QBManager

Application of Derivatives

85592 For all positive value of $x$ and $y$ , the value of
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{x y} is:$

1

\leq 9

2

< 9

3

\geq 9

4

> 9

Explanation:

(C) : Given,
For all positive value of $x$ and $y$ We have,
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{x y}$
We can write as,
$\frac{1 + x + x^{2}}{x y} = \frac{1}{x y} + \frac{1}{y} + \frac{x}{y}$

We know that,
A.M $\geq$ G.M
$= \frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq \sqrt[3]{\frac{1}{x y} \times \frac{1}{y} \times \frac{x}{y}}$
$= \frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq 3 {(\frac{1}{y^{3}})}^{\frac{1}{3}}$
$\frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq \frac{3}{y}$
Similarly,
$\frac{1 + y + y^{2}}{x y} = \frac{1}{x y} + \frac{1}{x} + \frac{y}{x}$
$\frac{\frac{1}{x y} + \frac{1}{x} + \frac{y}{x}}{3} \geq {(\frac{1}{x y} \times \frac{1}{x} \times \frac{y}{x})}^{\frac{1}{3}}$
$\frac{1}{x y} + \frac{1}{x} + \frac{y}{x} \geq \frac{3}{x}$
On multiplying equation (i) and (ii) we get-
$(\frac{1}{x y} + \frac{1}{y} + \frac{x}{y}) (\frac{1}{x y} + \frac{1}{x} + \frac{y}{x}) \geq \frac{9}{x y}$
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{(x y)^{2}} \geq \frac{9}{x y}$
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{(x y)} \geq 9$

Karnataka CET-2002

Application of Derivatives

85593 The maximum of the function $3 \cos x - 4 \sin x$ is:

1 2

2 3

3 4

4 5

Explanation:

(D) : Given,
$f (x) = 3 \cos x - 4 \sin x$
We know that
$- \sqrt{a^{2} + b^{2}} \leq a \cos x + b \sin x \leq \sqrt{a^{2} + b^{2}}$
$a = 3 and b = - 4$
For maximum
$3 \cos x - 4 \sin x \leq \sqrt{(3)^{2} + (- 4)^{2}}$
$3 \cos x - 4 \sin x \leq \sqrt{25}$
$3 \cos x - 4 \sin x \leq 5$

Karnataka CET-2004

Application of Derivatives

85594 The maximum slope of the curve
$y = - x^{3} + 3 x^{2} + 2 x - 27 i$

1 1

2 23

3 5

4 -23

Explanation:

(C) : Given curve,
$y = - x^{2} + 3 x^{2} + 2 x - 27$
On differentiating both sides w.r.t $x$ we get-
$\frac{d y}{d x} = - 3 x^{2} + 6 x + 2$
For maximum slope
$\frac{dy}{dx} = 0$
Slope $(m) = - 3 x^{2} + 6 x + 2 = 0$
$\frac{d (m)}{d x} = - 6 x + 6 \Rightarrow \frac{d^{2} (m)}{d x^{2}} = - 6$
For maximum,
$\frac{d (m)}{d x} = 0$
$- 6 x + 6 = 0$
$x = 1$
Slope $(m) = - 3 x^{2} + 6 x + 2$
Value of maximum slope at $x = 1$
$m = - 3 (1)^{2} + 6 (1) + 2$
$= - 3 + 6 + 2$
$m_{max} = 5$

Application of Derivatives

85595 The minimum value of $f (x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$ if $a^{2} > b^{2}$ , is

1

b^{2}

2

a^{2} - b^{2}

3

a^{2}

4

a^{2} + b^{2}

Explanation:

