QBManager

Limits, Continuity and Differentiability

79964 If $f (x) = \frac{x}{1 + x} + \frac{x}{(1 + x) (1 + 2 x)} + \frac{x}{(1 + 2 x) (1 + 3 x)}$ $+ \dots \infty$ , then

1

f (x)

is continuous for all

x

2

f (x)

is discontinuous for finite number of point

3

f (x)

is continuous for finite number of point

4 None of the above

Explanation:

(B) : Given,
$f (x) = \frac{x}{1 + x} + \frac{x}{(1 + x) (1 + 2 x)} + \frac{x}{(1 + 2 x) (1 + 3 x)} + \dots \dots \infty$
$f (0) = 0$
$lim_{x \to 0} f (x) = lim_{x \to 0} \frac{(x + 1) - 1}{1 + x} + \frac{(2 x + 1) - (x + 1)}{(x + 1) (2 x + 1)} + \frac{(3 x + 1) - (2 x + 1)}{(3 x + 1) (2 x + 1)} + \dots . \infty$
$= lim_{x \to 0} 1 - \frac{1}{1 + x} + \frac{1}{x + 1} - \frac{1}{2 x + 1} + \frac{1}{2 x + 1} - \frac{1}{3 x + 1} + \dots \dots \infty$
$= lim 1 = 1$
$∵ f (0) \neq lim_{x \to 0} f (x)$
$∴ f (x)$ is discontinuous for finite number of point

[JCECE-2013]

Limits, Continuity and Differentiability

79965 If $y = \frac{1}{t^{2} - t - 6}$ and $t = \frac{1}{x - 2}$ , then the value of $x$ which make the function $y$ discontinuous are

1

2, \frac{2}{3}, \frac{7}{3}

2

2, \frac{3}{2}, \frac{7}{3}

3

2, \frac{3}{2}, \frac{3}{7}

4 None of the above

Explanation:

(B) : Given,
$y = \frac{1}{t^{2} - t - 6} and t = \frac{1}{x - 2}$
$∵ t^{2} - t - 6 = 0$
$t^{2} - 3 t + 2 t - 6 = 0$
$t (t - 3) + 2 (t - 3) = 0$
$(t - 3) (t + 2) = 0$
$t = 3, - 2$
When, $t = - 2$
$\frac{1}{x - 2} = - 2 \Rightarrow x - 2 = - \frac{1}{2}$
$x = 2 - \frac{1}{2} \Rightarrow x = 3 / 2$
When, $t = 3$
$\frac{1}{x - 2} = 3 \Rightarrow x - 2 = \frac{1}{3}$
$x = 2 + \frac{1}{3} \Rightarrow x = 7 / 3$
When, $x - 2 = 0 \Rightarrow x = 2$
The value of $x$ which make the function $y$ discontinuous are $x = 2, 3 / 2$ and $7 / 3$ .

[JCECE-2012]

Limits, Continuity and Differentiability

79966 A function is defined as follows
$f (x) = {\begin{array}{cc} x^{m} \sin (\frac{1}{x}), x \neq 0 \\ 0, x = 0 \end{array}}$ what condition should be imposed on $m$ , so that $f (x)$ may be continuous at $x = 0$ ?

1

m > 0

2

m < 0

3

m = 0

4 any value of

m

Explanation:

(A) : Given, $f (x) = x^{m} \sin (\frac{1}{x})$
$f (0) = 0$
For $f (x)$ is continuous at $x = 0$
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} x^{m} \sin \frac{1}{x} = 0$
Which is possible only when $m > 0$
Hence, $f (x)$ is continuous when $m > 0$

[JCECE-2011]

Limits, Continuity and Differentiability

79968 Let $f (x) = {\begin{cases} \frac{\sin π x}{5 x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ if $f (x)$ is continuous at $x = 0$ , then $k$ is equal to

1

\frac{π}{5}

2

\frac{5}{π}

3 1

4 0

Explanation:

(A) : Given, $f (x) = \frac{\sin π x}{5 x}, x \neq 0$
$f (0) = k, x = 0$
$∵ f (x)$ is continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x) .$
$k = lim_{x \to 0} \frac{\sin π x}{5 x} \Rightarrow k = lim_{x \to 0} \frac{\sin π x}{5 π x} \times π$
$k = \frac{π}{5} lim_{x \to 0} \frac{\sin π x}{π x} [∵ lim_{x \to 0} \frac{\sin x}{x} = 1]$
$k = \frac{π}{5}$

