QBManager

Sets, Relation and Function

117023 For a suitable chosen real constant a, let a function $f : R - {a} \to R$ be defined by $f (x) =$ $\frac{a - x}{a + x}$ . Further suppose that for any real number $x \neq - a$ and $f (x) \neq - a$ , (fof)( $x) = x$ .
Then, $f (- \frac{1}{2})$ is equal to

1

\frac{1}{3}

2

- \frac{1}{3}

3 -3

4 3

Explanation:

D We have,
$f (x) = \frac{a - x}{a + x} [x \in R - {- a}]$
$f o f (x) = x$
$f [f (x)] = x$
$\frac{a - f (x)}{a + f (x)} = x$
$\frac{a - (\frac{a - x}{a + x})}{a + (\frac{a - x}{a + x})} = x$
$\frac{(a^{2} - a) + x (a + 1)}{(a^{2} + a) + x (a - 1)} = x$
$(a^{2} - a) + x (a + 1) = (a^{2} + a) x + x^{2} (a - 1)$
$x^{2} (a - 1) + x (a^{2} + a - a - 1) - a^{2} + a = 0$
$x^{2} (a - 1) + x (a^{2} - 1) - (a^{2} - a) = 0$
$x^{2} + x (a + 1) - a = 0$
$a = 1$
$f (x) = \frac{1 - x}{1 + x}$
$f (- \frac{1}{2}) = \frac{1 - (\frac{- 1}{2})}{1 + (\frac{- 1}{2})}$
$f (- \frac{1}{2}) = \frac{\frac{3}{2}}{\frac{1}{2}}$
$f (\frac{- 1}{2}) = 3$

JEE Main 06.09.2020 Shift-II

Sets, Relation and Function

117025 If $f (x) = \log (\frac{1 + x}{1 - x}), - 1 < x < 1$ , then $f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$ is

1

[f (x)]^{3}

2

[f (x)]^{2}

3

- f (x)

4

f (x)

5

3 f (x)

Explanation:

D Given $f (x) = \log [\frac{(1 + x)}{1 - x}]$
$f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$
$= \log (\frac{1 + \frac{(3 x + x^{3})}{(1 + 3 x^{2})}}{1 - (\frac{3 x + x^{3}}{1 + 3 x^{2}})}) - \log (\frac{1 + \frac{2 x}{1 + x^{2}}}{1 - \frac{2 x}{1 + x^{2}}})$
$= \log (\frac{1 + 3 x^{2} + 3 x + x^{3}}{1 + 3 x^{2} - 3 x - x^{3}}) - \log (\frac{1 + x^{2} + 2 x}{1 + x^{2} - 2 x})$
$= \log {(\frac{1 + x}{1 - x})}^{3} - \log {(\frac{1 + x}{1 - x})}^{2}$
$= 3 \log (\frac{1 + x}{1 - x}) - 2 \log (\frac{1 + x}{1 - x})$
$= \log {(\frac{1 + x}{1 - x})}^{2}$
$= f (x)$

Kerala CEE-2008

Sets, Relation and Function

117026 $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x),}$ solve for $x$ .

1 0

2 3

3 both (a) and (b)

4 0 and 6

Explanation:

A $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x)}$
$\log_{2} (9 - 2^{x}) = (3 - x) [∵ {alog}_{a}^{b} = b]$
$(9 - 2^{x}) = 2^{(3 - x)}$
$(9 - 2^{x}) = \frac{2^{3}}{2^{x}}$
$2^{x} (9 - 2^{x}) = 8$
$- 2^{2 x} + 2^{x} \cdot 9 = 8$
Let $2^{x} = y$ then-
$- y^{2} + 9 y - 8 = 0$
$y^{2} - 9 y + 8 = 0$
$(y - 8) (y - 1) = 0$
$y = 8 or y = 1$
$2^{x} = 2^{3} or 2^{x} = 2^{0}$
So, $x = 3$ or $x = 0$
But $x = 3$ does not satisfy the given equation, since log 0 is not defined.

