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Application of Derivatives

85689 Consider the following statements:
1.The function attains local minimum value at
$x = \frac{7}{5}$
2. $x = 2$ is the point of inflexion.
Which of the above statements is/are correct?

1 1 only

2 2 only

3 Both 1 and 2

4 Neither 1 nor 2

Explanation:

(C) :
$(1) f^{''} (\frac{7}{5}) = (\frac{7}{5} - 1) {(\frac{7}{5} - 2)}^{2} (5) + 0 + 0$
$= (\frac{2}{5}) (\frac{9}{25}) (5) > 0$
So, local minimum value at $x = \frac{7}{5}$
(2) $f^{''} (2) = 0$
So, both (1) and (2) are correct.

SCRA-2015

Application of Derivatives

85690 Let $f (x) = {\begin{matrix} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{matrix}$ . Then the set of all values of $b$ , for which $f (x)$ has maximum value at $x = 1$ , is:

1

(- 6, - 2)

2

(2, 6)

3

[- 6, - 2) \cup (2, 6]

4

[- \sqrt{6}, - 2) \cup (2, \sqrt{6}]

Explanation:

(C) : Given, $f (x) = {\begin{cases} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{cases}$
Then, for $x < 1, f (x) = x^{3} - x^{2} + 10 x - 7$
$f^{'} (x) = 3 x^{2} - 2 x + 10 > 0$
$\Rightarrow f (x)$ is increasing.
And, for $x > 1, f^{'} (x) < 0 \Rightarrow f (x)$ is decreasing
$lim_{x \to 1^{+}} f (x) = - 2 + \log_{2} (b^{2} - 4)$
From maximum value at $x = 1$
$\Rightarrow 3 \geq - 2 + \log_{2} (b^{2} - 4)$
$\log_{2} (b^{2} - 4) \leq 5$
$b^{2} - 4 > 0 and b^{2} - 4 \leq 32 or b^{2} - 36 \leq 0$
$b^{2} > 4 and b^{2} \leq 36$
$b > \pm 2 and b^{2} \leq 36$
$b > \pm 2 and b \leq 6$
So, $b \in [- 6, - 2] \cup (2, 6]$

JEE Main-2022-26.07.2022

Application of Derivatives

85691 Let $x = 2$ be a local minima of the function $f (x)$ $= 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$ . If $M$ is local maximum value of the function $f$ in $(- 4, 4)$ , then $M =$

1

18 \sqrt{6} - \frac{33}{2}

2

12 \sqrt{6} - \frac{33}{2}

3

12 \sqrt{6} - \frac{31}{2}

4

18 \sqrt{6} - \frac{31}{2}

Explanation:

(B) : $f (x) = 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$
Then, $f^{'} (x) = 8 x^{3} - 36 x + 8$
$f^{'} (x) = 4 (2 x^{3} - 9 x + 2)$
For maxima/minima $f^{'} (x) = 0$
$4 (2 x^{3} - 9 x + 2) = 0$
$2 x^{3} - 9 x + 2 = 0$
Then,

Since, $f^{'} (x)$ change sign from $+ v e$ to $- v e$ passing through $- 1 + \sqrt{\frac{3}{2}}$
Then, $x = - 1 + \sqrt{\frac{3}{2}}$ is point of local maxima.
So, $M = f (- 1 + \sqrt{\frac{3}{2}})$
$M = 2 {(- 1 + \sqrt{\frac{3}{2}})}^{4} - 18 {(- 1 + \sqrt{\frac{3}{2}})}^{2} + 8 (- 1 + \sqrt{\frac{3}{2}}) + 12$
$M = 2 (1 - 4 \sqrt{\frac{3}{2}} + 6 \times \frac{3}{2} - 4 \times \frac{3}{2} \sqrt{\frac{3}{2}} + \frac{3}{2} \times \frac{3}{2})$
$- 18 (1 - 2 \sqrt{\frac{3}{2}} + \frac{3}{2}) - 8 + 8 \sqrt{\frac{3}{2}} + 12$
$M = 2 - 4 \sqrt{6} + 18 - 6 \sqrt{6} + \frac{9}{2} - 18 + 18 \sqrt{6} - 27 - 8 + 4 \sqrt{6} + 12$
$M = 12 \sqrt{6} - \frac{33}{2}$

