RD Sharma Class 12 Ex 9.2 Solutions Chapter 9 Continuity

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	9
Exercise	9.2
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 9.2 Solutions Chapter 9 Continuity

Question 1. Prove that the function $f(x)=\begin{cases}\frac{sinx}{x} \ \ \ \ ,x\le0\\x+1\ \ \ ,x\le0\end{cases}$ is continuous everywhere.

Solution:

We know sin x/ x is continuous everywhere since it is the composite function of the functions sin x and x which are continuous.
When x > 0, we have f(x) = x + 1.
Given that $f(x)=\binom{\frac{sinx}{x}, x < 0}{x + 1, x \geq 0}$
Now, (LHL at x = 0) = lim_{{x ⇢ 0^–}}f(x)
= lim_{{h ⇢ 0}} f(0 – h)
= lim_{{h ⇢ 0}} f(-h)
= lim{h ⇢ 0} (sin (-h)/(-h))
= lim_{{h ⇢ 0}} (sin h/ h)
= 1
(RHL at x = 0) = lim_{{x ⇢ 0+}}f(x)
= lim_{{h ⇢ 0}} f(0 + h)
= lim_{{h ⇢ 0}}f(h)
= lim_{{h ⇢ 0}} (h + 1)
= 1
and, f(0) = 0 + 1 = 1.
We observe that: lim_{{x ⇢ 0^–}}f(x) = lim_{{x ⇢ 0⁺}} f(x) = f(0).
Therefore, f(x) is everywhere continuous.

Question 2. Discuss the continuity of the function $f(x)=\begin{cases}\frac{x}{\left| x \right|}, & x \neq 0 \\ 0 , & x = 0\end{cases}$ .

Solution:

We have,

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

Now: (LHL at x = 0) = lim_{{x ⇢ 0^–}} f(x)

= lim_{{h ⇢ 0}} f(0 – h)

= lim_{{h ⇢ 0}} f(–h)

= lim_{{h⇢ 0}} (–1)

= –1

(RHL at x = 0) = lim_{{x ⇢ 0⁺}} f(x)

= lim_{{h ⇢ 0}} f(0 + h)

= lim_{{h ⇢ 0}} (1)

= 1

We observe that, lim_{{x ⇢ 0^–}}f(x) ≠ lim_{{x ⇢ 0⁺}} f(x).

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

Question 3. Find the points of discontinuity, if any, of the following functions:

(i) $f(x)=\begin{cases}x^3 - x^2 + 2x - 2, & \text{ if }x \neq 1 \\ 4 , & \text{ if } x = 1\end{cases}$

Solution:

Since a polynomial function is everywhere continuous.
At x = 1, we have
(LHL at x = 1) = lim_{{x ⇢ 1^–}} f(x)
= lim_{{h ⇢ 0}} f(1 – h)
= lim_{{h ⇢ 0}} ((1 – h)³– (1 – h )² + 2(1 – h) – 2)
= 1 – 1 + 2 – 2
= 0
(RHL at x = 1) = lim_{{x ⇢ 1⁺}} f(x)
= lim_{{h ⇢ 0}} f(1 + h)
= lim{h ⇢ 0} ((1 + h)³ – (1 + h )² + 2(1 + h) – 2)
= 1 – 1 + 2 – 2
= 0
Also, f(1) = 4.
We observe that, lim_{{x ⇢ 0^–}} f(x) = lim_{{x ⇢ 0⁺}} f(x) ≠ f(1).
Therefore, f(x) is discontinuous only at x = 1.

(ii) $f( x ) =\begin{cases}\frac{x^4 - 16}{x - 2}, & \text{ if } x \neq 2 \\ 16 , & \text{ if } x = 2\end{cases}$

Solution:

When x ≠ 2 then
f(x) = $\frac{x^4 - 16}{x - 2}$
= $\frac{x^4 - 2^4}{x - 2}$
= $\frac{\left( x^2 + 4 \right)\left( x - 2 \right)\left( x + 2 \right)}{x - 2}$
= (x² + 4)(x + 2)
Since a polynomial function is everywhere continuous, (x² + 4) and (x + 2) are continuous everywhere.
So, the product function (x² + 4)(x + 2) is continuous.
Thus, f(x) is continuous at every x ≠ 2 .
We observe that lim_{x->2-}f(x) = lim_{x->2+}f(x) = f(2)
Therefore, f(x) is discontinuous only at x = 2.

(iii) $f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & \text{ if } x < 0 \\ 2x + 3, & x \geq 0\end{cases}$

Solution:

When x < 0, then f(x) = sin x/ x.
Since sin x as well as the identity function x are everywhere continuous, the quotient function sin x/x is continuous at each x < 0.
For x > 0, f(x) becomes a polynomial function. Therefore, f(x) is continuous at each x > 0.
We have: (LHL at x = 0) = lim_{x->0-}f(x)
= lim_{{h -> 0}} f(0 – h)
= lim_{{h -> 0}} f (-h)
= lim_{{h -> 0}} (sin(-h)/(-h))
= lim_{{h -> 0}} (sin h/h)
= 1
(RHL at x = 0) = lim_{_{x -> 0+}} f(x)
= lim_{{h -> 0}}f(0 + h)
= lim_{{h -> 0}} f(h)
= lim_{{h -> 0}} (2h + 3)
= 3
We observe that lim_{{x -> 0-}} f(x) ≠ lim_{{x -> 0+}} f(x)
Therefore, f(x) is discontinuous only at x = 0.

(iv) $f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, & \text{ if } x \neq 0 \\ 4 , & \text{ if } x = 0\end{cases}$

Solution:

At x ≠ 0, then f(x) = sin 3x/ x.
Since the functions sin 3x and x are everywhere continuous. So, the quotient function sin 3x/x is continuous at each x ≠ 0.
We have: (LHL at x = 0) = lim_{{x -> 0+}} f(x)
= lim_{{h -> 0}} f(0 + h)
= lim_{{h -> 0}} f(h)
= lim_{{h -> 0}} (sin 3h/h)
= lim_{{h -> 0}} 3 (sin h/h) = 3
(RHL at x = 0) = lim_{{x -> 0+}} f(x)
= lim_{{h -> 0}} f(0 + h)
= lim_{{h -> 0}} f(h)
= lim_{{h -> 0}} (sin 3h/h)
= lim_{{h -> 0}} 3 (sin 3h/h)
= 3
Also, f(0) = 4.
We observe that lim_{{x -> 0-}} f(x) = lim_{{x -> 0+}} f(x) = f(0)
Therefore, f(x) is discontinuous only at x = 0.