Manipal UGET-2013

Sets, Relation and Function

117027 The number of positive integral solutions of $x^{2} + 9 < (x + 3)^{2} < 8 x + 25$ , is

1 2

2 3

3 4

4 5

Explanation:

D We have, given inequality as
$x^{2} + 9 < (x + 3)^{2}$
$= x^{2} + 9 < x^{2} + 9 + 6 x$
$= 6 x > 0$
$= x > 0$
Again, $(x + 3)^{2} < 8 x + 25$
$= x^{2} + 9 + 6 x < 8 x + 25$
$= x^{2} - 2 x - 16 < 0$
$= (x - 1)^{2} - 17 < 0$
$= (x - 1)^{2} < 17$
$= x \in (1 - \sqrt{17}, 1 + \sqrt{17})$
Hence, integral value of $x$ are
$1, 2, 3, 4, 5 .$ The number of positive integral solution is 5 .

Manipal UGET-2015

Sets, Relation and Function

117028 The greatest value of the function $f (x) = {xe}^{- x}$ in $[0, \infty)$ , is

1 0

2

\frac{1}{e}

3

- e

4 e

Explanation:

B Given, $f (x) = {xe}^{- x}$
On differentiating both sides w.r.t. $x$ , we get
$f^{'} (x) = e^{- x} (1 - x)$
$f^{'} (x) = 0$
$x = 1$
Now, $f (0) = 0$
$lim_{x \to \infty} f (x) = lim_{x \to \infty} x e^{- x}$
$= lim_{x \to \infty} \frac{x}{e^{x}} = lim_{x \to \infty} \frac{1}{e^{x}} = 0$ Hence, the greatest value of $f (x)$ is $\frac{1}{e}$

Manipal UGET-2015

Sets, Relation and Function

117023 For a suitable chosen real constant a, let a function $f : R - {a} \to R$ be defined by $f (x) =$ $\frac{a - x}{a + x}$ . Further suppose that for any real number $x \neq - a$ and $f (x) \neq - a$ , (fof)( $x) = x$ .
Then, $f (- \frac{1}{2})$ is equal to

1

\frac{1}{3}

2

- \frac{1}{3}

3 -3

4 3

Explanation:

D We have,
$f (x) = \frac{a - x}{a + x} [x \in R - {- a}]$
$f o f (x) = x$
$f [f (x)] = x$
$\frac{a - f (x)}{a + f (x)} = x$
$\frac{a - (\frac{a - x}{a + x})}{a + (\frac{a - x}{a + x})} = x$
$\frac{(a^{2} - a) + x (a + 1)}{(a^{2} + a) + x (a - 1)} = x$
$(a^{2} - a) + x (a + 1) = (a^{2} + a) x + x^{2} (a - 1)$
$x^{2} (a - 1) + x (a^{2} + a - a - 1) - a^{2} + a = 0$
$x^{2} (a - 1) + x (a^{2} - 1) - (a^{2} - a) = 0$
$x^{2} + x (a + 1) - a = 0$
$a = 1$
$f (x) = \frac{1 - x}{1 + x}$
$f (- \frac{1}{2}) = \frac{1 - (\frac{- 1}{2})}{1 + (\frac{- 1}{2})}$
$f (- \frac{1}{2}) = \frac{\frac{3}{2}}{\frac{1}{2}}$
$f (\frac{- 1}{2}) = 3$

JEE Main 06.09.2020 Shift-II

Sets, Relation and Function

117025 If $f (x) = \log (\frac{1 + x}{1 - x}), - 1 < x < 1$ , then $f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$ is

1

[f (x)]^{3}

2

[f (x)]^{2}

3

- f (x)

4

f (x)

5

3 f (x)

Explanation:

D Given $f (x) = \log [\frac{(1 + x)}{1 - x}]$
$f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$
$= \log (\frac{1 + \frac{(3 x + x^{3})}{(1 + 3 x^{2})}}{1 - (\frac{3 x + x^{3}}{1 + 3 x^{2}})}) - \log (\frac{1 + \frac{2 x}{1 + x^{2}}}{1 - \frac{2 x}{1 + x^{2}}})$
$= \log (\frac{1 + 3 x^{2} + 3 x + x^{3}}{1 + 3 x^{2} - 3 x - x^{3}}) - \log (\frac{1 + x^{2} + 2 x}{1 + x^{2} - 2 x})$
$= \log {(\frac{1 + x}{1 - x})}^{3} - \log {(\frac{1 + x}{1 - x})}^{2}$
$= 3 \log (\frac{1 + x}{1 - x}) - 2 \log (\frac{1 + x}{1 - x})$
$= \log {(\frac{1 + x}{1 - x})}^{2}$
$= f (x)$

Kerala CEE-2008

Sets, Relation and Function

117026 $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x),}$ solve for $x$ .

1 0

2 3

3 both (a) and (b)

4 0 and 6

Explanation:

A $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x)}$
$\log_{2} (9 - 2^{x}) = (3 - x) [∵ {alog}_{a}^{b} = b]$
$(9 - 2^{x}) = 2^{(3 - x)}$
$(9 - 2^{x}) = \frac{2^{3}}{2^{x}}$
$2^{x} (9 - 2^{x}) = 8$
$- 2^{2 x} + 2^{x} \cdot 9 = 8$
Let $2^{x} = y$ then-
$- y^{2} + 9 y - 8 = 0$
$y^{2} - 9 y + 8 = 0$
$(y - 8) (y - 1) = 0$
$y = 8 or y = 1$
$2^{x} = 2^{3} or 2^{x} = 2^{0}$
So, $x = 3$ or $x = 0$
But $x = 3$ does not satisfy the given equation, since log 0 is not defined.

Manipal UGET-2013

Sets, Relation and Function

117027 The number of positive integral solutions of $x^{2} + 9 < (x + 3)^{2} < 8 x + 25$ , is

1 2

2 3

3 4

4 5

Explanation:

D We have, given inequality as
$x^{2} + 9 < (x + 3)^{2}$
$= x^{2} + 9 < x^{2} + 9 + 6 x$
$= 6 x > 0$
$= x > 0$
Again, $(x + 3)^{2} < 8 x + 25$
$= x^{2} + 9 + 6 x < 8 x + 25$
$= x^{2} - 2 x - 16 < 0$
$= (x - 1)^{2} - 17 < 0$
$= (x - 1)^{2} < 17$
$= x \in (1 - \sqrt{17}, 1 + \sqrt{17})$
Hence, integral value of $x$ are
$1, 2, 3, 4, 5 .$ The number of positive integral solution is 5 .

Manipal UGET-2015

Sets, Relation and Function

117028 The greatest value of the function $f (x) = {xe}^{- x}$ in $[0, \infty)$ , is

1 0

2

\frac{1}{e}

3

- e

4 e

Explanation:

B Given, $f (x) = {xe}^{- x}$
On differentiating both sides w.r.t. $x$ , we get
$f^{'} (x) = e^{- x} (1 - x)$
$f^{'} (x) = 0$
$x = 1$
Now, $f (0) = 0$
$lim_{x \to \infty} f (x) = lim_{x \to \infty} x e^{- x}$
$= lim_{x \to \infty} \frac{x}{e^{x}} = lim_{x \to \infty} \frac{1}{e^{x}} = 0$ Hence, the greatest value of $f (x)$ is $\frac{1}{e}$

Manipal UGET-2015

Sets, Relation and Function

117023 For a suitable chosen real constant a, let a function $f : R - {a} \to R$ be defined by $f (x) =$ $\frac{a - x}{a + x}$ . Further suppose that for any real number $x \neq - a$ and $f (x) \neq - a$ , (fof)( $x) = x$ .
Then, $f (- \frac{1}{2})$ is equal to

1

\frac{1}{3}

2

- \frac{1}{3}

3 -3

4 3

Explanation:

D We have,
$f (x) = \frac{a - x}{a + x} [x \in R - {- a}]$
$f o f (x) = x$
$f [f (x)] = x$
$\frac{a - f (x)}{a + f (x)} = x$
$\frac{a - (\frac{a - x}{a + x})}{a + (\frac{a - x}{a + x})} = x$
$\frac{(a^{2} - a) + x (a + 1)}{(a^{2} + a) + x (a - 1)} = x$
$(a^{2} - a) + x (a + 1) = (a^{2} + a) x + x^{2} (a - 1)$
$x^{2} (a - 1) + x (a^{2} + a - a - 1) - a^{2} + a = 0$
$x^{2} (a - 1) + x (a^{2} - 1) - (a^{2} - a) = 0$
$x^{2} + x (a + 1) - a = 0$
$a = 1$
$f (x) = \frac{1 - x}{1 + x}$
$f (- \frac{1}{2}) = \frac{1 - (\frac{- 1}{2})}{1 + (\frac{- 1}{2})}$
$f (- \frac{1}{2}) = \frac{\frac{3}{2}}{\frac{1}{2}}$
$f (\frac{- 1}{2}) = 3$

JEE Main 06.09.2020 Shift-II

Sets, Relation and Function

117025 If $f (x) = \log (\frac{1 + x}{1 - x}), - 1 < x < 1$ , then $f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$ is

1

[f (x)]^{3}

2

[f (x)]^{2}

3

- f (x)

4

f (x)

5

3 f (x)

Explanation:

D Given $f (x) = \log [\frac{(1 + x)}{1 - x}]$
$f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$
$= \log (\frac{1 + \frac{(3 x + x^{3})}{(1 + 3 x^{2})}}{1 - (\frac{3 x + x^{3}}{1 + 3 x^{2}})}) - \log (\frac{1 + \frac{2 x}{1 + x^{2}}}{1 - \frac{2 x}{1 + x^{2}}})$
$= \log (\frac{1 + 3 x^{2} + 3 x + x^{3}}{1 + 3 x^{2} - 3 x - x^{3}}) - \log (\frac{1 + x^{2} + 2 x}{1 + x^{2} - 2 x})$
$= \log {(\frac{1 + x}{1 - x})}^{3} - \log {(\frac{1 + x}{1 - x})}^{2}$
$= 3 \log (\frac{1 + x}{1 - x}) - 2 \log (\frac{1 + x}{1 - x})$
$= \log {(\frac{1 + x}{1 - x})}^{2}$
$= f (x)$

Kerala CEE-2008

Sets, Relation and Function

117026 $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x),}$ solve for $x$ .

1 0

2 3

3 both (a) and (b)

4 0 and 6

Explanation:

A $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x)}$
$\log_{2} (9 - 2^{x}) = (3 - x) [∵ {alog}_{a}^{b} = b]$
$(9 - 2^{x}) = 2^{(3 - x)}$
$(9 - 2^{x}) = \frac{2^{3}}{2^{x}}$
$2^{x} (9 - 2^{x}) = 8$
$- 2^{2 x} + 2^{x} \cdot 9 = 8$
Let $2^{x} = y$ then-
$- y^{2} + 9 y - 8 = 0$
$y^{2} - 9 y + 8 = 0$
$(y - 8) (y - 1) = 0$
$y = 8 or y = 1$
$2^{x} = 2^{3} or 2^{x} = 2^{0}$
So, $x = 3$ or $x = 0$
But $x = 3$ does not satisfy the given equation, since log 0 is not defined.

Manipal UGET-2013

Sets, Relation and Function

117027 The number of positive integral solutions of $x^{2} + 9 < (x + 3)^{2} < 8 x + 25$ , is

1 2

2 3

3 4

4 5

Explanation:

D We have, given inequality as
$x^{2} + 9 < (x + 3)^{2}$
$= x^{2} + 9 < x^{2} + 9 + 6 x$
$= 6 x > 0$
$= x > 0$
Again, $(x + 3)^{2} < 8 x + 25$
$= x^{2} + 9 + 6 x < 8 x + 25$
$= x^{2} - 2 x - 16 < 0$
$= (x - 1)^{2} - 17 < 0$
$= (x - 1)^{2} < 17$
$= x \in (1 - \sqrt{17}, 1 + \sqrt{17})$
Hence, integral value of $x$ are
$1, 2, 3, 4, 5 .$ The number of positive integral solution is 5 .

