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Application of Derivatives

85694 If the maximum value of a, for which the function $f_{a} (x) = \tan^{- 1} 2 x - 3 a x + 7$ is nondecreasing in $(- \frac{π}{6}, \frac{π}{6})$ , is $\overset{―}{a}$ , then $f_{\overset{―}{a}} (\frac{π}{8})$ is equal to

1

8 - \frac{9 π}{4 (9 + π^{2})}

2

8 - \frac{4 π}{9 (4 + π^{2})}

3

8 (\frac{1 + π^{2}}{9 + π^{2}})

4

8 - \frac{π}{4}

Explanation:

(A) : Given,
Function $f (x) = \tan^{- 1} 2 x - 3 a x + 7$
Then, $\int_{0}^{1} (x) = \frac{1 \times 2}{1 + (2 x)^{2}} - 3 a$
$\int_{0}^{1} (x) = \frac{2}{1 + 4 x^{2}} - 3 a$
For maxima/ minimum $\int_{0}^{1} (x) = 0$
$\frac{2}{1 + 4 x^{2}} - 3 a = 0$
$3 a = \frac{2}{1 + 4 x^{2}}$
$a = \frac{2}{3 (1 + 4 x^{2})}$
* Since, $f (x)$ is non-decreasing in $(\frac{- π}{6}, \frac{π}{6})$
Then, $a = \frac{2}{3 (1 + 4 \times \frac{π^{2}}{36})}$
$a = \frac{2}{3 (1 + \frac{π^{2}}{9})} = \frac{2 \times 9}{3 (9 + π^{2})}$
So, $\int_{a} (\frac{π}{8}) = \tan^{- 1} 2 \times \frac{π}{8} - 3 \times \frac{2 \times 9}{3 (9 + π^{2})} \times \frac{π}{8} + 7$
$= \tan^{- 1} \frac{π}{4} - \frac{9 π}{(9 + π^{2})} \times \frac{1}{4} + 7 = 1 - \frac{9 π}{4 (9 + π^{2})} + 7$
$\int_{a} (\frac{π}{8}) = 8 - \frac{9 π}{4 (9 + π^{2})}$

JEE Main-2022-26.07.2022

Application of Derivatives

85695 The sum of the absolute maximum and minimum values of the function $f (x) =∣ x^{2} - 5 x$ $+ 6 ∣ - 3 x + 2$ in the interval $[- 1, 3]$ is equal to

1 12

2 10

3 24

4 13

Explanation:

(B) : Given,
$f (x) = | x^{2} - 5 x + 6 | - 3 x + 2$
$f (x) = x^{2} - 8 x + 8; x \in [- 1, 2]$
and, $f (x) = - x^{2} + 2 x - 4; x \in [2, 3)$
$∴ f_{max} = f (- 1) = 17$
$f_{min} = f (3) = - 7$
Since, $f (x)$ is decreasing function in $[- 1, 3]$

So, $f_{max} + f_{min} = 17 - 7 = 10$

JEE Main-2023-01.02.2023

Application of Derivatives

85696 The maximum value of $3 \cos x + 4 \sin x + 5$ is

1 5

2 6

3 7

4 none of these

Explanation:

(D) : Maximum value of $3 \cos x + 4 \sin x + 5$
We know that,
$- \sqrt{a^{2} + b^{2}} \leq a \cos θ + b \sin θ \leq \sqrt{a^{2} + b^{2}}$
$- \sqrt{3^{2} + 4^{2}} \leq 3 \cos x + 4 \sin x \leq \sqrt{3^{2} + 4^{2}}$
$5 - \sqrt{25} \leq 3 \cos x + 4 \sin x + 5 \leq \sqrt{25} + 5$
$0 \leq 3 \cos x + 4 \sin x + 5 \leq 5 + 5$
$0 \leq 3 \cos x + 4 \sin x + 5 \leq 10$
The maximum value of
$3 \cos x + 4 \sin x + 5 = 10$

AMU-2009

Application of Derivatives

85697 If the maximum value of $y = a \cos x -$
$\frac{1}{3} \cos 3 x$ occurs when $x = \frac{π}{6}$ , then the value of a
is

