QBManager

Sets, Relation and Function

117305 The domain of the
function $f(x)=\frac{\sqrt{4-x^2}}{\cos ^{-1}(2-x)}$ is:

1 $[0,2]$

2 $[0,2)$

3 $[1,2)$

4 $[1,2]$

UPSEE-2015

Sets, Relation and Function

117306 Let $f=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}$ be a function from $R$ to $R$. Determine the range of $\boldsymbol f$ .

1

[0, 1)

2

[0, - 1]

3

[0, 2)

4 None of these

Explanation:

A Given,
$f (x) = {(x, \frac{x^{2}}{1 + x^{2}}) : x \in R}$ be a function from $R$ to $R$ .
Consider, $y = \frac{x^{2}}{1 + x^{2}}$
$y + y^{\frac{1}{2} + x^{2}} = x^{2}$
$y = x^{2} - y x^{2}$
$y = x^{2} (1 - y) \Rightarrow x^{2} = \frac{y}{1 - y}$
$x = \sqrt{\frac{y}{1 - y}}$
$∵ x$ is real, then -
$\frac{y}{1 - y} \geq 0$
$\frac{y (1 - y)}{(1 - y)^{2}} \geq 0$
Then,
$y (1 - y) \geq 0 and (1 - y)^{2} > 0$
$0 \leq y \leq 1$ and $- y > - 1$
$0 \leq y \leq 1$ and $y < 1$
original image
So, common interval $=$ range of $f = 0 \leq y \leq 1 = [0, 1)$

UPSEE-2014

Sets, Relation and Function

117307 The greatest value of
$f (x) = (x + 1)^{1 / 3} - (x - 1)^{1 / 3}$ on $[0, 1]$ is

1 1

2 2

3 3

4

\frac{1}{3}

Explanation:

B Given,
$f (x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$
Both side difference-
Then,
$f^{'} (x) = [\frac{1}{3} \times \frac{1}{(x + 1)^{\frac{2}{3}}} - \frac{1}{3} \times \frac{1}{(x + 1)^{\frac{2}{3}}}]$
$f^{'} (x) = \frac{1}{3} (x + 1)^{\frac{- 2}{3}} - \frac{1}{3} (x - 1)^{\frac{- 2}{3}}$
$= \frac{(x - 1)^{\frac{2}{3}} - (x + 1)^{\frac{2}{3}}}{3 (x + 1)^{\frac{2}{3}}}$
$f^{'} (x)$ does exist at $x = \pm 1$
$f^{'} (x) = 0$
$(x - 1)^{\frac{2}{3}} = (x + 1)^{\frac{2}{3}}$
$x = 0$
$∵ f^{'} (x) \neq 0$ for any other values of $x \in [0, 1]$ . the value of $f (x)$ at $x = 0$ is 2 .
Hence, the greatest value of $f (x)$ is 2 .

UPSEE-2010

Sets, Relation and Function

117308 The set of points where the function $f (x) = x | x |$ is differentiable is

1

(- \infty, \infty)

2

(- \infty, 0) \cup (0, \infty)

3

(0, \infty)

4

[0, \infty]

Explanation:

A Given,
UPSE 2010
$\begin{aligned} f (x) = x | x | \\ f (x) = x | x | = {\begin{array}{cc} x^{2}, & x \geq 0 \\ - x^{2}, & x < 0 \end{array} \end{aligned}$
$L.H.D = lim_{x \to \overset{―}{0}} \frac{f (x) - 0}{x - 0}$
$lim_{x \to \overset{―}{0}} \frac{f (- h)}{- h - 0} = \frac{h^{2}}{h} = 0$
$R.H.D = lim_{x \to \overset{―}{0}} \frac{f (h)}{h} = \frac{h^{2}}{h} = 0$

UPSEE-2010

Sets, Relation and Function

117305 The domain of the
function $f (x) = \frac{\sqrt{4 - x^{2}}}{\cos^{- 1} (2 - x)}$ is:

1

[0, 2]

2

[0, 2)

3

[1, 2)

4

[1, 2]

Explanation:

C Given, $f (x) = \frac{\sqrt{4 - x^{2}}}{\cos^{- 1} (2 - x)}$
For domain $f (x)$ be defined $4 - x^{2} \geq 0, - 1 \leq 2 - x \leq 1$ and $2 - x$
$x^{2} \leq 4, - 1 \leq 2 - x \leq 1$
$- 2 \leq x \leq 2$ and $- 3 \leq - x \leq - 1$
$- 2 \leq x \leq 2$ and $1 \leq x \leq 3$
Then,
original image
So, common interval $=$ domain of $f (x) = [1, 2)$

UPSEE-2015

Sets, Relation and Function

117306 Let $f = {(x, \frac{x^{2}}{1 + x^{2}}) : x \in R}$ be a function from $R$ to $R$ . Determine the range of $\boldsymbol f$ .