(v) $f\left( x \right) = \begin{cases}\frac{\sin x}{x} + \cos x, & \text{ if } x \neq 0 \\ 5 , & \text { if } x = 0\end{cases}$

Solution:

When x ≠ 0, then f(x) = sin x/ x + cos x.
We know that sin x as well as cos x are everywhere continuous. Thus, the given function is continuous at each x ≠ 0.
Let us consider the point x = 0.
Given: $f\left( x \right) = \binom{\frac{\text{ sin } x}{x} + \ \text{ cos } x, \text{ if } x \neq 0}{5, \text{ if } x = 0 }$
We have: (LHL at x = 0) = lim_{{x -> 0-}} f(x)
= lim_{{h -> 0}} f(0 – h)
= lim_{{h -> 0}} f(-h)
= lim_{{h -> 0}} [(sin (-h)/(-h)) + cos (-h)]
= lim_{{h -> 0}} sin(-h)/(-h) + lim_{{h -> 0}} cos(-h)
= 1 + 1
= 2
(RHL at x = 0) = lim_{{x -> 0+}} f(x)
= lim_{{h -> 0}} f(0 + h)
= lim_{{h -> 0}} f(h)
= lim_{{h -> 0}} [(sin h/h) + cos (-h)]
= lim_{{h -> 0}} sin h/h + lim_{{h -> 0}} cos(-h)
= 1 + 1
= 2
Also, f(0) = 5.
We observe that lim_{{x -> 0-}} f(x) = lim_{{x -> 0^+}} f(x) ≠ f(0)
Therefore, f(x) is discontinuous only at x = 0.

(vi) $f\left( x \right) = \begin{cases}\frac{x^4 + x^3 + 2 x^2}{\tan^{- 1} x}, & \text{ if } x \neq 0 \\ 10 , & \text{ if } x = 0\end{cases}$

Solution:

When x ≠ 0, then $f( x) = \frac{x^4 + x^3 + 2 x^2}{tan^{- 1} x}$
x⁴ + x³ + 2x² being a polynomial function is continuous everywhere.
Also, tan^-1x is everywhere continuous.
Let us consider the point x = 0.
We have:
(LHL at x = 0) = lim_{{x -> 0-}}f(x)
= lim_{{h -> 0}} f(0 – h)
= lim_{{h -> 0}} f(-h)
= $\lim_{h \to 0} \left( \frac{\left( - h \right)^4 + \left( - h \right)^3 + 2 \left( - h \right)^2}{\tan^{- 1} \left( - h \right)} \right)$
= $\lim_{h \to 0} \left( \frac{\left( h \right)^3 - \left( h \right)^2 + 2\left( h \right)}{- \frac{\tan^{- 1} \left( h \right)}{h}} \right)$
= 0
(RHL at x = 0) = lim_{{x -> 0+}} f(x)
= lim_{{h -> 0}} f(0 + h)
= lim_{{h -> 0}} f(h)
= $\lim_{h \to 0} \left( \frac{\left( h \right)^4 + \left( h \right)^3 + 2 \left( h \right)^2}{\tan^{- 1} \left( h \right)} \right)$
= $\lim_{h \to 0} \left( \frac{\left( h \right)^3 + \left( h \right)^2 + 2\left( h \right)}{\frac{\tan^{- 1} \left( h \right)}{h}} \right)$
= 0
Also, f(0) = 10.
We observe that lim_{{x -> 0-}} f(x) = lim_{{x -> 0⁺}} f(x) ≠ f(0)
Therefore, f(x) is discontinuous only at x = 0.

(vii) $f\left( x \right) = \begin{cases}\frac{e^x - 1}{\log_e (1 + 2x)}, & \text{ if }x \neq 0 \\ 7 , & \text{ if } x = 0\end{cases}$

Solution:

We have,

$\lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \frac{e^x - 1}{\log_e \left( 1 + 2x \right)}$

= $\lim_{x\to0}\frac{\left( \frac{e^x - 1}{x} \right)}{\left( \frac{2 \log_e \left( 1 + 2x \right)}{2x} \right)}$

= $\frac{1}{2} \times \frac{\lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)}{\lim_{x \to 0} \left( \frac{\log_e \left( 1 + 2x \right)}{2x} \right)}$

= 1/2

It is given that f(0) = 7.

We observe that lim_{{x -> 0}} f(x) ≠ f(0)

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

(viii) $f\left( x \right) = \begin{cases}\left| x - 3 \right|, & \text{ if } x \geq 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & \text{ if } x < 1\end{cases}$

Solution:

At x > 1, f(x) being a modulus function is continuous for each x > 1.
When x < 1, then f( x ) being a composite of polynomial and continuous functions would be continuous.
At x = 1, we have
(LHL at x = 1) = lim_{{x -> 1-}} f(x)
= lim_{{h -> 0}} f(1 – h)
= $\lim_{h\to0} \left[ \frac{\left( 1 - h \right)^2}{4} - \frac{3\left( 1 - h \right)}{2} + \frac{13}{4} \right]$
= 1/4 – 3/2 + 13/4
= 2
(RHL at x = 1) = lim_{{x -> 1+}} f(x)
= lim_{{h -> 0}} f(1 + h)
= lim_{{h -> 0}} |1 + h – 3|
= |-2|
= 2
Also f(1) = |1 – 3| = |- 2| = 2
We observe that, lim_{{x -> 1-}} f(x) = lim_{{x -> 1+}} f(x) = f(1)
Therefore, the given function is everywhere continuous.

(ix) $f\left( x \right) = \begin{cases}\left| x \right| + 3 , & \text{ if } x \leq - 3 \\ - 2x , & \text { if } - 3 < x < 3 \\ 6x + 2 , & \text{ if } x > 3\end{cases}” height=”111″ width=”398″><span class=$

Solution:

f(x) being a modulus function is continuous for each x ≤ – 3.
At – 3 < x < 3 f(x) being a polynomial function is continuous.
At x > 3, f(x) being a polynomial function is continuous.
At x = 3,
We have: (LHL at x = 3) = lim_{{x -> 3-}} f(x)
= lim_{{h -> 0}} f(3 – h)
= lim_{{h -> 0}} -2(3 – h)
= -6
(RHL at x = 3) = lim_{{x -> 3+}} f(x)
= lim_{{h -> 0}} f(3 + h)
= lim_{{h -> 0}} 6(3 + h) + 2
= 20
We observe that lim_{{x -> 3-}} f(x) ≠ lim_{{x -> 3+}} f(x)
Therefore, f(x) is discontinuous only at x = 3.