JEE Main-2023-25.01.2023

Application of Derivatives

85692 Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p$ -9) $x - 6$ have a maxima for some value of $x <$ 0 and a minima for some value of $x > 0$ . Then, the set of all values of $p$ is

1

(\frac{9}{2}, \infty)

2

(0, \frac{9}{2})

3

(- \frac{9}{2}, \frac{9}{2})

4

(- \infty, \frac{9}{2})

Explanation:

(D) : Given,
$f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$
Then, $f^{'} (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)$
From equation, $f^{'} (0) < 0$
$3 (2 p - 9) < 0$
$p < \frac{9}{2}$
So,
$p \in (- \infty, \frac{9}{2})$

JEE Main-2023-25.01.2023

Application of Derivatives

85693 If the function $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$ and
$g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b$ have
common extreme point, then $a + 2 b + 7$ is equal to :

1

\frac{3}{2}

2 3

3 6

4 4

Explanation:

(C) : Given, $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$
and, $g (x) = \frac{x^{3}}{3} + a x + b x^{2}$
Then, $f^{'} (x) = x^{2} + 2 b + ax$
And,
$g^{'} (x) = x^{2} + a + 2 b x$
Then, $f^{'} (x) - g^{'} (x) = 0$ (for find $x)$
$x^{2} + 2 b + a x - x^{2} - a - 2 b x = 0$
$(2 b - a) + a x - 2 b x = 0$
$(2 b - a) - x (2 b - a) = 0$
$(2 b - a) (1 - x) = 0$
$2 b = a$ or $x = 1$
But given $a \neq 2 b$ then $x = 1$
Put, $x = 1$ in $f^{'} (x) = 0$ or $g^{'} (x) = 0$
$1 + 2 b + a = 0$
$7 + 2 b + a = 6$
So, $a + 2 b + 7 = 6$

JEE Main-2023-30.01.2023

Application of Derivatives

85689 Consider the following statements:
1.The function attains local minimum value at
$x = \frac{7}{5}$
2. $x = 2$ is the point of inflexion.
Which of the above statements is/are correct?

1 1 only

2 2 only

3 Both 1 and 2

4 Neither 1 nor 2

Explanation:

(C) :
$(1) f^{''} (\frac{7}{5}) = (\frac{7}{5} - 1) {(\frac{7}{5} - 2)}^{2} (5) + 0 + 0$
$= (\frac{2}{5}) (\frac{9}{25}) (5) > 0$
So, local minimum value at $x = \frac{7}{5}$
(2) $f^{''} (2) = 0$
So, both (1) and (2) are correct.

SCRA-2015

Application of Derivatives

85690 Let $f (x) = {\begin{matrix} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{matrix}$ . Then the set of all values of $b$ , for which $f (x)$ has maximum value at $x = 1$ , is:

1

(- 6, - 2)

2

(2, 6)

3

[- 6, - 2) \cup (2, 6]

4

[- \sqrt{6}, - 2) \cup (2, \sqrt{6}]

Explanation:

(C) : Given, $f (x) = {\begin{cases} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{cases}$
Then, for $x < 1, f (x) = x^{3} - x^{2} + 10 x - 7$
$f^{'} (x) = 3 x^{2} - 2 x + 10 > 0$
$\Rightarrow f (x)$ is increasing.
And, for $x > 1, f^{'} (x) < 0 \Rightarrow f (x)$ is decreasing
$lim_{x \to 1^{+}} f (x) = - 2 + \log_{2} (b^{2} - 4)$
From maximum value at $x = 1$
$\Rightarrow 3 \geq - 2 + \log_{2} (b^{2} - 4)$
$\log_{2} (b^{2} - 4) \leq 5$
$b^{2} - 4 > 0 and b^{2} - 4 \leq 32 or b^{2} - 36 \leq 0$
$b^{2} > 4 and b^{2} \leq 36$
$b > \pm 2 and b^{2} \leq 36$
$b > \pm 2 and b \leq 6$
So, $b \in [- 6, - 2] \cup (2, 6]$

