QBManager

Sets, Relation and Function

116902 If $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2 x - 1}$ then the value of $x$ is

1

7 / 2

2

5 / 2

3

1 / 2

4

3 / 2

Explanation:

D $4^{x} - 3^{x - 1 / 2} = 3^{x + 1 / 2} - 2^{2 x - 1}$
$4^{x} - 3^{x} \times 3^{- 1 / 2} = 3^{x} \times 3^{1 / 2} - 2^{2 x} \times 2^{- 1}$
$2^{2 x} + 2^{2 x} \times 2^{- 1} = 3^{x} \times 3^{1 / 2} + 3^{x} 3^{- 1 / 2}$
$2^{2 x} (1 + \frac{1}{2}) = 3^{x} (\sqrt{3} + \frac{1}{\sqrt{3}})$
$2^{2 x} (\frac{3}{2}) = 3^{x} (\frac{3 + 1}{\sqrt{3}})$
$\frac{2^{2 x}}{3^{x}} = \frac{4}{\sqrt{3}} \times \frac{2}{3}$
$\frac{2^{2 x}}{3^{x}} = \frac{8}{3 \sqrt{3}}$
$\frac{4^{x}}{3^{x}} = \frac{4^{\frac{3}{2}}}{3^{\frac{3}{2}}}$
${(\frac{4}{3})}^{x} = {(\frac{4}{3})}^{3 / 2}$
On comparing both side, we get-
$x = \frac{3}{2}$

Shift-I

Sets, Relation and Function

116904 If $x = \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}}$ , then $x^{2} (x - 4)^{2}$ is equal to:

1 7

2 4

3 2

4 1

Explanation:

D Given, $x = \sqrt{\frac{(2 + \sqrt{3})}{(2 - \sqrt{3})}}$
On simplifying, we gets -
$x = \sqrt{\frac{(2 + \sqrt{3}) (2 + \sqrt{3})}{(2 - \sqrt{3}) (2 + \sqrt{3})}}$
$x = \sqrt{\frac{(2 + \sqrt{3})^{2}}{4 - 3}}$
$x = 2 + \sqrt{3}$
Then, $x - 4 = - 2 + \sqrt{3}$
$∴ x (x - 4) = (2 + \sqrt{3}) (- 2 + \sqrt{3})$
$= 3 - 4$
$= - 1$ So, $x^{2} (x - 4)^{2} = (- 1)^{2} = 1$

AP EAMCET-2006

Sets, Relation and Function

116905 If $f (x) = \log (x + \sqrt{x^{2} + 1})$ , then $f (x)$ is

1 even function

2 odd function

3 periodic function

4 none of these

Explanation:

B Given, $f (x) = \log (x + \sqrt{x^{2} + 1})$
We know that, for odd function-
$f (- x) = - f (x) \Rightarrow f (x) + f (- x) = 0$
For even function-
$f (- x) = f (x) \Rightarrow f (x) + f (- x) = 2 f (x)$
We have,
$f (- x) = \log (- x + \sqrt{(- x)^{2} + 1})$
$f (- x) = \log (- x + \sqrt{x^{2} + 1})$
$f (x) + f (- x) = \log (x + \sqrt{x^{2} + 1}) + \log (- x + \sqrt{x^{2} + 1})$
$= \log (x + \sqrt{x^{2} + 1}) (- x + \sqrt{x^{2} + 1})$
$= \log ({(\sqrt{x^{2} + 1})}^{2} - x^{2})$
$= \log (x^{2} - 1 - x^{2})$
$= \log (1)$
$= 0$ Hence it is odd function

AIEEE-2003

Sets, Relation and Function

116908 If $f (x) = \cos (\log_{e} x)$ , then
$f (x) f (y) - \frac{1}{2} [f (\frac{y}{x}) + f (x y)]$ has the value

1 1

2

1 / 2

3 -2

4 0

Explanation:

D Given,
$f (x) = \cos (\log_{e} x), f (\frac{y}{x}) = \cos (\log_{e} \frac{y}{x})$
$F (x y) = \cos (\log_{e} x y)$
$∴ f (x) f (y) - \frac{1}{2} [f (\frac{y}{x}) + f (x y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \frac{1}{2} [\cos (\log_{e} \frac{y}{x}) + \cos (\log_{e} x y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) -$
$\frac{1}{2} [\cos (\log_{e} y - \log_{e} x) + \cos (\log_{e} x + \log_{e} y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \frac{1}{2} [2 \cos (\log_{e} x) \cos (\log_{e} y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \cos (\log_{e} y) \cos (\log_{e} x) = 0$