(x) $f\left( x \right) = \begin{cases}x^{10} - 1, & \text{ if } x \leq 1 \\ x^2 , & \text{ if } x > 1\end{cases}” height=”79″ width=”321″><span class=$

Solution:

According to the question it is given that function f is defined at all the points of the real line.
Let us considered c be a point on the real line.
Case I: If c< 1, then f(c) = c¹⁰ −1 and
lim_{{x-> c}} f(x) = lim_{x->c}(x¹⁰ – 1)
= c¹⁰ −1.
∴ lim_{x->c}f(x) = f(c)
Hence, f is continuous for all x < 1.
Case II: If c = 1, then the left hand limit of f at x = 1.
The right hand limit of f at x = 1 is, lim_(x->1)f(x) = lim_(x->1)(x²) = 1² = 1
So we conclude that the left and right hand limit of f at x = 1 do not coincide. So, f is not continuous at x = 1.
Case III: If c>1, then f(c) = c²
lim_(x->c)f(x) = lim_(x->c) f(c) = c²
∴ lim_(x->c)f(x) = f(c)
Therefore, f(x) is discontinuous only at x = 1.

(xi) $f\left( x \right) = \begin{cases}2x , & \text{ if } & x < 0 \\ 0 , & \text{ if } & 0 \leq x \leq 1 \\ 4x , & \text{ if } & x > 1\end{cases}” height=”237″ width=”277″><span class=$

Solution:

Let us considered a be a point on the real line.
Case I: if a < 0, then f(c) = 2a.
lim_{x->a}(a) = 2a.
∴ lim_{{N -> 0}}f(x) = f(a)
So, f is continuous at all points such that x < 0.
Case II: If 0 < a < 1 then f(x) and lim_{x->a}f(x)=lim_{x->a}(0)=0 .
∴ lim_{x->a}f(x)=f(a)
So, f is continuous at all points of the interval (0, 1).
Case III: If a =1 then f(a) = f(1) = 0.
The left hand limit of f at x = 1 is,
lim_{x->1}f(x) = lim_{x->1}f(1)
The right hand limit of f at x = 1 is,
lim_{x->1}f(x) = lim_{x->1}(4x) = 4(1) = 4
So we conclude that LHL ≠ RHL. Thus, f is not continuous at x = 1.
Case IV: If a > 1, then f(a) = 4a and lim_{x->a}f(4x) = 4a .
∴ lim_{x->a}f(x) = f(a)
So, f(x) is discontinuous only at x = 1.

(xii) $f\left( x \right) = \begin{cases}\sin x - \cos x , & \text{ if } x \neq 0 \\ - 1 , & \text{ if } x = 0\end{cases}$

Solution:

It is evident that f is defined at all points of the real line. Let p be a real number.
Case I: if p ≠ 0 , then f (p) = sin p – cos p
lim_{{x → p}}f(x) = lim_{x→p}( sin x – cos x ) = sin p – cos p
∴ lim_{{x →p}}f(x) = f(p)
Therefore, f is continuous at all points x, such that x ≠ 0.
Case II: if p = 0 , then f (0) = – 1.
lim_{{x →0-}}f(x) = lim_{{x →0^-}}(sin x – cos x) = sin 0 – cos 0 = 0 – 1 = -1
lim_{{x →0+}}f(x) = lim_{{x →0}}(sin x – cos x) = sin 0 – cos 0 = 0 – 1 = -1
We observe that: lim_{{x →0-}} f (x) = lim_{{x →0+}}f (x)= f(0)
Therefore, f is a continuous function everywhere.

(xiii) $f\left( x \right) = \begin{cases}- 2 , & \text{ if }& x \leq - 1 \\ 2x , & \text{ if } & - 1 < x < 1 \\ 2 , & \text{ if } & x \geq 1\end{cases}$

Solution:

The given function is defined at all points of the real line. Let us considered a be a point on the real line.
Case I: If a < -1 then f(a)= -2 and lim_{x->a}(x) = lim_{x->a}(-2) = -2
∴ lim_{x->a}f(x) = f(a)
f is continuous for all x < −1.
Case II: If a =1 then f(a) = f(-1) = -2
LHL = lim_{x->-1}f(x) = lim_{x->-1}f(-2) = -2
RHL = lim_{x->-1}f(x) = lim_{x->-1}f(2x) = 2(-1) = -2
We observe that lim_{x->-1}f(x) = f(-1)
Therefore, f is continuous at x = −1.
Case III: if -1 < a < 1,then f(a) = 2a
lim_{x->a}f(x) = lim_{x->a}f(2x) = 2a
∴ lim_{x->a}f(x) = f(a)
Therefore, f is continuous at all points of the interval (−1, 1).
Case IV: if a = 1, then f(c) = f(1) = 2(1) = 2.
LHL = lim_{x->1}f(x) = lim_{x->1}2 = 2
RHL = lim_{x->1}f(x) = lim_{x->1}2 = 2
We observe that: lim_{x->1}f(x) = lim_{x->1}f(c)
Therefore, f is continuous at x = 2.
Therefore, f is a continuous function everywhere.