JEE Main-2022-26.07.2022

Application of Derivatives

85691 Let $x = 2$ be a local minima of the function $f (x)$ $= 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$ . If $M$ is local maximum value of the function $f$ in $(- 4, 4)$ , then $M =$

1

18 \sqrt{6} - \frac{33}{2}

2

12 \sqrt{6} - \frac{33}{2}

3

12 \sqrt{6} - \frac{31}{2}

4

18 \sqrt{6} - \frac{31}{2}

Explanation:

(B) : $f (x) = 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$
Then, $f^{'} (x) = 8 x^{3} - 36 x + 8$
$f^{'} (x) = 4 (2 x^{3} - 9 x + 2)$
For maxima/minima $f^{'} (x) = 0$
$4 (2 x^{3} - 9 x + 2) = 0$
$2 x^{3} - 9 x + 2 = 0$
Then,

Since, $f^{'} (x)$ change sign from $+ v e$ to $- v e$ passing through $- 1 + \sqrt{\frac{3}{2}}$
Then, $x = - 1 + \sqrt{\frac{3}{2}}$ is point of local maxima.
So, $M = f (- 1 + \sqrt{\frac{3}{2}})$
$M = 2 {(- 1 + \sqrt{\frac{3}{2}})}^{4} - 18 {(- 1 + \sqrt{\frac{3}{2}})}^{2} + 8 (- 1 + \sqrt{\frac{3}{2}}) + 12$
$M = 2 (1 - 4 \sqrt{\frac{3}{2}} + 6 \times \frac{3}{2} - 4 \times \frac{3}{2} \sqrt{\frac{3}{2}} + \frac{3}{2} \times \frac{3}{2})$
$- 18 (1 - 2 \sqrt{\frac{3}{2}} + \frac{3}{2}) - 8 + 8 \sqrt{\frac{3}{2}} + 12$
$M = 2 - 4 \sqrt{6} + 18 - 6 \sqrt{6} + \frac{9}{2} - 18 + 18 \sqrt{6} - 27 - 8 + 4 \sqrt{6} + 12$
$M = 12 \sqrt{6} - \frac{33}{2}$

JEE Main-2023-25.01.2023

Application of Derivatives

85692 Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p$ -9) $x - 6$ have a maxima for some value of $x <$ 0 and a minima for some value of $x > 0$ . Then, the set of all values of $p$ is

1

(\frac{9}{2}, \infty)

2

(0, \frac{9}{2})

3

(- \frac{9}{2}, \frac{9}{2})

4

(- \infty, \frac{9}{2})

Explanation:

(D) : Given,
$f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$
Then, $f^{'} (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)$
From equation, $f^{'} (0) < 0$
$3 (2 p - 9) < 0$
$p < \frac{9}{2}$
So,
$p \in (- \infty, \frac{9}{2})$

JEE Main-2023-25.01.2023

Application of Derivatives

85693 If the function $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$ and
$g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b$ have
common extreme point, then $a + 2 b + 7$ is equal to :

1

\frac{3}{2}

2 3

3 6

4 4

Explanation:

(C) : Given, $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$
and, $g (x) = \frac{x^{3}}{3} + a x + b x^{2}$
Then, $f^{'} (x) = x^{2} + 2 b + ax$
And,
$g^{'} (x) = x^{2} + a + 2 b x$
Then, $f^{'} (x) - g^{'} (x) = 0$ (for find $x)$
$x^{2} + 2 b + a x - x^{2} - a - 2 b x = 0$
$(2 b - a) + a x - 2 b x = 0$
$(2 b - a) - x (2 b - a) = 0$
$(2 b - a) (1 - x) = 0$
$2 b = a$ or $x = 1$
But given $a \neq 2 b$ then $x = 1$
Put, $x = 1$ in $f^{'} (x) = 0$ or $g^{'} (x) = 0$
$1 + 2 b + a = 0$
$7 + 2 b + a = 6$
So, $a + 2 b + 7 = 6$

JEE Main-2023-30.01.2023

Application of Derivatives

85689 Consider the following statements:
1.The function attains local minimum value at
$x = \frac{7}{5}$
2. $x = 2$ is the point of inflexion.
Which of the above statements is/are correct?

