RD Sharma Solutions Class 12 Mathematics Chapter 8 MCQ

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RD Sharma Solutions Class 12 Mathematics Chapter 8 MCQ

Updated on Jan 27, 2022 01:56 PM IST

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RD Sharma Class 12 Solutions Chapter 8 MCQ Continuity - Other Exercise
Continuity Excercise: MCQ
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter 8 MCQ Continuity - Other Exercise

Continuity Excercise: MCQ

Continuity exercise multiple choice question 1

Answer:
The correct option is (c)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
$f (x) = \frac{4 - x^{2}}{4 x - x^{3}}$
Solution:
Since a function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a} f (x) = f (a) \\ f (x) = \frac{4 - x^{2}}{4 x - x^{3}} \\ f (x) = \frac{4 - x^{2}}{x (4 - x^{2})} \\ f (x) = \frac{1}{x} \\ Where x \neq 0, \pm 2 \end{aligned}

f(x) is continuous for all real numbers except (0,

\pm

2).
Therefore, it is continuous exactly at three points (0,

\pm

2)
So, the correct option is (c)

Continuity exercise multiple choice question 2

Answer:
The correct option is (a) and (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \\ (ii) lim_{h \to 0} {\frac{f (a + h) - f (a)}{h}} = f^{'} (a^{+}) for right hand derivative \\ lim_{h \to 0} {\frac{f (a - h) - f (a)}{- h}} = f^{'} (a^{-}) for left hand derivative \end{aligned}$
Given:
$f (x) = | x - a | ϕ (x)$
Solution:
Using formula (ii)
$\begin{aligned} = lim_{h \to 0} {\frac{f (a + h) - f (a)}{h}} \\ = lim_{h \to 0} {\frac{| h + a - a | ϕ (a + h) - | a - a | ϕ (a)}{h}} \end{aligned}$
$\begin{aligned} = lim_{h \to 0} \frac{h ϕ (a + h)}{h} \\ = lim_{h \to 0} ϕ (a + h) = ϕ (a) = f^{'} (a^{+}) . . . . (i) \end{aligned}$
For Left Hand Derivative
$\begin{aligned} = lim_{h \to 0} {\frac{f (a - h) - f (a)}{- h}} \\ = lim_{h \to 0} {\frac{| a - h - a | ϕ (a - h) - | a - a | ϕ (a)}{h}} \\ = lim_{h \to 0} \frac{| - h | ϕ (a - h)}{- h} \\ = lim_{h \to 0} - ϕ (a - h) \end{aligned}$
Therefore

$\begin{aligned} = lim_{h \to 0} - ϕ (a - h) \\ = - ϕ (a) \\ = f^{'} (a^{-}) . . . . . (i i) \end{aligned}$
From (i) and (ii), we get
$f^{'} (a^{+}) \neq f^{'} (a^{-})$
So, correct option is (a) and (b)

Continuity exercise multiple choice question 3

Answer:
The correct option is (a) and (d)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \\ (ii) lim_{h \to 0} {\frac{f (a + h) - f (a)}{h}} = f^{'} (a^{+}) for right hand derivative \\ lim_{h \to 0} {\frac{f (a - h) - f (a)}{- h}} = f^{'} (a^{-}) for left hand derivative \end{aligned}

Given:
$f (x) = | \log_{10} x |$
Solution:
$f (x) = | \log_{10} x |$

= | \frac{\log x}{\log_{e} 10} | = | \log_{e} x \log_{10} e |

Calculate right hand derivative
Using limit at x = 1
Apply limit at x = 1

\begin{aligned} = lim_{h \to 0} {\frac{f (1 + h) - f (1)}{h}} \\ = lim_{h \to 0} {\frac{| \log_{10} e \log (1 + h) | - | \log_{10} e \log 1 |}{h}} \\ = lim_{h \to 0} \frac{\log_{10} e | \log (1 + h) |}{h} \end{aligned}

Therefore applying L-Hospital rule

\begin{aligned} = \log_{10} e \times 1 \\ = f^{'} (1^{+}) \end{aligned}

For left hand limit

= lim_{h \to 0} {\frac{f (1 - h) - f (1)}{- h}}

\begin{aligned} = lim_{h \to 0} {\frac{| \log_{10} e \log (1 - h) | - | \log_{10} e \log 1 |}{- h}} \\ = lim_{h \to 0} \frac{- \log_{10} e | \log (1 - h) |}{h} \\ - \log_{10} e = f^{'} (1^{-}) \end{aligned}

\begin{aligned} The function f (x) = | \log_{10} x | is continuous and in the right hand derivative \\ f^{'} (1^{-}) = - \log_{10} e \end{aligned}

So the correct option is (a) and (d)

Continuity exercise multiple choice question 4

Answer:
The correct option is (C)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \\ (ii) lim_{x \to a} {\frac{f (x)}{g (x)}} = \frac{1}{m} where lim_{x \to a} f (x) = 1, lim_{x \to a} g (x) = m \\ (iii) lim_{x \to 0} \frac{\sin x}{x} = 1 \end{aligned}

Given:
$f (x) = {\begin{cases} \frac{36^{x} - 9^{x} - 4^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}} \\ k \end{cases}$ $\begin{aligned} , & x \neq 0 \\ , & x = 0 \end{aligned}$
Solution:
Function f(x) is continuous at x = 0
To calculate the value of k, understand that function is continuous at x = 0

\begin{aligned} \Rightarrow lim_{x \to 0} (\frac{36^{x} - 9^{x} - 4^{x} + 1}{\sqrt{2} - \sqrt{1 + \cos x}}) = k \\ \Rightarrow lim_{x \to 0} (\frac{(9^{x} - 1) (4^{x} - 1)}{\sqrt{2} - \sqrt{1 + \cos x}}) = k \\ \Rightarrow lim_{x \to 0} (\frac{(9^{x} - 1) (4^{x} - 1)}{\sqrt{2} [2 \sin^{2} (\frac{x}{4})]}) = k \end{aligned}

\Rightarrow lim_{x \to 0} (\frac{(9^{x} - 1) (4^{x} - 1)}{2 \sqrt{2} x^{2} \cdot \frac{\frac{\sin^{2} x}{4}}{\frac{x^{2}}{16} \times 16}}) = k

\Rightarrow lim_{x \to 0} {\frac{(9^{x} - 1) (4^{x} - 1)}{\frac{\sqrt{2}}{8} x^{2} \times \frac{\sin^{2} x y}{\frac{x^{2}}{16}}}} = k

Using formula (ii) and (iii)

\begin{aligned} \Rightarrow \frac{8 \ln 9 \ln 4}{\sqrt{2}} = k [understand that lim_{x \to 0} (\frac{a^{x} - 1}{a^{x}}) = \ln a] \\ \Rightarrow \frac{32 \ln 3 \ln 2}{\sqrt{2}} = k \\ \Rightarrow k = 16 \sqrt{2} \ln 2 \ln 3 \end{aligned}

So correct option is (C)

Continuity exercise multiple choice question 5

Answer:
The correct option is (d)
Hint:
Use the given formula:

\begin{aligned} (i) If lim_{x \to 0^{+}} f (x) \neq lim_{x \to 0^{-}} f (x) then f (x) is discontinuous at x = 0 . \\ (ii)If lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{-}} f (x) = f (0) then f (x) is continuous at x = 0 \end{aligned}

\begin{aligned} (iii) A function f (x) is said to be continuous at a point x = a of its domain, if \\ lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \end{aligned}

Given:
$f (x) = {\begin{array}{lr} \frac{| x^{2} - x |}{x^{2} - x} \\ 1 \\ - 1 \end{array}$ $\begin{array}{r} , x \neq 0, 1 \\ , x = 0 \\ , x = 1 \end{array}$
Solution:
Simplify the given function
$f (x) = {\begin{array}{lr} \frac{| x^{2} - x |}{x^{2} - x} \\ 1 \\ - 1 \end{array}$ $\begin{array}{r} , x \neq 0, 1 \\ , x = 0 \\ , x = 1 \end{array}$
$\begin{aligned} f (x) = {\begin{aligned} 1, x > 1 \\ 1, x \leq 0 \\ - 1, 0 \leq x \leq 1 \end{aligned} \\ Using RHL lim_{x \to 0^{+}} f (x) = lim_{h \to 0} f (h) = - 1 \\ Using LHL lim_{x \to 0^{-}} f (x) = lim_{h \to 0} f (h) = 1 \end{aligned}$
f(x) is discontinuous at x =0
Again, Using RHL,

\begin{aligned} lim_{x \to 1^{+}} f (x) = lim_{h \to 0} f (1 + h) = 1 \end{aligned}

Using LHL,

\begin{aligned} lim_{x \to 1^{-}} f (x) = lim_{h \to 0} f (1 - h) = - 1 \end{aligned}

f(x) is discontinuous at x = 1
Therefore,
f(x) is continuous for all except at x = 0 and x = 1
So, the correct option is (d)

Continuity exercise multiple choice question 6

Answer:
The correct option is (C)
Hint:
Use the given formula:

(i) lim_{x \to 0} \frac{\log (1 - x)}{x} = 1 and lim_{x \to 0} \frac{\sin x}{x} = 1

(ii) A function f(x) is said to be continuous at a point x=a of its domain, if

\begin{aligned} lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \\ (iii) lim_{x \to a} {f (x) g (x)} = 1 m, where lim_{x \to a} f (x) = 1, lim_{x \to a} g (x) = m \end{aligned}

Given:
$f (x) = {\begin{cases} \frac{1 - \sin x}{(π - 2 x)^{2}} \frac{\log \sin x}{(\log (1 + π^{2} - 4 π x + 4 x^{2}))} \\ k \end{cases}$ $\begin{aligned} , & x \neq \frac{π}{2} \\ , & x = \frac{π}{2} \end{aligned}$
Solution:
Function f(x) is continuous at

x = \frac{π}{2}

lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})

Using substitution method

\begin{aligned} Substitute \frac{π}{2} - x = t \\ \Rightarrow lim_{t \to 0} (\frac{π}{2} - t) = f (\frac{π}{2}) \end{aligned}

\Rightarrow lim_{t \to 0} {\frac{1 - \sin (\frac{π}{2} - t)}{4 t^{2}} \frac{\log \sin (\frac{π}{2} - t)}{\log (1 + π^{2} - 4 π (\frac{π}{2} - t) + 4 {(\frac{π}{2} - t)}^{2}}} = k

\Rightarrow lim_{t \to 0} \frac{(2 \sin^{2} t / 2) \log \cos 1}{4 t^{2} \cdot \log (1 + π^{2} - 2 π^{2} + 4 π t + π^{2} + 4 t^{2} - 4 π t)} = k

\begin{aligned} \Rightarrow lim_{t \to 0} \frac{2 \sin^{2} t / 2 \log \cos t}{4 t^{2} \log (1 + 4 t^{2})} = k \\ \Rightarrow lim_{t \to 0} \frac{2 \sin^{2} t / 2 \log \sqrt{1 - \sin^{2} t}}{4 \times 4 \frac{t^{2}}{4}} \frac{1}{4 t^{2} \cdot \frac{\log (1 + 4 t^{2})}{4 t^{2}}} = k [∵ \cos^{2} x = 1 - \sin^{2} x] \end{aligned}

