QBManager

Limits, Continuity and Differentiability

79959 The value of $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$ is

1

\frac{1}{3}

2 2

3 1

4

\frac{1}{2}

Explanation:

(C) : Given, $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$
Applying L-Hospital rule,
$= lim_{x \to 0} \frac{\sinh x + \sin x}{x \cdot \cos x + \sin x}$
Again applying L-Hospital rule,
$= lim_{x \to 0} \frac{\cosh x + \cos x}{x \cdot (- \sin x) + \cos x + \cos x} = \frac{\cosh 0^{\circ} + \cos 0^{\circ}}{0 + 2 \cos 0^{\circ}} = \frac{1 + 1}{2} = 1$

UPSEE-2016

Limits, Continuity and Differentiability

79960 Let $f (x) = {\begin{array}{rr} x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x = 0 \end{array}}$ ,
Then, $f (x)$ is continuous but not differentiable at $x = 0$ , if

1

n \in (0, 1)

2

n \in [1, \infty)

3

n \in (- \infty, 0)

4

n = 0

Explanation:

(A) : Given,
$f (x) = x^{n} \sin \frac{1}{x}$
$f (0) = 0$
$∵ f (x)$ is continuous at $x = 0$ ,
Hence, $f (x)$ will not be differentiable if $n \leq 1$ but
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} x^{n} \sin \frac{1}{x} = 0$
$f^{'} (0) = \frac{f (x) - f (0)}{x - 0}$
$f^{'} (0) = \frac{x^{n} \sin \frac{1}{x} - 0}{x}$
$f^{'} (0) = x^{n - 1} \sin \frac{1}{x}$
continuous at $x = 0$
$∴$ Possible values of $n$ is $n \in (0, 1)$

UPSEE-2014

Limits, Continuity and Differentiability

79961 $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ is equal to

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$
$lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{+}} \frac{x}{x + x^{2}} [∵ x > 0]$
$= lim_{x \to 0^{+}} \frac{x}{x (1 + x)} = lim_{x \to 0^{+}} \frac{1}{1 + x} = \frac{1}{1 + 0} = 1$
$lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{-}} \frac{x}{- x + x^{2}} {∵ x < 0}$
$= lim_{x \to 0^{-}} \frac{x}{x (- 1 + x)} = lim_{x \to 0^{-}} \frac{1}{- 1 + x} = \frac{1}{- 1} = - 1$
$∵ lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} \neq lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}}$
Therefore, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ does not exist

UPSEE-2012

Limits, Continuity and Differentiability

79962 If $a_{1} = 1$ and $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}, n \geq 1$ and if $lim_{n \to \infty} a_{n} = a$ , then the value of $a$ is

1

\sqrt{2}

2

- \sqrt{2}

3 2

4 None of these

Explanation:

(A) : Given, $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$ , and $lim_{n \to \infty} a_{n} = a$
On taking both side $lim_{x \to \infty}$ ,
$lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$
$a = \frac{4 + 3 a}{3 + 2 a}$
$3 a + 2 a^{2} = 4 + 3 a$
$a^{2} = 2 \Rightarrow a = \sqrt{2}$

UPSEE-2012

Limits, Continuity and Differentiability

79963 If $f (x) = {\begin{cases} - x^{2}, When x \leq 0 \\ 5 x - 4, When 0 < x \leq 1 \\ 4 x^{2} - 3 x, When 1 < x < 2 \\ 3 x + 4, When x \geq 2 \end{cases}$
then

1

f (x)

is continuous at

x = 0

2

f (x)

is continuous at

x = 2

3

f (x)

is discontinuous at

x = 1

4 None of the above

Explanation:

(B) : Given,
$f (x) = - x^{2}, x \leq 0$
$= 5 x - 4, 0 < x \leq 1$
$= 4 x^{2} - 3 x, 1 < X < 2$
$= 3 x + 4, x \geq 2$
$At, x = 0$
$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (- x^{2}) = 0$
$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} 5 x - 4 = 0 - 4 = - 4$
$\begin{array}{ll} ∵ lim_{x \to 0^{-}} f (x) \neq lim_{x \to 0^{+}} f (x) \\ ∵ f (x) is discontinuous \end{array}$
$∵ f (x)$ is discontinuous at $x = 0$
At, $x = 1$
$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (5 x - 4)$
$= 5 \times 1 - 4 = 1$
$\begin{array}{r} lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} 4 x^{2} - 3 x \\ = 4 - 3 = 1 \end{array}$
$∵ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x)$
$∴ f (x)$ is continuous at $x = 1$
At, $x = 2$
$lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} 4 x^{2} - 3 x$
$= 4 \times 2^{2} - 3 \times 2$
$= 16 - 6 = 10$
$lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} 3 x + 4$
$= 3 \times 2 + 4 = 10$
$∵ lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{+}} f (x)$
Therefore, $f (x)$ continuous at $x = 2$

[JCECE-2014]

Limits, Continuity and Differentiability

79959 The value of $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$ is

1

\frac{1}{3}

2 2

3 1

4

\frac{1}{2}

Explanation:

(C) : Given, $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$
Applying L-Hospital rule,
$= lim_{x \to 0} \frac{\sinh x + \sin x}{x \cdot \cos x + \sin x}$
Again applying L-Hospital rule,
$= lim_{x \to 0} \frac{\cosh x + \cos x}{x \cdot (- \sin x) + \cos x + \cos x} = \frac{\cosh 0^{\circ} + \cos 0^{\circ}}{0 + 2 \cos 0^{\circ}} = \frac{1 + 1}{2} = 1$

UPSEE-2016

Limits, Continuity and Differentiability

79960 Let $f (x) = {\begin{array}{rr} x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x = 0 \end{array}}$ ,
Then, $f (x)$ is continuous but not differentiable at $x = 0$ , if

1

n \in (0, 1)

2

n \in [1, \infty)

3

n \in (- \infty, 0)

4

n = 0

Explanation:

(A) : Given,
$f (x) = x^{n} \sin \frac{1}{x}$
$f (0) = 0$
$∵ f (x)$ is continuous at $x = 0$ ,
Hence, $f (x)$ will not be differentiable if $n \leq 1$ but
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} x^{n} \sin \frac{1}{x} = 0$
$f^{'} (0) = \frac{f (x) - f (0)}{x - 0}$
$f^{'} (0) = \frac{x^{n} \sin \frac{1}{x} - 0}{x}$
$f^{'} (0) = x^{n - 1} \sin \frac{1}{x}$
continuous at $x = 0$
$∴$ Possible values of $n$ is $n \in (0, 1)$

UPSEE-2014

Limits, Continuity and Differentiability

79961 $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ is equal to

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$
$lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{+}} \frac{x}{x + x^{2}} [∵ x > 0]$
$= lim_{x \to 0^{+}} \frac{x}{x (1 + x)} = lim_{x \to 0^{+}} \frac{1}{1 + x} = \frac{1}{1 + 0} = 1$
$lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{-}} \frac{x}{- x + x^{2}} {∵ x < 0}$
$= lim_{x \to 0^{-}} \frac{x}{x (- 1 + x)} = lim_{x \to 0^{-}} \frac{1}{- 1 + x} = \frac{1}{- 1} = - 1$
$∵ lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} \neq lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}}$
Therefore, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ does not exist

UPSEE-2012

Limits, Continuity and Differentiability

79962 If $a_{1} = 1$ and $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}, n \geq 1$ and if $lim_{n \to \infty} a_{n} = a$ , then the value of $a$ is

1

\sqrt{2}

2

- \sqrt{2}

3 2

4 None of these

Explanation:

(A) : Given, $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$ , and $lim_{n \to \infty} a_{n} = a$
On taking both side $lim_{x \to \infty}$ ,
$lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$
$a = \frac{4 + 3 a}{3 + 2 a}$
$3 a + 2 a^{2} = 4 + 3 a$
$a^{2} = 2 \Rightarrow a = \sqrt{2}$

UPSEE-2012

Limits, Continuity and Differentiability

79963 If $f (x) = {\begin{cases} - x^{2}, When x \leq 0 \\ 5 x - 4, When 0 < x \leq 1 \\ 4 x^{2} - 3 x, When 1 < x < 2 \\ 3 x + 4, When x \geq 2 \end{cases}$
then

