QBManager

Sets, Relation and Function

116907 If $f (x) = 2 x^{2}$ , find $\frac{f (3.8) - f (4)}{3.8 - 4}$

1 156

2 0.156

3 1.56

4 15.6

Explanation:

D Given, $f (x) = 2 x^{2}$
Then find $\frac{f (3.8) - f (4)}{(3.8 - 4)} =$ ?
So, $\frac{f (3.8) - f (4)}{3.8 - 4} = \frac{2 \times (3.8)^{2} - 4^{2} \times 2}{3.8 - 4}$
$= \frac{2 [(3.8)^{2} - 4^{2}]}{(3.8 - 4)}$
$= \frac{- 2 [4^{2} - (3.8)^{2}]}{- (4 - 3.8)}$
$= \frac{2 (4 - 3.8) (4 + 3.8)}{(4 - 3.8)} = 2 \times 7.8 = 15.6$

Karnataka CET-2015

Sets, Relation and Function

116975 If $y = 3^{x - 1} + 3^{- x - 1}$ (x real), then the least value of $y$ is

1 2

2 6

3

2 / 3

4 None of these

Explanation:

C Given that,
$y = 3^{x - 1} + 3^{- x - 1}$
Now we know that-
$A.M \geq G \cdot M$
$\frac{3^{x - 1} + 3^{- x - 1}}{2} \geq {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq 2 {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq \frac{2}{3}$

Jamia Millia Islamia-2006

Sets, Relation and Function

117024 If $p$ and $q$ are positive real numbers such that $p^{2} + q^{2} = 1$ , then the maximum value of $(p + q)$ is

1 2

2

\frac{1}{2}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

D Given,
$p^{2} + q^{2} = 1$
Applying $AM \geq GM$ inequality
$\frac{p^{2} + q^{2}}{2} \geq \sqrt{p^{2} q^{2}}$
$p^{2} + q^{2} \geq 2 p q$
$\frac{1}{2} \geq p q$
$pq \leq \frac{1}{2}$
Now we know that -
$(p + q)^{2} = p^{2} + q^{2} = p q$
$(p + q)^{2} = 1 + 2 p q$
$(p + q)^{2} \leq 1 + 1$
$(p + q) \leq \sqrt{2}$
Hence maximum value of
$p + q = \sqrt{2}$

AIEEE-2007

Sets, Relation and Function

116859 If $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$ , then

1

2.5 < p < 3

2

p > 3

3

1.5 < p < 2

4

2 <

p

< 2.5

Explanation:

B Given, $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$
$p = \log_{π} 3 + \log_{π} 4 + 1$
$p = \log_{π} (3 \times 4) + 1$
$p = \log_{π} (12) + 1$
We know that -
Then,
$12 > (π^{2}) = (3.14)^{2} = 9.8596$
$12 > π^{2}$
$\log_{π} 12 > \log_{π} π^{2}$
$\log_{π} 12 > 2$ So, $p > 3$

UPSEE-2016

Sets, Relation and Function

116906 If $f : R \to R$ , such that $f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$ , then $f$ is

1 an odd function

2 a neither even nor odd function

3 an even function

4 a periodic function

Explanation:

A Given,
$f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$
$f (x) = \frac{e^{x} + \frac{1}{e^{x}}}{e^{x} - \frac{1}{e^{x}}}$
$f (x) = \frac{e^{2 x} + 1}{e^{2 x} - 1}$
And $f (- x) = \frac{e^{- x} + e^{x}}{e^{- x} - e^{x}}$
$f (- x) = \frac{\frac{1}{e^{x}} + e^{x}}{\frac{1}{e^{x}} - e^{x}}$
$f (- x) = \frac{1 + e^{2 x}}{1 - e^{2 x}}$
$f (- x) = - (\frac{e^{2 x} + 1}{e^{2 x} - 1})$
$f (- x) = - f (x)$
[ From equation (i)]
$∴ f (x)$ is odd function