\Rightarrow lim_{t \to 0} \frac{2}{16} {(\frac{\sin t / 2}{t / 2})}^{2} \cdot \frac{\frac{1}{2} \log (1 - \sin 2 t)}{4 t^{2} \cdot \frac{\log (1 + 4 t^{2})}{4 t^{2}}} = k

\Rightarrow \frac{1}{8} lim_{t \to 0} {(\frac{\sin t / 2}{t / 2})}^{2} \cdot \frac{- \sin^{2} f}{2} \cdot \frac{\log (1 - \sin^{2} t)}{4 t^{2} (\sin^{2} t)} \times \frac{1}{\log (1 + 4 t^{2})} = k

\Rightarrow \frac{- 1}{8} lim_{t \to 0} {(\frac{\sin t / 2}{t / 2})}^{2} \cdot \frac{\sin^{2} f}{8 t^{2}} \cdot \frac{\log (1 - \sin 2 t)}{(- \sin^{2} t - 1)} \times \frac{1}{\frac{\log (1 + 4 + 2)}{4 t^{2}}} = k

\Rightarrow - \frac{1}{8 \times 8} lim_{t \to 0} {(\frac{\sin t / 2}{t / 2})}^{2} \cdot lim_{t \to 0} {(\frac{\sin t}{t})}^{2} \cdot lim_{t \to 0} \frac{\log (1 - \sin^{2} t)}{- \sin 2 t} \cdot lim_{d \to 0} \frac{1}{\frac{\log (1 + 4 t^{2})}{4 t^{2}}} = k

Using formula (iii)

\Rightarrow \frac{1}{64} lim_{t \to 0} {\frac{\sin^{2} \frac{t}{2} \frac{\log (1 - \sin^{2} t)}{8 t^{2}}}{{(\frac{t}{2})}^{2}} \frac{\log (1 + 4 t^{2})}{4 t^{2}}} = k

\Rightarrow - \frac{1}{64} lim_{t \to 0} {(\frac{\sin t}{t})}^{2} lim_{t \to 0} \log \frac{1 - \sin^{2} t}{- \sin^{2} t} = k

[∵ lim_{x \to 0} \frac{\sin x}{x} = 4, lim_{x \to 0} \frac{\log (1 + x)}{x} = 1]

Using standard limit formula (i)

\Rightarrow k = - \frac{1}{64}

So, option (c) is correct.

Continuity exercise multiple choice question 7

Answer:
The correct option is (C)
Hint:
Use the given formula:

(i) lim_{x \to 0} \frac{\log (1 - x)}{x} = 1 and lim_{x \to 0} \frac{\tan x}{x} = 1

(ii) A function f(x) is said to be continuous at a point x = a of its domain

\begin{aligned} lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \\ (iii) lim_{x \to a} {f (x) g (x)} = 1 m, where lim_{x \to a} f (x) = 1, lim_{x \to a} g (x) = m \end{aligned}

Given:
$f (x) = (x + 1)^{\cot x}$
Solution:

f (x) = (x + 1)^{\cot x}

taking log on both sides

\log f (x) = (\cot x) (\log (x + 1))

lim_{x \to 0} \log f (x) = lim_{x \to 0} (\frac{\frac{\log (x + 1)}{x}}{\frac{\tan x}{x}})

Using formula (iii)

lim_{x \to 0} \log f (x) = \frac{lim_{x \to 0} \frac{\log (x + 1)}{x}}{lim_{x \to 0} \frac{\tan x}{x}}

Using standard limit formula (i)

\begin{aligned} lim_{x \to 0} f (x) = e \\ f (0) = e \end{aligned}

So, the correct option is (C)

Continuity exercise multiple choice question 8

Answer:
The correct option is (b)
Hint:
Use the given formula:
$(i) Standard limit lim_{x \to 0} \frac{\log (1 + x)}{x} = 1$
(ii) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \\ (iii) lim_{x \to a} {f (x) \pm g (x)} = 1 \pm m, where lim_{x \to a} f (x) = 1, lim_{x \to a} g (x) = m \end{aligned}

Given:
$f (x) = {\begin{cases} \frac{\log (1 + a x) - \log (1 - b x)}{x} \\ k \end{cases}$ $\begin{aligned} , x & \neq 0 \\ , x & = 0 \end{aligned}$
And f(x) is continuous at x = 0
Solution:

∵

f(x) is continuous at x = 0

\Rightarrow lim_{x \to 0} \frac{\log (1 + a x) - \log (1 - b x)}{x} = k

Using formula (ii)

\Rightarrow a lim_{x \to 0} \frac{\log (1 + a x)}{a x} + b lim_{x \to 0} \frac{\log (1 - b x)}{- b x} = k

Using formula (i)

\Rightarrow a + b = k

So, the correct option is (b)

Continuity exercise multiple choice question 9

Answer:
The correct option is (b)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
$f (x) = {\begin{cases} \frac{e^{1 / x} - 1}{e^{1 / x} + 1} \\ 0 \end{cases}$ $\begin{aligned} , x & \neq 0 \\ , x & = 0 \end{aligned}$
Solution:
Using substitution method,

Let e^{1 / x} = t so, x \to 0, t \to \infty

\begin{aligned} lim_{t \to \infty} f (x) = lim_{t \to \infty} (\frac{t - 1}{t + 1}) \\ = \frac{1 - 0}{1 + 0} = 1 [L - Hospital rule] \end{aligned}

And,

\begin{aligned} f (0) = 0 \\ Therefore, lim_{x \to 0} f (x) \neq f (0) \end{aligned}

Hence, f(x) is discontinuous at x = 0
So, option (b) is correct.

Continuity exercise multiple choice question 10

Answer:
The correct option is (d)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:

f (x) = {\begin{cases} \frac{x - 4}{| x - 4 |} + a \\ a + b \\ \frac{x - 4}{| x - 4 |} + b \end{cases}

\begin{aligned} , & x < 4 \\ , & x = 4 \\ , & x > 4 \end{aligned}

and f(x) is continuous at x = 4
Solution:
Using RHL at x = 4

\begin{aligned} = lim_{x \to 4^{+}} f (x) \\ = lim_{h \to 0} f (4 + h) \end{aligned}

\begin{aligned} = lim_{h \to 0} (\frac{4 + h - 4}{| 4 + h - 4 |} + b) \\ = lim_{h \to 0} (\frac{h}{| h |} + b) \\ = 1 + b \end{aligned}

Using LHL

\begin{aligned} lim_{x \to 4^{-}} f (x) = lim_{h \to 0} f (4 - h) \\ = lim_{h \to 0} (\frac{4 - h - 4}{| 4 - h - 4 |} + a) \\ = lim_{h \to \infty} (\frac{- h}{| - h |} + a) \\ = a - 1 \end{aligned}

f(x) will be continuous at x = 4, if

\begin{aligned} lim_{x \to 4^{+}} f (x) = lim_{x \to 4^{-}} f (x) = f (4) \\ a - 1 = b + 1 = a + b \\ b = - 1 and a = 1 \end{aligned}

So, the correct option is (d)

Continuity exercise multiple choice question 11

Answer:
The correct option is (b)
Hint:
Use the given formula:
$\begin{aligned} (i) lim_{x \to 0} f (x) g (x) = e lim_{x \to 0} (f (x) - 1) \cdot g (x) \\ Where lim_{x \to 0} f (x) = 1 \end{aligned}$
$\begin{aligned} And lim_{x \to 0} g (x) = 0 \\ (ii) lim_{x \to 0} {\frac{\cos x - 1}{x}} = - 1 \end{aligned}$
(iii) A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:

f (x) = {\begin{cases} (\cos x)^{\frac{1}{x}} \\ k \end{cases}

\begin{aligned} , x & \neq 0 \\ , x & = 0 \end{aligned}

And f(x) is continuous at x = 0
Solution:

lim_{x \to 0} f (x) = f (0)

Calculate the value of k

lim_{x \to 0} (\cos x)^{\frac{1}{x}} = k

Using formula (i)

lim_{x \to 0} f (x)^{g (x)} = e lim_{x \to 0} {\frac{\cos x - 1}{x}} = k

Apply formula (ii)

\begin{aligned} e^{0} = k \\ k = 1 \end{aligned}

So, option (b) is correct.

Continuity exercise multiple choice question 12

Answer:
The correct option is (a)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
$f (x) = | x | + | x - 1 |$
Solution:
Both are polynomial function and we know that the polynomial functions are continuous everywhere,
So, according to option,
f(x) is continuous at x = 0, as well as at x = 1
So, option is (a) is correct.

Continuity exercise multiple choice question 13

Answer:
The correct option is (d)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
$f (x) = {\begin{cases} \frac{x^{4} - 5 x^{2} + 4}{| (x - 1) (x - 2) |} \\ 6 \\ 12 \end{cases}$ $\begin{aligned} , & x \neq 1, 2 \\ , & x = 1 \\ , & x = 2 \end{aligned}$
Solution:
we write

\begin{aligned} x^{4} - 5 x^{2} + 4 & = x^{4} - x^{2} - 4 x^{2} + 4 = x^{2} (x^{2} - 1) - 4 (x^{2} - 1) \\ = (x^{2} - 4) (x^{2} - 1) \\ = (x + 2) (x - 2) (x + 1) (x - 1) \end{aligned}

∴

we have

f (x) = {\begin{matrix} \frac{(x + 2) (x - 2) (x + 1) (x - 1)}{| (x - 1) (x - 2) |} \\ 6 \\ 12 \end{matrix}

$\begin{aligned} , & x \neq 1, 2 \\ , & x = 1 \\ , & x = 2 \end{aligned}$

f (x) = {\begin{matrix} \frac{x^{4} - 5 x^{2} + 4}{(x - 1) (x - 2)} \\ 6 \\ 12 \end{matrix}

$\begin{aligned} , & x \neq 1, 2 \\ , & x = 1 \\ , & x = 2 \end{aligned}$
$= {\begin{matrix} \int (x + 1) (x + 2), x < 1 \\ - (x + 1) (x + 2), 1 < x < 2 \\ (x + 1) (x + 2), x > 2 \\ 6, x = 1 \\ 12, x = 2 \end{matrix}$
Using RHL at x = 1

\begin{aligned} lim_{x \to 1^{+}} f (x) = lim_{h \to 0} f (1 + h) \\ = - (2) \cdot (3) = - 6 \end{aligned}

Using LHL at x = 1

\begin{aligned} lim_{x \to 1^{-}} f (x) = lim_{h \to 0} f (1 - h) \\ = (2) (3) = 6 \end{aligned}

And using RHL at x = 2

lim_{x \to 2^{+}} f (x) = lim_{h \to 0} f (1 + h) = 12

Using LHL at x = 2

lim_{x \to 2^{-}} f (x) = lim_{h \to 0} f (1 - h) = - 12

f(x) is discontinuous at x=1 and x=2 for f(x) continuous at R[1, 2]
So, the correct option is (d)

Continuity exercise multiple choice question 14

Answer:
The correct option is (c)
Hint:

Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \\ (ii) lim_{x \to 0} \frac{\sin x}{x} = 1 \end{aligned}$
Given:
f(x) is continuous at x = 0 and
$f (x) = {\begin{array}{cc} \frac{\sin (a + 1) x + \sin x}{x} & , x < 0 \\ C & , x = 0 \\ \frac{\sqrt{x + b x^{2}} - \sqrt{x}}{b x \sqrt{x}} & , x > 0 \end{array}$
Solution:
Using RHL
$\begin{matrix} lim_{x \to 0^{+}} f (x) = lim_{h \to 0} f (0 + h) \\ = lim_{h \to 0} f (h) \\ = lim_{h \to 0} \frac{\sqrt{h + b h^{2}} - \sqrt{h}}{b h \sqrt{h}} \end{matrix}$
$\begin{aligned} = lim_{h \to 0} \frac{\sqrt{h} (\sqrt{1 + b h}) - \sqrt{n}}{b h \sqrt{h}} \\ = lim_{h \to 0} \frac{\sqrt{1 + b n} - 1}{b h} \\ = lim_{h \to 0} \frac{\sqrt{1 + b h} - 1}{b h} \times \frac{\sqrt{1 + b h} + 1}{\sqrt{1 + b h} + 1} \end{aligned}$
$\begin{aligned} = lim_{h \to 0} \frac{(1 + b h) - 1}{b h (\sqrt{1 + b h} + 1)} \\ = lim_{h \to 0} \frac{b h}{b h (\sqrt{1 + b h} + 1)} \\ = lim_{h \to 0} \frac{1}{\sqrt{1 + b h} + 1} \end{aligned}$
$\begin{aligned} = \frac{1}{\sqrt{1 + 6 \cdot 0} + 1} \\ = \frac{1}{1 + 1} \\ = \frac{1}{2} \end{aligned}$
Using L.H.L
$\begin{aligned} lim_{x \to 0^{-}} f (x) & = lim_{h \to 0} f (0 - h) \\ = lim_{h \to 0} f (- h) \\ = lim_{n \to 0} \frac{\sin (a + 1) (- h) + \sin (- h)}{(- h)} \end{aligned}$
$\begin{aligned} = lim_{h \to 0} \frac{- \sin (a + 1) \cdot h - \sinh}{- h} \\ = lim_{h \to 0} {- \frac{\sin (a + 1) \cdot h}{- h} + \frac{\sin h}{h}} [∵ \sin (- θ) = - \sin θ] \end{aligned}$
$\begin{aligned} = lim_{h \to 0} {\frac{\sin (a + 1)}{(a + 1)} \times (a + 1) + \frac{\sinh}{h}} \\ = (a + 1) lim_{h \to 0} \frac{\sin (a + 1)}{a + 1} + lim_{h \to 0} \frac{\sin h}{n} \end{aligned}$
$\begin{aligned} = (a + 1) \cdot 1 + 1 [: lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = a + 1 + 1 \\ = a + 2 \end{aligned}$
Since function f(x) is continuous at x = 0
$\begin{aligned} lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x) \\ = f (0) \\ \Rightarrow a + 2 = \frac{1}{2} \\ \Rightarrow a + 2 = \frac{1}{2} \\ or c = \frac{1}{2} \end{aligned}$
$\begin{aligned} \begin{aligned} \Rightarrow a & = \frac{1}{2} - 2 \\ = - \frac{3}{2} or \\ c = \frac{1}{2} \end{aligned} \\ and from f (x), \\ b \in R - {0} \end{aligned}$
So, the correct option is (c).

Continuity exercise 5 multiple choice question 15

Answer:
The correct option is (c)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
f(x) is continuous at x = 0 and

f (x) = {\begin{cases} m x + 1, x \leq \frac{π}{2} \\ \sin x + n, x > \frac{π}{2} \end{cases}

Solution:
Using RHL

\begin{aligned} lim_{x \to \frac{π}{2}} f (x) = lim_{h \to 0} f (\frac{π}{2} + h) \\ = lim_{h \to 0} [\sin (\frac{π}{2} + h) + n] \\ = lim_{h \to 0} (\cos h + n) = 1 + n \end{aligned}

Using LHL

$\begin{aligned} lim_{x \to \frac{π}{2}} f (x) = lim_{h \to 0} f (\frac{π}{2} - h) \\ = lim_{h \to 0} m (\frac{π}{2} - h) + 1 = \frac{m π}{2} + 1 \end{aligned}$

$\begin{aligned} Function f (x) is continuous at x = \frac{π}{2} \\ lim_{x \to \frac{π^{+}}{2}} f (x) = lim_{x \to \frac{π}{2}} f (x) \\ \frac{m π}{2} + 1 = n + 1 \\ \frac{m π}{2} = n \end{aligned}$
So, option (c) is correct

Continuity exercise multiple choice question 16

Answer:
The correct option is (c)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
$f (x) = \frac{\sqrt{a^{2} - a x + x^{2}} - \sqrt{a^{2} + a x + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}}$
Solution:
Using rationalization method,

lim_{x \to 0} f (x) = lim_{x \to 0} \frac{(a^{2} - a x + x^{2}) - (a^{2} + a x + x^{2})}{(\sqrt{a + x} - \sqrt{a - x}) (\sqrt{a^{2} - a x + x^{2}} + \sqrt{a^{2} + a x + x^{2}})}

lim_{x \to 0} \frac{- 2 a x (\sqrt{a + x} + \sqrt{a - x})}{2 x (\sqrt{a^{2} - a x + x^{2}} + \sqrt{a^{2} + a x + x^{2}})}

lim_{x \to 0} \frac{- a (\sqrt{a + x} + \sqrt{a - x})}{(\sqrt{a^{2} - a x + x^{2}} + \sqrt{a^{2} + a x + x^{2}}})

= lim_{x \to 0} \frac{- 2 a (\sqrt{a})}{2 a} = - \sqrt{a}

Understand that, f(x) is continuous for all x, then it is continuous at x = 0

\begin{aligned} lim_{x \to 0} f (x) = 0 \\ lim_{x \to 0} - \sqrt{a} = f (0) \end{aligned}

So, the correct option is (c)

Continuity exercise multiple choice question 17

Answer:
The correct option is (c)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
$f (x) = {\begin{cases} 1, | x | \geq 1 \\ \frac{1}{n^{2}}, \frac{1}{n} < | x | < \frac{1}{n - 1}, n = 2, 3, \dots \\ 0, x = 0 \end{cases}$
Solution:
Step 1: Simplify the given function

f (x) = {\begin{cases} 1, | x | \geq 1 \\ \frac{1}{n^{2}}, \frac{1}{n} < | x | < \frac{1}{n - 1}, n = 2, 3, \dots \\ 0, x = 0 \end{cases}

Case 2: Calculate the RHL

\begin{matrix} lim_{x \to {\frac{1}{n}}^{+}} f (x) = lim_{h \to 0} (\frac{1}{n} + h) \\ = {(\frac{1}{h})}^{2} \end{matrix}

Case 3: Calculate the RHL

\begin{aligned} lim_{x \to {\frac{1}{n}}^{-}} f (x) = lim_{h \to 0} f (\frac{1}{n} - h) = {(\frac{1}{n - 1})}^{2} \\ Therefore, lim_{x \to {\frac{1}{n}}^{+}} f (x) \neq lim_{x \to {\frac{1}{n}}^{-}} f (x) \end{aligned}

Hence, the correct answer is option (c)
Hence, f(x) is discontinuous only at

x = \pm \frac{1}{n^{'}} n \in Z - {0} and x = 0

Continuity exercise multiple choice question 18

Answer:
The correct option is (c)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
$f (x) = \frac{(27 - 2 x)^{\frac{1}{3}} - 3}{9 - 3 (243 + 5 x)^{\frac{1}{5}}}, (x \neq 0)$
Solution:
Step 1: Understand that, if f(x) to be continuous at x = 0 then,

lim_{x \to 0} f (x) = f (0)

Therefore,

= f (0) = lim_{x \to 0} \frac{(27 - 2 x)^{\frac{1}{3}} - 3}{9 - 3 (243 + 5 x)^{\frac{1}{5}}}

= f (0) = lim_{x \to 0} \frac{(27 - 2 x)^{\frac{1}{3}} - (27)^{\frac{1}{3}}}{3 [3 - (243 + 5 x)^{\frac{1}{5}}]}

= lim_{x \to 0} \frac{(27 - 2 x)^{1 / 3} - (27)^{4 / 3}}{3 {(243)^{15} - (243 + 5 x)^{1 / 5}}}

= \frac{1}{3} lim_{x \to 0} \frac{x {(27 - 2 x)^{1 / 3} - (27)^{4 / 3}}}{x (243)^{1 / 5} - (243 + 5 x)^{1 / 5}}}

= \frac{1}{3} lim_{x \to 0} \frac{\frac{(27 - 2 x)^{1 / 3} - (2 + 1^{1 / 3}}{x}}{\frac{(243)^{1 / 5} - (243 + 5 x)^{1 / 5}}{x}}

= \frac{2}{3} lim_{x \to 0} \frac{\frac{(27 - 2 x)^{1 / 3} - (27)^{1 / 3}}{5}}{5 \frac{(243 + 5 x)^{1 / 5} - (243)^{1 / 5}}{5 x}}

= \frac{2}{15} lim_{x \to 0} \frac{\frac{(27 - 2 x)^{1 / 3} - (27)^{1 / 3}}{27 - 2 x - 27}}{\frac{(243 + 5 x)^{1 / 5} - (243)^{1 / 5}}{243 + 5 x - 243}}

= f (0) = \frac{2}{15} \times \frac{\frac{1}{3} \times 27^{- \frac{2}{3}}}{\frac{1}{5} \times 243^{- \frac{4}{5}}}

= f (0) = \frac{2}{15} \times \frac{\frac{1}{3} \times \frac{1}{27^{\frac{2}{3}}}}{\frac{1}{5} \times \frac{1}{243^{\frac{4}{5}}}}

\begin{aligned} \begin{aligned} = \frac{2}{9} \frac{(243)^{\frac{4}{5}}}{(27)^{\frac{2}{3}}} \\ = & \frac{2}{9} \frac{{(3^{5})}^{4 / 5}}{{(3^{3})}^{2 / 3}} \\ ∴ f (0) & = \frac{2}{9} \times \frac{3^{4}}{3^{2}} \\ = 2 . \end{aligned} \\ Therefore, f (0) = 2 \end{aligned}

Hence, the correct option is (c)

Continuity exercise multiple choice question 19

Answer:
The correct option is (d)
Hint:
Use the given formula :
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \\ (ii) lim_{x \to a} {\frac{f (x)}{g (x)}} = \frac{1}{m}, \end{aligned}

Where, lim_{x \to a} f (x) = 1, lim_{x \to a} g (x) = m

Given:
$f (x) = \frac{2 - (256 - 7 x)^{\frac{1}{8}}}{(5 x + 32)^{\frac{1}{5}} - 2} is continuous everywhere$
Solution:
$f (x) = \frac{2 - (256 - 7 x)^{\frac{1}{8}}}{(5 x + 32)^{\frac{1}{5}} - 2}$
Now,

lim_{x \to 0} f (x) = lim_{x \to 0} \frac{2 - (256 - 7 x)^{\frac{1}{8}}}{(5 x + 32)^{\frac{1}{5}} - 2}

Using factorization method

\Rightarrow lim_{x \to 0} \frac{(256)^{1 / 8} - (256 - 7 x)^{1 / 8}}{(5 x + 32)^{1 / 5} - (32)^{1 / 5}} = x (0)

\Rightarrow lim_{x \to 0} \frac{x}{x} {\frac{(256)^{1 / 8} - (256 - 7 x)^{1 / 8}}{(5 x + 32)^{1 / 5} - (32)^{1 / 5}}} = f (0)