1

f (x)

is continuous at

x = 0

2

f (x)

is continuous at

x = 2

3

f (x)

is discontinuous at

x = 1

4 None of the above

Explanation:

(B) : Given,
$f (x) = - x^{2}, x \leq 0$
$= 5 x - 4, 0 < x \leq 1$
$= 4 x^{2} - 3 x, 1 < X < 2$
$= 3 x + 4, x \geq 2$
$At, x = 0$
$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (- x^{2}) = 0$
$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} 5 x - 4 = 0 - 4 = - 4$
$\begin{array}{ll} ∵ lim_{x \to 0^{-}} f (x) \neq lim_{x \to 0^{+}} f (x) \\ ∵ f (x) is discontinuous \end{array}$
$∵ f (x)$ is discontinuous at $x = 0$
At, $x = 1$
$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (5 x - 4)$
$= 5 \times 1 - 4 = 1$
$\begin{array}{r} lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} 4 x^{2} - 3 x \\ = 4 - 3 = 1 \end{array}$
$∵ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x)$
$∴ f (x)$ is continuous at $x = 1$
At, $x = 2$
$lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} 4 x^{2} - 3 x$
$= 4 \times 2^{2} - 3 \times 2$
$= 16 - 6 = 10$
$lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} 3 x + 4$
$= 3 \times 2 + 4 = 10$
$∵ lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{+}} f (x)$
Therefore, $f (x)$ continuous at $x = 2$

[JCECE-2014]

Limits, Continuity and Differentiability

79959 The value of $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$ is

1

\frac{1}{3}

2 2

3 1

4

\frac{1}{2}

Explanation:

(C) : Given, $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$
Applying L-Hospital rule,
$= lim_{x \to 0} \frac{\sinh x + \sin x}{x \cdot \cos x + \sin x}$
Again applying L-Hospital rule,
$= lim_{x \to 0} \frac{\cosh x + \cos x}{x \cdot (- \sin x) + \cos x + \cos x} = \frac{\cosh 0^{\circ} + \cos 0^{\circ}}{0 + 2 \cos 0^{\circ}} = \frac{1 + 1}{2} = 1$

UPSEE-2016

Limits, Continuity and Differentiability

79960 Let $f (x) = {\begin{array}{rr} x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x = 0 \end{array}}$ ,
Then, $f (x)$ is continuous but not differentiable at $x = 0$ , if

1

n \in (0, 1)

2

n \in [1, \infty)

3

n \in (- \infty, 0)

4

n = 0

Explanation:

(A) : Given,
$f (x) = x^{n} \sin \frac{1}{x}$
$f (0) = 0$
$∵ f (x)$ is continuous at $x = 0$ ,
Hence, $f (x)$ will not be differentiable if $n \leq 1$ but
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} x^{n} \sin \frac{1}{x} = 0$
$f^{'} (0) = \frac{f (x) - f (0)}{x - 0}$
$f^{'} (0) = \frac{x^{n} \sin \frac{1}{x} - 0}{x}$
$f^{'} (0) = x^{n - 1} \sin \frac{1}{x}$
continuous at $x = 0$
$∴$ Possible values of $n$ is $n \in (0, 1)$

UPSEE-2014

Limits, Continuity and Differentiability

79961 $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ is equal to

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$
$lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{+}} \frac{x}{x + x^{2}} [∵ x > 0]$
$= lim_{x \to 0^{+}} \frac{x}{x (1 + x)} = lim_{x \to 0^{+}} \frac{1}{1 + x} = \frac{1}{1 + 0} = 1$
$lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{-}} \frac{x}{- x + x^{2}} {∵ x < 0}$
$= lim_{x \to 0^{-}} \frac{x}{x (- 1 + x)} = lim_{x \to 0^{-}} \frac{1}{- 1 + x} = \frac{1}{- 1} = - 1$
$∵ lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} \neq lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}}$
Therefore, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ does not exist

UPSEE-2012

Limits, Continuity and Differentiability

79962 If $a_{1} = 1$ and $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}, n \geq 1$ and if $lim_{n \to \infty} a_{n} = a$ , then the value of $a$ is