UPSEE-2007

Limits, Continuity and Differentiability

79964 If $f (x) = \frac{x}{1 + x} + \frac{x}{(1 + x) (1 + 2 x)} + \frac{x}{(1 + 2 x) (1 + 3 x)}$ $+ \dots \infty$ , then

1

f (x)

is continuous for all

x

2

f (x)

is discontinuous for finite number of point

3

f (x)

is continuous for finite number of point

4 None of the above

Explanation:

(B) : Given,
$f (x) = \frac{x}{1 + x} + \frac{x}{(1 + x) (1 + 2 x)} + \frac{x}{(1 + 2 x) (1 + 3 x)} + \dots \dots \infty$
$f (0) = 0$
$lim_{x \to 0} f (x) = lim_{x \to 0} \frac{(x + 1) - 1}{1 + x} + \frac{(2 x + 1) - (x + 1)}{(x + 1) (2 x + 1)} + \frac{(3 x + 1) - (2 x + 1)}{(3 x + 1) (2 x + 1)} + \dots . \infty$
$= lim_{x \to 0} 1 - \frac{1}{1 + x} + \frac{1}{x + 1} - \frac{1}{2 x + 1} + \frac{1}{2 x + 1} - \frac{1}{3 x + 1} + \dots \dots \infty$
$= lim 1 = 1$
$∵ f (0) \neq lim_{x \to 0} f (x)$
$∴ f (x)$ is discontinuous for finite number of point

[JCECE-2013]

Limits, Continuity and Differentiability

79965 If $y = \frac{1}{t^{2} - t - 6}$ and $t = \frac{1}{x - 2}$ , then the value of $x$ which make the function $y$ discontinuous are

1

2, \frac{2}{3}, \frac{7}{3}

2

2, \frac{3}{2}, \frac{7}{3}

3

2, \frac{3}{2}, \frac{3}{7}

4 None of the above

Explanation:

(B) : Given,
$y = \frac{1}{t^{2} - t - 6} and t = \frac{1}{x - 2}$
$∵ t^{2} - t - 6 = 0$
$t^{2} - 3 t + 2 t - 6 = 0$
$t (t - 3) + 2 (t - 3) = 0$
$(t - 3) (t + 2) = 0$
$t = 3, - 2$
When, $t = - 2$
$\frac{1}{x - 2} = - 2 \Rightarrow x - 2 = - \frac{1}{2}$
$x = 2 - \frac{1}{2} \Rightarrow x = 3 / 2$
When, $t = 3$
$\frac{1}{x - 2} = 3 \Rightarrow x - 2 = \frac{1}{3}$
$x = 2 + \frac{1}{3} \Rightarrow x = 7 / 3$
When, $x - 2 = 0 \Rightarrow x = 2$
The value of $x$ which make the function $y$ discontinuous are $x = 2, 3 / 2$ and $7 / 3$ .

[JCECE-2012]

Limits, Continuity and Differentiability

79966 A function is defined as follows
$f (x) = {\begin{array}{cc} x^{m} \sin (\frac{1}{x}), x \neq 0 \\ 0, x = 0 \end{array}}$ what condition should be imposed on $m$ , so that $f (x)$ may be continuous at $x = 0$ ?

1

m > 0

2

m < 0

3

m = 0

4 any value of

m

Explanation:

(A) : Given, $f (x) = x^{m} \sin (\frac{1}{x})$
$f (0) = 0$
For $f (x)$ is continuous at $x = 0$
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} x^{m} \sin \frac{1}{x} = 0$
Which is possible only when $m > 0$
Hence, $f (x)$ is continuous when $m > 0$

[JCECE-2011]

Limits, Continuity and Differentiability

79968 Let $f (x) = {\begin{cases} \frac{\sin π x}{5 x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ if $f (x)$ is continuous at $x = 0$ , then $k$ is equal to

1

\frac{π}{5}

2

\frac{5}{π}

3 1

4 0

Explanation:

(A) : Given, $f (x) = \frac{\sin π x}{5 x}, x \neq 0$
$f (0) = k, x = 0$
$∵ f (x)$ is continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x) .$
$k = lim_{x \to 0} \frac{\sin π x}{5 x} \Rightarrow k = lim_{x \to 0} \frac{\sin π x}{5 π x} \times π$
$k = \frac{π}{5} lim_{x \to 0} \frac{\sin π x}{π x} [∵ lim_{x \to 0} \frac{\sin x}{x} = 1]$
$k = \frac{π}{5}$