Question 4. In the following, determine the value of constant involved in the definition so that the given function is continuous:

(i) $f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, & \text{ if } x \neq 0 \\ 3k , & \text{ if } x = 0\end{cases}$

Solution:

If f( x ) is continuous at x = 0, then
⇒ lim{x -> 0} f(x) = f(0)
⇒ lim_{{x -> 0}} sin 2x/5x = f(0)
⇒ lim_{{x -> 0}} 2 sin 2x/10x = f(0)
⇒ 2/5 lim_{{x -> 0}} sin 2x/2x = f(0)
⇒ k = 2/15

(ii) $f\left( x \right) = \begin{cases}kx + 5, & \text{ if } x \leq 2 \\ x - 1, & \text{ if } x > 2\end{cases}” height=”79″ width=”315″><span class=$

Solution:

If f(x) is continuous at x = 2, then
lim_{{x -> 2-}} f(x) = lim_{{x -> 2+}} f(x)
⇒ lim_{{h -> 0}} (k (2 – h) + 5) = lim_{{h -> 0}} (2 + h -1)
⇒ lim_{{h -> 0}} f(2 – h) = lim_{{h -> 0}} f(2 + h)
⇒ 2k + 5 = 1
⇒ 2k = – 4
⇒ k = – 2

(iii) $f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if } x < 0 \\ \cos 2x , & \text{ if } x \geq 0\end{cases}$

Solution:

If f(x) is continuous at x = 0, then
lim_{{x -> 0-}} f(x) = lim_{{x -> 0+}} f(x)
⇒ lim_{{h -> 0}} f(-h) = lim_{{h -> 0}} f(h)
⇒ $\lim_{h -> 0} \left( k\left( \left( – h \right)^2 – 3h \right) \right) = lim_{h -> 0} \left( \cos 2h \right)” height=”47″ width=”527″>⇒ 0 = 1, which is not possibleTherefore, the given function is not continuous for any value of k.</blockquote><h3 class=$ (iv) $f\left( x \right) = \begin{cases}2 , & \text{ if } x \leq 3 \\ ax + b, & \text{ if } 3 < x < 5 \\ 9 , & \text{ if } x \geq 5\end{cases}$
Solution:
If f(x) is continuous at x = 3 and 5, then
lim_{{x -> 3-}} f(x) = lim_{{x -> 3+}} f(x)
and lim_{{x -> 5-}} f(x) = lim_{{x -> 5+}} f(x)
⇒ lim_{{h -> 0}} f(3 – h) = lim_{{h -> 0}} f(3 + h)
and lim_{{h -> 0}} f(5 – h) = lim_{{h -> 0}} f(5 + h)
⇒ 2 = 3a + b and 5a + b = 9
⇒ 2 = 3a + b and 5a + b = 9
⇒ a = 7/2 and b = -17/2.
(v) $f\left( x \right) = \begin{cases}4 , & \text{ if } x \leq - 1 \\ a x^2 + b, & \text{ if } - 1 < x < 0 \\ \cos x, &\text{ if }x \geq 0\end{cases}$
Solution:
If f(x) is continuous at x = −1 and 0, then
lim_{{x -> -1-}} f(x) = lim_{{x -> – 1+}} f(x) and lim_{{x -> 0-}} f(x) = lim_{{x -> 0+}} f(x)
⇒ lim_{{h -> 0}} f(-1 – h) = lim_{{h -> 0}} f(-1 + h) and lim_{{h -> 0}} f(-h) = lim_{{h -> 0}} f(h)
⇒ lim_{{h -> 0}} (4) = lim_{{h -> 0}} (a (-1 + h)² + b)
Also, $\lim_{h -> 0} \left( a \left( – h \right)^2 + b \right) = \lim_{h -> 0} \left( \cos h \right)” height=”47″ width=”461″>⇒ 4 = a + b and b = 1⇒ a = 3 and b = 1.</blockquote><h3 class=$ (vi) $f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & \text{ if } - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & \text{ if } 0 \leq x \leq 1\end{cases}$
Solution:
If f(x) is continuous at x = 0, then $\lim_{x \to 0^-} f( x ) = \lim_{x \to 0^+} f( x )$
⇒ $\lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} f\left( h \right)$
⇒ $\lim_{h \to 0} \left( \frac{\sqrt{1 - ph} - \sqrt{1 + ph}}{- h} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)$
⇒ $\lim_{h \to 0} \left( \frac{\left( \sqrt{1 - ph} - \sqrt{1 + ph} \right)\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)$
⇒ $\lim_{h \to 0} \left( \frac{\left( 1 - ph - 1 - ph \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)$
⇒ $\lim_{h \to 0} \left( \frac{\left( - 2ph \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)$
⇒ $\lim_{h \to 0} \left( \frac{\left( 2p \right)}{\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)$
⇒ 2p/2 = -1/2
⇒ p = -1/2.
(vii) $f\left( x \right) = \begin{cases}5 , & \text{ if } & x \leq 2 \\ ax + b, & \text{ if } & 2 < x < 10 \\ 21 , & \text{ if } & x \geq 10\end{cases}$
Solution:
If f(x) is continuous at x = 2 and x = 10, then
lim_{{x -> 2-}} f(x) = lim_{{x -> 2+}} f(x) and lim_{{x -> {10}^–}} f(x) = lim_{{x -> {10}+}} f(x)
⇒ lim_{{h -> 0}} f(2 – h) = lim_{{h -> 0}} f(2 + h) and lim_{{h -> 0}} f(10 – h) = lim_{{h -> 0}} f(10 + h)
⇒ lim_{{h -> 0}} (5) = lim_{{h -> 0}} (a (2 + h) + b)
And lim_{{h -> 0}} a (10 – h) + b = lim_{{h -> 0}} (21)
On solving equations, we get,
⇒ a = 2 and b = 1.
(viii) $f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x} , & x < \frac{\pi}{2} \\ 3 , & x = \frac{\pi}{2} \\ \frac{3 \tan 2x}{2x - \pi}, & x > \frac{\pi}{2}\end{cases}” height=”113″ width=”287″> π/2-}} f(x) = f(π/2)
⇒ lim_{{h -> 0}} f(π/2 – h) = f(π/2)
⇒ lim_{{h -> 0}} f(π/2 – h) = 3
⇒ $\lim_{h \to 0} \left[ \frac{k \cos \left( \frac{\pi}{2} - h \right)}{\pi - 2\left( \frac{\pi}{2} - h \right)} \right] = 3$
⇒ $\lim_{h \to 0} \left[ \frac{k \sin h}{\pi - \pi + 2h} \right] = 3$
⇒ lim_{{h -> 0}} (k sin h/2h) = 3
⇒ k/2 lim_{{h -> 0}}sin h/h =3
⇒ k/2 = 3
⇒ k = 6
Question 5. The function $f\left( x \right) = \begin{cases}x^2 /a , & \text{ if } 0 \leq x < 1 \\ a , & \text{ if } 1 \leq x < \sqrt{2} \\ \frac{2 b^2 - 4b}{x^2}, & \text{ if } \sqrt{2} \leq x < \infty\end{cases}$ is continuous on (0, ∞), then find the most suitable values of a and b.
Solution:
Given that f is continuous on ( 0, ∞ ).
So, f is continuous at x = 1 and x = √2.
At x = 1, we have lim_{{x -> 1-}} f(x)
= lim_{{h -> 0}}f(1 – h)
= lim_{{h -> 0}} [(1 – h)²/a]
= 1/a
At x = √2, we have
lim_{{x -> √2-}} f(x) = lim_{{h -> 0}} f(√2 + h)
= lim_{{h -> 0}} (a)
= a
$\lim_{x \to \sqrt{2}^+} f\left( x \right) = \lim_{h \to 0} f\left( \sqrt{2} + h \right) = \lim_{h \to 0} \left[ \frac{2 b^2 - 4b}{\left( \sqrt{2} + h \right)^2} \right] = \frac{2 b^2 - 4b}{2} = b^2 - 2b$
f is continuous at x = 1 and √2.
⇒ 1/a = a and b² – 2b = a
⇒ a²= 1 and b² – 2b = a
⇒ a = ±1 and b² – 2b = a . . . (1)
If a = 1, then b² – 2b = 1
⇒ b² – 2b – 1 = 0
⇒ b = $\frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2}$ = 1 ± √2
If a = −1, then b² – 2b = – 1
⇒ b² – 2b + 1 = 0
⇒ b = 1
Therefore, a = −1, b = 1 or a = 1, b = √2.are the most suitable values of a and b.
Question 6. Find the values of a and b so that the function f(x) defined by $f\left( x \right) = \begin{cases}x + a\sqrt{2}\sin x , & \text{ if }0 \leq x < \pi/4 \\ 2x \cot x + b , & \text{ if } \pi/4 \leq x < \pi/2 \\ a \cos 2x - b \sin x, & \text{ if } \pi/2 \leq x \leq \pi\end{cases}$ becomes continuous on [0, π].
Solution:
f is continuous at x = π.
At x = π/4, we have
lim_{{x -> π/4-}} f(x) = lim_{{h -> 0}} f(π/4 – h)
= lim{h -> 0} [(π/4 – h) + √2a sin (π/4 – h)]
= π/4 + √2a sin π/4
= π/4 + a
= lim_{{h -> 0}} [2 (π/4 + h) cot (π/4 + h) + b]
= [2 π/4 cot π/4 + b]
= π/2 + b