COMEDK-2019

Sets, Relation and Function

116902 If $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2 x - 1}$ then the value of $x$ is

1

7 / 2

2

5 / 2

3

1 / 2

4

3 / 2

Explanation:

D $4^{x} - 3^{x - 1 / 2} = 3^{x + 1 / 2} - 2^{2 x - 1}$
$4^{x} - 3^{x} \times 3^{- 1 / 2} = 3^{x} \times 3^{1 / 2} - 2^{2 x} \times 2^{- 1}$
$2^{2 x} + 2^{2 x} \times 2^{- 1} = 3^{x} \times 3^{1 / 2} + 3^{x} 3^{- 1 / 2}$
$2^{2 x} (1 + \frac{1}{2}) = 3^{x} (\sqrt{3} + \frac{1}{\sqrt{3}})$
$2^{2 x} (\frac{3}{2}) = 3^{x} (\frac{3 + 1}{\sqrt{3}})$
$\frac{2^{2 x}}{3^{x}} = \frac{4}{\sqrt{3}} \times \frac{2}{3}$
$\frac{2^{2 x}}{3^{x}} = \frac{8}{3 \sqrt{3}}$
$\frac{4^{x}}{3^{x}} = \frac{4^{\frac{3}{2}}}{3^{\frac{3}{2}}}$
${(\frac{4}{3})}^{x} = {(\frac{4}{3})}^{3 / 2}$
On comparing both side, we get-
$x = \frac{3}{2}$

Shift-I

Sets, Relation and Function

116904 If $x = \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}}$ , then $x^{2} (x - 4)^{2}$ is equal to:

1 7

2 4

3 2

4 1

Explanation:

D Given, $x = \sqrt{\frac{(2 + \sqrt{3})}{(2 - \sqrt{3})}}$
On simplifying, we gets -
$x = \sqrt{\frac{(2 + \sqrt{3}) (2 + \sqrt{3})}{(2 - \sqrt{3}) (2 + \sqrt{3})}}$
$x = \sqrt{\frac{(2 + \sqrt{3})^{2}}{4 - 3}}$
$x = 2 + \sqrt{3}$
Then, $x - 4 = - 2 + \sqrt{3}$
$∴ x (x - 4) = (2 + \sqrt{3}) (- 2 + \sqrt{3})$
$= 3 - 4$
$= - 1$ So, $x^{2} (x - 4)^{2} = (- 1)^{2} = 1$

AP EAMCET-2006

Sets, Relation and Function

116905 If $f (x) = \log (x + \sqrt{x^{2} + 1})$ , then $f (x)$ is

1 even function

2 odd function

3 periodic function

4 none of these

Explanation:

B Given, $f (x) = \log (x + \sqrt{x^{2} + 1})$
We know that, for odd function-
$f (- x) = - f (x) \Rightarrow f (x) + f (- x) = 0$
For even function-
$f (- x) = f (x) \Rightarrow f (x) + f (- x) = 2 f (x)$
We have,
$f (- x) = \log (- x + \sqrt{(- x)^{2} + 1})$
$f (- x) = \log (- x + \sqrt{x^{2} + 1})$
$f (x) + f (- x) = \log (x + \sqrt{x^{2} + 1}) + \log (- x + \sqrt{x^{2} + 1})$
$= \log (x + \sqrt{x^{2} + 1}) (- x + \sqrt{x^{2} + 1})$
$= \log ({(\sqrt{x^{2} + 1})}^{2} - x^{2})$
$= \log (x^{2} - 1 - x^{2})$
$= \log (1)$
$= 0$ Hence it is odd function

AIEEE-2003

Sets, Relation and Function

116908 If $f (x) = \cos (\log_{e} x)$ , then
$f (x) f (y) - \frac{1}{2} [f (\frac{y}{x}) + f (x y)]$ has the value

1 1

2

1 / 2

3 -2

4 0

Explanation:

D Given,
$f (x) = \cos (\log_{e} x), f (\frac{y}{x}) = \cos (\log_{e} \frac{y}{x})$
$F (x y) = \cos (\log_{e} x y)$
$∴ f (x) f (y) - \frac{1}{2} [f (\frac{y}{x}) + f (x y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \frac{1}{2} [\cos (\log_{e} \frac{y}{x}) + \cos (\log_{e} x y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) -$
$\frac{1}{2} [\cos (\log_{e} y - \log_{e} x) + \cos (\log_{e} x + \log_{e} y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \frac{1}{2} [2 \cos (\log_{e} x) \cos (\log_{e} y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \cos (\log_{e} y) \cos (\log_{e} x) = 0$