UPSEE-2007

Limits, Continuity and Differentiability

79964 If $f (x) = \frac{x}{1 + x} + \frac{x}{(1 + x) (1 + 2 x)} + \frac{x}{(1 + 2 x) (1 + 3 x)}$ $+ \dots \infty$ , then

1

f (x)

is continuous for all

x

2

f (x)

is discontinuous for finite number of point

3

f (x)

is continuous for finite number of point

4 None of the above

Explanation:

(B) : Given,
$f (x) = \frac{x}{1 + x} + \frac{x}{(1 + x) (1 + 2 x)} + \frac{x}{(1 + 2 x) (1 + 3 x)} + \dots \dots \infty$
$f (0) = 0$
$lim_{x \to 0} f (x) = lim_{x \to 0} \frac{(x + 1) - 1}{1 + x} + \frac{(2 x + 1) - (x + 1)}{(x + 1) (2 x + 1)} + \frac{(3 x + 1) - (2 x + 1)}{(3 x + 1) (2 x + 1)} + \dots . \infty$
$= lim_{x \to 0} 1 - \frac{1}{1 + x} + \frac{1}{x + 1} - \frac{1}{2 x + 1} + \frac{1}{2 x + 1} - \frac{1}{3 x + 1} + \dots \dots \infty$
$= lim 1 = 1$
$∵ f (0) \neq lim_{x \to 0} f (x)$
$∴ f (x)$ is discontinuous for finite number of point

[JCECE-2013]

Limits, Continuity and Differentiability

79965 If $y = \frac{1}{t^{2} - t - 6}$ and $t = \frac{1}{x - 2}$ , then the value of $x$ which make the function $y$ discontinuous are

1

2, \frac{2}{3}, \frac{7}{3}

2

2, \frac{3}{2}, \frac{7}{3}

3

2, \frac{3}{2}, \frac{3}{7}

4 None of the above

Explanation:

(B) : Given,
$y = \frac{1}{t^{2} - t - 6} and t = \frac{1}{x - 2}$
$∵ t^{2} - t - 6 = 0$
$t^{2} - 3 t + 2 t - 6 = 0$
$t (t - 3) + 2 (t - 3) = 0$
$(t - 3) (t + 2) = 0$
$t = 3, - 2$
When, $t = - 2$
$\frac{1}{x - 2} = - 2 \Rightarrow x - 2 = - \frac{1}{2}$
$x = 2 - \frac{1}{2} \Rightarrow x = 3 / 2$
When, $t = 3$
$\frac{1}{x - 2} = 3 \Rightarrow x - 2 = \frac{1}{3}$
$x = 2 + \frac{1}{3} \Rightarrow x = 7 / 3$
When, $x - 2 = 0 \Rightarrow x = 2$
The value of $x$ which make the function $y$ discontinuous are $x = 2, 3 / 2$ and $7 / 3$ .

[JCECE-2012]

Limits, Continuity and Differentiability

79966 A function is defined as follows
$f (x) = {\begin{array}{cc} x^{m} \sin (\frac{1}{x}), x \neq 0 \\ 0, x = 0 \end{array}}$ what condition should be imposed on $m$ , so that $f (x)$ may be continuous at $x = 0$ ?

1

m > 0

2

m < 0

3

m = 0

4 any value of

m

Explanation:

(A) : Given, $f (x) = x^{m} \sin (\frac{1}{x})$
$f (0) = 0$
For $f (x)$ is continuous at $x = 0$
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} x^{m} \sin \frac{1}{x} = 0$
Which is possible only when $m > 0$
Hence, $f (x)$ is continuous when $m > 0$

[JCECE-2011]

Limits, Continuity and Differentiability

79968 Let $f (x) = {\begin{cases} \frac{\sin π x}{5 x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ if $f (x)$ is continuous at $x = 0$ , then $k$ is equal to

1

\frac{π}{5}

2

\frac{5}{π}

3 1

4 0

Explanation:

(A) : Given, $f (x) = \frac{\sin π x}{5 x}, x \neq 0$
$f (0) = k, x = 0$
$∵ f (x)$ is continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x) .$
$k = lim_{x \to 0} \frac{\sin π x}{5 x} \Rightarrow k = lim_{x \to 0} \frac{\sin π x}{5 π x} \times π$
$k = \frac{π}{5} lim_{x \to 0} \frac{\sin π x}{π x} [∵ lim_{x \to 0} \frac{\sin x}{x} = 1]$
$k = \frac{π}{5}$