MHT-CET 2020

Sets, Relation and Function

116907 If $f (x) = 2 x^{2}$ , find $\frac{f (3.8) - f (4)}{3.8 - 4}$

1 156

2 0.156

3 1.56

4 15.6

Explanation:

D Given, $f (x) = 2 x^{2}$
Then find $\frac{f (3.8) - f (4)}{(3.8 - 4)} =$ ?
So, $\frac{f (3.8) - f (4)}{3.8 - 4} = \frac{2 \times (3.8)^{2} - 4^{2} \times 2}{3.8 - 4}$
$= \frac{2 [(3.8)^{2} - 4^{2}]}{(3.8 - 4)}$
$= \frac{- 2 [4^{2} - (3.8)^{2}]}{- (4 - 3.8)}$
$= \frac{2 (4 - 3.8) (4 + 3.8)}{(4 - 3.8)} = 2 \times 7.8 = 15.6$

Karnataka CET-2015

Sets, Relation and Function

116975 If $y = 3^{x - 1} + 3^{- x - 1}$ (x real), then the least value of $y$ is

1 2

2 6

3

2 / 3

4 None of these

Explanation:

C Given that,
$y = 3^{x - 1} + 3^{- x - 1}$
Now we know that-
$A.M \geq G \cdot M$
$\frac{3^{x - 1} + 3^{- x - 1}}{2} \geq {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq 2 {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq \frac{2}{3}$

Jamia Millia Islamia-2006

Sets, Relation and Function

117024 If $p$ and $q$ are positive real numbers such that $p^{2} + q^{2} = 1$ , then the maximum value of $(p + q)$ is

1 2

2

\frac{1}{2}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

D Given,
$p^{2} + q^{2} = 1$
Applying $AM \geq GM$ inequality
$\frac{p^{2} + q^{2}}{2} \geq \sqrt{p^{2} q^{2}}$
$p^{2} + q^{2} \geq 2 p q$
$\frac{1}{2} \geq p q$
$pq \leq \frac{1}{2}$
Now we know that -
$(p + q)^{2} = p^{2} + q^{2} = p q$
$(p + q)^{2} = 1 + 2 p q$
$(p + q)^{2} \leq 1 + 1$
$(p + q) \leq \sqrt{2}$
Hence maximum value of
$p + q = \sqrt{2}$

AIEEE-2007

Sets, Relation and Function

116859 If $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$ , then

1

2.5 < p < 3

2

p > 3

3

1.5 < p < 2

4

2 <

p

< 2.5

Explanation:

B Given, $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$
$p = \log_{π} 3 + \log_{π} 4 + 1$
$p = \log_{π} (3 \times 4) + 1$
$p = \log_{π} (12) + 1$
We know that -
Then,
$12 > (π^{2}) = (3.14)^{2} = 9.8596$
$12 > π^{2}$
$\log_{π} 12 > \log_{π} π^{2}$
$\log_{π} 12 > 2$ So, $p > 3$

UPSEE-2016

Sets, Relation and Function

116906 If $f : R \to R$ , such that $f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$ , then $f$ is

1 an odd function

2 a neither even nor odd function

3 an even function

4 a periodic function

Explanation:

A Given,
$f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$
$f (x) = \frac{e^{x} + \frac{1}{e^{x}}}{e^{x} - \frac{1}{e^{x}}}$
$f (x) = \frac{e^{2 x} + 1}{e^{2 x} - 1}$
And $f (- x) = \frac{e^{- x} + e^{x}}{e^{- x} - e^{x}}$
$f (- x) = \frac{\frac{1}{e^{x}} + e^{x}}{\frac{1}{e^{x}} - e^{x}}$
$f (- x) = \frac{1 + e^{2 x}}{1 - e^{2 x}}$
$f (- x) = - (\frac{e^{2 x} + 1}{e^{2 x} - 1})$
$f (- x) = - f (x)$
[ From equation (i)]
$∴ f (x)$ is odd function