COMEDK-2019

Sets, Relation and Function

116902 If $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2 x - 1}$ then the value of $x$ is

1

7 / 2

2

5 / 2

3

1 / 2

4

3 / 2

Explanation:

D $4^{x} - 3^{x - 1 / 2} = 3^{x + 1 / 2} - 2^{2 x - 1}$
$4^{x} - 3^{x} \times 3^{- 1 / 2} = 3^{x} \times 3^{1 / 2} - 2^{2 x} \times 2^{- 1}$
$2^{2 x} + 2^{2 x} \times 2^{- 1} = 3^{x} \times 3^{1 / 2} + 3^{x} 3^{- 1 / 2}$
$2^{2 x} (1 + \frac{1}{2}) = 3^{x} (\sqrt{3} + \frac{1}{\sqrt{3}})$
$2^{2 x} (\frac{3}{2}) = 3^{x} (\frac{3 + 1}{\sqrt{3}})$
$\frac{2^{2 x}}{3^{x}} = \frac{4}{\sqrt{3}} \times \frac{2}{3}$
$\frac{2^{2 x}}{3^{x}} = \frac{8}{3 \sqrt{3}}$
$\frac{4^{x}}{3^{x}} = \frac{4^{\frac{3}{2}}}{3^{\frac{3}{2}}}$
${(\frac{4}{3})}^{x} = {(\frac{4}{3})}^{3 / 2}$
On comparing both side, we get-
$x = \frac{3}{2}$

Shift-I

Sets, Relation and Function

116904 If $x = \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}}$ , then $x^{2} (x - 4)^{2}$ is equal to:

1 7

2 4

3 2

4 1

Explanation:

D Given, $x = \sqrt{\frac{(2 + \sqrt{3})}{(2 - \sqrt{3})}}$
On simplifying, we gets -
$x = \sqrt{\frac{(2 + \sqrt{3}) (2 + \sqrt{3})}{(2 - \sqrt{3}) (2 + \sqrt{3})}}$
$x = \sqrt{\frac{(2 + \sqrt{3})^{2}}{4 - 3}}$
$x = 2 + \sqrt{3}$
Then, $x - 4 = - 2 + \sqrt{3}$
$∴ x (x - 4) = (2 + \sqrt{3}) (- 2 + \sqrt{3})$
$= 3 - 4$
$= - 1$ So, $x^{2} (x - 4)^{2} = (- 1)^{2} = 1$

AP EAMCET-2006

Sets, Relation and Function

116905 If $f (x) = \log (x + \sqrt{x^{2} + 1})$ , then $f (x)$ is

1 even function

2 odd function

3 periodic function

4 none of these

Explanation:

B Given, $f (x) = \log (x + \sqrt{x^{2} + 1})$
We know that, for odd function-
$f (- x) = - f (x) \Rightarrow f (x) + f (- x) = 0$
For even function-
$f (- x) = f (x) \Rightarrow f (x) + f (- x) = 2 f (x)$
We have,
$f (- x) = \log (- x + \sqrt{(- x)^{2} + 1})$
$f (- x) = \log (- x + \sqrt{x^{2} + 1})$
$f (x) + f (- x) = \log (x + \sqrt{x^{2} + 1}) + \log (- x + \sqrt{x^{2} + 1})$
$= \log (x + \sqrt{x^{2} + 1}) (- x + \sqrt{x^{2} + 1})$
$= \log ({(\sqrt{x^{2} + 1})}^{2} - x^{2})$
$= \log (x^{2} - 1 - x^{2})$
$= \log (1)$
$= 0$ Hence it is odd function

AIEEE-2003

Sets, Relation and Function

116908 If $f (x) = \cos (\log_{e} x)$ , then
$f (x) f (y) - \frac{1}{2} [f (\frac{y}{x}) + f (x y)]$ has the value

1 1

2

1 / 2

3 -2

4 0

Explanation:

D Given,
$f (x) = \cos (\log_{e} x), f (\frac{y}{x}) = \cos (\log_{e} \frac{y}{x})$
$F (x y) = \cos (\log_{e} x y)$
$∴ f (x) f (y) - \frac{1}{2} [f (\frac{y}{x}) + f (x y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \frac{1}{2} [\cos (\log_{e} \frac{y}{x}) + \cos (\log_{e} x y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) -$
$\frac{1}{2} [\cos (\log_{e} y - \log_{e} x) + \cos (\log_{e} x + \log_{e} y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \frac{1}{2} [2 \cos (\log_{e} x) \cos (\log_{e} y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \cos (\log_{e} y) \cos (\log_{e} x) = 0$

