L.H.L ≠ R.H.L
Thus, f(x) is not continuous at x = 0.
Hence f(x) is continuous at each point except x = 0.

Question 7.
(i) f(x) = \(\left\{\begin{array}{ccc}
x^{10}-1 & , \text { if } & x \leq 1 \\
x^2 & \text {, if } & x>1
\end{array}\right.\) (NCERT)
(ii) f(x) = tan x, 0 ≤ x ≤ \(\frac{\pi}{4}\).
Solution:
(i) When x < 1,
f(x) = x¹⁰ – 1,
which is polynomial in x and hence continuous for each x < 1. When x > 1,
f(x) = x²,
which is polynomial in x and hence continuous for each x > 1.

at x = 1
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) x¹⁰
= (1)¹⁰ – 1
= 1 – 1 = 0
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) x²
= 1² = 1
and f(1) = 1¹⁰ – 1 = 0
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)
Hence f(x) is discontinuous at x = 1.
Thus, the function f(x) is continuous everywhere except x = 1.

(ii) f(x) = tan x
given D_f = [0, \(\frac{\pi}{4}\)]
Let c ∈ D_f be any arbitrary point
Here, \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) tan x
= tan c
= f(c)
∴ f is continuous at x = c but c is any arbitrary point.
Hence f is continuous at every point of its domain.

Locate the points discontinuity (if any) of the following (8 to 10) functions:

Question 8.
(i) f(x) = \(\left\{\begin{array}{l}
2 x+3, \text { if } x \leq 2 \\
2 x-3, \text { if } x>2
\end{array}\right.\) (NCERT)
(ii) f(x) = \(\begin{cases}x^3-3, & \text { if } x \leq 2 \\ x^2+1, & \text { if } x>2\end{cases}\) (NCERT)
Solution:
(i) Given f(x) = \(\left\{\begin{array}{l}
2 x+3, \text { if } x \leq 2 \\
2 x-3, \text { if } x>2
\end{array}\right.\)
Here D_f = R
Since examine the continuity of f at all x ∈ R.
Let c ∈ R be any arbitrary point.

Case – I:
When c < 2,
f(x) = 2c + 3
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (2x + 3)
= 2c + 3
= f(c)
∴ f is continuous for all c < 2.

Case – II:
When c > 2,
f(c) = 2c + 3
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2x – 3
= 2c – 3
= f(c)
∴ f is continuous for all c > 2.

Case – III:
When c = 2,
f(c) = 2c – 3
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) (2x – 3)
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [2 (2 + h) – 3]
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [2h + 1]
= 1
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) (2x + 3)
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [2 (2 – h) + 3]
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [7 – 2h]
= 7
∴ \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x)
Thus f is discontinuous at x = 2.
Thus on combining all three cases, f is continuous for all x = c except x = 2 but c is arbitrary.
Hence f is continuous everywhere on R – {2} and has one point of discontinuity at x = 2.

(ii) Given f(x) = \(\begin{cases}x^2-3 & ; \quad x \leq 2 \\ x^2+1 & ; \quad x>2\end{cases}\)
Here D_f = R, so we examine the continuity of function f at all x ∈ R.
Let c ∈ R be any arbitrary point.
Three cases arises.

Case – I :
When c < 2, f(c) = c² – 3
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x² – 3
= c² – 3
= f(c)
∴ f is continuous for all c < 2. Case – II : When c > 2,
f(c) = c² + 1
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (x² + 1)
= c² + 1
= f(c)
∴ f is continuous for all x = c > 2.

Case – III:
When c = 2
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\)
= \(\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}\) [(2 + h)² + 1] = 5
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) (x³ – 3)
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) [(2 – h)³ – 3]
= 8 – 3 = 5
∴ \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = f(2)
Hence f is continuous everywhere on R.
∴ f is continuous everywhere on R.
Hence it has no point of discontinuity.