Manipal UGET-2015

Sets, Relation and Function

117028 The greatest value of the function $f (x) = {xe}^{- x}$ in $[0, \infty)$ , is

1 0

2

\frac{1}{e}

3

- e

4 e

Explanation:

B Given, $f (x) = {xe}^{- x}$
On differentiating both sides w.r.t. $x$ , we get
$f^{'} (x) = e^{- x} (1 - x)$
$f^{'} (x) = 0$
$x = 1$
Now, $f (0) = 0$
$lim_{x \to \infty} f (x) = lim_{x \to \infty} x e^{- x}$
$= lim_{x \to \infty} \frac{x}{e^{x}} = lim_{x \to \infty} \frac{1}{e^{x}} = 0$ Hence, the greatest value of $f (x)$ is $\frac{1}{e}$

Manipal UGET-2015

Sets, Relation and Function

117023 For a suitable chosen real constant a, let a function $f : R - {a} \to R$ be defined by $f (x) =$ $\frac{a - x}{a + x}$ . Further suppose that for any real number $x \neq - a$ and $f (x) \neq - a$ , (fof)( $x) = x$ .
Then, $f (- \frac{1}{2})$ is equal to

1

\frac{1}{3}

2

- \frac{1}{3}

3 -3

4 3

Explanation:

D We have,
$f (x) = \frac{a - x}{a + x} [x \in R - {- a}]$
$f o f (x) = x$
$f [f (x)] = x$
$\frac{a - f (x)}{a + f (x)} = x$
$\frac{a - (\frac{a - x}{a + x})}{a + (\frac{a - x}{a + x})} = x$
$\frac{(a^{2} - a) + x (a + 1)}{(a^{2} + a) + x (a - 1)} = x$
$(a^{2} - a) + x (a + 1) = (a^{2} + a) x + x^{2} (a - 1)$
$x^{2} (a - 1) + x (a^{2} + a - a - 1) - a^{2} + a = 0$
$x^{2} (a - 1) + x (a^{2} - 1) - (a^{2} - a) = 0$
$x^{2} + x (a + 1) - a = 0$
$a = 1$
$f (x) = \frac{1 - x}{1 + x}$
$f (- \frac{1}{2}) = \frac{1 - (\frac{- 1}{2})}{1 + (\frac{- 1}{2})}$
$f (- \frac{1}{2}) = \frac{\frac{3}{2}}{\frac{1}{2}}$
$f (\frac{- 1}{2}) = 3$

JEE Main 06.09.2020 Shift-II

Sets, Relation and Function

117025 If $f (x) = \log (\frac{1 + x}{1 - x}), - 1 < x < 1$ , then $f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$ is

1

[f (x)]^{3}

2

[f (x)]^{2}

3

- f (x)

4

f (x)

5

3 f (x)

Explanation:

D Given $f (x) = \log [\frac{(1 + x)}{1 - x}]$
$f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$
$= \log (\frac{1 + \frac{(3 x + x^{3})}{(1 + 3 x^{2})}}{1 - (\frac{3 x + x^{3}}{1 + 3 x^{2}})}) - \log (\frac{1 + \frac{2 x}{1 + x^{2}}}{1 - \frac{2 x}{1 + x^{2}}})$
$= \log (\frac{1 + 3 x^{2} + 3 x + x^{3}}{1 + 3 x^{2} - 3 x - x^{3}}) - \log (\frac{1 + x^{2} + 2 x}{1 + x^{2} - 2 x})$
$= \log {(\frac{1 + x}{1 - x})}^{3} - \log {(\frac{1 + x}{1 - x})}^{2}$
$= 3 \log (\frac{1 + x}{1 - x}) - 2 \log (\frac{1 + x}{1 - x})$
$= \log {(\frac{1 + x}{1 - x})}^{2}$
$= f (x)$

Kerala CEE-2008

Sets, Relation and Function

117026 $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x),}$ solve for $x$ .