⇒ – b – a = b and π/4 + a = π/2 + b
⇒ a = π/6 and b = -π/12
Question 7. The function f(x) is defined as follows: $[f\left( x \right) = \begin{cases}x^2 + ax + b , & 0 \leq x < 2 \\ 3x + 2 , & 2 \leq x \leq 4 \\ 2ax + 5b , & 4 < x \leq 8\end{cases}$ . If f is continuous on [0, 8], find the values of a and b.
Solution:
Given that f is continuous on [0, 8].
So, f is continuous at x = 2 and x = 4
At x = 2,
lim_{{x -> 2-}} f(x) = lim_{{h -> 0}} f(2 – h)
= lim_{{h -> 0}} (2 – h)² + a(2 – h) + b
= 4 + 2a + b
lim_{{x -> 2+}}f(x) = lim_{{h -> 0}} f(2 + h)
= lim_{{h -> 0}} [3(2 + h) + h]
= 8
At x = 4,
lim_{{x -> 4-}} f(x) = lim_{{h -> 0}} f(4 – h)
= lim_{h -> 0} [3(4 – h) + 2]
= 14
lim_{{x -> 4+}} f(x) = lim_{{h -> 0}} f(4 + h)
= lim_{{h -> 0}} [2a(4 + h) + 5b]
= 8a + 5b
So, f is continuous at x = 2 and x = 4.
lim_{{x -> 2-}} f(x) = lim_{{x -> 2+}} f(x)
and, lim_{{x -> 4-}} f(x) = lim_{{x -> 4+}} f(x)
⇒ 4 + 2a + b = 8 and 8a + 5b = 14
⇒ 2a + b = 4 and 8a + 5b = 14
On solving, we get
a = 3 and b = -2
Question 8. If $f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}$ for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].
Solution:
If x ≠ π/4, tan (π/4 – x) and cot2x are continuous in [0, π/2]. Then the function $\frac{\tan \left( \frac{\pi}{4} - x \right)}{\cot 2x}$ is continuous for each x ≠ π/4.
Now, let us assume that the point x = π/4.
We have,
(LHL at x = π/4) = lim_{{x -> π/4-}}f(x)
= lim_{{h -> 0}} f(π/4 – h)
= $\lim_{h \to 0} \left( \frac{\tan\left( \frac{\pi}{4} - \frac{\pi}{4} + h \right)}{\cot\left( \frac{\pi}{2} - 2h \right)} \right)$
= lim_{{h -> 0}} tan h/tan 2h
= $\lim_{h \to 0} \left( \frac{\frac{\tan \left( h \right)}{h}}{\frac{2 \tan \left( 2h \right)}{2h}} \right)$
= $\frac{1}{2}\left( \frac{\lim_{h \to 0} \frac{\tan \left( h \right)}{h}}{\lim_{h \to 0} \frac{\tan \left( 2h \right)}{2h}} \right)$
= 1/2
(RHL at x = π/4) = lim_{{x -> π/4+}} f(x)
= lim_{{h -> 0}} f(π/4 + h)
= $\lim_{h \to 0} \left( \frac{\tan \left( \frac{\pi}{4} - \frac{\pi}{4} - h \right)}{\cot \left( \frac{\pi}{2} + 2h \right)} \right)$
= lim{h -> 0} tan (-h)/tan (-2h)
= lim_{{h -> 0}}tan h/tan 2h
= $\lim_{h \to 0} \left( \frac{\frac{\tan \left( h \right)}{h}}{\frac{2 \tan \left( 2h \right)}{2h}} \right)$
= $\frac{1}{2}\left( \frac{\lim_{h \to 0} \frac{\tan \left( h \right)}{h}}{\lim_{h \to 0} \frac{\tan \left( 2h \right)}{2h}} \right)$
= 1/2
If f(x) is continuous at x = π/4 then
lim_{{x -> π/4-}} f(x) = lim_{{x -> π/4+}}f(x) = f(π/4)
∴ f(π/4) = 1/2
Hence, the function will be everywhere continuous.
Question 9. Discuss the continuity of the function $f\left( x \right) = \begin{cases}2x - 1 , & \text { if } x < 2 \\ \frac{3x}{2} , & \text{ if } x \geq 2\end{cases}$ .
Solution:
When x < 2, f(x) being a polynomial function is continuous.
When x > 2, f(x) being a polynomial and continuous function is continuous.
At x = 2, we have:
(LHL at x = 2) = lim_{{x -> 2-}} f(x)
= lim_{{h -> 0}}f(2 – h)
= lim_{{h -> 0}}(2(2 – h) – 1)
= 4 – 1
= 3
(RHL at x = 2) = lim_{{x -> 2+}} f(x)
= lim_{{h -> 0}} f(2 + h)
= lim_{{h -> 0}} 3 (h + 2)/2
= 3
Also, f(2) = 3(2)/2 = 3
∴ $\lim_{x \to 2^-} f\left( x \right) = \lim_{x \to 2^+} f\left( x \right) = f\left( 2 \right)$
So, f(x) is continuous at x = 2.
Question 10. Discuss the continuity of f(x) = sin |x|.
Solution:
f is clearly the composition of two functions, f = h o g, where g (x) = |x| and h (x) = sin x
Since, hog(x) = h(g(x)) = h(|x|) = \sin|x|
g(x)=|x| being a modulus function must be continuous for all real numbers.
Let us assume that a be a real number.
Case 1:
If a > 0 then g(a) = a
lim_{{x -> c}}(g(x)) = lim_{{x -> c}}(x) = a
So, lim_{{x -> c}}(g(x)) = g(a)
So, g is the continuous on all the points, i.e., x > 0
Case 2:
If a < 0 then g(a) = -a
lim_{{x -> c}}(g(x)) = lim_{{x -> c}}(-x) = -a
So, lim_{{x -> c}}(g(x)) = g(a)
So, g is the continuous on all the points x < 0
Case 3:
If a = 0 then g(a) = g(0) = 0
lim_{{x -> 0^–}}(g(x)) = lim_{{x -> 0^–}}(-x) = 0
lim_{{x -> 0⁺}}(g(x)) = lim_{{x -> 0⁺}}(x) = 0
So, lim_{{x -> 0^–}}(g(x)) = lim_{{x -> 0⁺}}(g(x)) = g(0)
So, g is continus at point x = 0
So, lim_{{x -> c}}(g(x)) = g(a)
So we conclude that h(x) = sinx is defined for every real number.
Let us considered b be the real number. Now put x = b + k
If x->b , then k ->0
So, h(b) = sin b
lim_{{x -> b}}(h(x)) = lim_{{x -> b}}sin x
= lim_{{k -> 0}}sin (b + k)
= lim_{{k -> 0}}(sinb cos k + cos b sink)
= lim_{{k -> 0}}(sinb cos k) + lim_{{k -> 0}}(cos b sink)
= sinb cos 0 + cos b sin 0
= sin b + 0
= sin b
Hence, lim_{{x -> c}}h(x) = g(c)
So, h is continuous function
Hence, f(x) = hog(x) = h(g(x)) = h(|x|) = sin|x|
Question 11. Prove that $f\left( x \right) = \begin{cases}\frac{\sin x}{x} , & x < 0 \\ x + 1 , & x \geq 0\end{cases}$ is everywhere continuous.
Solution:
When x < 0, sin x/x being the composite of two continuous functions is continuous.
When x > 0, we have f(x) being a polynomial function is continuous.
At x = 0:
(LHL at x = 0) = lim_{{x -> 0^–}} f(x)
= lim_{{h -> 0}} f(0 – h)
= lim_{{h -> 0}} f(-h)
= lim_{{h -> 0}} (sin (-h)/(-h))
= lim_{{h -> 0}} (sin h/h)
= 1
(RHL at x = 0) = lim_{{x -> 0⁺}} f(x)
= lim_{{h -> 0}} f(0 + h)
= lim_{{h -> 0}}f(h)
= lim_{{h -> 0}} (h + 1)
= 1
Also, f(x) = 0 + 1 = 1
So we conclude that lim_{{x -> 0^–}} f(x) = lim_{{x -> 0⁺}} f(x) = f(0)
So, f(x) is everywhere continuous.
Question 12. Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.