Question 9.
(i) f(x) = \(\left\{\begin{array}{cc}
\frac{x^4-16}{x-2}, & x \neq 2 \\
16, & x=2
\end{array}\right.\)
(ii) f(x) = \(\left\{\begin{array}{cc}
\frac{\sin 3 x}{x}+\cos x & , x>0 \\
4-3 x & , x \leq 0
\end{array}\right.\)
Solution:
(i) We note that D_f = R, then we have to examine f for continuity for all x ∈ R and c ∈ R be any arbitrary element.
\(\ {Lt}_{x \rightarrow 2} \frac{x^4-16}{x-2}\)
= \(\ {Lt}_{x \rightarrow 2} \frac{\left(x^2-4\right)\left(x^2+4\right)}{x-2}\)
= \(\underset{x \rightarrow 2}{\mathrm{Lt}}\) (x + 2) (x² + 4)
= 4 . 8
= 32
also f(2) = 16
∴ \(\underset{x \rightarrow 2}{\mathrm{Lt}}\) f(x) ≠ f(2).
Thus f is discontinuous at x = 2.
at c ≠ 2 be any arbitrary point, then f(c) = \(\frac{c^4-16}{c-2}\)
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{x^4-16}{x-2}\)
= \(\frac{c^4-16}{c-2}\)
= f(c)
∴ f is continuous at x = c ≠ 2
f(x) has only one point of discontinuity at x = 2.

(ii) Here D_f = R, so we examine f for continuity at all x ∈ R
Let c ∈ R be any real number. Three cases arise.

Case – I:
When c < 0, f(c) = 4 – 3c
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 4 – 3x
= 4 – 3c
= f(c)
∴ f is continuous for all c < 0

Case – II:
When c > 0,
f(c) = \(\frac{\sin 3 c}{c}\) + cos c
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) \(\frac{\sin 3 x}{x}\) + cos x
= \(\frac{\sin 3 c}{c}\) + cos c
= f(c)
∴ f is continuous for all c > 0

Case – III:
When c = 0
Then f(c) = f(0)
= 4 – 3 × 0
= 4
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (\(\frac{\sin 3 x}{x}\) + cos x)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) 3 (\(\frac{\sin 3 x}{x}\)) + \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) cos x
= 3 × 1 + 1 = 4
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\sin \theta}{\theta}\) = 1]
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) 4 – 3x
= 4 – 3 × 0 = 4
also f(0) = 4
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
Thus f is continuous at x = 0
Hence on combining all three cases, f is continuous for all c ∈ R i.e. for all x ∈ R.
Thus f be a continuous function.
Hence there is no point for which f is discontinuous.

Question 10.
(i) f(x) = \(\left\{\begin{array}{ccc}
2 x & , \text { if } & x<0 \\ 0 & , \text { if } & 0 \leq x \leq 1 \\ 4 x & , \text { if } & x>1
\end{array}\right.\) (NCERT)
(ii) f(x) = \(\left\{\begin{array}{cc}
x+2, & x \leq 1 \\
x-2, & 1<x<2 \\
0, & x \geq 2
\end{array}\right.\)
Solution:
(i) When x < 0 and x > 1
f(x) = 2x and 4x, which is polynomial in x and hence continuous everywhere.
∴ f(x) is continuous for each x < 0 and x > 1.
When 0 < x < 1 ; f(x) = 0, which is a constant function.
and hence continuous everywhere.
∴ f(x) is continuous for each x ∈ (0, 1).

at x = 0 ;
L.H.L. = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) 2x = 0
R.H.L. = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = 0
and f(0) = 0
∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = f(0)
Thus f(x) is continuous at x = 0.

at x = 1 ;
L.H.L. = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) 0 = 0

R.H.L. = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\)
= \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) 4x
= 4 × 1 = 4
∴ L.H.L. ≠ R.H.L.
Thus f(x) is continuous at x = 1.

(ii) D_f = R;
So we examine for continuity for all values of x ∈ R.
Let c ∈ R be any real number.
Five cases arise.

Case – I :
When c < 1
then f(c) = c + 2
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) x + 2
= c + 2
= f(c)
∴ f is continuous for all c < 1.