(A) : Given,
$f (x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$
On differentiating both sides w.r.t. $x$ , we get-
$f^{'} (x) = a^{2} 2 \cos x \cdot (- \sin x) + b^{2} 2 \sin x \cos x$
$= - a^{2} \sin 2 x + b^{2} \sin 2 x$
$= (b^{2} - a^{2}) \sin 2 x$
For minimum $f^{'} (x) = 0$
$(b^{2} - a^{2}) \sin 2 x = 0$
$\sin 2 x = 0$
$2 x = 0$
$x = 0$
$f (0) = a^{2} (1)^{2} + b^{2} (0)$
$f (0) = a^{2} (maxima)$
Again on differentiating both sides w.r.t. $x$ , we get-
$f^{''} (x) = (b^{2} - a^{2}) 2 \cos 2 x$
$f^{''} (0) = (b^{2} - a^{2}) 2 < 0$
$f (\frac{π}{2}) = a^{2} (0) + b^{2} (1)$
$f (\frac{π}{2}) = b^{2} (minima)$
$f^{''} (\frac{π}{2}) = 2 (a^{2} - b^{2})$
$> 0$
$f (\frac{π}{2}) = b^{2}$
Hence, minimum value is $b^{2}$

MHT CET-2020

Application of Derivatives

85592 For all positive value of $x$ and $y$ , the value of
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{x y} is:$

1

\leq 9

2

< 9

3

\geq 9

4

> 9

Explanation:

(C) : Given,
For all positive value of $x$ and $y$ We have,
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{x y}$
We can write as,
$\frac{1 + x + x^{2}}{x y} = \frac{1}{x y} + \frac{1}{y} + \frac{x}{y}$

We know that,
A.M $\geq$ G.M
$= \frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq \sqrt[3]{\frac{1}{x y} \times \frac{1}{y} \times \frac{x}{y}}$
$= \frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq 3 {(\frac{1}{y^{3}})}^{\frac{1}{3}}$
$\frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq \frac{3}{y}$
Similarly,
$\frac{1 + y + y^{2}}{x y} = \frac{1}{x y} + \frac{1}{x} + \frac{y}{x}$
$\frac{\frac{1}{x y} + \frac{1}{x} + \frac{y}{x}}{3} \geq {(\frac{1}{x y} \times \frac{1}{x} \times \frac{y}{x})}^{\frac{1}{3}}$
$\frac{1}{x y} + \frac{1}{x} + \frac{y}{x} \geq \frac{3}{x}$
On multiplying equation (i) and (ii) we get-
$(\frac{1}{x y} + \frac{1}{y} + \frac{x}{y}) (\frac{1}{x y} + \frac{1}{x} + \frac{y}{x}) \geq \frac{9}{x y}$
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{(x y)^{2}} \geq \frac{9}{x y}$
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{(x y)} \geq 9$

Karnataka CET-2002

Application of Derivatives

85593 The maximum of the function $3 \cos x - 4 \sin x$ is:

1 2

2 3

3 4

4 5

Explanation:

(D) : Given,
$f (x) = 3 \cos x - 4 \sin x$
We know that
$- \sqrt{a^{2} + b^{2}} \leq a \cos x + b \sin x \leq \sqrt{a^{2} + b^{2}}$
$a = 3 and b = - 4$
For maximum
$3 \cos x - 4 \sin x \leq \sqrt{(3)^{2} + (- 4)^{2}}$
$3 \cos x - 4 \sin x \leq \sqrt{25}$
$3 \cos x - 4 \sin x \leq 5$

Karnataka CET-2004

Application of Derivatives

85594 The maximum slope of the curve
$y = - x^{3} + 3 x^{2} + 2 x - 27 i$

1 1

2 23

3 5

4 -23

Explanation:

(C) : Given curve,
$y = - x^{2} + 3 x^{2} + 2 x - 27$
On differentiating both sides w.r.t $x$ we get-
$\frac{d y}{d x} = - 3 x^{2} + 6 x + 2$
For maximum slope
$\frac{dy}{dx} = 0$
Slope $(m) = - 3 x^{2} + 6 x + 2 = 0$
$\frac{d (m)}{d x} = - 6 x + 6 \Rightarrow \frac{d^{2} (m)}{d x^{2}} = - 6$
For maximum,
$\frac{d (m)}{d x} = 0$
$- 6 x + 6 = 0$
$x = 1$
Slope $(m) = - 3 x^{2} + 6 x + 2$
Value of maximum slope at $x = 1$
$m = - 3 (1)^{2} + 6 (1) + 2$
$= - 3 + 6 + 2$
$m_{max} = 5$