1 1 only

2 2 only

3 Both 1 and 2

4 Neither 1 nor 2

Explanation:

(C) :
$(1) f^{''} (\frac{7}{5}) = (\frac{7}{5} - 1) {(\frac{7}{5} - 2)}^{2} (5) + 0 + 0$
$= (\frac{2}{5}) (\frac{9}{25}) (5) > 0$
So, local minimum value at $x = \frac{7}{5}$
(2) $f^{''} (2) = 0$
So, both (1) and (2) are correct.

SCRA-2015

Application of Derivatives

85690 Let $f (x) = {\begin{matrix} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{matrix}$ . Then the set of all values of $b$ , for which $f (x)$ has maximum value at $x = 1$ , is:

1

(- 6, - 2)

2

(2, 6)

3

[- 6, - 2) \cup (2, 6]

4

[- \sqrt{6}, - 2) \cup (2, \sqrt{6}]

Explanation:

(C) : Given, $f (x) = {\begin{cases} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{cases}$
Then, for $x < 1, f (x) = x^{3} - x^{2} + 10 x - 7$
$f^{'} (x) = 3 x^{2} - 2 x + 10 > 0$
$\Rightarrow f (x)$ is increasing.
And, for $x > 1, f^{'} (x) < 0 \Rightarrow f (x)$ is decreasing
$lim_{x \to 1^{+}} f (x) = - 2 + \log_{2} (b^{2} - 4)$
From maximum value at $x = 1$
$\Rightarrow 3 \geq - 2 + \log_{2} (b^{2} - 4)$
$\log_{2} (b^{2} - 4) \leq 5$
$b^{2} - 4 > 0 and b^{2} - 4 \leq 32 or b^{2} - 36 \leq 0$
$b^{2} > 4 and b^{2} \leq 36$
$b > \pm 2 and b^{2} \leq 36$
$b > \pm 2 and b \leq 6$
So, $b \in [- 6, - 2] \cup (2, 6]$

JEE Main-2022-26.07.2022

Application of Derivatives

85691 Let $x = 2$ be a local minima of the function $f (x)$ $= 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$ . If $M$ is local maximum value of the function $f$ in $(- 4, 4)$ , then $M =$

1

18 \sqrt{6} - \frac{33}{2}

2

12 \sqrt{6} - \frac{33}{2}

3

12 \sqrt{6} - \frac{31}{2}

4

18 \sqrt{6} - \frac{31}{2}

Explanation:

(B) : $f (x) = 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$
Then, $f^{'} (x) = 8 x^{3} - 36 x + 8$
$f^{'} (x) = 4 (2 x^{3} - 9 x + 2)$
For maxima/minima $f^{'} (x) = 0$
$4 (2 x^{3} - 9 x + 2) = 0$
$2 x^{3} - 9 x + 2 = 0$
Then,

Since, $f^{'} (x)$ change sign from $+ v e$ to $- v e$ passing through $- 1 + \sqrt{\frac{3}{2}}$
Then, $x = - 1 + \sqrt{\frac{3}{2}}$ is point of local maxima.
So, $M = f (- 1 + \sqrt{\frac{3}{2}})$
$M = 2 {(- 1 + \sqrt{\frac{3}{2}})}^{4} - 18 {(- 1 + \sqrt{\frac{3}{2}})}^{2} + 8 (- 1 + \sqrt{\frac{3}{2}}) + 12$
$M = 2 (1 - 4 \sqrt{\frac{3}{2}} + 6 \times \frac{3}{2} - 4 \times \frac{3}{2} \sqrt{\frac{3}{2}} + \frac{3}{2} \times \frac{3}{2})$
$- 18 (1 - 2 \sqrt{\frac{3}{2}} + \frac{3}{2}) - 8 + 8 \sqrt{\frac{3}{2}} + 12$
$M = 2 - 4 \sqrt{6} + 18 - 6 \sqrt{6} + \frac{9}{2} - 18 + 18 \sqrt{6} - 27 - 8 + 4 \sqrt{6} + 12$
$M = 12 \sqrt{6} - \frac{33}{2}$