\Rightarrow lim_{x \to 0} \frac{\frac{(256)^{1 / 8} - (256 - 7 x)^{1 / 8}}{x}}{\frac{(5 x + 32)^{1 / 5} - (32)^{1 / 5}}{5}} = f (0)

\Rightarrow + \frac{7}{5} lim_{x \to 0} \frac{\frac{(256 - 7 x)^{1 / 8} - (256)^{1 / 8}}{- 7 x}}{\frac{(5 x + 32)^{1 / 5} - (32)^{1 / 5}}{5 x}} = f (0)

\Rightarrow \frac{7}{5} lim_{x \to 0} \frac{\frac{(256 - 7 x)^{1 / 8} - (256)^{1 / 8}}{256 - 7 x - 256}}{\frac{(5 x + 32)^{1 / 5 - (32)^{1 / 5}}}{32 + 5 x - 32}} = f (0)

\Rightarrow \frac{7 \frac{1}{8} (256)^{1 / 8 - 1}}{5} \frac{1}{\frac{1}{5} (32)^{1 / 5 - 1}} = f (0) [∴ lim_{x \to a} \frac{x^{n} - a^{4}}{x - a} = n a^{n - 1}]

\begin{aligned} \Rightarrow \frac{7}{5} \times \frac{1}{8} \times \frac{5}{1} \frac{(256)^{- 7 / 8}}{(32)^{- 4 / 5}} = f (0) \\ \Rightarrow \frac{7}{8} \frac{(32)^{\frac{4}{5}}}{(256)^{- \frac{1}{8}}} = f (0) \end{aligned}

\begin{aligned} \Rightarrow f (0) & = \frac{7}{8} \times \frac{{72^{5})}^{\frac{4}{5}}}{{(2^{8})}^{\frac{7}{8}}} \\ = \frac{7}{2^{3}} \times \frac{2^{4}}{2^{7}} \\ = \frac{7}{2^{6}} = \frac{7}{64} \end{aligned}

So the correct-option is (d)

Continuity exercise multiple choice question 20

Answer:
The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
$f (x) = {\begin{matrix} \frac{\sqrt{1 + p x} - \sqrt{1 - p x}}{x}, - 1 \leq x < 0 \\ \frac{2 x + 1}{x - 2}, 0 \leq x \leq 1 \end{matrix}$
Step 1 : Rationalize the function

∴ lim_{x \to 0^{-}} f (x) = lim_{h \to 0} f (0 - h) = lim_{h \to 0} f (- h)

lim_{h \to 0} \frac{\sqrt{1 - p h} - \sqrt{1 + p h}}{- h}

\begin{aligned} = lim_{h \to 0} \frac{(\sqrt{1 - p h} - \sqrt{1 + p h}) \times (\sqrt{1 - p h} + \sqrt{1 + p h})}{(- h) (\sqrt{1 - p h} + \sqrt{1 + p h})} \\ = lim_{h \to 0} \frac{((- p h) - (1 + p h)}{(- h) \sqrt{1 - p h} + \sqrt{1 + p^{n}}} \end{aligned}

\begin{aligned} = lim_{h \to 0} \frac{- 2 p h}{- n (\sqrt{1 - p h} + \sqrt{1 + p h})} \\ = lim_{h \to 0} \frac{+ 2 p}{\sqrt{1 - p h} + \sqrt{1 + p h}} \end{aligned}

\begin{aligned} = \frac{2 p}{\sqrt{1} + \sqrt{1}} \\ = \frac{2 p}{1 + 1} \\ = \frac{2 p}{2} \\ = p \end{aligned}

\begin{aligned} And lim_{x \to 0^{+}} f (x) = lim_{h \to 0} f (0 + h) = lim_{h \to 0} f (h) \\ = lim_{n \to 0} \frac{2 h + 1}{h - 2} \\ = \frac{2 \times 0 + 1}{0 - 2} \\ = \frac{1}{- 2} \end{aligned}

\begin{aligned} And lim_{x \to 0^{+}} f (x) = lim_{h \to 0} f (0 + h) = lim_{h \to 0} f (h) \\ = lim_{n \to 0} \frac{2 h + 1}{h - 2} \\ = \frac{2 \times 0 + 1}{0 - 2} \\ = \frac{1}{- 2} \end{aligned}

\begin{aligned} = \frac{- 1}{2} \end{aligned}

Also,

\begin{aligned} f (x) = \frac{2 \cdot 0 + 1}{0 - 2} \\ = \frac{- 1}{2} \end{aligned}

From (i)

\begin{aligned} p = - \frac{1}{2} = - \frac{1}{2} \\ \Rightarrow p = \frac{- 1}{2} \end{aligned}

So, the correct option is (b)

Continuity exercise multiple choice question 21

Answer:
The correct option is (c)
Hint:
If a function is continuous at x=2, then

lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)

Given:
The function f(x)

= {\begin{cases} x^{2} a, 0 \leq x < 1 \\ a, 1 \leq x < \sqrt{2} \\ \frac{2 b^{2} - 4 b}{x^{2}}, \sqrt{2 \leq x < \infty} \end{cases} is continuous for 0 \leq x < \infty

Solution:
Step 1: Calculate f(x) is continuous for

x = 1, \sqrt{2}

Calculate for f(x) is continuous at x = 1

\begin{aligned} lim_{x \to 1^{-}} f (x) = lim_{x \to 1 +} f (x) = f (1) \dots (i) \\ lim_{x \to 1^{-}} f (x) = lim_{h \to 0} f (1 - h) = lim_{h \to 0} \frac{(1 - h)^{2}}{a} \end{aligned}

\begin{aligned} = lim_{h \to 0} \frac{(1 + h^{2} - 2 h)}{a} \cdot [∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b] \\ = lim_{h \to 0} \frac{1}{a} + \frac{1}{a} h^{2} - \frac{2 h}{a} \end{aligned}

\begin{aligned} = \frac{1}{a} + 0 - 2 \frac{1}{a} \times 0 \\ = \frac{3}{a} \end{aligned}

f(a) = a

\begin{aligned} lim_{x \to 1^{+}} f (x) = lim_{h \to 0} f (1 + n) = lim_{n \to 0} a = a \\ from (i) \Rightarrow \frac{1}{a} = a = a \end{aligned}

\begin{aligned} \Rightarrow \frac{1}{a} = a \\ \Rightarrow a^{2} = 1 \\ \Rightarrow a = \pm 1 \end{aligned}

Step 2: Calculate for f(x) is continuous at

\begin{array}{r} x = \sqrt{2} \end{array}

Therefore,

\begin{aligned} lim_{x \to \sqrt{2}} - f (x) = lim_{x \to \sqrt{2}} f (x) = f (\sqrt{2}) - (ii) \\ \Rightarrow lim_{x \to \sqrt{2}} - f (x) = lim_{h \to 0} f (\sqrt{2} - h) = lim_{h \to 0} a = a \end{aligned}

\begin{aligned} and lim_{x \to \sqrt{2} +} f (x) = lim_{h \to 0} f (\sqrt{2} + h) \\ = lim_{h \to 0} \frac{2 b^{2} - 4 b}{(\sqrt{2} + h)^{2}} \\ = (2 b^{2} - 4 b) lim_{h \to 0} \frac{1}{(\sqrt{2} + 4)^{2}} \end{aligned}

\begin{aligned} = (2 b^{2} - 4 b) \cdot \frac{1}{(\sqrt{2} + 0)^{2}} \\ = (2 b^{2} - 4 b) \cdot \frac{1}{(\sqrt{2})^{2}} \\ = \frac{2 b^{2} - 4 b}{2} \\ also, f (\sqrt{2}) = \frac{2 b^{2} - 4 b}{(\sqrt{2})^{2}} \end{aligned}

\begin{aligned} = \frac{2 b^{2} - 4 b}{2} \end{aligned}

from(ii), we have

\begin{aligned} \Rightarrow a = \frac{2 b^{2} - 4 b}{2} \\ = \frac{2 b^{2} - 4 b}{2} \\ \Rightarrow a = \frac{2 b^{2} - 4 b}{2} \\ \Rightarrow 2 a - 2 b^{2} + 4 b = 0 \\ \Rightarrow b^{2} - a - 2 b = 0 \end{aligned}

\begin{aligned} When - a = 1 then \\ \Rightarrow b^{2} - 2 b - 1 = 0 \\ \Rightarrow b = \frac{2 \pm \sqrt{4 - (- 4)}}{2} \\ = \frac{2 \pm \sqrt{4 + 4}}{2} \\ = \frac{2 \pm \sqrt{8}}{2} \\ = 1 \pm \sqrt{2} \end{aligned}

\begin{aligned} Also, when a = - 1 then \\ \Rightarrow b^{2} - 2 b + 1 = 0 \\ \Rightarrow b = \frac{2 \pm \sqrt{4 - 4}}{2} \\ = \frac{2}{2} \\ = 1 \\ Thus a = - 1 and b = 1 \end{aligned}

The correct option is (c)

Continuity exercise multiple choice question 22

Answer:
The correct option is (a)
Hint:

lim_{h \to 0} f ((\frac{π}{2}) - h)

Given:
$\begin{aligned} The function f (x) = \frac{1 - \sin x}{(π - 2 x)^{2}}, when x \neq \frac{π}{2} and f (\frac{π}{2}) = λ, then f (x) will be continuous function \\ for x = \frac{π}{2} \end{aligned}$
Solution:
Step 1: Calculate, f(x) is continuous for

x = \frac{π}{2}

Therefore,

\begin{aligned} ∴ lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2}) \\ = lim_{x \to \frac{π}{2} (π - 2 x)^{2}} = f (\frac{π}{2}) \end{aligned}

\begin{aligned} Assume, (\frac{π}{2} - x) = t and substitute in equation (i) \\ ∴ lim_{t \to 0} [\frac{1 - \sin (\frac{π}{2} - t)}{(2 t)^{2}}] = f (\frac{π}{2}) \end{aligned}

\begin{aligned} lim_{t \to 0} (\frac{1 - \cos t}{4 t^{2}}) = f (\frac{π}{2}) \\ \frac{1}{4} lim_{t \to 0} [\frac{2 \sin^{2} (\frac{t}{2})}{t^{2}}] = f (\frac{π}{2}) \\ \frac{1}{4} lim_{i \to 0} [\frac{\frac{2}{4} \sin^{2} (\frac{t}{2})}{\frac{t^{2}}{4}}] = f (\frac{π}{2}) \end{aligned}

Simplify further,

\begin{aligned} = \frac{1}{8} lim_{t \to 0} [\frac{\sin^{2} (\frac{t}{2})}{\frac{t^{2}}{4}}] = f (\frac{π}{2}) \\ = \frac{1}{8} lim_{t \to 0} {[\frac{\sin (\frac{t}{2})}{\frac{t}{2}}]}^{2} = f (\frac{π}{2}) \end{aligned}

Apply formula (i) i.e

lim_{h \to 0} f ((\frac{π}{2}) - h)

\begin{aligned} = \frac{1}{8} \times (1)^{2} = f (\frac{π}{2}) \\ = f (\frac{π}{2}) = \frac{1}{8} \\ ∴ f (\frac{π}{2}) = λ \\ ∴\Rightarrow λ = \frac{1}{8} \end{aligned}