Application of Derivatives

85595 The minimum value of $f (x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$ if $a^{2} > b^{2}$ , is

1

b^{2}

2

a^{2} - b^{2}

3

a^{2}

4

a^{2} + b^{2}

Explanation:

(A) : Given,
$f (x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$
On differentiating both sides w.r.t. $x$ , we get-
$f^{'} (x) = a^{2} 2 \cos x \cdot (- \sin x) + b^{2} 2 \sin x \cos x$
$= - a^{2} \sin 2 x + b^{2} \sin 2 x$
$= (b^{2} - a^{2}) \sin 2 x$
For minimum $f^{'} (x) = 0$
$(b^{2} - a^{2}) \sin 2 x = 0$
$\sin 2 x = 0$
$2 x = 0$
$x = 0$
$f (0) = a^{2} (1)^{2} + b^{2} (0)$
$f (0) = a^{2} (maxima)$
Again on differentiating both sides w.r.t. $x$ , we get-
$f^{''} (x) = (b^{2} - a^{2}) 2 \cos 2 x$
$f^{''} (0) = (b^{2} - a^{2}) 2 < 0$
$f (\frac{π}{2}) = a^{2} (0) + b^{2} (1)$
$f (\frac{π}{2}) = b^{2} (minima)$
$f^{''} (\frac{π}{2}) = 2 (a^{2} - b^{2})$
$> 0$
$f (\frac{π}{2}) = b^{2}$
Hence, minimum value is $b^{2}$

MHT CET-2020

Application of Derivatives

85592 For all positive value of $x$ and $y$ , the value of
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{x y} is:$

1

\leq 9

2

< 9

3

\geq 9

4

> 9

Explanation:

(C) : Given,
For all positive value of $x$ and $y$ We have,
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{x y}$
We can write as,
$\frac{1 + x + x^{2}}{x y} = \frac{1}{x y} + \frac{1}{y} + \frac{x}{y}$

We know that,
A.M $\geq$ G.M
$= \frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq \sqrt[3]{\frac{1}{x y} \times \frac{1}{y} \times \frac{x}{y}}$
$= \frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq 3 {(\frac{1}{y^{3}})}^{\frac{1}{3}}$
$\frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq \frac{3}{y}$
Similarly,
$\frac{1 + y + y^{2}}{x y} = \frac{1}{x y} + \frac{1}{x} + \frac{y}{x}$
$\frac{\frac{1}{x y} + \frac{1}{x} + \frac{y}{x}}{3} \geq {(\frac{1}{x y} \times \frac{1}{x} \times \frac{y}{x})}^{\frac{1}{3}}$
$\frac{1}{x y} + \frac{1}{x} + \frac{y}{x} \geq \frac{3}{x}$
On multiplying equation (i) and (ii) we get-
$(\frac{1}{x y} + \frac{1}{y} + \frac{x}{y}) (\frac{1}{x y} + \frac{1}{x} + \frac{y}{x}) \geq \frac{9}{x y}$
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{(x y)^{2}} \geq \frac{9}{x y}$
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{(x y)} \geq 9$

Karnataka CET-2002

Application of Derivatives

85593 The maximum of the function $3 \cos x - 4 \sin x$ is:

1 2

2 3

3 4

4 5

Explanation:

(D) : Given,
$f (x) = 3 \cos x - 4 \sin x$
We know that
$- \sqrt{a^{2} + b^{2}} \leq a \cos x + b \sin x \leq \sqrt{a^{2} + b^{2}}$
$a = 3 and b = - 4$
For maximum
$3 \cos x - 4 \sin x \leq \sqrt{(3)^{2} + (- 4)^{2}}$
$3 \cos x - 4 \sin x \leq \sqrt{25}$
$3 \cos x - 4 \sin x \leq 5$