JEE Main-2023-25.01.2023

Application of Derivatives

85692 Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p$ -9) $x - 6$ have a maxima for some value of $x <$ 0 and a minima for some value of $x > 0$ . Then, the set of all values of $p$ is

1

(\frac{9}{2}, \infty)

2

(0, \frac{9}{2})

3

(- \frac{9}{2}, \frac{9}{2})

4

(- \infty, \frac{9}{2})

Explanation:

(D) : Given,
$f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$
Then, $f^{'} (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)$
From equation, $f^{'} (0) < 0$
$3 (2 p - 9) < 0$
$p < \frac{9}{2}$
So,
$p \in (- \infty, \frac{9}{2})$

JEE Main-2023-25.01.2023

Application of Derivatives

85693 If the function $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$ and
$g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b$ have
common extreme point, then $a + 2 b + 7$ is equal to :

1

\frac{3}{2}

2 3

3 6

4 4

Explanation:

(C) : Given, $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$
and, $g (x) = \frac{x^{3}}{3} + a x + b x^{2}$
Then, $f^{'} (x) = x^{2} + 2 b + ax$
And,
$g^{'} (x) = x^{2} + a + 2 b x$
Then, $f^{'} (x) - g^{'} (x) = 0$ (for find $x)$
$x^{2} + 2 b + a x - x^{2} - a - 2 b x = 0$
$(2 b - a) + a x - 2 b x = 0$
$(2 b - a) - x (2 b - a) = 0$
$(2 b - a) (1 - x) = 0$
$2 b = a$ or $x = 1$
But given $a \neq 2 b$ then $x = 1$
Put, $x = 1$ in $f^{'} (x) = 0$ or $g^{'} (x) = 0$
$1 + 2 b + a = 0$
$7 + 2 b + a = 6$
So, $a + 2 b + 7 = 6$

JEE Main-2023-30.01.2023

Application of Derivatives

85689 Consider the following statements:
1.The function attains local minimum value at
$x = \frac{7}{5}$
2. $x = 2$ is the point of inflexion.
Which of the above statements is/are correct?

1 1 only

2 2 only

3 Both 1 and 2

4 Neither 1 nor 2

Explanation:

(C) :
$(1) f^{''} (\frac{7}{5}) = (\frac{7}{5} - 1) {(\frac{7}{5} - 2)}^{2} (5) + 0 + 0$
$= (\frac{2}{5}) (\frac{9}{25}) (5) > 0$
So, local minimum value at $x = \frac{7}{5}$
(2) $f^{''} (2) = 0$
So, both (1) and (2) are correct.

SCRA-2015

Application of Derivatives

85690 Let $f (x) = {\begin{matrix} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{matrix}$ . Then the set of all values of $b$ , for which $f (x)$ has maximum value at $x = 1$ , is:

1

(- 6, - 2)

2

(2, 6)

3

[- 6, - 2) \cup (2, 6]

4

[- \sqrt{6}, - 2) \cup (2, \sqrt{6}]

Explanation:

(C) : Given, $f (x) = {\begin{cases} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{cases}$
Then, for $x < 1, f (x) = x^{3} - x^{2} + 10 x - 7$
$f^{'} (x) = 3 x^{2} - 2 x + 10 > 0$
$\Rightarrow f (x)$ is increasing.
And, for $x > 1, f^{'} (x) < 0 \Rightarrow f (x)$ is decreasing
$lim_{x \to 1^{+}} f (x) = - 2 + \log_{2} (b^{2} - 4)$
From maximum value at $x = 1$
$\Rightarrow 3 \geq - 2 + \log_{2} (b^{2} - 4)$
$\log_{2} (b^{2} - 4) \leq 5$
$b^{2} - 4 > 0 and b^{2} - 4 \leq 32 or b^{2} - 36 \leq 0$
$b^{2} > 4 and b^{2} \leq 36$
$b > \pm 2 and b^{2} \leq 36$
$b > \pm 2 and b \leq 6$
So, $b \in [- 6, - 2] \cup (2, 6]$