Solution:
g is defined at all integral points. Let n be an integer. Then,
g(n) = n − [n]
= n − n
= 0
At x = n, we have:
LHL = lim_{{x -> n^–}}g(x) = lim_{x->n^–}(x – [x])
= lim_{x->n^–}(x) – lim_{_{x->n^–}}[x]
= n − (n − 1)
= 1
RHL = lim_{x->n⁺}g(x) = lim_{x->n⁺}(x – [x])
= lim_{{x-> n⁺}}(x) – lim_{x->n⁺}[x]
= n − n
= 0
So we conclude that lim_{{x -> 0^–}} f( x ) ≠ lim_{{x -> 0⁺}}f(x)
So, g is discontinuous at all integral points.
Question 13. Discuss the continuity of the following functions:
(i) f(x) = sin x + cos x
(ii) f(x) = sin x − cos x
(iii) f(x) = sin x cos x
Solution:
We know that if g and h are two continuous functions, then g + h, g − h and g o h are also continuous.
Let g (x) = sin x, defined for every real number.
Let a be a real number. Put x = a + h
If x → a, then h → 0 g(a)=sin a.
lim_{x->a}g(x) = lim_{x->a} sina
= lim_{h->0} sin (a+h)
= lim_{h->0}[sin a cos h + cos a sin h]
= lim_{h->0}(sin a cos h )+lim_{h->0}(cos a sin h)
= sin a cos 0 + cos a sin 0
= sin (a + 0)
= sin a
∴ lim_{x->c}g(x) = g(c)
Similarly, cos x can be proved as a continuous function.
So we conclude that
(i) f (x) = g (x) + h (x) = sin x + cos x is a continuous function.
(ii) f (x) = g (x) − h (x) = sin x − cos x is a continuous function.
(iii) f (x) = g (x) h (x) = sin x cos x is a continuous function.
Question 14. Show that f (x) = cos x² is a continuous function.
Solution:
f can be written as the composition of two functions as f = g o h, where g (x) = cos x and h (x) = x²
∵ (g o h)(x) = g(h (x)) = g(x²) = cos(x²) = f(x)
Let c be a real number.
Then, g(c) = cos c
If x-> c , then h->0 and lim_{x->c} g(x)
= lim_{x->c}cos c
= cos c
∴ lim_{x->c}g(x) = g(c)
So, g(x) = cos x is a continuous function.
Now, h(x) = x²
Let k be a real number, then h(k) = k²
lim_x->h(x) = lim_{x->k} x²= k²
∴ lim_{x->k}h(x) = h(k)
So, h is a continuous function.
So, f(x) being a composite of two continuous functions is a continuous function.
Question 15. Show that f (x) = |cos x| is a continuous function.
Solution:
f is the composition of two functions as, f = g o h, where g(x) = |x| and h(x) = cos x
(g o h)(x) = g(h(x)) = g(cos x) = |cos x| = f(x)
Clearly, g(x) being a modulus function would be continuous at all points.
Now, h (x) = cos x. We know that h (x) = cos x is defined for every real number.
Let c be a real number.
Put x = c + h.
If x → c, then h → 0.
⇒ h(c) = cos c
So, h (x) = cos x is a continuous function.
Therefore, f(x) being a composite of two continuous functions is a continuous function.
Question 16. Find all the points of discontinuity of f defined by f (x) = |x| − |x + 1|.
Solution:
f is the composition of two functions as f(x) = g(x) – h(x), where g(x) = |x| and h(x) = |x + 1|.
Let c be a real number.
Case I: If c < 0 , then g(c) = -c and lim_{x->c}g(x) = lim_{x->c} = -c
∴ lim_{x->c}g(x) = g(c)
So, g is continuous at all points x < 0.
Case II: If c < 0 , then g(c) = -c and lim_(x->c)g(x) = lim_(x->c)(-x) = -c
∴ lim_(x->c)g(x) = g(c)
So, g is continuous at all points x > 0.
Case III: if c = 0 , then g (c) = g(0) = 0
lim_{x->0-}g(x) = lim_{x->0-}(- x) = 0
lim_{x->0+}g(x) = lim_{x->0+}(x) = 0
∴ lim_{x->0+}g(x) = lim_{x->0+}(x) = g(0)
So, g is continuous everywhere.
Clearly, h is defined for every real number. Let c be a real number.
Case I: if c < – 1, then h (c) = – (c + 1)
lim_{x->c}h (x) = lim_{x->c}[-(x + 1)]
= -(c + 1)
∴ lim_{{x-> c}}h (x) = h(c)
Therefore, f being a composite of two continuous functions is a continuous function.
Question 17. Determine if $f\left( x \right) = \begin{cases}x^2 \sin\frac{1}{x} , & x \neq 0 \\ 0 , & x = 0\end{cases}$ is a continuous function?
Solution:
Let us assume that c be a real number.
Case I: If c ≠ 0 , then f(c)= c² sin (1/c)
lim_{x->c}f(x) = lim_{x->c}(x² sin 1/x)
= (lim_{x->c}x²) (lim_{x->c}sin 1/x)
= c² sin (1/c)
lim_{x->c}f(x) = f(c)
So, f is continuous at all points such that x ≠ 0
Case II: If c = 0 then f(0) = 0
lim_{{x -> 0^–}} f(x) = lim_{{x -> 0^–}} (x² sin 1/x) = lim_{{x -> 0}} (x² sin 1/x)
So, -1 ≤ sin 1/x ≤ 1, x ≠ 0
-x² ≤ x²sin 1/x ≤ x²
lim_{{x -> 0}} (-x²) ≤ lim_{{x -> 0}} (x²sin 1/x) ≤ lim_{{x -> 0}} x²
0 ≤ lim_{{x -> 0}} (x²sin 1/x) ≤ 0
lim_{{x -> 0}} (x²sin 1/x) = 0
lim_{{x -> 0^–}} f(x) = 0
Similarly, lim_{{x -> 0⁺}} f(x) = lim_{{x -> 0⁺}} (x² sin 1/x) = lim_{{x -> 0}} (x² sin 1/x) = 0
Thus, f is a continuous function.
Question 18. Given the function $f\left( x \right) = \frac{1}{x + 2}+2$ . Find the points of discontinuity of the function f(f(x)).
Solution:
Here, $f[f(x)]=\frac{1}{\frac{1}{x+2}+2}=\frac{x+2}{2x+5}$
We observe that f(f( x )) is not defined at x + 2 = 0 and 2x + 5 = 0.
If x + 2 = 0, then x = – 2 and if 2x + 5 = 0, then x = -5/2
Hence, the function is discontinuous at x = -5/2 and – 2.
Question 19. Find all point of discontinuity of the function f(t) = $\frac{1}{t^2 + t - 2}$ , where t = 1/(x – 1).
Solution:
Here, $f\left( t \right) = \frac{1}{t^2 + t - 2}$
Now, let u = 1/(x – 1)
Therefore f( u ) = $\frac{1}{u^2 + 2u - u - 2}$
= $\frac{1}{\left( u + 2 \right)\left( u - 1 \right)}$
So, f (u ) is not defined at u = -2 and u = 1.
If u = – 2, then -2 = 1/(x – 1)
⇒ 2x = 1
⇒ x = 1/2
If u = 1, then 1 = 1/(x – 1)
⇒ x = 2
Hence, the function is discontinuous at x = 1/2 , 2.
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RD Sharma Class 12 Ex 9.2 Solutions Chapter 9 Continuity