Karnataka CET-2004

Application of Derivatives

85594 The maximum slope of the curve
$y = - x^{3} + 3 x^{2} + 2 x - 27 i$

1 1

2 23

3 5

4 -23

Explanation:

(C) : Given curve,
$y = - x^{2} + 3 x^{2} + 2 x - 27$
On differentiating both sides w.r.t $x$ we get-
$\frac{d y}{d x} = - 3 x^{2} + 6 x + 2$
For maximum slope
$\frac{dy}{dx} = 0$
Slope $(m) = - 3 x^{2} + 6 x + 2 = 0$
$\frac{d (m)}{d x} = - 6 x + 6 \Rightarrow \frac{d^{2} (m)}{d x^{2}} = - 6$
For maximum,
$\frac{d (m)}{d x} = 0$
$- 6 x + 6 = 0$
$x = 1$
Slope $(m) = - 3 x^{2} + 6 x + 2$
Value of maximum slope at $x = 1$
$m = - 3 (1)^{2} + 6 (1) + 2$
$= - 3 + 6 + 2$
$m_{max} = 5$

Application of Derivatives

85595 The minimum value of $f (x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$ if $a^{2} > b^{2}$ , is

1

b^{2}

2

a^{2} - b^{2}

3

a^{2}

4

a^{2} + b^{2}

Explanation:

(A) : Given,
$f (x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$
On differentiating both sides w.r.t. $x$ , we get-
$f^{'} (x) = a^{2} 2 \cos x \cdot (- \sin x) + b^{2} 2 \sin x \cos x$
$= - a^{2} \sin 2 x + b^{2} \sin 2 x$
$= (b^{2} - a^{2}) \sin 2 x$
For minimum $f^{'} (x) = 0$
$(b^{2} - a^{2}) \sin 2 x = 0$
$\sin 2 x = 0$
$2 x = 0$
$x = 0$
$f (0) = a^{2} (1)^{2} + b^{2} (0)$
$f (0) = a^{2} (maxima)$
Again on differentiating both sides w.r.t. $x$ , we get-
$f^{''} (x) = (b^{2} - a^{2}) 2 \cos 2 x$
$f^{''} (0) = (b^{2} - a^{2}) 2 < 0$
$f (\frac{π}{2}) = a^{2} (0) + b^{2} (1)$
$f (\frac{π}{2}) = b^{2} (minima)$
$f^{''} (\frac{π}{2}) = 2 (a^{2} - b^{2})$
$> 0$
$f (\frac{π}{2}) = b^{2}$
Hence, minimum value is $b^{2}$

MHT CET-2020

Application of Derivatives

85592 For all positive value of $x$ and $y$ , the value of
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{x y} is:$

1

\leq 9

2

< 9

3

\geq 9

4

> 9

Explanation:

(C) : Given,
For all positive value of $x$ and $y$ We have,
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{x y}$
We can write as,
$\frac{1 + x + x^{2}}{x y} = \frac{1}{x y} + \frac{1}{y} + \frac{x}{y}$

We know that,
A.M $\geq$ G.M
$= \frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq \sqrt[3]{\frac{1}{x y} \times \frac{1}{y} \times \frac{x}{y}}$
$= \frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq 3 {(\frac{1}{y^{3}})}^{\frac{1}{3}}$
$\frac{1}{x y} + \frac{1}{y} + \frac{x}{y} \geq \frac{3}{y}$
Similarly,
$\frac{1 + y + y^{2}}{x y} = \frac{1}{x y} + \frac{1}{x} + \frac{y}{x}$
$\frac{\frac{1}{x y} + \frac{1}{x} + \frac{y}{x}}{3} \geq {(\frac{1}{x y} \times \frac{1}{x} \times \frac{y}{x})}^{\frac{1}{3}}$
$\frac{1}{x y} + \frac{1}{x} + \frac{y}{x} \geq \frac{3}{x}$
On multiplying equation (i) and (ii) we get-
$(\frac{1}{x y} + \frac{1}{y} + \frac{x}{y}) (\frac{1}{x y} + \frac{1}{x} + \frac{y}{x}) \geq \frac{9}{x y}$
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{(x y)^{2}} \geq \frac{9}{x y}$
$\frac{(1 + x + x^{2}) (1 + y + y^{2})}{(x y)} \geq 9$