UPSEE-2007

Limits, Continuity and Differentiability

79964 If $f (x) = \frac{x}{1 + x} + \frac{x}{(1 + x) (1 + 2 x)} + \frac{x}{(1 + 2 x) (1 + 3 x)}$ $+ \dots \infty$ , then

1

f (x)

is continuous for all

x

2

f (x)

is discontinuous for finite number of point

3

f (x)

is continuous for finite number of point

4 None of the above

Explanation:

(B) : Given,
$f (x) = \frac{x}{1 + x} + \frac{x}{(1 + x) (1 + 2 x)} + \frac{x}{(1 + 2 x) (1 + 3 x)} + \dots \dots \infty$
$f (0) = 0$
$lim_{x \to 0} f (x) = lim_{x \to 0} \frac{(x + 1) - 1}{1 + x} + \frac{(2 x + 1) - (x + 1)}{(x + 1) (2 x + 1)} + \frac{(3 x + 1) - (2 x + 1)}{(3 x + 1) (2 x + 1)} + \dots . \infty$
$= lim_{x \to 0} 1 - \frac{1}{1 + x} + \frac{1}{x + 1} - \frac{1}{2 x + 1} + \frac{1}{2 x + 1} - \frac{1}{3 x + 1} + \dots \dots \infty$
$= lim 1 = 1$
$∵ f (0) \neq lim_{x \to 0} f (x)$
$∴ f (x)$ is discontinuous for finite number of point

[JCECE-2013]

Limits, Continuity and Differentiability

79965 If $y = \frac{1}{t^{2} - t - 6}$ and $t = \frac{1}{x - 2}$ , then the value of $x$ which make the function $y$ discontinuous are

1

2, \frac{2}{3}, \frac{7}{3}

2

2, \frac{3}{2}, \frac{7}{3}

3

2, \frac{3}{2}, \frac{3}{7}

4 None of the above

Explanation:

(B) : Given,
$y = \frac{1}{t^{2} - t - 6} and t = \frac{1}{x - 2}$
$∵ t^{2} - t - 6 = 0$
$t^{2} - 3 t + 2 t - 6 = 0$
$t (t - 3) + 2 (t - 3) = 0$
$(t - 3) (t + 2) = 0$
$t = 3, - 2$
When, $t = - 2$
$\frac{1}{x - 2} = - 2 \Rightarrow x - 2 = - \frac{1}{2}$
$x = 2 - \frac{1}{2} \Rightarrow x = 3 / 2$
When, $t = 3$
$\frac{1}{x - 2} = 3 \Rightarrow x - 2 = \frac{1}{3}$
$x = 2 + \frac{1}{3} \Rightarrow x = 7 / 3$
When, $x - 2 = 0 \Rightarrow x = 2$
The value of $x$ which make the function $y$ discontinuous are $x = 2, 3 / 2$ and $7 / 3$ .

[JCECE-2012]

Limits, Continuity and Differentiability

79966 A function is defined as follows
$f (x) = {\begin{array}{cc} x^{m} \sin (\frac{1}{x}), x \neq 0 \\ 0, x = 0 \end{array}}$ what condition should be imposed on $m$ , so that $f (x)$ may be continuous at $x = 0$ ?

1

m > 0

2

m < 0

3

m = 0

4 any value of

m

Explanation:

(A) : Given, $f (x) = x^{m} \sin (\frac{1}{x})$
$f (0) = 0$
For $f (x)$ is continuous at $x = 0$
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} x^{m} \sin \frac{1}{x} = 0$
Which is possible only when $m > 0$
Hence, $f (x)$ is continuous when $m > 0$

[JCECE-2011]

Limits, Continuity and Differentiability

79968 Let $f (x) = {\begin{cases} \frac{\sin π x}{5 x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ if $f (x)$ is continuous at $x = 0$ , then $k$ is equal to

1

\frac{π}{5}

2

\frac{5}{π}

3 1

4 0

Explanation:

(A) : Given, $f (x) = \frac{\sin π x}{5 x}, x \neq 0$
$f (0) = k, x = 0$
$∵ f (x)$ is continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x) .$
$k = lim_{x \to 0} \frac{\sin π x}{5 x} \Rightarrow k = lim_{x \to 0} \frac{\sin π x}{5 π x} \times π$
$k = \frac{π}{5} lim_{x \to 0} \frac{\sin π x}{π x} [∵ lim_{x \to 0} \frac{\sin x}{x} = 1]$
$k = \frac{π}{5}$

UPSEE-2007