1

\sqrt{2}

2

- \sqrt{2}

3 2

4 None of these

Explanation:

(A) : Given, $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$ , and $lim_{n \to \infty} a_{n} = a$
On taking both side $lim_{x \to \infty}$ ,
$lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$
$a = \frac{4 + 3 a}{3 + 2 a}$
$3 a + 2 a^{2} = 4 + 3 a$
$a^{2} = 2 \Rightarrow a = \sqrt{2}$

UPSEE-2012

Limits, Continuity and Differentiability

79963 If $f (x) = {\begin{cases} - x^{2}, When x \leq 0 \\ 5 x - 4, When 0 < x \leq 1 \\ 4 x^{2} - 3 x, When 1 < x < 2 \\ 3 x + 4, When x \geq 2 \end{cases}$
then

1

f (x)

is continuous at

x = 0

2

f (x)

is continuous at

x = 2

3

f (x)

is discontinuous at

x = 1

4 None of the above

Explanation:

(B) : Given,
$f (x) = - x^{2}, x \leq 0$
$= 5 x - 4, 0 < x \leq 1$
$= 4 x^{2} - 3 x, 1 < X < 2$
$= 3 x + 4, x \geq 2$
$At, x = 0$
$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (- x^{2}) = 0$
$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} 5 x - 4 = 0 - 4 = - 4$
$\begin{array}{ll} ∵ lim_{x \to 0^{-}} f (x) \neq lim_{x \to 0^{+}} f (x) \\ ∵ f (x) is discontinuous \end{array}$
$∵ f (x)$ is discontinuous at $x = 0$
At, $x = 1$
$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (5 x - 4)$
$= 5 \times 1 - 4 = 1$
$\begin{array}{r} lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} 4 x^{2} - 3 x \\ = 4 - 3 = 1 \end{array}$
$∵ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x)$
$∴ f (x)$ is continuous at $x = 1$
At, $x = 2$
$lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} 4 x^{2} - 3 x$
$= 4 \times 2^{2} - 3 \times 2$
$= 16 - 6 = 10$
$lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} 3 x + 4$
$= 3 \times 2 + 4 = 10$
$∵ lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{+}} f (x)$
Therefore, $f (x)$ continuous at $x = 2$

[JCECE-2014]

Limits, Continuity and Differentiability

79959 The value of $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$ is

1

\frac{1}{3}

2 2

3 1

4

\frac{1}{2}

Explanation:

(C) : Given, $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$
Applying L-Hospital rule,
$= lim_{x \to 0} \frac{\sinh x + \sin x}{x \cdot \cos x + \sin x}$
Again applying L-Hospital rule,
$= lim_{x \to 0} \frac{\cosh x + \cos x}{x \cdot (- \sin x) + \cos x + \cos x} = \frac{\cosh 0^{\circ} + \cos 0^{\circ}}{0 + 2 \cos 0^{\circ}} = \frac{1 + 1}{2} = 1$

UPSEE-2016

Limits, Continuity and Differentiability

79960 Let $f (x) = {\begin{array}{rr} x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x = 0 \end{array}}$ ,
Then, $f (x)$ is continuous but not differentiable at $x = 0$ , if

1

n \in (0, 1)

2

n \in [1, \infty)

3

n \in (- \infty, 0)

4

n = 0

Explanation:

(A) : Given,
$f (x) = x^{n} \sin \frac{1}{x}$
$f (0) = 0$
$∵ f (x)$ is continuous at $x = 0$ ,
Hence, $f (x)$ will not be differentiable if $n \leq 1$ but
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} x^{n} \sin \frac{1}{x} = 0$
$f^{'} (0) = \frac{f (x) - f (0)}{x - 0}$
$f^{'} (0) = \frac{x^{n} \sin \frac{1}{x} - 0}{x}$
$f^{'} (0) = x^{n - 1} \sin \frac{1}{x}$
continuous at $x = 0$
$∴$ Possible values of $n$ is $n \in (0, 1)$