1 -2

2 2

3

\frac{2}{\sqrt{3}}

4

- \frac{2}{\sqrt{3}}

Explanation:

(B) : Given,
$y = a \cos x - \frac{1}{3} \cos 3 x$
Then, $\frac{d y}{d x} = - a \sin x + \frac{1}{3} \sin 3 x \times 3$
$\frac{d y}{d x} = - a \sin x + \sin 3 x$
For maxim of minima, $\frac{dy}{dx} = 0$
$- a \sin x = - \sin 3 x \Rightarrow a = \frac{\sin 3 x}{\sin x}$
So, when $x = \frac{π}{6}$ , then the value of $a$ is
$a = \frac{\sin 3 \times \frac{π}{6}}{\sin \frac{π}{6}} = \frac{\sin \frac{π}{2}}{\sin \frac{π}{6}} = \frac{\sin 90^{\circ}}{\sin 30^{\circ}} \Rightarrow a = \frac{1}{1 / 2} \Rightarrow a = 2$

Application of Derivatives

85694 If the maximum value of a, for which the function $f_{a} (x) = \tan^{- 1} 2 x - 3 a x + 7$ is nondecreasing in $(- \frac{π}{6}, \frac{π}{6})$ , is $\overset{―}{a}$ , then $f_{\overset{―}{a}} (\frac{π}{8})$ is equal to

1

8 - \frac{9 π}{4 (9 + π^{2})}

2

8 - \frac{4 π}{9 (4 + π^{2})}

3

8 (\frac{1 + π^{2}}{9 + π^{2}})

4

8 - \frac{π}{4}

Explanation:

(A) : Given,
Function $f (x) = \tan^{- 1} 2 x - 3 a x + 7$
Then, $\int_{0}^{1} (x) = \frac{1 \times 2}{1 + (2 x)^{2}} - 3 a$
$\int_{0}^{1} (x) = \frac{2}{1 + 4 x^{2}} - 3 a$
For maxima/ minimum $\int_{0}^{1} (x) = 0$
$\frac{2}{1 + 4 x^{2}} - 3 a = 0$
$3 a = \frac{2}{1 + 4 x^{2}}$
$a = \frac{2}{3 (1 + 4 x^{2})}$
* Since, $f (x)$ is non-decreasing in $(\frac{- π}{6}, \frac{π}{6})$
Then, $a = \frac{2}{3 (1 + 4 \times \frac{π^{2}}{36})}$
$a = \frac{2}{3 (1 + \frac{π^{2}}{9})} = \frac{2 \times 9}{3 (9 + π^{2})}$
So, $\int_{a} (\frac{π}{8}) = \tan^{- 1} 2 \times \frac{π}{8} - 3 \times \frac{2 \times 9}{3 (9 + π^{2})} \times \frac{π}{8} + 7$
$= \tan^{- 1} \frac{π}{4} - \frac{9 π}{(9 + π^{2})} \times \frac{1}{4} + 7 = 1 - \frac{9 π}{4 (9 + π^{2})} + 7$
$\int_{a} (\frac{π}{8}) = 8 - \frac{9 π}{4 (9 + π^{2})}$

JEE Main-2022-26.07.2022

Application of Derivatives

85695 The sum of the absolute maximum and minimum values of the function $f (x) =∣ x^{2} - 5 x$ $+ 6 ∣ - 3 x + 2$ in the interval $[- 1, 3]$ is equal to

1 12

2 10

3 24

4 13

Explanation:

(B) : Given,
$f (x) = | x^{2} - 5 x + 6 | - 3 x + 2$
$f (x) = x^{2} - 8 x + 8; x \in [- 1, 2]$
and, $f (x) = - x^{2} + 2 x - 4; x \in [2, 3)$
$∴ f_{max} = f (- 1) = 17$
$f_{min} = f (3) = - 7$
Since, $f (x)$ is decreasing function in $[- 1, 3]$

So, $f_{max} + f_{min} = 17 - 7 = 10$

JEE Main-2023-01.02.2023

Application of Derivatives

85696 The maximum value of $3 \cos x + 4 \sin x + 5$ is

1 5

2 6

3 7

4 none of these

Explanation:

(D) : Maximum value of $3 \cos x + 4 \sin x + 5$
We know that,
$- \sqrt{a^{2} + b^{2}} \leq a \cos θ + b \sin θ \leq \sqrt{a^{2} + b^{2}}$
$- \sqrt{3^{2} + 4^{2}} \leq 3 \cos x + 4 \sin x \leq \sqrt{3^{2} + 4^{2}}$
$5 - \sqrt{25} \leq 3 \cos x + 4 \sin x + 5 \leq \sqrt{25} + 5$
$0 \leq 3 \cos x + 4 \sin x + 5 \leq 5 + 5$
$0 \leq 3 \cos x + 4 \sin x + 5 \leq 10$
The maximum value of
$3 \cos x + 4 \sin x + 5 = 10$

AMU-2009

Application of Derivatives

85697 If the maximum value of $y = a \cos x -$
$\frac{1}{3} \cos 3 x$ occurs when $x = \frac{π}{6}$ , then the value of a
is

1 -2

2 2

3

\frac{2}{\sqrt{3}}

4

- \frac{2}{\sqrt{3}}

Explanation:

(B) : Given,
$y = a \cos x - \frac{1}{3} \cos 3 x$
Then, $\frac{d y}{d x} = - a \sin x + \frac{1}{3} \sin 3 x \times 3$
$\frac{d y}{d x} = - a \sin x + \sin 3 x$
For maxim of minima, $\frac{dy}{dx} = 0$
$- a \sin x = - \sin 3 x \Rightarrow a = \frac{\sin 3 x}{\sin x}$
So, when $x = \frac{π}{6}$ , then the value of $a$ is
$a = \frac{\sin 3 \times \frac{π}{6}}{\sin \frac{π}{6}} = \frac{\sin \frac{π}{2}}{\sin \frac{π}{6}} = \frac{\sin 90^{\circ}}{\sin 30^{\circ}} \Rightarrow a = \frac{1}{1 / 2} \Rightarrow a = 2$

Application of Derivatives

85694 If the maximum value of a, for which the function $f_{a} (x) = \tan^{- 1} 2 x - 3 a x + 7$ is nondecreasing in $(- \frac{π}{6}, \frac{π}{6})$ , is $\overset{―}{a}$ , then $f_{\overset{―}{a}} (\frac{π}{8})$ is equal to

1

8 - \frac{9 π}{4 (9 + π^{2})}

2

8 - \frac{4 π}{9 (4 + π^{2})}

3

8 (\frac{1 + π^{2}}{9 + π^{2}})

4

8 - \frac{π}{4}

Explanation:

(A) : Given,
Function $f (x) = \tan^{- 1} 2 x - 3 a x + 7$
Then, $\int_{0}^{1} (x) = \frac{1 \times 2}{1 + (2 x)^{2}} - 3 a$
$\int_{0}^{1} (x) = \frac{2}{1 + 4 x^{2}} - 3 a$
For maxima/ minimum $\int_{0}^{1} (x) = 0$
$\frac{2}{1 + 4 x^{2}} - 3 a = 0$
$3 a = \frac{2}{1 + 4 x^{2}}$
$a = \frac{2}{3 (1 + 4 x^{2})}$
* Since, $f (x)$ is non-decreasing in $(\frac{- π}{6}, \frac{π}{6})$
Then, $a = \frac{2}{3 (1 + 4 \times \frac{π^{2}}{36})}$
$a = \frac{2}{3 (1 + \frac{π^{2}}{9})} = \frac{2 \times 9}{3 (9 + π^{2})}$
So, $\int_{a} (\frac{π}{8}) = \tan^{- 1} 2 \times \frac{π}{8} - 3 \times \frac{2 \times 9}{3 (9 + π^{2})} \times \frac{π}{8} + 7$
$= \tan^{- 1} \frac{π}{4} - \frac{9 π}{(9 + π^{2})} \times \frac{1}{4} + 7 = 1 - \frac{9 π}{4 (9 + π^{2})} + 7$
$\int_{a} (\frac{π}{8}) = 8 - \frac{9 π}{4 (9 + π^{2})}$