JEE Main-2022-26.07.2022

Application of Derivatives

85691 Let $x = 2$ be a local minima of the function $f (x)$ $= 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$ . If $M$ is local maximum value of the function $f$ in $(- 4, 4)$ , then $M =$

1

18 \sqrt{6} - \frac{33}{2}

2

12 \sqrt{6} - \frac{33}{2}

3

12 \sqrt{6} - \frac{31}{2}

4

18 \sqrt{6} - \frac{31}{2}

Explanation:

(B) : $f (x) = 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$
Then, $f^{'} (x) = 8 x^{3} - 36 x + 8$
$f^{'} (x) = 4 (2 x^{3} - 9 x + 2)$
For maxima/minima $f^{'} (x) = 0$
$4 (2 x^{3} - 9 x + 2) = 0$
$2 x^{3} - 9 x + 2 = 0$
Then,

Since, $f^{'} (x)$ change sign from $+ v e$ to $- v e$ passing through $- 1 + \sqrt{\frac{3}{2}}$
Then, $x = - 1 + \sqrt{\frac{3}{2}}$ is point of local maxima.
So, $M = f (- 1 + \sqrt{\frac{3}{2}})$
$M = 2 {(- 1 + \sqrt{\frac{3}{2}})}^{4} - 18 {(- 1 + \sqrt{\frac{3}{2}})}^{2} + 8 (- 1 + \sqrt{\frac{3}{2}}) + 12$
$M = 2 (1 - 4 \sqrt{\frac{3}{2}} + 6 \times \frac{3}{2} - 4 \times \frac{3}{2} \sqrt{\frac{3}{2}} + \frac{3}{2} \times \frac{3}{2})$
$- 18 (1 - 2 \sqrt{\frac{3}{2}} + \frac{3}{2}) - 8 + 8 \sqrt{\frac{3}{2}} + 12$
$M = 2 - 4 \sqrt{6} + 18 - 6 \sqrt{6} + \frac{9}{2} - 18 + 18 \sqrt{6} - 27 - 8 + 4 \sqrt{6} + 12$
$M = 12 \sqrt{6} - \frac{33}{2}$

JEE Main-2023-25.01.2023

Application of Derivatives

85692 Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p$ -9) $x - 6$ have a maxima for some value of $x <$ 0 and a minima for some value of $x > 0$ . Then, the set of all values of $p$ is

1

(\frac{9}{2}, \infty)

2

(0, \frac{9}{2})

3

(- \frac{9}{2}, \frac{9}{2})

4

(- \infty, \frac{9}{2})

Explanation:

(D) : Given,
$f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$
Then, $f^{'} (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)$
From equation, $f^{'} (0) < 0$
$3 (2 p - 9) < 0$
$p < \frac{9}{2}$
So,
$p \in (- \infty, \frac{9}{2})$

JEE Main-2023-25.01.2023

Application of Derivatives

85693 If the function $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$ and
$g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b$ have
common extreme point, then $a + 2 b + 7$ is equal to :

1

\frac{3}{2}

2 3

3 6

4 4

Explanation:

(C) : Given, $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$
and, $g (x) = \frac{x^{3}}{3} + a x + b x^{2}$
Then, $f^{'} (x) = x^{2} + 2 b + ax$
And,
$g^{'} (x) = x^{2} + a + 2 b x$
Then, $f^{'} (x) - g^{'} (x) = 0$ (for find $x)$
$x^{2} + 2 b + a x - x^{2} - a - 2 b x = 0$
$(2 b - a) + a x - 2 b x = 0$
$(2 b - a) - x (2 b - a) = 0$
$(2 b - a) (1 - x) = 0$
$2 b = a$ or $x = 1$
But given $a \neq 2 b$ then $x = 1$
Put, $x = 1$ in $f^{'} (x) = 0$ or $g^{'} (x) = 0$
$1 + 2 b + a = 0$
$7 + 2 b + a = 6$
So, $a + 2 b + 7 = 6$

JEE Main-2023-30.01.2023

Application of Derivatives

85689 Consider the following statements:
1.The function attains local minimum value at
$x = \frac{7}{5}$
2. $x = 2$ is the point of inflexion.
Which of the above statements is/are correct?