Hence, the value of

\begin{aligned} λ is \frac{1}{8} \end{aligned}

Hence, the correct answer is option (a)

Continuity exercise multiple choice question 23

Answer:
The correct option is (d)
Hint:
If a function f is continuous at x = a, then

lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)

(i) lim_{x \to 0} (\frac{\sin x}{x}) = 1

(ii) lim_{x \to 0} \frac{\log (1 + x)}{x} = 1

(iii) lim_{x \to 0} (\frac{a^{x} - 1}{x}) = \log a

Given:
The function

f (x) = {\begin{cases} \frac{{(4^{x} - 1)}^{3}}{\sin (x a) \log (1 + x^{2} 3)} & , x \neq 0 \\ 12 (\log 4)^{3} & , x = 0 \end{cases}

Solution:
Step 1: Calculate, f(x) is continuous for x = 0

\begin{aligned} ∴ lim_{x \to 0} f (x) = f (0) \\ lim_{x \to 0} [\frac{{(4^{x} - 1)}^{3}}{\sin \frac{x}{a} \log (1 + \frac{x^{2}}{3})}] = 12 (\log 4)^{3} \end{aligned}

lim_{x \to 0} [\frac{\frac{{(4^{x} - 1)}^{3}}{x^{3}}}{\frac{\sin \frac{x}{a} \log (1 + \frac{x^{2}}{3})}{x^{3}}}] = 12 (\log 4)^{3}

lim_{x \to 0} [\frac{a {(\frac{4^{x} - 1}{x})}^{3}}{(\frac{\sin \frac{x}{a}}{\frac{x}{a}}) \log (1 + \frac{x^{2}}{3})} x^{2}] = 12 (\log 4)^{3}

3 a lim_{x \to 0} [\frac{{(\frac{4^{x} - 1}{x})}^{3}}{(\frac{\sin \frac{x}{a}}{\frac{x}{a}}) \frac{\log (1 + \frac{x^{2}}{3})}{(\frac{x^{2}}{3})}}] = 12 (\log 4)^{3}

3 a [\frac{lim_{x \to 0} {(\frac{4^{x} - 1}{x})}^{3}}{lim_{x \to 0} (\frac{\sin \frac{x}{a}}{\frac{x}{a}}) lim_{x \to 0} \frac{\log (1 + \frac{x^{2}}{3})}{(\frac{x^{2}}{3})}}] = 12 (\log 4)^{3}

Apply formula (i), formula (ii) and formula (iii)

\begin{aligned} ∴ 3 a (\log 4)^{3} = 12 (\log 4)^{3} \\ a = 4 \end{aligned}

Hence, the correct answer is option (d)

Continuity exercise multiple choice question 24

Answer:
The correct option is (C)
Hint:
If a function f is continuous at x = a, then

lim_{x \to a -} f (x) = lim_{x \to a^{+}} f (x) = f (a)

Given:
the function

f (x) = t a n x

Solution:
Step 1: Check continuity of the function

f (x) = \tan x on the set {n π : n \in Z}

∴ f (x) = \tan n π

is defined at the integral point
Check continuity of the function

f (x) = \tan x on the set {2 n π : n \in Z}

f (x) = \tan 2 n π

is defined at the integral point
Check continuity of the function

f (x) = \tan x on the set {(2 n + 1) \frac{π}{2} : n \in Z}

f (x) = \tan (2 n + 1) \frac{π}{2}

is defined at the integral point

\begin{aligned} f (x) = \tan (n π + \frac{π}{2}) \\ f (x) = - \cot (n π) \end{aligned}

is not defined at the integral point
Check continuity of the function

f (x) = \tan x on the set {\frac{n π}{2} : n \in Z}

f (x) = \tan \frac{n π}{2}

is defined at the integral point
Hence, the correct answer is option (c)

Continuity exercise multiple choice question 25

Answer:
The correct option is (b)
Hint:
$lim_{x \to 0} \frac{\sin x}{x} = 1$
Given:
$f (x) = {\begin{cases} \frac{\sin 3 x}{x} & , x \neq 0 \\ \frac{k}{2} & , x \neq 0 \end{cases}$
Solution:
is continuous at x = 0, then
LHL = RHL

\begin{aligned} lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x) \\ lim_{x \to 0^{-}} \frac{\sin 3 x}{x} = f (0) \end{aligned}

\begin{aligned} \Rightarrow 3 lim_{x \to 0} \frac{\sin 3 x}{3 x} = \frac{k}{2} \\ \Rightarrow 3.1 = \frac{k}{2} [\frac{\sin x}{x} = 1 ∴ \frac{\sin 3 x}{3 x} = 1] \end{aligned}

Hence, k = 6
So, the correct option is (b)

Continuity exercise multiple choice question 26

Answer:
The correct option is (b)
Hint:
If a function f is continuous at x = a, then

lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)

(i) lim_{x \to 0} (\frac{\sin^{- 1} x}{x}) = 1

(ii) lim_{x \to 0} (\frac{\tan^{- 1} x}{x}) = 1

Given:
the function

f (x) = \frac{2 x - \sin^{- 1} x}{2 x + \tan^{- 1} x}

is continuous at each point of its domain
Solution:
Step 1: Understand that according to question, f(x) is continuous at x = 0

\begin{aligned} ∴ lim f (x) = f (0) \\ lim_{x \to 0} \frac{2 x - \sin^{- 1} x}{2 x + \tan^{- 1} x} = f (0) \end{aligned}

lim_{x \to \infty} \frac{x (2 - \frac{\sin^{- 1} x}{x})}{x (2 + \frac{\tan^{- 1} x}{x})} = f (0)

Simplify further,

lim_{x \to 0} \frac{2 - \frac{\sin^{- 1} x}{x}}{2 + \frac{\tan^{- 1} x}{x}} = f (0)

\frac{2 - lim_{x \to 0} (\frac{\sin^{- 1} x}{x})}{2 + lim_{x \to 0} (\frac{\tan^{- 1} x}{x})} = f (0)

Apply formula (i) and formula (ii)

\begin{aligned} \frac{2 - 1}{2 + 1} = f (0) \\ f (0) = \frac{1}{3} \end{aligned}

Hence, the correct answer is option (b)

Continuity exercise multiple choice question 27

Answer:
The correct option is (a)
Hint:
If a function is continuous at x = a, then

lim_{x \to t} f (x) = lim_{x \to 0^{+}} f (x) = f (a)

Solution: the function

is continuous at every point of its domain
Step 1: Understand that according to question, f(x) is continuous at every point of its domain.
Therefore, it is continuous at x = 1

\begin{aligned} ∴ lim_{x \to 1} f (x) = f (1) \\ lim_{h \to 0} f (1 + h) = f (1) \\ lim_{h \to 0} [4 (1 + h)^{2} + 3 b (1 + h)] = 5 (1) - 4 \\ 4 + 3 b = 1 \end{aligned}

Simplify further,

\begin{aligned} 3 b = - 3 \\ b = - 1 \end{aligned}

Hence, the correct answer is option (a)

Continuity exercise multiple choice question 28

Answer:
The correct option is (b)
Hint:
If a function f is continuous at x = a then

lim_{x \to 0^{-}} f (a) = lim_{x \to 0^{+}} f (x) = f (a)

Given:
The function

f (x) = \frac{1}{1 - x}

Step 1: Understand that

{f : R - {1} \to R}

\begin{aligned} Therefore f (f (x)) = f (\frac{x - 1}{x}) \\ \begin{array}{c} f (f (x)) = f (\frac{1}{1 - (\frac{1}{x - 1})}) \\ = \frac{x - 1}{x} \end{array} \end{aligned}

Step 2: Understand that ∴ f \circ f : R - {0, 1} \to R

∴ f (f (f (x))) = f (\frac{x - 1}{x})

\begin{aligned} f (f (f (x))) = f (\frac{x - 1}{1 - (\frac{x - 1}{x})}) \\ = x \end{aligned}

\begin{aligned} Step 3: Understand that ∴ f \circ f \circ f : R - {0, 1} \to R \\ Therefore, f (f (f (x))) is not defined at x = 0, 1 \\ Hence, f (f (f (x))) is discontinuous at {0, 1} \\ The set of point discontinuous of the function f (f (f (x))) is {0, 1} \end{aligned}

Hence, the correct answer is option (b)

Continuity exercise multiple choice question 29

Answer:
The correct option is (b)
Hint:
If a function f is continuous at x = a, then

lim_{x \to 4 -} f (x) = lim_{x \to 0 +} f (x) = f (a)

(i) lim_{x \to 0} (\frac{\tan x}{x}) = 1

Given:
The function

f (x) = \frac{\tan (\frac{π}{4} x)}{\cot 2 x}, x \neq \frac{π}{4}

Step 1: Understand that, if f(x) is continuous a

x = \frac{π}{4}

Understand that, if

\frac{π}{4} - x = y

Therefore,

x \to \frac{π}{4} and y \to 0

\begin{aligned} lim_{x \to \frac{π}{4}} \frac{\tan (\frac{π}{4} - x)}{\cot 2 x} = f (\frac{π}{4}) \\ ∴ lim_{y \to 0} (\frac{\tan y}{\cot 2 (\frac{π}{4} - y)}) = f (\frac{π}{4}) \end{aligned}

lim_{y \to 0} (\frac{\tan y}{\cot (\frac{π}{2} - 2 y)}) = f (\frac{π}{4})

\begin{aligned} lim_{y \to 0} (\frac{\tan y}{\tan 2 y}) = f (\frac{π}{4}) \\ \frac{1}{2} lim_{y \to 0} (\frac{\frac{\tan y}{y}}{\frac{\tan 2 y}{2 y}}) = f (\frac{π}{4}) \end{aligned}

Apply formula (i)

\begin{aligned} ∴ \frac{1}{2} (\frac{1}{1}) = f (\frac{π}{4}) \\ f (\frac{π}{4}) = \frac{1}{2} \end{aligned}

Hence, the correct answer is option (b)

Continuity exercise multiple choice question 30

Answer:
The correct option is (a)
Hint:
If a function f is continuous at x = a, then

lim_{x \to a +} f (x) = lim_{x \to a -} f (x) = f (a)

Given:
the function

f (x) = \frac{x^{3} + x^{2} - 16 x + 20}{x - 2}

is not defined for x = 2. In order to make f(x) continuous at x = 2
Step 1: Take

f (x) = x^{3} + x^{2} - 16 x + 20

and calculate further

∴ f (x) = x^{3} + x^{2} - 16 x + 20

\begin{aligned} = x^{2} (x - 2) + 3 x (x - 2) - 10 (x - 2) \end{aligned}

Simplify further,

\begin{aligned} = (x - 2) (x - 2) (x + 5) \\ = (x - 2)^{2} (x + 5) \end{aligned}

Therefore, the function can be rewritten as

\begin{aligned} ∴ f (x) = \frac{(x - 2)^{2} (x + 5)}{(x - 2)} \\ f (x) = (x - 2) (x + 5) \end{aligned}

Step 2: Understand that if f(x) is continuous at x = 2
Therefore,

\begin{aligned} ∴ lim_{h \to 0} f (x) = f (x) \\ = lim_{h \to 0} (x - 2) (x + 5) = f (x) \\ f (2) = 0 \end{aligned}

Hence, the correct answer is option (a)