1 0

2 3

3 both (a) and (b)

4 0 and 6

Explanation:

A $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x)}$
$\log_{2} (9 - 2^{x}) = (3 - x) [∵ {alog}_{a}^{b} = b]$
$(9 - 2^{x}) = 2^{(3 - x)}$
$(9 - 2^{x}) = \frac{2^{3}}{2^{x}}$
$2^{x} (9 - 2^{x}) = 8$
$- 2^{2 x} + 2^{x} \cdot 9 = 8$
Let $2^{x} = y$ then-
$- y^{2} + 9 y - 8 = 0$
$y^{2} - 9 y + 8 = 0$
$(y - 8) (y - 1) = 0$
$y = 8 or y = 1$
$2^{x} = 2^{3} or 2^{x} = 2^{0}$
So, $x = 3$ or $x = 0$
But $x = 3$ does not satisfy the given equation, since log 0 is not defined.

Manipal UGET-2013

Sets, Relation and Function

117027 The number of positive integral solutions of $x^{2} + 9 < (x + 3)^{2} < 8 x + 25$ , is

1 2

2 3

3 4

4 5

Explanation:

D We have, given inequality as
$x^{2} + 9 < (x + 3)^{2}$
$= x^{2} + 9 < x^{2} + 9 + 6 x$
$= 6 x > 0$
$= x > 0$
Again, $(x + 3)^{2} < 8 x + 25$
$= x^{2} + 9 + 6 x < 8 x + 25$
$= x^{2} - 2 x - 16 < 0$
$= (x - 1)^{2} - 17 < 0$
$= (x - 1)^{2} < 17$
$= x \in (1 - \sqrt{17}, 1 + \sqrt{17})$
Hence, integral value of $x$ are
$1, 2, 3, 4, 5 .$ The number of positive integral solution is 5 .

Manipal UGET-2015

Sets, Relation and Function

117028 The greatest value of the function $f (x) = {xe}^{- x}$ in $[0, \infty)$ , is

1 0

2

\frac{1}{e}

3

- e

4 e

Explanation:

B Given, $f (x) = {xe}^{- x}$
On differentiating both sides w.r.t. $x$ , we get
$f^{'} (x) = e^{- x} (1 - x)$
$f^{'} (x) = 0$
$x = 1$
Now, $f (0) = 0$
$lim_{x \to \infty} f (x) = lim_{x \to \infty} x e^{- x}$
$= lim_{x \to \infty} \frac{x}{e^{x}} = lim_{x \to \infty} \frac{1}{e^{x}} = 0$ Hence, the greatest value of $f (x)$ is $\frac{1}{e}$

Manipal UGET-2015

Sets, Relation and Function

117023 For a suitable chosen real constant a, let a function $f : R - {a} \to R$ be defined by $f (x) =$ $\frac{a - x}{a + x}$ . Further suppose that for any real number $x \neq - a$ and $f (x) \neq - a$ , (fof)( $x) = x$ .
Then, $f (- \frac{1}{2})$ is equal to

1

\frac{1}{3}

2

- \frac{1}{3}

3 -3

4 3

Explanation:

D We have,
$f (x) = \frac{a - x}{a + x} [x \in R - {- a}]$
$f o f (x) = x$
$f [f (x)] = x$
$\frac{a - f (x)}{a + f (x)} = x$
$\frac{a - (\frac{a - x}{a + x})}{a + (\frac{a - x}{a + x})} = x$
$\frac{(a^{2} - a) + x (a + 1)}{(a^{2} + a) + x (a - 1)} = x$
$(a^{2} - a) + x (a + 1) = (a^{2} + a) x + x^{2} (a - 1)$
$x^{2} (a - 1) + x (a^{2} + a - a - 1) - a^{2} + a = 0$
$x^{2} (a - 1) + x (a^{2} - 1) - (a^{2} - a) = 0$
$x^{2} + x (a + 1) - a = 0$
$a = 1$
$f (x) = \frac{1 - x}{1 + x}$
$f (- \frac{1}{2}) = \frac{1 - (\frac{- 1}{2})}{1 + (\frac{- 1}{2})}$
$f (- \frac{1}{2}) = \frac{\frac{3}{2}}{\frac{1}{2}}$
$f (\frac{- 1}{2}) = 3$