1

[0, 1)

2

[0, - 1]

3

[0, 2)

4 None of these

Explanation:

A Given,
$f (x) = {(x, \frac{x^{2}}{1 + x^{2}}) : x \in R}$ be a function from $R$ to $R$ .
Consider, $y = \frac{x^{2}}{1 + x^{2}}$
$y + y^{\frac{1}{2} + x^{2}} = x^{2}$
$y = x^{2} - y x^{2}$
$y = x^{2} (1 - y) \Rightarrow x^{2} = \frac{y}{1 - y}$
$x = \sqrt{\frac{y}{1 - y}}$
$∵ x$ is real, then -
$\frac{y}{1 - y} \geq 0$
$\frac{y (1 - y)}{(1 - y)^{2}} \geq 0$
Then,
$y (1 - y) \geq 0 and (1 - y)^{2} > 0$
$0 \leq y \leq 1$ and $- y > - 1$
$0 \leq y \leq 1$ and $y < 1$
original image
So, common interval $=$ range of $f = 0 \leq y \leq 1 = [0, 1)$

UPSEE-2014

Sets, Relation and Function

117307 The greatest value of
$f (x) = (x + 1)^{1 / 3} - (x - 1)^{1 / 3}$ on $[0, 1]$ is

1 1

2 2

3 3

4

\frac{1}{3}

Explanation:

B Given,
$f (x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$
Both side difference-
Then,
$f^{'} (x) = [\frac{1}{3} \times \frac{1}{(x + 1)^{\frac{2}{3}}} - \frac{1}{3} \times \frac{1}{(x + 1)^{\frac{2}{3}}}]$
$f^{'} (x) = \frac{1}{3} (x + 1)^{\frac{- 2}{3}} - \frac{1}{3} (x - 1)^{\frac{- 2}{3}}$
$= \frac{(x - 1)^{\frac{2}{3}} - (x + 1)^{\frac{2}{3}}}{3 (x + 1)^{\frac{2}{3}}}$
$f^{'} (x)$ does exist at $x = \pm 1$
$f^{'} (x) = 0$
$(x - 1)^{\frac{2}{3}} = (x + 1)^{\frac{2}{3}}$
$x = 0$
$∵ f^{'} (x) \neq 0$ for any other values of $x \in [0, 1]$ . the value of $f (x)$ at $x = 0$ is 2 .
Hence, the greatest value of $f (x)$ is 2 .

UPSEE-2010

Sets, Relation and Function

117308 The set of points where the function $f (x) = x | x |$ is differentiable is

1

(- \infty, \infty)

2

(- \infty, 0) \cup (0, \infty)

3

(0, \infty)

4

[0, \infty]

Explanation:

A Given,
UPSE 2010
$\begin{aligned} f (x) = x | x | \\ f (x) = x | x | = {\begin{array}{cc} x^{2}, & x \geq 0 \\ - x^{2}, & x < 0 \end{array} \end{aligned}$
$L.H.D = lim_{x \to \overset{―}{0}} \frac{f (x) - 0}{x - 0}$
$lim_{x \to \overset{―}{0}} \frac{f (- h)}{- h - 0} = \frac{h^{2}}{h} = 0$
$R.H.D = lim_{x \to \overset{―}{0}} \frac{f (h)}{h} = \frac{h^{2}}{h} = 0$

UPSEE-2010

Sets, Relation and Function

117305 The domain of the
function $f (x) = \frac{\sqrt{4 - x^{2}}}{\cos^{- 1} (2 - x)}$ is:

1

[0, 2]

2

[0, 2)

3

[1, 2)

4

[1, 2]

Explanation:

C Given, $f (x) = \frac{\sqrt{4 - x^{2}}}{\cos^{- 1} (2 - x)}$
For domain $f (x)$ be defined $4 - x^{2} \geq 0, - 1 \leq 2 - x \leq 1$ and $2 - x$
$x^{2} \leq 4, - 1 \leq 2 - x \leq 1$
$- 2 \leq x \leq 2$ and $- 3 \leq - x \leq - 1$
$- 2 \leq x \leq 2$ and $1 \leq x \leq 3$
Then,
original image
So, common interval $=$ domain of $f (x) = [1, 2)$

UPSEE-2015

Sets, Relation and Function

117306 Let $f = {(x, \frac{x^{2}}{1 + x^{2}}) : x \in R}$ be a function from $R$ to $R$ . Determine the range of $\boldsymbol f$ .

1

[0, 1)

2

[0, - 1]

3

[0, 2)

4 None of these

Explanation:

A Given,
$f (x) = {(x, \frac{x^{2}}{1 + x^{2}}) : x \in R}$ be a function from $R$ to $R$ .
Consider, $y = \frac{x^{2}}{1 + x^{2}}$
$y + y^{\frac{1}{2} + x^{2}} = x^{2}$
$y = x^{2} - y x^{2}$
$y = x^{2} (1 - y) \Rightarrow x^{2} = \frac{y}{1 - y}$
$x = \sqrt{\frac{y}{1 - y}}$
$∵ x$ is real, then -
$\frac{y}{1 - y} \geq 0$
$\frac{y (1 - y)}{(1 - y)^{2}} \geq 0$
Then,
$y (1 - y) \geq 0 and (1 - y)^{2} > 0$
$0 \leq y \leq 1$ and $- y > - 1$
$0 \leq y \leq 1$ and $y < 1$
original image
So, common interval $=$ range of $f = 0 \leq y \leq 1 = [0, 1)$

UPSEE-2014

Sets, Relation and Function

117307 The greatest value of
$f (x) = (x + 1)^{1 / 3} - (x - 1)^{1 / 3}$ on $[0, 1]$ is

1 1

2 2

3 3

4

\frac{1}{3}

Explanation:

B Given,
$f (x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$
Both side difference-
Then,
$f^{'} (x) = [\frac{1}{3} \times \frac{1}{(x + 1)^{\frac{2}{3}}} - \frac{1}{3} \times \frac{1}{(x + 1)^{\frac{2}{3}}}]$
$f^{'} (x) = \frac{1}{3} (x + 1)^{\frac{- 2}{3}} - \frac{1}{3} (x - 1)^{\frac{- 2}{3}}$
$= \frac{(x - 1)^{\frac{2}{3}} - (x + 1)^{\frac{2}{3}}}{3 (x + 1)^{\frac{2}{3}}}$
$f^{'} (x)$ does exist at $x = \pm 1$
$f^{'} (x) = 0$
$(x - 1)^{\frac{2}{3}} = (x + 1)^{\frac{2}{3}}$
$x = 0$
$∵ f^{'} (x) \neq 0$ for any other values of $x \in [0, 1]$ . the value of $f (x)$ at $x = 0$ is 2 .
Hence, the greatest value of $f (x)$ is 2 .

UPSEE-2010

Sets, Relation and Function

117308 The set of points where the function $f (x) = x | x |$ is differentiable is

1

(- \infty, \infty)

2

(- \infty, 0) \cup (0, \infty)

3

(0, \infty)

4

[0, \infty]

Explanation:

A Given,
UPSE 2010
$\begin{aligned} f (x) = x | x | \\ f (x) = x | x | = {\begin{array}{cc} x^{2}, & x \geq 0 \\ - x^{2}, & x < 0 \end{array} \end{aligned}$
$L.H.D = lim_{x \to \overset{―}{0}} \frac{f (x) - 0}{x - 0}$
$lim_{x \to \overset{―}{0}} \frac{f (- h)}{- h - 0} = \frac{h^{2}}{h} = 0$
$R.H.D = lim_{x \to \overset{―}{0}} \frac{f (h)}{h} = \frac{h^{2}}{h} = 0$

UPSEE-2010

Sets, Relation and Function

117305 The domain of the
function $f (x) = \frac{\sqrt{4 - x^{2}}}{\cos^{- 1} (2 - x)}$ is:

1

[0, 2]