Continuity exercise multiple choice question 31

Answer:
The correct option is (a)
Hint:
f(x) is continuous at x = 0

(LHL of f (x) at x = 0) = (RHL of f (x) at x = 0) = f (0)

lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x) = f (0) . . . . . (i)

\begin{aligned} Now, & lim_{x \to 0^{-}} f (x) \\ = lim_{x \to 0} a \sin \frac{π}{2} (x + 1) \end{aligned}

\begin{aligned} = lim_{x \to 0} a \sin (\frac{π}{2} + \frac{π}{2} x) \\ = lim_{x \to 0} a \cos \frac{π x}{2} = a \end{aligned}

\begin{aligned} [∴ f (x) = a \sin \frac{π}{2} (x + 1), if x \leq 0] \\ [\cos 0 = 1] \end{aligned}

\begin{aligned} Again, & lim_{x \to 0^{+}} f (x) \end{aligned}

= lim_{x \to 0} \frac{\tan x - \sin x}{x^{3}} [∵ f (x) = \frac{\tan x - \sin x}{x^{3}} if x > 0]

\begin{aligned} = lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - \sin x}{x^{3}} \\ = lim_{x \to 0} \frac{\sin x - \sin x \cos x}{\cos x \times x^{3}} = lim_{x \to 0} \frac{\sin x (1 - \cos x)}{\cos x \times x^{3}} \end{aligned}

= lim_{x \to 0} \frac{1}{\cos x} lim_{x \to 0} \frac{\sin x}{x} \times \frac{2 \sin^{2} \frac{x}{2}}{\frac{x^{2}}{4} \times 4} [∵ 1 - \cos x = 2 \sin^{2} \frac{x}{2}]

= \frac{1}{1} \times 1 \times \frac{1}{2} lim_{x \to 0} {(\frac{\sin \frac{x}{2}}{\frac{x}{2}})}^{2}

= \frac{1}{2} [lim_{\frac{x}{2} \to 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}] = \frac{1}{2} \times 1 = \frac{1}{2}

\begin{matrix} Also, f (0) = a \sin \frac{π}{2} (0 + 1) \\ = a \sin \frac{π}{2} = a \end{matrix}

Putting above values in (i) we get,

\begin{matrix} a = \frac{1}{2} \end{matrix}

So, the correct option is (a)

Continuity exercise multiple choice question 32

Answer:
The correct option is (d)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a)

Given:
$f (x) = {\begin{cases} a x^{2} + b & , 0 \leq x < 1 \\ 4 & , x = 1 \\ x + 3 & , 1 < x \leq 2 \end{cases}$
Step 1: Understand that, f(x) is continuous at x = 1
Therefore,

\begin{aligned} ∴ lim_{h \to 0} f (x) = f (1) \\ lim_{h \to 0} f (1 - h) = 4 \\ lim_{h \to 0} a (1 - h)^{2} + b = 4 \end{aligned}

Therefore, the possible values for (a, b) can be (2, 2), (3, 1), (4, 0) but (a,b) ≠ (5, 2)
Hence, the correct answer is option (d).

Continuity exercise multiple choice question 33

Answer:
The correct option is (b)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if

lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)

(i) Standard limits

lim_{x \to 0} \frac{\sin x}{x} = 1, lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1

Given:
$f (x) = {\begin{array}{cc} \frac{\log (1 + 3 x) - \log (1 - 2 x)}{x}, x \neq 0 \\ k, x = 0 \end{array}$
Step 1: Understand that, if f(x) is continuous at x = 0

\begin{aligned} ∴ lim_{x \to 0} f (x) = f (0) \\ lim_{x \to 0} f (x) = lim_{x \to 0} \frac{\log (1 + 3 x) - \log (1 - 2 x)}{x} = k \end{aligned}

lim_{x \to 0} \frac{3 \log (1 + 3 x)}{3 x} - \frac{2 \log (1 - 2 x)}{2 x} = k

Simplifying further,

3 lim_{x \to 0} \frac{\log (1 + 3 x)}{3 x} + 2 lim_{x \to 0} \frac{\log (1 - 2 x)}{- 2 x} = k

Applying formula (i)

\begin{aligned} ∴ 3 \times 1 + 2 \times 1 = k \\ k = 3 + 2 \\ k = 5 \end{aligned}

So, correct option is (b)

Continuity exercise multiple choice question 34

Answer:
The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a} f (x) = f (a) \\ lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \end{aligned}

(ii) Standard limits

\begin{aligned} lim_{x \to a} \frac{s i n x}{x} = 1 \end{aligned}

Given:
$f (x) = {\begin{cases} \frac{1 - \cos 10 x}{x^{2}} & , x < 0 \\ a & , x = 0 \\ \frac{\sqrt{x}}{\sqrt{625 + x} - 25} & , x > 0 \end{cases}$
then the value of a so that f(x) may be continuous at x = 0
Using LHL

\begin{aligned} lim_{x \to 0^{-}} f (x) = lim_{h \to 0} f (- h) \\ = lim_{h \to 0} \frac{1 - \cos (- 10 h)}{(- h)^{2}} \\ = 50 lim_{h \to 0} \frac{(\sin 5 h)^{2}}{(5 h)^{2}} \\ = 50 (Using standard limits) \end{aligned}

Function f(x) is continuous at x=0
50 = a
So, the correct option is (b).

Continuity exercise multiple choice question 35

Answer:
The correct option is (a)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a} f (x) = f (a) \\ lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \end{aligned}

(ii) Standard limits

\begin{aligned} lim_{x \to 0} \frac{\sin x}{x} = 1 \\ (iii) lim_{x \to 0} {f (x) \pm g (x)} = 1 \pm m, \end{aligned}

\begin{aligned} Where, lim_{x \to a} f (x) = 1, lim_{x \to a} g (x) = m \end{aligned}

Given:
$\begin{aligned} f (x) = x \sin \frac{1}{x} \end{aligned}$
Solution:

\begin{aligned} f (x) = x \sin \frac{1}{x} \end{aligned}

= lim_{x \to 0} f (x) = lim_{x \to 0} x \sin \frac{1}{x} = 0

Function f(x) is continuous at x = 0

\begin{aligned} = lim_{x \to 0} f (x) = f (0) \\ = 0 = f (0) \end{aligned}

So, option (a) is correct.

Continuity exercise multiple choice question 36

Answer:
The correct option is (d)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a} f (x) = f (a) \\ lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \end{aligned}

(ii) Standard limits

\begin{aligned} lim_{x \to 0} \frac{\sin x}{x} = 1 \end{aligned}

Given:
$\begin{array}{r} f (x) = \sin \frac{1}{x} \end{array}$
Solution:
$\begin{array}{r} f (x) = \sin \frac{1}{x} \end{array}$

lim_{x \to 0} f (x) = lim_{x \to 0} \sin \frac{1}{x}

Function f(x) is continuous at x = 0

\begin{aligned} lim_{x \to 0} f (x) = f (0) \\ lim_{x \to 0} \sin \frac{1}{x} = k \end{aligned}

Value does not exist for fraction to be continuous.
So, the option (d) is correct.

Continuity exercise multiple choice question 37

Answer:
The correct option is (e)
Hint:
Use this definition of continuity to solve this problem. A function f(x) is said to be continuous at x=a
if

lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)

Given:
$f (x) = {\begin{array}{cl} (1 + a x)^{1 / x} & , x < 0 \\ b & , x = 0 \\ \frac{(x + c)^{1 / 3} - 1}{(x + 1)^{1 / 2 - 1}} & , x > 0 \end{array}$
may be continuous at x = 0
Solution:
Now at x = 1

f (x) = {\begin{array}{cl} (1 + a x)^{1 / x} & , x < 0 \\ b & , x = 0 \\ \frac{(x + c)^{1 / 3} - 1}{(x + 1)^{1 / 2 - 1}} & , x > 0 \end{array}

To find the value of a, b and c we will use definition of continuous function
So we know if f(x) is continuous at x = 0 then we have,

lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x) = f (0) - (1)

Using equation (1) We have,

lim_{x \to 0^{-}} f (x) = f (0) \Rightarrow lim_{h \to 0} f (0 - h) = f (0)

\begin{aligned} \Rightarrow lim_{h \to 0} {1 + a (- h)}^{\frac{1}{h}} = f (0) [∵ f (x) = (1 + a x)^{1 / x}, x < 0] \\ \Rightarrow lim_{h \to 0} (1 - a h)^{- \frac{1}{h}} = f (0) = b [∵ f (x) = b, x = 0] \end{aligned}

Taking log both sides, then

\begin{aligned} \Rightarrow lim_{h \to 0} \log (1 - a b)^{- 1 / n} = \log b \\ \Rightarrow lim_{h \to 0} - \frac{1}{n} \log (1 - a h) = \log b [∵ \log a^{m} = m l o g a] \end{aligned}

\Rightarrow lim_{h \to 0} \frac{a \log (1 - a h)}{- a h} = \log b [multiply and divide a on L.H.S]

\begin{aligned} \Rightarrow a lim_{h \to 0} \frac{\log {1 + (- a b)}}{(- a b)} = \log b \\ \Rightarrow a \cdot 1 = \log b & [∴ lim_{x \to 0} \frac{\log (1 + x)}{x} = 1] \\ \Rightarrow a = \log b - (2) \end{aligned}

Again, using equation (1) we get,

\begin{aligned} lim_{x \to 0 +} f (x) = {(0) \\ \Rightarrow lim_{h \to 0} f (0 + h) = f (0) \end{aligned}

\begin{aligned} \Rightarrow lim_{h \to 0} f (n) = f^{'} (0)) \\ \Rightarrow lim_{h \to 0} (\frac{(h + c)^{1 / 3} - 1}{(h + 1)^{1 / 2} - 1}) = f (0) [∵ f (x) = \frac{(x + c)^{1 / 3} - 4}{(x + 1)^{1 / 2} - 1}, x > 0] \end{aligned}

\Rightarrow lim_{h \to 0} (\frac{(h + c)^{1 / 3} - 1}{(n + 1)^{1 / 2} - 1} \times \frac{(n + 1)^{1 / 2} + 1}{(n + 1)^{1 / 2} + 1}) = b [∵ f (x) = b, x = 0]

\Rightarrow lim_{h \to 0} \frac{((h + r)^{4} 3 - 1) \times ((n + 1)^{42} + 1)}{((h + 1)^{1 / 2} - 1) ((h + 1)^{1 / 2} + 1)} = b

\Rightarrow lim_{h \to 0} \frac{((h + c)^{1 / 3} - 1) ((n + 1)^{4 / 2} + 1)}{{((h + 1)^{1 / 2})}^{2} - (1)^{2}} = b [∵ (a + b) (a - b) = a^{2} - b^{2}]

\begin{aligned} \Rightarrow lim_{h \to 0} \frac{((h + c)^{1 / 3} - 1) ((h + 1)^{42} + 1)}{h + 1 - 1} = b \\ \Rightarrow lim_{h \to 0} \frac{((h + c)^{1 / 3} - 1) (n + 1)^{4 / 2} + 1)}{h} = b \end{aligned}