MHT-CET 2020

Sets, Relation and Function

116907 If $f (x) = 2 x^{2}$ , find $\frac{f (3.8) - f (4)}{3.8 - 4}$

1 156

2 0.156

3 1.56

4 15.6

Explanation:

D Given, $f (x) = 2 x^{2}$
Then find $\frac{f (3.8) - f (4)}{(3.8 - 4)} =$ ?
So, $\frac{f (3.8) - f (4)}{3.8 - 4} = \frac{2 \times (3.8)^{2} - 4^{2} \times 2}{3.8 - 4}$
$= \frac{2 [(3.8)^{2} - 4^{2}]}{(3.8 - 4)}$
$= \frac{- 2 [4^{2} - (3.8)^{2}]}{- (4 - 3.8)}$
$= \frac{2 (4 - 3.8) (4 + 3.8)}{(4 - 3.8)} = 2 \times 7.8 = 15.6$

Karnataka CET-2015

Sets, Relation and Function

116975 If $y = 3^{x - 1} + 3^{- x - 1}$ (x real), then the least value of $y$ is

1 2

2 6

3

2 / 3

4 None of these

Explanation:

C Given that,
$y = 3^{x - 1} + 3^{- x - 1}$
Now we know that-
$A.M \geq G \cdot M$
$\frac{3^{x - 1} + 3^{- x - 1}}{2} \geq {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq 2 {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq \frac{2}{3}$

Jamia Millia Islamia-2006

Sets, Relation and Function

117024 If $p$ and $q$ are positive real numbers such that $p^{2} + q^{2} = 1$ , then the maximum value of $(p + q)$ is

1 2

2

\frac{1}{2}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

D Given,
$p^{2} + q^{2} = 1$
Applying $AM \geq GM$ inequality
$\frac{p^{2} + q^{2}}{2} \geq \sqrt{p^{2} q^{2}}$
$p^{2} + q^{2} \geq 2 p q$
$\frac{1}{2} \geq p q$
$pq \leq \frac{1}{2}$
Now we know that -
$(p + q)^{2} = p^{2} + q^{2} = p q$
$(p + q)^{2} = 1 + 2 p q$
$(p + q)^{2} \leq 1 + 1$
$(p + q) \leq \sqrt{2}$
Hence maximum value of
$p + q = \sqrt{2}$

AIEEE-2007

Sets, Relation and Function

116859 If $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$ , then

1

2.5 < p < 3

2

p > 3

3

1.5 < p < 2

4

2 <

p

< 2.5

Explanation:

B Given, $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$
$p = \log_{π} 3 + \log_{π} 4 + 1$
$p = \log_{π} (3 \times 4) + 1$
$p = \log_{π} (12) + 1$
We know that -
Then,
$12 > (π^{2}) = (3.14)^{2} = 9.8596$
$12 > π^{2}$
$\log_{π} 12 > \log_{π} π^{2}$
$\log_{π} 12 > 2$ So, $p > 3$

UPSEE-2016

Sets, Relation and Function

116906 If $f : R \to R$ , such that $f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$ , then $f$ is

1 an odd function

2 a neither even nor odd function

3 an even function

4 a periodic function

Explanation:

A Given,
$f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$
$f (x) = \frac{e^{x} + \frac{1}{e^{x}}}{e^{x} - \frac{1}{e^{x}}}$
$f (x) = \frac{e^{2 x} + 1}{e^{2 x} - 1}$
And $f (- x) = \frac{e^{- x} + e^{x}}{e^{- x} - e^{x}}$
$f (- x) = \frac{\frac{1}{e^{x}} + e^{x}}{\frac{1}{e^{x}} - e^{x}}$
$f (- x) = \frac{1 + e^{2 x}}{1 - e^{2 x}}$
$f (- x) = - (\frac{e^{2 x} + 1}{e^{2 x} - 1})$
$f (- x) = - f (x)$
[ From equation (i)]
$∴ f (x)$ is odd function

MHT-CET 2020

Sets, Relation and Function

116907 If $f (x) = 2 x^{2}$ , find $\frac{f (3.8) - f (4)}{3.8 - 4}$

1 156

2 0.156

3 1.56

4 15.6

Explanation:

D Given, $f (x) = 2 x^{2}$
Then find $\frac{f (3.8) - f (4)}{(3.8 - 4)} =$ ?
So, $\frac{f (3.8) - f (4)}{3.8 - 4} = \frac{2 \times (3.8)^{2} - 4^{2} \times 2}{3.8 - 4}$
$= \frac{2 [(3.8)^{2} - 4^{2}]}{(3.8 - 4)}$
$= \frac{- 2 [4^{2} - (3.8)^{2}]}{- (4 - 3.8)}$
$= \frac{2 (4 - 3.8) (4 + 3.8)}{(4 - 3.8)} = 2 \times 7.8 = 15.6$

Karnataka CET-2015

Sets, Relation and Function

116975 If $y = 3^{x - 1} + 3^{- x - 1}$ (x real), then the least value of $y$ is

1 2

2 6

3

2 / 3

4 None of these

Explanation:

C Given that,
$y = 3^{x - 1} + 3^{- x - 1}$
Now we know that-
$A.M \geq G \cdot M$
$\frac{3^{x - 1} + 3^{- x - 1}}{2} \geq {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq 2 {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq \frac{2}{3}$

Jamia Millia Islamia-2006

Sets, Relation and Function

117024 If $p$ and $q$ are positive real numbers such that $p^{2} + q^{2} = 1$ , then the maximum value of $(p + q)$ is

1 2

2

\frac{1}{2}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

D Given,
$p^{2} + q^{2} = 1$
Applying $AM \geq GM$ inequality
$\frac{p^{2} + q^{2}}{2} \geq \sqrt{p^{2} q^{2}}$
$p^{2} + q^{2} \geq 2 p q$
$\frac{1}{2} \geq p q$
$pq \leq \frac{1}{2}$
Now we know that -
$(p + q)^{2} = p^{2} + q^{2} = p q$
$(p + q)^{2} = 1 + 2 p q$
$(p + q)^{2} \leq 1 + 1$
$(p + q) \leq \sqrt{2}$
Hence maximum value of
$p + q = \sqrt{2}$

AIEEE-2007

Sets, Relation and Function

116859 If $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$ , then

1

2.5 < p < 3

2

p > 3

3

1.5 < p < 2

4

2 <

p

< 2.5

Explanation:

B Given, $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$
$p = \log_{π} 3 + \log_{π} 4 + 1$
$p = \log_{π} (3 \times 4) + 1$
$p = \log_{π} (12) + 1$
We know that -
Then,
$12 > (π^{2}) = (3.14)^{2} = 9.8596$
$12 > π^{2}$
$\log_{π} 12 > \log_{π} π^{2}$
$\log_{π} 12 > 2$ So, $p > 3$

UPSEE-2016

Sets, Relation and Function

116906 If $f : R \to R$ , such that $f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$ , then $f$ is

1 an odd function

2 a neither even nor odd function

3 an even function

4 a periodic function

Explanation:

A Given,
$f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$
$f (x) = \frac{e^{x} + \frac{1}{e^{x}}}{e^{x} - \frac{1}{e^{x}}}$
$f (x) = \frac{e^{2 x} + 1}{e^{2 x} - 1}$
And $f (- x) = \frac{e^{- x} + e^{x}}{e^{- x} - e^{x}}$
$f (- x) = \frac{\frac{1}{e^{x}} + e^{x}}{\frac{1}{e^{x}} - e^{x}}$
$f (- x) = \frac{1 + e^{2 x}}{1 - e^{2 x}}$
$f (- x) = - (\frac{e^{2 x} + 1}{e^{2 x} - 1})$
$f (- x) = - f (x)$
[ From equation (i)]
$∴ f (x)$ is odd function

MHT-CET 2020

Sets, Relation and Function

116907 If $f (x) = 2 x^{2}$ , find $\frac{f (3.8) - f (4)}{3.8 - 4}$

1 156

2 0.156

3 1.56

4 15.6

Explanation:

D Given, $f (x) = 2 x^{2}$
Then find $\frac{f (3.8) - f (4)}{(3.8 - 4)} =$ ?
So, $\frac{f (3.8) - f (4)}{3.8 - 4} = \frac{2 \times (3.8)^{2} - 4^{2} \times 2}{3.8 - 4}$
$= \frac{2 [(3.8)^{2} - 4^{2}]}{(3.8 - 4)}$
$= \frac{- 2 [4^{2} - (3.8)^{2}]}{- (4 - 3.8)}$
$= \frac{2 (4 - 3.8) (4 + 3.8)}{(4 - 3.8)} = 2 \times 7.8 = 15.6$

Karnataka CET-2015

Sets, Relation and Function

116975 If $y = 3^{x - 1} + 3^{- x - 1}$ (x real), then the least value of $y$ is

1 2

2 6

3

2 / 3

4 None of these

Explanation:

C Given that,
$y = 3^{x - 1} + 3^{- x - 1}$
Now we know that-
$A.M \geq G \cdot M$
$\frac{3^{x - 1} + 3^{- x - 1}}{2} \geq {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq 2 {(3^{x - 1} \cdot 3^{- x - 1})}^{\frac{1}{2}}$
$3^{x - 1} + 3^{- x - 1} \geq \frac{2}{3}$

Jamia Millia Islamia-2006

Sets, Relation and Function

117024 If $p$ and $q$ are positive real numbers such that $p^{2} + q^{2} = 1$ , then the maximum value of $(p + q)$ is

1 2

2

\frac{1}{2}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

D Given,
$p^{2} + q^{2} = 1$
Applying $AM \geq GM$ inequality
$\frac{p^{2} + q^{2}}{2} \geq \sqrt{p^{2} q^{2}}$
$p^{2} + q^{2} \geq 2 p q$
$\frac{1}{2} \geq p q$
$pq \leq \frac{1}{2}$
Now we know that -
$(p + q)^{2} = p^{2} + q^{2} = p q$
$(p + q)^{2} = 1 + 2 p q$
$(p + q)^{2} \leq 1 + 1$
$(p + q) \leq \sqrt{2}$
Hence maximum value of
$p + q = \sqrt{2}$

AIEEE-2007

Sets, Relation and Function

116859 If $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$ , then

1

2.5 < p < 3

2

p > 3

3

1.5 < p < 2

4

2 <

p

< 2.5

Explanation:

B Given, $p = \frac{1}{\log_{3} π} + \frac{1}{\log_{4} π} + 1$
$p = \log_{π} 3 + \log_{π} 4 + 1$
$p = \log_{π} (3 \times 4) + 1$
$p = \log_{π} (12) + 1$
We know that -
Then,
$12 > (π^{2}) = (3.14)^{2} = 9.8596$
$12 > π^{2}$
$\log_{π} 12 > \log_{π} π^{2}$
$\log_{π} 12 > 2$ So, $p > 3$

UPSEE-2016

Sets, Relation and Function

116906 If $f : R \to R$ , such that $f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$ , then $f$ is

1 an odd function

2 a neither even nor odd function

3 an even function

4 a periodic function

Explanation:

A Given,
$f (x) = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$
$f (x) = \frac{e^{x} + \frac{1}{e^{x}}}{e^{x} - \frac{1}{e^{x}}}$
$f (x) = \frac{e^{2 x} + 1}{e^{2 x} - 1}$
And $f (- x) = \frac{e^{- x} + e^{x}}{e^{- x} - e^{x}}$
$f (- x) = \frac{\frac{1}{e^{x}} + e^{x}}{\frac{1}{e^{x}} - e^{x}}$
$f (- x) = \frac{1 + e^{2 x}}{1 - e^{2 x}}$
$f (- x) = - (\frac{e^{2 x} + 1}{e^{2 x} - 1})$
$f (- x) = - f (x)$
[ From equation (i)]
$∴ f (x)$ is odd function

MHT-CET 2020