2

[0, 2)

3

[1, 2)

4

[1, 2]

Explanation:

C Given, $f (x) = \frac{\sqrt{4 - x^{2}}}{\cos^{- 1} (2 - x)}$
For domain $f (x)$ be defined $4 - x^{2} \geq 0, - 1 \leq 2 - x \leq 1$ and $2 - x$
$x^{2} \leq 4, - 1 \leq 2 - x \leq 1$
$- 2 \leq x \leq 2$ and $- 3 \leq - x \leq - 1$
$- 2 \leq x \leq 2$ and $1 \leq x \leq 3$
Then,
original image
So, common interval $=$ domain of $f (x) = [1, 2)$

UPSEE-2015

Sets, Relation and Function

117306 Let $f = {(x, \frac{x^{2}}{1 + x^{2}}) : x \in R}$ be a function from $R$ to $R$ . Determine the range of $\boldsymbol f$ .

1

[0, 1)

2

[0, - 1]

3

[0, 2)

4 None of these

Explanation:

A Given,
$f (x) = {(x, \frac{x^{2}}{1 + x^{2}}) : x \in R}$ be a function from $R$ to $R$ .
Consider, $y = \frac{x^{2}}{1 + x^{2}}$
$y + y^{\frac{1}{2} + x^{2}} = x^{2}$
$y = x^{2} - y x^{2}$
$y = x^{2} (1 - y) \Rightarrow x^{2} = \frac{y}{1 - y}$
$x = \sqrt{\frac{y}{1 - y}}$
$∵ x$ is real, then -
$\frac{y}{1 - y} \geq 0$
$\frac{y (1 - y)}{(1 - y)^{2}} \geq 0$
Then,
$y (1 - y) \geq 0 and (1 - y)^{2} > 0$
$0 \leq y \leq 1$ and $- y > - 1$
$0 \leq y \leq 1$ and $y < 1$
original image
So, common interval $=$ range of $f = 0 \leq y \leq 1 = [0, 1)$

UPSEE-2014

Sets, Relation and Function

117307 The greatest value of
$f (x) = (x + 1)^{1 / 3} - (x - 1)^{1 / 3}$ on $[0, 1]$ is

1 1

2 2

3 3

4

\frac{1}{3}

Explanation:

B Given,
$f (x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$
Both side difference-
Then,
$f^{'} (x) = [\frac{1}{3} \times \frac{1}{(x + 1)^{\frac{2}{3}}} - \frac{1}{3} \times \frac{1}{(x + 1)^{\frac{2}{3}}}]$
$f^{'} (x) = \frac{1}{3} (x + 1)^{\frac{- 2}{3}} - \frac{1}{3} (x - 1)^{\frac{- 2}{3}}$
$= \frac{(x - 1)^{\frac{2}{3}} - (x + 1)^{\frac{2}{3}}}{3 (x + 1)^{\frac{2}{3}}}$
$f^{'} (x)$ does exist at $x = \pm 1$
$f^{'} (x) = 0$
$(x - 1)^{\frac{2}{3}} = (x + 1)^{\frac{2}{3}}$
$x = 0$
$∵ f^{'} (x) \neq 0$ for any other values of $x \in [0, 1]$ . the value of $f (x)$ at $x = 0$ is 2 .
Hence, the greatest value of $f (x)$ is 2 .

UPSEE-2010

Sets, Relation and Function

117308 The set of points where the function $f (x) = x | x |$ is differentiable is

1

(- \infty, \infty)

2

(- \infty, 0) \cup (0, \infty)

3

(0, \infty)

4

[0, \infty]

Explanation:

A Given,
UPSE 2010
$\begin{aligned} f (x) = x | x | \\ f (x) = x | x | = {\begin{array}{cc} x^{2}, & x \geq 0 \\ - x^{2}, & x < 0 \end{array} \end{aligned}$
$L.H.D = lim_{x \to \overset{―}{0}} \frac{f (x) - 0}{x - 0}$
$lim_{x \to \overset{―}{0}} \frac{f (- h)}{- h - 0} = \frac{h^{2}}{h} = 0$
$R.H.D = lim_{x \to \overset{―}{0}} \frac{f (h)}{h} = \frac{h^{2}}{h} = 0$

UPSEE-2010

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