\Rightarrow lim_{h \to 0} \frac{((h + c)^{1 / 3} - 1)}{h} \times lim_{h \to 0} ((n + 1)^{1 / 2} + 1) = b [\begin{array}{ll} ∵ lim_{x \to a} f (x) \cdot g (x) = & lim_{x \to a} f (x) \\ x lim_{x \to a} g (x) \end{array}]

\begin{aligned} \Rightarrow lim_{h \to 0} (\frac{(h + c)^{1 / 3} - 1^{4 / 3}}{h}) \times (0 + 1)^{1 / 2} + 1 = b \\ \Rightarrow lim_{h \to 0} \frac{(h + c)^{1 / 3} - 1^{1 / 3}}{(h + c) - c} \times 1^{1 / 2} + 1 = b \\ \Rightarrow g \times 2 = b \end{aligned}

\begin{aligned} where g ∴= lim_{h \to 0} \frac{(h + c)^{1 / 3} - 1^{1 / 3}}{(h + c) - c} - (a) \\ \Rightarrow g \dots = \frac{6}{2} - (3) \end{aligned}

To find the value of g, we will use

lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - L}

But this formula is applicable on g only if c=L
So we take c = L then we have,

g = lim_{h \to 0} \frac{(h + 1)^{1 / 3} - 1^{1 / 3}}{(h + 1) - 1}

[∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}]

\begin{aligned} = \frac{1}{3} \cdot 1^{1 / 3 - 1} \\ = \frac{1}{3} 1^{- 2 / 3} = \frac{1}{3} \cdot 1 = \frac{1}{3} \\ ∴ g = \frac{1}{3} \end{aligned}

Equation (3)

\begin{aligned} \Rightarrow \frac{1}{3} = \frac{b}{2} \\ \Rightarrow b = \frac{2}{3} \end{aligned}

Then equation 2 becomes,

\begin{aligned} \Rightarrow a = \log \frac{2}{3} \\ ∴ a = \frac{\log 2}{3}, b = \frac{2}{3}, c = 1 \end{aligned}

Continuity exercise multiple choice question 38

Answer:
The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a} f (x) = f (a) \\ lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \end{aligned}

Given:
$f (x) = {\begin{cases} 2 \sqrt{x} & , 0 \leq x \leq 1 \\ 4 - 2 x & , 1 < x < \frac{5}{2} \\ 2 x - 7 & , \frac{5}{2} \leq x \leq 4 \end{cases}$
Solution:
Now at x = 1

\begin{aligned} lim_{x \to 1^{+}} f (x) = lim_{h \to 0} f (1 + h) \\ = lim_{h \to 0} f (4 - 2 (1 + h)) \\ = 2 \end{aligned}

Again,

\begin{aligned} lim_{x \to 1^{-}} f (x) = lim_{h \to 0} f (1 - h) \\ = lim_{h \to 0} (2 \sqrt{1 - h}) \\ = 2 \\ = lim_{x^{+} \to 0} f (x) = f (0) = lim_{x \to 0^{-}} f (x) \end{aligned}

Therefore, function is continuous at x = 0
Now we will check the continuity of f(x) at

\begin{array}{r} x = \frac{5}{2} \end{array}

\begin{aligned} So lim_{x \to \frac{5}{2}} f (x) = lim_{h \to 0} f (\frac{5}{2} - h) \\ = lim_{h \to 0} (4 - 2 (\frac{5}{2} - h)) [∵ f (x) = 4 - 2 x, x < \frac{5}{2}] \\ = 4 - 2 (\frac{5}{2} - 0) \end{aligned}

\begin{aligned} = 4 - 2 \times \frac{5}{2} \\ = 4 - 5 \\ = - 1 \end{aligned}

\begin{aligned} And lim_{x \to {\frac{5}{2}}^{+}} f (x) & = lim_{h \to 0} f (\frac{5}{2} + h) \\ = lim_{h \to 0} 2 (\frac{5}{2} + h) - 7 [∵ f (x) = 2 x - 7, x > \frac{5}{2}] \\ = 2 (\frac{5}{2} + 0) - 7 \end{aligned}

\begin{aligned} = 2 \times \frac{5}{2} - 7 \\ = 5 - 7 \\ = - 2 \\ lim_{x \to \frac{5}{2}} f (x) & \neq lim_{x \to \frac{5}{2}} f (x) \\ le - 1 \neq - 2 \end{aligned}

Therefore the function f(x) is discontinuous at

\begin{array}{r} x = \frac{5}{2} \end{array}

So, correct option is (b)

Continuity exercise multiple choice question 39

Answer:
The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a} f (x) = f (a) \\ lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \end{aligned}

(ii) Standard limits

lim_{x \to 0} \frac{\sin x}{x} = 1

Given:
$f (x) = {\begin{cases} \frac{1 - \sin^{2} x}{3 \cos^{2} x} & , x < \frac{π}{2} \\ a & , x = \frac{π}{2} \\ \frac{b (1 - \sin x)}{(π - 2 x)^{2}} & , x > \frac{π}{2} \end{cases}$
Solution:
$f (x) = {\begin{cases} \frac{1 - \sin^{2} x}{3 \cos^{2} x} & , x < \frac{π}{2} \\ a & , x = \frac{π}{2} \\ \frac{b (1 - \sin x)}{(π - 2 x)^{2}} & , x > \frac{π}{2} \end{cases}$
So it is given that the function f(x) is continuous at

x = \frac{π}{2}

So, we have

\begin{aligned} lim_{x \to \frac{π}{2}} + f (x) = lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2}) - (1) \\ L.H.L at x = \frac{π}{2} \\ i.e lim_{x \to \frac{π}{2}} f (x) = lim_{h \to 0} f (\frac{π}{2} - h) \end{aligned}

= lim_{h \to 0} \frac{1 - \sin^{2} (π / 2 - h)}{3 \cos^{2} (π / 2 - h)} [∴ f (x) = \frac{1 - \sin^{2} x}{3 \cos^{2} x}, x < π / 2]

= lim_{h \to 0} \frac{1 - {\sin (π / 2 - h)}^{2}}{3 {\cos (π / 2 - h)}^{2}}

= lim_{h \to 0} \frac{1 - (\cos h)^{2}}{3 (\sin h)^{2}} [\begin{matrix} ∵ \sin (π / 2 - θ) = \cos θ, \\ \cos (π / 2 - θ) = \sin θ \end{matrix}]

= lim_{h \to 0} \frac{1 - \cos^{2} h}{3 \sin^{2} h}

\begin{aligned} = lim_{n \to 0} \frac{\sin^{2} h}{3 \sin^{2} h} [∵ \cos^{2} θ + \sin^{2} θ =⊥\Rightarrow 1 - \cos^{2} θ = \sin^{2} θ] \\ = \frac{1}{3} lim_{h \to 0} 1 = \frac{1}{3} . - (2) \end{aligned}

\begin{aligned} R.H.I at x = π / 2 \\ i.e lim_{x \to π / 2^{+}} f (x) = lim_{h \to 0} f (π / 2 + h) \end{aligned}

= lim_{h \to 0} \frac{b {1 - \sin (π / 2 + h)}}{{π - 2 (π / 2 + h)}^{2}} [∵ f (x) = \frac{b (1 - \sin x)}{(π - 2 x)^{2}}, x > π / 2]

= lim_{h \to 0} \frac{b {1 - \cosh}}{{π - 2 (\frac{π}{2}) - 2 h}^{2}} [∵ \sin (π / 2 + θ) = \cos θ]

= lim_{h \to 0} \frac{b {2 \sin^{2} h / 2}}{(π - π - 2 h)^{2}} [∵ 1 - \cos 2 θ = 2 \sin^{2} θ]

\begin{aligned} = lim_{h \to 0} \frac{\frac{2 b \sin^{2} h}{2}}{(- 2 h)^{2}} \\ = lim_{h \to 0} \frac{\frac{2 b \sin^{2} h}{2}}{4 h^{2}} \\ = \frac{26}{4} lim_{h \to 0} \frac{\frac{\sin^{2} n}{2}}{h^{2}} \end{aligned}

\begin{aligned} = \frac{b}{2} lim_{h \to 0} \frac{\frac{\sin^{2} h}{2}}{\frac{h^{2}}{4} x^{4}} \\ = \frac{b}{8} lim_{h \to 0} {(\frac{\frac{\sin h}{2}}{\frac{h}{2}})}^{2} \\ = \frac{6}{8} {(lim_{h \to 0} \frac{\sin n / 2}{h / 2})}^{2} \end{aligned}

\begin{aligned} = \frac{b}{8} \times 1^{2} [∵ lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = \frac{b}{8} \times 1 \\ = \frac{b}{8} - (3) \end{aligned}

\begin{aligned} also, f (x) a t - x = π / 2 \\ i.e f (π / 2) = a - (4) & [∵ f (x) = a, x = π / 2] \end{aligned}

Putting the value of equation 2 ,3 , 4 in (1)
we get ,

\begin{aligned} - \frac{1}{3} = \frac{b}{8} = a \\ Sor a = \frac{1}{3} or \frac{b}{8} = \frac{1}{3} \Rightarrow b = \frac{8}{3} \\ ∴ a = \frac{1}{3}, b = \frac{8}{2} \end{aligned}

Continuity exercise multiple choice question 40

Answer:
The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a} f (x) = f (a) \\ lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \end{aligned}

Given:
$f (x) = {\begin{cases} \frac{1}{5} (2 x^{2} + 3) & , x \leq 1 \\ 6 - 5 x & , 1 < x < 3 \\ x - 3 & , x \geq 3 \end{cases}$
Solution:
Now at x = 1

\begin{aligned} lim_{x \to 1^{+}} f (x) = lim_{h \to 0} f (1 + h) \\ lim_{h \to 0} [6 - 5 (1 + h)] = 1 \\ lim_{h \to 1^{+}} f (x) = lim_{x \to 1^{-}} f (x) = f (1) \end{aligned}

Therefore continuous at x = 1
Now, at x = 3

\begin{aligned} lim_{x \to 3^{+}} f (x) = lim_{h \to 0} f (3 + h) \\ lim_{h \to 0} [(3 + h) - 3] \\ = 0 \end{aligned}

For left hand limit,

\begin{aligned} = lim_{x \to 3^{-}} f (x) = lim_{h \to 0} f (3 - h) \\ = lim_{h \to 0} [6 - 5 (3 - h)] = - 9 \end{aligned}

We have

= lim_{x \to 3^{+}} f (x) \neq lim_{x \to 3^{-}} f (x)

Therefore discontinuous at x = 3
So, the correct option is (b).