Casae – II :
When 1 < c < 2
then f(c) = c – 2
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (x – 2)
= c – 2
= f(c)
∴ f is continuous for all c where 1 < c < 2.

Case – III:
When c > 2 then f(c) = 0
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 0
= 0 = f(c)
∴ f is continuous for all c where c < 2.

Case – IV:
at c = 1 then
f(c) = f(1)
= 1 + 2 = 3
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) x – 2
= 1 – 2 = – 1
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) x + 2
= 1 + 2 = 3
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) Thus f is discontinuous at x = c = 1.

Case – V:
at x = c = 2 then f(c) = f(2) = 0 \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) x – 2
= 2 – 2 = 0
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) 0 = 0
∴ \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x)
= f(0) = 0
Thus f is continuous at x = c = 2.
On combining all five cases, f is discontinuous at x = 1.

Question 11.
Find the value of the constant k so that the function f defined by f(x) = \(\begin{cases}k x^2 & , x \leq 2 \\ x-3, & x>2\end{cases}\) may be continuous.Solution:
at x = 2:
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) x – 3
= 2 – 3
= – 1
and \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) kx²
= \(\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}\) k (2 – h)²
= 4k
As f(x) be a continuous function.
So f(x) is also continuous at x = 2.
∴ \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = f(2)
⇒ – 1 = 4k
⇒ k = \(\frac{1}{4}\).

Question 12.
For what choice a and b are the following functions continuous:
(i) f(x) = \(\left\{\begin{array}{ccc}
-5 & \text {, if } & x \leq-1 \\
a x-b & \text {, if } & -1<x<3 \\ 7 & \text {, if } & x \geq 3 \end{array}\right.\) (ii) f(x) = \(\left\{\begin{array}{cc} x^2 & , x \leq 0 \\ a x+b & , x>0
\end{array}\right.\)
Solution:
(i) Given f(x) = \(\left\{\begin{array}{ccc}
-5 & \text {, if } & x \leq-1 \\
a x-b & \text {, if } & -1<x<3 \\ 7 & \text {, if } & x \geq 3 \end{array}\right.
\) Since f be a continuous function.
Thus f must be continuous at x = – 1 and x = 3.

at x = – 1 :
\(\underset{x \rightarrow-1^{-}}{\ {Lt}}\) f(x) = \(\underset{x \rightarrow-1^{-}}{\ {Lt}}\) – 5
= – 5 \(\underset{x \rightarrow-1^{+}}{\ {Lt}}\) f(x)
= \(\underset{x \rightarrow-1^{+}}{\ {Lt}}\) ax – b
= – a – b
also f (- 1) = – a – b
Since f is continuous at x = – 1
∴ \(\underset{x \rightarrow-1^{-}}{\ {Lt}}\) f(x) = \(\underset{x \rightarrow-1^{+}}{\ {Lt}}\)
f(x) = f(- 1)
∴ – 5 = – a – b ⇒ a + b = 5 …………(1)

at x = 3 :

\(\underset{x \rightarrow 3^{+}}{\ {Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\ {Lt}}\) 7
= 7 \(\underset{x \rightarrow 3^{-}}{\ {Lt}}\) f(x)
= \(\underset{x \rightarrow 3^{-}}{\ {Lt}}\) ax – b
= 3a – b
Also f(3) = 7
Since f is continuous at x = 3.
∴ \(\underset{x \rightarrow 3^{-}}{\ {Lt}}\) f(x)
= \(\underset{x \rightarrow 3^{-}}{\ {Lt}}\) f(x) = f(3)
∴ 3a – b = 7 ……….(2)
On solving (1) and (2)
simultaneously, we get a = 3 and b = 2.

(ii) Given, f(x) = \(\left\{\begin{array}{cc} x^2 & , x \leq 0 \\ a x+b & , x>0
\end{array}\right.\)
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (ax + b)
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) ah + b = b
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x²
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) (- h)² = 0
Also f(0) = 0
Since the function f(x) is continuous everywhere.
∴ function is continous at x = 0
Thus \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
∴ b = 0 and a can have any value.