UPSEE-2014

Limits, Continuity and Differentiability

79961 $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ is equal to

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$
$lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{+}} \frac{x}{x + x^{2}} [∵ x > 0]$
$= lim_{x \to 0^{+}} \frac{x}{x (1 + x)} = lim_{x \to 0^{+}} \frac{1}{1 + x} = \frac{1}{1 + 0} = 1$
$lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{-}} \frac{x}{- x + x^{2}} {∵ x < 0}$
$= lim_{x \to 0^{-}} \frac{x}{x (- 1 + x)} = lim_{x \to 0^{-}} \frac{1}{- 1 + x} = \frac{1}{- 1} = - 1$
$∵ lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} \neq lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}}$
Therefore, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ does not exist

UPSEE-2012

Limits, Continuity and Differentiability

79962 If $a_{1} = 1$ and $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}, n \geq 1$ and if $lim_{n \to \infty} a_{n} = a$ , then the value of $a$ is

1

\sqrt{2}

2

- \sqrt{2}

3 2

4 None of these

Explanation:

(A) : Given, $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$ , and $lim_{n \to \infty} a_{n} = a$
On taking both side $lim_{x \to \infty}$ ,
$lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$
$a = \frac{4 + 3 a}{3 + 2 a}$
$3 a + 2 a^{2} = 4 + 3 a$
$a^{2} = 2 \Rightarrow a = \sqrt{2}$

UPSEE-2012

Limits, Continuity and Differentiability

79963 If $f (x) = {\begin{cases} - x^{2}, When x \leq 0 \\ 5 x - 4, When 0 < x \leq 1 \\ 4 x^{2} - 3 x, When 1 < x < 2 \\ 3 x + 4, When x \geq 2 \end{cases}$
then

1

f (x)

is continuous at

x = 0

2

f (x)

is continuous at

x = 2

3

f (x)

is discontinuous at

x = 1

4 None of the above

Explanation:

(B) : Given,
$f (x) = - x^{2}, x \leq 0$
$= 5 x - 4, 0 < x \leq 1$
$= 4 x^{2} - 3 x, 1 < X < 2$
$= 3 x + 4, x \geq 2$
$At, x = 0$
$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (- x^{2}) = 0$
$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} 5 x - 4 = 0 - 4 = - 4$
$\begin{array}{ll} ∵ lim_{x \to 0^{-}} f (x) \neq lim_{x \to 0^{+}} f (x) \\ ∵ f (x) is discontinuous \end{array}$
$∵ f (x)$ is discontinuous at $x = 0$
At, $x = 1$
$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (5 x - 4)$
$= 5 \times 1 - 4 = 1$
$\begin{array}{r} lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} 4 x^{2} - 3 x \\ = 4 - 3 = 1 \end{array}$
$∵ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x)$
$∴ f (x)$ is continuous at $x = 1$
At, $x = 2$
$lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} 4 x^{2} - 3 x$
$= 4 \times 2^{2} - 3 \times 2$
$= 16 - 6 = 10$
$lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} 3 x + 4$
$= 3 \times 2 + 4 = 10$
$∵ lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{+}} f (x)$
Therefore, $f (x)$ continuous at $x = 2$

[JCECE-2014]

Limits, Continuity and Differentiability

79959 The value of $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$ is

1

\frac{1}{3}

2 2

3 1

4

\frac{1}{2}

Explanation:

(C) : Given, $lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x}$
Applying L-Hospital rule,
$= lim_{x \to 0} \frac{\sinh x + \sin x}{x \cdot \cos x + \sin x}$
Again applying L-Hospital rule,
$= lim_{x \to 0} \frac{\cosh x + \cos x}{x \cdot (- \sin x) + \cos x + \cos x} = \frac{\cosh 0^{\circ} + \cos 0^{\circ}}{0 + 2 \cos 0^{\circ}} = \frac{1 + 1}{2} = 1$

UPSEE-2016

Limits, Continuity and Differentiability

79960 Let $f (x) = {\begin{array}{rr} x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x = 0 \end{array}}$ ,
Then, $f (x)$ is continuous but not differentiable at $x = 0$ , if

1

n \in (0, 1)

2

n \in [1, \infty)

3

n \in (- \infty, 0)

4

n = 0

Explanation:

(A) : Given,
$f (x) = x^{n} \sin \frac{1}{x}$
$f (0) = 0$
$∵ f (x)$ is continuous at $x = 0$ ,
Hence, $f (x)$ will not be differentiable if $n \leq 1$ but
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} x^{n} \sin \frac{1}{x} = 0$
$f^{'} (0) = \frac{f (x) - f (0)}{x - 0}$
$f^{'} (0) = \frac{x^{n} \sin \frac{1}{x} - 0}{x}$
$f^{'} (0) = x^{n - 1} \sin \frac{1}{x}$
continuous at $x = 0$
$∴$ Possible values of $n$ is $n \in (0, 1)$

UPSEE-2014

Limits, Continuity and Differentiability

79961 $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ is equal to

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$
$lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{+}} \frac{x}{x + x^{2}} [∵ x > 0]$
$= lim_{x \to 0^{+}} \frac{x}{x (1 + x)} = lim_{x \to 0^{+}} \frac{1}{1 + x} = \frac{1}{1 + 0} = 1$
$lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}} = lim_{x \to 0^{-}} \frac{x}{- x + x^{2}} {∵ x < 0}$
$= lim_{x \to 0^{-}} \frac{x}{x (- 1 + x)} = lim_{x \to 0^{-}} \frac{1}{- 1 + x} = \frac{1}{- 1} = - 1$
$∵ lim_{x \to 0^{+}} \frac{x}{| x | + x^{2}} \neq lim_{x \to 0^{-}} \frac{x}{| x | + x^{2}}$
Therefore, $lim_{x \to 0} \frac{x}{| x | + x^{2}}$ does not exist

UPSEE-2012

Limits, Continuity and Differentiability

79962 If $a_{1} = 1$ and $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}, n \geq 1$ and if $lim_{n \to \infty} a_{n} = a$ , then the value of $a$ is

1

\sqrt{2}

2

- \sqrt{2}

3 2

4 None of these

Explanation:

(A) : Given, $a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$ , and $lim_{n \to \infty} a_{n} = a$
On taking both side $lim_{x \to \infty}$ ,
$lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \frac{4 + 3 a_{n}}{3 + 2 a_{n}}$
$a = \frac{4 + 3 a}{3 + 2 a}$
$3 a + 2 a^{2} = 4 + 3 a$
$a^{2} = 2 \Rightarrow a = \sqrt{2}$

UPSEE-2012

Limits, Continuity and Differentiability

79963 If $f (x) = {\begin{cases} - x^{2}, When x \leq 0 \\ 5 x - 4, When 0 < x \leq 1 \\ 4 x^{2} - 3 x, When 1 < x < 2 \\ 3 x + 4, When x \geq 2 \end{cases}$
then

1

f (x)

is continuous at

x = 0

2

f (x)

is continuous at

x = 2

3

f (x)

is discontinuous at

x = 1

4 None of the above

Explanation:

(B) : Given,
$f (x) = - x^{2}, x \leq 0$
$= 5 x - 4, 0 < x \leq 1$
$= 4 x^{2} - 3 x, 1 < X < 2$
$= 3 x + 4, x \geq 2$
$At, x = 0$
$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (- x^{2}) = 0$
$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} 5 x - 4 = 0 - 4 = - 4$
$\begin{array}{ll} ∵ lim_{x \to 0^{-}} f (x) \neq lim_{x \to 0^{+}} f (x) \\ ∵ f (x) is discontinuous \end{array}$
$∵ f (x)$ is discontinuous at $x = 0$
At, $x = 1$
$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (5 x - 4)$
$= 5 \times 1 - 4 = 1$
$\begin{array}{r} lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} 4 x^{2} - 3 x \\ = 4 - 3 = 1 \end{array}$
$∵ lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x)$
$∴ f (x)$ is continuous at $x = 1$
At, $x = 2$
$lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} 4 x^{2} - 3 x$
$= 4 \times 2^{2} - 3 \times 2$
$= 16 - 6 = 10$
$lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} 3 x + 4$
$= 3 \times 2 + 4 = 10$
$∵ lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{+}} f (x)$
Therefore, $f (x)$ continuous at $x = 2$

[JCECE-2014]

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