COMEDK-2019

Sets, Relation and Function

116902 If $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2 x - 1}$ then the value of $x$ is

1

7 / 2

2

5 / 2

3

1 / 2

4

3 / 2

Explanation:

D $4^{x} - 3^{x - 1 / 2} = 3^{x + 1 / 2} - 2^{2 x - 1}$
$4^{x} - 3^{x} \times 3^{- 1 / 2} = 3^{x} \times 3^{1 / 2} - 2^{2 x} \times 2^{- 1}$
$2^{2 x} + 2^{2 x} \times 2^{- 1} = 3^{x} \times 3^{1 / 2} + 3^{x} 3^{- 1 / 2}$
$2^{2 x} (1 + \frac{1}{2}) = 3^{x} (\sqrt{3} + \frac{1}{\sqrt{3}})$
$2^{2 x} (\frac{3}{2}) = 3^{x} (\frac{3 + 1}{\sqrt{3}})$
$\frac{2^{2 x}}{3^{x}} = \frac{4}{\sqrt{3}} \times \frac{2}{3}$
$\frac{2^{2 x}}{3^{x}} = \frac{8}{3 \sqrt{3}}$
$\frac{4^{x}}{3^{x}} = \frac{4^{\frac{3}{2}}}{3^{\frac{3}{2}}}$
${(\frac{4}{3})}^{x} = {(\frac{4}{3})}^{3 / 2}$
On comparing both side, we get-
$x = \frac{3}{2}$

Shift-I

Sets, Relation and Function

116904 If $x = \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}}$ , then $x^{2} (x - 4)^{2}$ is equal to:

1 7

2 4

3 2

4 1

Explanation:

D Given, $x = \sqrt{\frac{(2 + \sqrt{3})}{(2 - \sqrt{3})}}$
On simplifying, we gets -
$x = \sqrt{\frac{(2 + \sqrt{3}) (2 + \sqrt{3})}{(2 - \sqrt{3}) (2 + \sqrt{3})}}$
$x = \sqrt{\frac{(2 + \sqrt{3})^{2}}{4 - 3}}$
$x = 2 + \sqrt{3}$
Then, $x - 4 = - 2 + \sqrt{3}$
$∴ x (x - 4) = (2 + \sqrt{3}) (- 2 + \sqrt{3})$
$= 3 - 4$
$= - 1$ So, $x^{2} (x - 4)^{2} = (- 1)^{2} = 1$

AP EAMCET-2006

Sets, Relation and Function

116905 If $f (x) = \log (x + \sqrt{x^{2} + 1})$ , then $f (x)$ is

1 even function

2 odd function

3 periodic function

4 none of these

Explanation:

B Given, $f (x) = \log (x + \sqrt{x^{2} + 1})$
We know that, for odd function-
$f (- x) = - f (x) \Rightarrow f (x) + f (- x) = 0$
For even function-
$f (- x) = f (x) \Rightarrow f (x) + f (- x) = 2 f (x)$
We have,
$f (- x) = \log (- x + \sqrt{(- x)^{2} + 1})$
$f (- x) = \log (- x + \sqrt{x^{2} + 1})$
$f (x) + f (- x) = \log (x + \sqrt{x^{2} + 1}) + \log (- x + \sqrt{x^{2} + 1})$
$= \log (x + \sqrt{x^{2} + 1}) (- x + \sqrt{x^{2} + 1})$
$= \log ({(\sqrt{x^{2} + 1})}^{2} - x^{2})$
$= \log (x^{2} - 1 - x^{2})$
$= \log (1)$
$= 0$ Hence it is odd function

AIEEE-2003

Sets, Relation and Function

116908 If $f (x) = \cos (\log_{e} x)$ , then
$f (x) f (y) - \frac{1}{2} [f (\frac{y}{x}) + f (x y)]$ has the value

1 1

2

1 / 2

3 -2

4 0

Explanation:

D Given,
$f (x) = \cos (\log_{e} x), f (\frac{y}{x}) = \cos (\log_{e} \frac{y}{x})$
$F (x y) = \cos (\log_{e} x y)$
$∴ f (x) f (y) - \frac{1}{2} [f (\frac{y}{x}) + f (x y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \frac{1}{2} [\cos (\log_{e} \frac{y}{x}) + \cos (\log_{e} x y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) -$
$\frac{1}{2} [\cos (\log_{e} y - \log_{e} x) + \cos (\log_{e} x + \log_{e} y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \frac{1}{2} [2 \cos (\log_{e} x) \cos (\log_{e} y)]$
$= \cos (\log_{e} x) \cos (\log_{e} y) - \cos (\log_{e} y) \cos (\log_{e} x) = 0$

COMEDK-2019

Question Bank Series

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