Question 1. Prove that the function is continuous everywhere.

Question 2. Discuss the continuity of the function .

Question 3. Find the points of discontinuity, if any, of the following functions:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

Question 4. In the following, determine the value of constant involved in the definition so that the given function is continuous:

(i)

(ii)

(iii)

(v)

(vii)

(viii)

Question 5. The function is continuous on (0, ∞), then find the most suitable values of a and b.

Question 6. Find the values of a and b so that the function f(x) defined by becomes continuous on [0, π].

Question 7. The function f(x) is defined as follows: . If f is continuous on [0, 8], find the values of a and b.

Question 8. If for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].

Question 9. Discuss the continuity of the function .

Question 10. Discuss the continuity of f(x) = sin |x|.

Question 11. Prove that is everywhere continuous.

Question 12. Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.

Question 13. Discuss the continuity of the following functions:

(i) f(x) = sin x + cos x

(ii) f(x) = sin x − cos x

(iii) f(x) = sin x cos x

Question 14. Show that f (x) = cos x2 is a continuous function.

Question 15. Show that f (x) = |cos x| is a continuous function.

Question 16. Find all the points of discontinuity of f defined by f (x) = |x| − |x + 1|.

Question 17. Determine if is a continuous function?

Question 18. Given the function . Find the points of discontinuity of the function f(f(x)).

Question 19. Find all point of discontinuity of the function f(t) = , where t = 1/(x – 1).

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Question 1. Prove that the function $f(x)=\begin{cases}\frac{sinx}{x} \ \ \ \ ,x\le0\\x+1\ \ \ ,x\le0\end{cases}$ is continuous everywhere.

Question 2. Discuss the continuity of the function $f(x)=\begin{cases}\frac{x}{\left| x \right|}, & x \neq 0 \\ 0 , & x = 0\end{cases}$ .

(i) $f(x)=\begin{cases}x^3 - x^2 + 2x - 2, & \text{ if }x \neq 1 \\ 4 , & \text{ if } x = 1\end{cases}$

(ii) $f( x ) =\begin{cases}\frac{x^4 - 16}{x - 2}, & \text{ if } x \neq 2 \\ 16 , & \text{ if } x = 2\end{cases}$

(iii) $f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & \text{ if } x < 0 \\ 2x + 3, & x \geq 0\end{cases}$

(iv) $f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, & \text{ if } x \neq 0 \\ 4 , & \text{ if } x = 0\end{cases}$

(v) $f\left( x \right) = \begin{cases}\frac{\sin x}{x} + \cos x, & \text{ if } x \neq 0 \\ 5 , & \text { if } x = 0\end{cases}$

(vi) $f\left( x \right) = \begin{cases}\frac{x^4 + x^3 + 2 x^2}{\tan^{- 1} x}, & \text{ if } x \neq 0 \\ 10 , & \text{ if } x = 0\end{cases}$

(vii) $f\left( x \right) = \begin{cases}\frac{e^x - 1}{\log_e (1 + 2x)}, & \text{ if }x \neq 0 \\ 7 , & \text{ if } x = 0\end{cases}$

(viii) $f\left( x \right) = \begin{cases}\left| x - 3 \right|, & \text{ if } x \geq 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & \text{ if } x < 1\end{cases}$

(ix) $f\left( x \right) = \begin{cases}\left| x \right| + 3 , & \text{ if } x \leq - 3 \\ - 2x , & \text { if } - 3 < x < 3 \\ 6x + 2 , & \text{ if } x > 3\end{cases}” height=”111″ width=”398″><span class=$

(x) $f\left( x \right) = \begin{cases}x^{10} - 1, & \text{ if } x \leq 1 \\ x^2 , & \text{ if } x > 1\end{cases}” height=”79″ width=”321″><span class=$

(xi) $f\left( x \right) = \begin{cases}2x , & \text{ if } & x < 0 \\ 0 , & \text{ if } & 0 \leq x \leq 1 \\ 4x , & \text{ if } & x > 1\end{cases}” height=”237″ width=”277″><span class=$

(xii) $f\left( x \right) = \begin{cases}\sin x - \cos x , & \text{ if } x \neq 0 \\ - 1 , & \text{ if } x = 0\end{cases}$

(xiii) $f\left( x \right) = \begin{cases}- 2 , & \text{ if }& x \leq - 1 \\ 2x , & \text{ if } & - 1 < x < 1 \\ 2 , & \text{ if } & x \geq 1\end{cases}$

(i) $f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, & \text{ if } x \neq 0 \\ 3k , & \text{ if } x = 0\end{cases}$

(ii) $f\left( x \right) = \begin{cases}kx + 5, & \text{ if } x \leq 2 \\ x - 1, & \text{ if } x > 2\end{cases}” height=”79″ width=”315″><span class=$

(iii) $f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if } x < 0 \\ \cos 2x , & \text{ if } x \geq 0\end{cases}$

(v) $f\left( x \right) = \begin{cases}4 , & \text{ if } x \leq - 1 \\ a x^2 + b, & \text{ if } - 1 < x < 0 \\ \cos x, &\text{ if }x \geq 0\end{cases}$

(vii) $f\left( x \right) = \begin{cases}5 , & \text{ if } & x \leq 2 \\ ax + b, & \text{ if } & 2 < x < 10 \\ 21 , & \text{ if } & x \geq 10\end{cases}$

(viii) $f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x} , & x < \frac{\pi}{2} \\ 3 , & x = \frac{\pi}{2} \\ \frac{3 \tan 2x}{2x - \pi}, & x > \frac{\pi}{2}\end{cases}” height=”113″ width=”287″><span class=$

Question 5. The function $f\left( x \right) = \begin{cases}x^2 /a , & \text{ if } 0 \leq x < 1 \\ a , & \text{ if } 1 \leq x < \sqrt{2} \\ \frac{2 b^2 - 4b}{x^2}, & \text{ if } \sqrt{2} \leq x < \infty\end{cases}$ is continuous on (0, ∞), then find the most suitable values of a and b.

Question 7. The function f(x) is defined as follows: $[f\left( x \right) = \begin{cases}x^2 + ax + b , & 0 \leq x < 2 \\ 3x + 2 , & 2 \leq x \leq 4 \\ 2ax + 5b , & 4 < x \leq 8\end{cases}$ . If f is continuous on [0, 8], find the values of a and b.

Question 8. If $f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}$ for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].

Question 9. Discuss the continuity of the function $f\left( x \right) = \begin{cases}2x - 1 , & \text { if } x < 2 \\ \frac{3x}{2} , & \text{ if } x \geq 2\end{cases}$ .

Question 11. Prove that $f\left( x \right) = \begin{cases}\frac{\sin x}{x} , & x < 0 \\ x + 1 , & x \geq 0\end{cases}$ is everywhere continuous.

Question 14. Show that f (x) = cos x² is a continuous function.

Question 17. Determine if $f\left( x \right) = \begin{cases}x^2 \sin\frac{1}{x} , & x \neq 0 \\ 0 , & x = 0\end{cases}$ is a continuous function?

Question 18. Given the function $f\left( x \right) = \frac{1}{x + 2}+2$ . Find the points of discontinuity of the function f(f(x)).

Question 19. Find all point of discontinuity of the function f(t) = $\frac{1}{t^2 + t - 2}$ , where t = 1/(x – 1).