JEE Main 06.09.2020 Shift-II

Sets, Relation and Function

117025 If $f (x) = \log (\frac{1 + x}{1 - x}), - 1 < x < 1$ , then $f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$ is

1

[f (x)]^{3}

2

[f (x)]^{2}

3

- f (x)

4

f (x)

5

3 f (x)

Explanation:

D Given $f (x) = \log [\frac{(1 + x)}{1 - x}]$
$f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$
$= \log (\frac{1 + \frac{(3 x + x^{3})}{(1 + 3 x^{2})}}{1 - (\frac{3 x + x^{3}}{1 + 3 x^{2}})}) - \log (\frac{1 + \frac{2 x}{1 + x^{2}}}{1 - \frac{2 x}{1 + x^{2}}})$
$= \log (\frac{1 + 3 x^{2} + 3 x + x^{3}}{1 + 3 x^{2} - 3 x - x^{3}}) - \log (\frac{1 + x^{2} + 2 x}{1 + x^{2} - 2 x})$
$= \log {(\frac{1 + x}{1 - x})}^{3} - \log {(\frac{1 + x}{1 - x})}^{2}$
$= 3 \log (\frac{1 + x}{1 - x}) - 2 \log (\frac{1 + x}{1 - x})$
$= \log {(\frac{1 + x}{1 - x})}^{2}$
$= f (x)$

Kerala CEE-2008

Sets, Relation and Function

117026 $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x),}$ solve for $x$ .

1 0

2 3

3 both (a) and (b)

4 0 and 6

Explanation:

A $\log_{2} (9 - 2^{x}) = 10^{\log (3 - x)}$
$\log_{2} (9 - 2^{x}) = (3 - x) [∵ {alog}_{a}^{b} = b]$
$(9 - 2^{x}) = 2^{(3 - x)}$
$(9 - 2^{x}) = \frac{2^{3}}{2^{x}}$
$2^{x} (9 - 2^{x}) = 8$
$- 2^{2 x} + 2^{x} \cdot 9 = 8$
Let $2^{x} = y$ then-
$- y^{2} + 9 y - 8 = 0$
$y^{2} - 9 y + 8 = 0$
$(y - 8) (y - 1) = 0$
$y = 8 or y = 1$
$2^{x} = 2^{3} or 2^{x} = 2^{0}$
So, $x = 3$ or $x = 0$
But $x = 3$ does not satisfy the given equation, since log 0 is not defined.

Manipal UGET-2013

Sets, Relation and Function

117027 The number of positive integral solutions of $x^{2} + 9 < (x + 3)^{2} < 8 x + 25$ , is

1 2

2 3

3 4

4 5

Explanation:

D We have, given inequality as
$x^{2} + 9 < (x + 3)^{2}$
$= x^{2} + 9 < x^{2} + 9 + 6 x$
$= 6 x > 0$
$= x > 0$
Again, $(x + 3)^{2} < 8 x + 25$
$= x^{2} + 9 + 6 x < 8 x + 25$
$= x^{2} - 2 x - 16 < 0$
$= (x - 1)^{2} - 17 < 0$
$= (x - 1)^{2} < 17$
$= x \in (1 - \sqrt{17}, 1 + \sqrt{17})$
Hence, integral value of $x$ are
$1, 2, 3, 4, 5 .$ The number of positive integral solution is 5 .

Manipal UGET-2015

Sets, Relation and Function

117028 The greatest value of the function $f (x) = {xe}^{- x}$ in $[0, \infty)$ , is

1 0

2

\frac{1}{e}

3

- e

4 e

Explanation:

B Given, $f (x) = {xe}^{- x}$
On differentiating both sides w.r.t. $x$ , we get
$f^{'} (x) = e^{- x} (1 - x)$
$f^{'} (x) = 0$
$x = 1$
Now, $f (0) = 0$
$lim_{x \to \infty} f (x) = lim_{x \to \infty} x e^{- x}$
$= lim_{x \to \infty} \frac{x}{e^{x}} = lim_{x \to \infty} \frac{1}{e^{x}} = 0$ Hence, the greatest value of $f (x)$ is $\frac{1}{e}$

Manipal UGET-2015