Karnataka CET-2002

Application of Derivatives

85593 The maximum of the function $3 \cos x - 4 \sin x$ is:

1 2

2 3

3 4

4 5

Explanation:

(D) : Given,
$f (x) = 3 \cos x - 4 \sin x$
We know that
$- \sqrt{a^{2} + b^{2}} \leq a \cos x + b \sin x \leq \sqrt{a^{2} + b^{2}}$
$a = 3 and b = - 4$
For maximum
$3 \cos x - 4 \sin x \leq \sqrt{(3)^{2} + (- 4)^{2}}$
$3 \cos x - 4 \sin x \leq \sqrt{25}$
$3 \cos x - 4 \sin x \leq 5$

Karnataka CET-2004

Application of Derivatives

85594 The maximum slope of the curve
$y = - x^{3} + 3 x^{2} + 2 x - 27 i$

1 1

2 23

3 5

4 -23

Explanation:

(C) : Given curve,
$y = - x^{2} + 3 x^{2} + 2 x - 27$
On differentiating both sides w.r.t $x$ we get-
$\frac{d y}{d x} = - 3 x^{2} + 6 x + 2$
For maximum slope
$\frac{dy}{dx} = 0$
Slope $(m) = - 3 x^{2} + 6 x + 2 = 0$
$\frac{d (m)}{d x} = - 6 x + 6 \Rightarrow \frac{d^{2} (m)}{d x^{2}} = - 6$
For maximum,
$\frac{d (m)}{d x} = 0$
$- 6 x + 6 = 0$
$x = 1$
Slope $(m) = - 3 x^{2} + 6 x + 2$
Value of maximum slope at $x = 1$
$m = - 3 (1)^{2} + 6 (1) + 2$
$= - 3 + 6 + 2$
$m_{max} = 5$

Application of Derivatives

85595 The minimum value of $f (x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$ if $a^{2} > b^{2}$ , is

1

b^{2}

2

a^{2} - b^{2}

3

a^{2}

4

a^{2} + b^{2}

Explanation:

(A) : Given,
$f (x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$
On differentiating both sides w.r.t. $x$ , we get-
$f^{'} (x) = a^{2} 2 \cos x \cdot (- \sin x) + b^{2} 2 \sin x \cos x$
$= - a^{2} \sin 2 x + b^{2} \sin 2 x$
$= (b^{2} - a^{2}) \sin 2 x$
For minimum $f^{'} (x) = 0$
$(b^{2} - a^{2}) \sin 2 x = 0$
$\sin 2 x = 0$
$2 x = 0$
$x = 0$
$f (0) = a^{2} (1)^{2} + b^{2} (0)$
$f (0) = a^{2} (maxima)$
Again on differentiating both sides w.r.t. $x$ , we get-
$f^{''} (x) = (b^{2} - a^{2}) 2 \cos 2 x$
$f^{''} (0) = (b^{2} - a^{2}) 2 < 0$
$f (\frac{π}{2}) = a^{2} (0) + b^{2} (1)$
$f (\frac{π}{2}) = b^{2} (minima)$
$f^{''} (\frac{π}{2}) = 2 (a^{2} - b^{2})$
$> 0$
$f (\frac{π}{2}) = b^{2}$
Hence, minimum value is $b^{2}$

MHT CET-2020