Continuity exercise multiple choice question 41

Answer:
The correct option is (d)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a} f (x) = f (a) \\ lim_{x \to a} f (x) = lim_{x \to a^{-}} f (x) = f (a) \end{aligned}

Given:
$f (x) = {\begin{cases} 5 x - 4 & , 0 < x \leq 1 \\ 4 x^{2} + 3 a x & , 1 < x < 2 \end{cases}$
is continuous at every point of domain
Solution:

f (x) = {\begin{cases} 5 x - 4 & , 0 < x \leq 1 \\ 4 x^{2} + 3 a x & , 1 < x < 2 \end{cases}

At x = 1 f(x)=1
So,

\begin{aligned} lim_{x \to 1^{-}} f (x) = lim_{h \to 0} f (1 - h) \\ = lim_{h \to 0} s (1 - h) - 4 [∵ f (x) = 5 x - 4 when x < 1] \end{aligned}

\begin{aligned} = lim_{h \to 0} 5 - 5 h - 4 \\ = lim_{h \to 0} 1 - 5 h \\ = 1 - 5 \times 0 \\ = 1 \end{aligned}

\begin{aligned} lim_{x \to 1^{+}} f (x) & = lim_{h \to 0} f (1 + h) \\ = lim_{h \to 0} {4 (1 + h)^{2} + 3 a (1 + n)} & [∵ f (x) = 4 x^{2} + 3 a x when x > 1] \end{aligned}

\begin{aligned} = lim_{h \to 0} {4 (1 + h^{2} + 2 b) + 3 a + 3 a h} [∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b] \\ = lim_{h \to 0} (4 + 4 h^{2} + 8 h + 3 a + 3 a h) \end{aligned}

\begin{aligned} = 4 + 4 \times 0^{2} + 8 x 0 + 3 a + 3 a \times 0 \\ = 4 + 0 + 0 + 3 a + 0 = 4 + 3 a - (2) \\ Also, f (1) = 5 \cdot 1 - 4 \\ = 5 - 4 \\ = 1 - (3) [∵ f (x) = 5 x - 4 when x = 1] \end{aligned}

It is given that the function f(x) is continuous at x=1

\begin{aligned} \Rightarrow lim_{x \to 1^{-}} f (x) & = lim_{x \to 1^{+}} f (x) \\ = f (1) \\ \Rightarrow 1 = 4 + 3 a \end{aligned}

\begin{aligned} = 1 [using equation (1), (2) and (3) \\ So, \Rightarrow 4 + 3 a = 1 \\ \Rightarrow 3 a & = 1 - 4 \\ = - 3 \\ \Rightarrow a & = \frac{- 3}{3} = - 1 \\ ∴ a = - 1 \end{aligned}

Continuity exercise multiple choice question 42

Answer:
The correct option is (a)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} lim_{x \to a} f (x) = f (a) \\ lim_{x \to a^{+}} f (a + h) = lim_{x \to a^{-}} f (a - h) = f (a) \end{aligned}

(ii) Standard limits

\begin{aligned} lim_{x \to 0} \frac{\sin x}{x} = 1 \\ (iii) lim_{x \to 0} {f (x) \pm g (x)} = 1 \pm m, \\ Where, lim_{x \to a} f (x) = 1, lim_{x \to a} g (x) = m \end{aligned}

Given:
$f (x) = {\begin{cases} \frac{(\sin (\cos x) - \cos x)}{(π - 2 x)^{2}} & , x \neq \frac{π}{2} \\ k & , x < \frac{π}{2} \end{cases}$
Solution:

f (x) = {\begin{cases} \frac{(\sin (\cos x) - \cos x)}{(π - 2 x)^{2}} & , x \neq \frac{π}{2} \\ k & , x < \frac{π}{2} \end{cases} - (1)

\begin{aligned} At x = \frac{π}{2} \\ lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2}) \\ \Rightarrow lim_{x \to π / 2} \frac{\sin (\cos x) - \cos x)}{(π - 2 x)^{2}} = k & [using 1] \end{aligned}

Let

\begin{aligned} π - 2 x = 2 y \\ \Rightarrow y = \frac{π - 2 x}{2} \\ \Rightarrow y = \frac{π}{2} - x \\ Since, when x \to \frac{π}{2} then y \to 0 . \end{aligned}

\Rightarrow lim_{y \to 0} \frac{\sin (\cos x) - \cos x}{(2 y)^{2}} = k [∵ π - 2 x = y]

\Rightarrow lim_{y \to 0} \frac{\sin {\cos (π / 2 - y)} - \cos (π / 2 - y)}{4 y^{2}} = k [∵ y = \frac{π}{2} - x \Rightarrow x = \frac{π}{2} - y]

\Rightarrow lim_{y \to 0} \frac{\sin (\sin y) - \sin y}{4 y^{2}} = k [∵ \cos (\frac{π}{2} - θ) = \sin θ]

\Rightarrow lim_{y \to 0} \frac{2 \sin (\frac{\sin y - y}{2}) \cdot \cos (\frac{\sin y + y}{2})}{4 y^{2}} = k [∵ \sin c - \sin D = 2 \sin (\frac{C - D}{2}) \cdot \cos (\frac{c + D}{2})]

\Rightarrow \frac{1}{2} lim_{y \to 0} \frac{(\frac{\sin y - y}{2}) \cdot \sin (\frac{\sin y - y}{2})}{y^{2} \cdot (\frac{\sin y - y}{2})} \cdot \cos (\frac{\sin y + y}{2})

\Rightarrow \frac{1}{2} lim_{y \to 0} (\frac{\frac{\sin y - y}{2}}{y}) \cdot {\frac{\sin (\frac{\sin y - y}{2})}{(\frac{\sin y - y}{2})} ∣ \cdot {\frac{\cos (\frac{\sin y + y}{2})}{y}} = k

\Rightarrow \frac{1}{2} lim_{y \to 0} \frac{\sin y - y}{2 y} \times lim_{y \to 0} \frac{\sin (\frac{\sin y - y}{2})}{(\frac{\sin y - y}{2})} \times lim_{y \to 0} \frac{\cos (\frac{\sin y + y}{2})}{y} = k

[∵ lim_{x \to a} f (x) \cdot g (x) \cdot h (x) = lim_{x \to a} f (x) lim_{x \to a} g (x) lim_{x \to a} h (x)]

\Rightarrow \frac{1}{4} (1 - 1) \times 1 \times lim_{y \to 0} \frac{\cos (\frac{\sin y + y}{2})}{y} = k [∵ lim_{x \to 0} \frac{\sin x}{x} = 1]

\begin{aligned} \Rightarrow \frac{1}{4} \times 0 \times 1 \times lim_{y \to 0} \frac{\cos (\frac{\sin y + y}{2})}{y^{\circ}} = R \\ \Rightarrow 0 = k \end{aligned}

\begin{array}{r} ∴ k = 0 \end{array}

So, the correct option is (a)

Continuity exercise multiple choice question 43

Answer:
The correct option is (d)
Hint:
use this to get

\frac{f (x)}{g (x)}

Given:
$\begin{aligned} f (x) = 2 x \\ g (x) = \frac{x^{2}}{2} + 1 \end{aligned}$
For the discontinuous function

\frac{f (x)}{g (x)} \to g (x) \neq 0

If g(x) = 0, then it is a discontinuous function
So, the correct option is (d)

Continuity exercise multiple choice question 44

Answer:
The correct option is (a)
Given:
We have

f (x) = \cot x = \frac{\cos x}{\sin x}

Now, f(x) is discontinuous when sinx = 0 we know that

\sin x = 0 at x = n π, n \in Z

f(x) = cot x is discontinuous on the set

{x : x = n π : n \in Z}

So, the correct option is (a)

Continuity exercise multiple choice question 45

Answer:
The correct option is (a)
Given:
The given function f is

f (x) = x^{2} \sin \frac{1}{x}, where x \neq 0

It is evident that f is defined at all points of the real line.
let C be a real number.

\begin{aligned} Case 1 : If c \neq 0, then f (c) = c^{2} \sin \frac{1}{c} \\ lim_{x \to c} f (x) = lim_{x \to c} (x^{2} \sin \frac{1}{x}) = (lim_{x \to c} x^{2}) (lim_{x \to c} \sin \frac{1}{x}) = c^{2} \sin \frac{1}{c} \end{aligned}

∴ lim_{x \to c} f (x) = f (c)

Therefore, f is continuous at all points x≠0
Case 2: If c=0, then f(0)=0

lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (x^{2} \sin \frac{1}{x}) = lim_{x \to 0^{-}} (x^{2} \sin \frac{1}{x}) = c^{2} \sin \frac{1}{c}

It is known that

\begin{aligned} - 1 \leq \sin \frac{1}{x} \leq 1, x \neq 0 \\ = - x^{2} \leq \sin \frac{1}{x} \leq x^{2} \end{aligned}

\begin{aligned} = lim_{x \to 0} (- x^{2}) \leq lim_{x \to 0} (x^{2} \sin \frac{1}{x}) \leq lim_{x \to 0} x^{2} \\ = 0 \leq lim_{x \to 0} (x^{2} \sin \frac{1}{x}) \leq 0 \end{aligned}

\begin{aligned} = lim_{x \to 0} (x^{2} \sin \frac{1}{x}) = 0 \\ ∴ lim_{x \to 0^{-}} f (x) = 0 \end{aligned}

Similarly,

lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} (x^{2} \sin \frac{1}{x}) = lim_{x \to 0^{+}} (x^{2} \sin \frac{1}{x}) = 0

∴ lim_{x \to 0^{-}} f (x) = f (0) = lim_{x \to 0^{+}} f (x)

There fore f is continuous at x = 0
So, the correct option is (a)

Continuity exercise multiple choice question 46

Answer:
The correct option is (d)
$f (x) = \frac{1}{x - [x]}$
The greatest integer function becomes discontinuous at every integer value of x. So, the whole function will become discontinuous at each integer value of x.
It has infinite number of points of discontinuity.
So, the correct option is (d)

Continuity exercise multiple choice question 47

Answer:
The correct option is (d)
The function f(x) = [x] is continuous for all x except all integral values of x
Hence, it is continuous at x = 1.5, which is not an integer
So, the correct option is (d)

Continuity exercise multiple choice question 48

Answer:
The correct option is (b)
Hint:
Standard limit:

\begin{array}{r} (i) lim_{x \to 0} \frac{\sin x}{x} = 1 \\ (i i) lim_{x \to 0} \cos x = 1 \end{array}

Given:
$f (x) = {\begin{array}{cl} \frac{\sin x}{x} + \cos x & , if x \neq 0 \\ k & , if x = 0 \end{array}$
f(x) is continuous at x = 0.
So,

\begin{aligned} = lim_{x \to 0} f (x) = f (0) \\ = lim_{x \to 0} (\frac{\sin x}{x} + \cos x) = k \\ = lim_{x \to 0} \frac{\sin x}{x} + lim_{x \to 0} \cos x = k \\ = 1 + 1 = k \\ = k = 2 \end{aligned}

RD Sharma class 12 solution of Continuity ex MCQ contains questions of the chapter Continuity. There are about 48 MCQs in this exercise that are extremely basic and simple to solve if you are aware of the fundamentals of this chapter. However, if you find any question tricky to solve, you can refer to the RD Sharma class 12th exercise MCQ to get a good understanding of all the concepts in this chapter.

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The essential concepts covered in the RD Sharma class 12th exercise MCQ are:-

The intuitive notion of continuity
Testing the continuity of a function
Continuity of a point
Continuity of an interval
Continuity of an open interval
Continuity of an closed interval
Definition and meaning of continuous function
Properties of continuous function

The questions prepared in the RD Sharma class 12th exercise MCQ are created by experts from academics across the country. The experts also provide helpful tips and tricks to solve the questions quickly and alternately, which will be less time-consuming. RD Sharma solutions Therefore, using Class 12 RD Sharma exercise MCQ solution can help prepare for the exams efficiently as the questions mentioned in the book are frequently asked in the board exams. The level of questions in the RD Sharma class 12th exercise MCQ is exactly the same as in the NCERT that are helpful in the preparation of public examinations as well.

The latest version of the RD Sharma class 12th exercise MCQ is available at the Career360 website. As the new version is released, the older materials are replaced. The solutions by Career360 are updated to the latest version. Also, the RD Sharma class 12th exercise MCQ can be downloaded free of cost from the Career360 website, which means it will not cost you a single penny to own the RD Sharma class 12 chapter 8 exercise MCQ.

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