JEE Main-2022-26.07.2022

Application of Derivatives

85695 The sum of the absolute maximum and minimum values of the function $f (x) =∣ x^{2} - 5 x$ $+ 6 ∣ - 3 x + 2$ in the interval $[- 1, 3]$ is equal to

1 12

2 10

3 24

4 13

Explanation:

(B) : Given,
$f (x) = | x^{2} - 5 x + 6 | - 3 x + 2$
$f (x) = x^{2} - 8 x + 8; x \in [- 1, 2]$
and, $f (x) = - x^{2} + 2 x - 4; x \in [2, 3)$
$∴ f_{max} = f (- 1) = 17$
$f_{min} = f (3) = - 7$
Since, $f (x)$ is decreasing function in $[- 1, 3]$

So, $f_{max} + f_{min} = 17 - 7 = 10$

JEE Main-2023-01.02.2023

Application of Derivatives

85696 The maximum value of $3 \cos x + 4 \sin x + 5$ is

1 5

2 6

3 7

4 none of these

Explanation:

(D) : Maximum value of $3 \cos x + 4 \sin x + 5$
We know that,
$- \sqrt{a^{2} + b^{2}} \leq a \cos θ + b \sin θ \leq \sqrt{a^{2} + b^{2}}$
$- \sqrt{3^{2} + 4^{2}} \leq 3 \cos x + 4 \sin x \leq \sqrt{3^{2} + 4^{2}}$
$5 - \sqrt{25} \leq 3 \cos x + 4 \sin x + 5 \leq \sqrt{25} + 5$
$0 \leq 3 \cos x + 4 \sin x + 5 \leq 5 + 5$
$0 \leq 3 \cos x + 4 \sin x + 5 \leq 10$
The maximum value of
$3 \cos x + 4 \sin x + 5 = 10$

AMU-2009

Application of Derivatives

85697 If the maximum value of $y = a \cos x -$
$\frac{1}{3} \cos 3 x$ occurs when $x = \frac{π}{6}$ , then the value of a
is

1 -2

2 2

3

\frac{2}{\sqrt{3}}

4

- \frac{2}{\sqrt{3}}

Explanation:

(B) : Given,
$y = a \cos x - \frac{1}{3} \cos 3 x$
Then, $\frac{d y}{d x} = - a \sin x + \frac{1}{3} \sin 3 x \times 3$
$\frac{d y}{d x} = - a \sin x + \sin 3 x$
For maxim of minima, $\frac{dy}{dx} = 0$
$- a \sin x = - \sin 3 x \Rightarrow a = \frac{\sin 3 x}{\sin x}$
So, when $x = \frac{π}{6}$ , then the value of $a$ is
$a = \frac{\sin 3 \times \frac{π}{6}}{\sin \frac{π}{6}} = \frac{\sin \frac{π}{2}}{\sin \frac{π}{6}} = \frac{\sin 90^{\circ}}{\sin 30^{\circ}} \Rightarrow a = \frac{1}{1 / 2} \Rightarrow a = 2$

Application of Derivatives

85694 If the maximum value of a, for which the function $f_{a} (x) = \tan^{- 1} 2 x - 3 a x + 7$ is nondecreasing in $(- \frac{π}{6}, \frac{π}{6})$ , is $\overset{―}{a}$ , then $f_{\overset{―}{a}} (\frac{π}{8})$ is equal to

1

8 - \frac{9 π}{4 (9 + π^{2})}

2

8 - \frac{4 π}{9 (4 + π^{2})}

3

8 (\frac{1 + π^{2}}{9 + π^{2}})

4

8 - \frac{π}{4}

Explanation:

(A) : Given,
Function $f (x) = \tan^{- 1} 2 x - 3 a x + 7$
Then, $\int_{0}^{1} (x) = \frac{1 \times 2}{1 + (2 x)^{2}} - 3 a$
$\int_{0}^{1} (x) = \frac{2}{1 + 4 x^{2}} - 3 a$
For maxima/ minimum $\int_{0}^{1} (x) = 0$
$\frac{2}{1 + 4 x^{2}} - 3 a = 0$
$3 a = \frac{2}{1 + 4 x^{2}}$
$a = \frac{2}{3 (1 + 4 x^{2})}$
* Since, $f (x)$ is non-decreasing in $(\frac{- π}{6}, \frac{π}{6})$
Then, $a = \frac{2}{3 (1 + 4 \times \frac{π^{2}}{36})}$
$a = \frac{2}{3 (1 + \frac{π^{2}}{9})} = \frac{2 \times 9}{3 (9 + π^{2})}$
So, $\int_{a} (\frac{π}{8}) = \tan^{- 1} 2 \times \frac{π}{8} - 3 \times \frac{2 \times 9}{3 (9 + π^{2})} \times \frac{π}{8} + 7$
$= \tan^{- 1} \frac{π}{4} - \frac{9 π}{(9 + π^{2})} \times \frac{1}{4} + 7 = 1 - \frac{9 π}{4 (9 + π^{2})} + 7$
$\int_{a} (\frac{π}{8}) = 8 - \frac{9 π}{4 (9 + π^{2})}$

JEE Main-2022-26.07.2022

Application of Derivatives

85695 The sum of the absolute maximum and minimum values of the function $f (x) =∣ x^{2} - 5 x$ $+ 6 ∣ - 3 x + 2$ in the interval $[- 1, 3]$ is equal to

1 12

2 10

3 24

4 13

Explanation:

(B) : Given,
$f (x) = | x^{2} - 5 x + 6 | - 3 x + 2$
$f (x) = x^{2} - 8 x + 8; x \in [- 1, 2]$
and, $f (x) = - x^{2} + 2 x - 4; x \in [2, 3)$
$∴ f_{max} = f (- 1) = 17$
$f_{min} = f (3) = - 7$
Since, $f (x)$ is decreasing function in $[- 1, 3]$

So, $f_{max} + f_{min} = 17 - 7 = 10$

JEE Main-2023-01.02.2023

Application of Derivatives

85696 The maximum value of $3 \cos x + 4 \sin x + 5$ is

1 5

2 6

3 7

4 none of these

Explanation:

(D) : Maximum value of $3 \cos x + 4 \sin x + 5$
We know that,
$- \sqrt{a^{2} + b^{2}} \leq a \cos θ + b \sin θ \leq \sqrt{a^{2} + b^{2}}$
$- \sqrt{3^{2} + 4^{2}} \leq 3 \cos x + 4 \sin x \leq \sqrt{3^{2} + 4^{2}}$
$5 - \sqrt{25} \leq 3 \cos x + 4 \sin x + 5 \leq \sqrt{25} + 5$
$0 \leq 3 \cos x + 4 \sin x + 5 \leq 5 + 5$
$0 \leq 3 \cos x + 4 \sin x + 5 \leq 10$
The maximum value of
$3 \cos x + 4 \sin x + 5 = 10$

AMU-2009

Application of Derivatives

85697 If the maximum value of $y = a \cos x -$
$\frac{1}{3} \cos 3 x$ occurs when $x = \frac{π}{6}$ , then the value of a
is

1 -2

2 2

3

\frac{2}{\sqrt{3}}

4

- \frac{2}{\sqrt{3}}

Explanation:

(B) : Given,
$y = a \cos x - \frac{1}{3} \cos 3 x$
Then, $\frac{d y}{d x} = - a \sin x + \frac{1}{3} \sin 3 x \times 3$
$\frac{d y}{d x} = - a \sin x + \sin 3 x$
For maxim of minima, $\frac{dy}{dx} = 0$
$- a \sin x = - \sin 3 x \Rightarrow a = \frac{\sin 3 x}{\sin x}$
So, when $x = \frac{π}{6}$ , then the value of $a$ is
$a = \frac{\sin 3 \times \frac{π}{6}}{\sin \frac{π}{6}} = \frac{\sin \frac{π}{2}}{\sin \frac{π}{6}} = \frac{\sin 90^{\circ}}{\sin 30^{\circ}} \Rightarrow a = \frac{1}{1 / 2} \Rightarrow a = 2$

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