1 1 only

2 2 only

3 Both 1 and 2

4 Neither 1 nor 2

Explanation:

(C) :
$(1) f^{''} (\frac{7}{5}) = (\frac{7}{5} - 1) {(\frac{7}{5} - 2)}^{2} (5) + 0 + 0$
$= (\frac{2}{5}) (\frac{9}{25}) (5) > 0$
So, local minimum value at $x = \frac{7}{5}$
(2) $f^{''} (2) = 0$
So, both (1) and (2) are correct.

SCRA-2015

Application of Derivatives

85690 Let $f (x) = {\begin{matrix} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{matrix}$ . Then the set of all values of $b$ , for which $f (x)$ has maximum value at $x = 1$ , is:

1

(- 6, - 2)

2

(2, 6)

3

[- 6, - 2) \cup (2, 6]

4

[- \sqrt{6}, - 2) \cup (2, \sqrt{6}]

Explanation:

(C) : Given, $f (x) = {\begin{cases} x^{3} - x^{2} + 10 x - 7, x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4), x > 1 \end{cases}$
Then, for $x < 1, f (x) = x^{3} - x^{2} + 10 x - 7$
$f^{'} (x) = 3 x^{2} - 2 x + 10 > 0$
$\Rightarrow f (x)$ is increasing.
And, for $x > 1, f^{'} (x) < 0 \Rightarrow f (x)$ is decreasing
$lim_{x \to 1^{+}} f (x) = - 2 + \log_{2} (b^{2} - 4)$
From maximum value at $x = 1$
$\Rightarrow 3 \geq - 2 + \log_{2} (b^{2} - 4)$
$\log_{2} (b^{2} - 4) \leq 5$
$b^{2} - 4 > 0 and b^{2} - 4 \leq 32 or b^{2} - 36 \leq 0$
$b^{2} > 4 and b^{2} \leq 36$
$b > \pm 2 and b^{2} \leq 36$
$b > \pm 2 and b \leq 6$
So, $b \in [- 6, - 2] \cup (2, 6]$

JEE Main-2022-26.07.2022

Application of Derivatives

85691 Let $x = 2$ be a local minima of the function $f (x)$ $= 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$ . If $M$ is local maximum value of the function $f$ in $(- 4, 4)$ , then $M =$

1

18 \sqrt{6} - \frac{33}{2}

2

12 \sqrt{6} - \frac{33}{2}

3

12 \sqrt{6} - \frac{31}{2}

4

18 \sqrt{6} - \frac{31}{2}

Explanation:

(B) : $f (x) = 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$
Then, $f^{'} (x) = 8 x^{3} - 36 x + 8$
$f^{'} (x) = 4 (2 x^{3} - 9 x + 2)$
For maxima/minima $f^{'} (x) = 0$
$4 (2 x^{3} - 9 x + 2) = 0$
$2 x^{3} - 9 x + 2 = 0$
Then,

Since, $f^{'} (x)$ change sign from $+ v e$ to $- v e$ passing through $- 1 + \sqrt{\frac{3}{2}}$
Then, $x = - 1 + \sqrt{\frac{3}{2}}$ is point of local maxima.
So, $M = f (- 1 + \sqrt{\frac{3}{2}})$
$M = 2 {(- 1 + \sqrt{\frac{3}{2}})}^{4} - 18 {(- 1 + \sqrt{\frac{3}{2}})}^{2} + 8 (- 1 + \sqrt{\frac{3}{2}}) + 12$
$M = 2 (1 - 4 \sqrt{\frac{3}{2}} + 6 \times \frac{3}{2} - 4 \times \frac{3}{2} \sqrt{\frac{3}{2}} + \frac{3}{2} \times \frac{3}{2})$
$- 18 (1 - 2 \sqrt{\frac{3}{2}} + \frac{3}{2}) - 8 + 8 \sqrt{\frac{3}{2}} + 12$
$M = 2 - 4 \sqrt{6} + 18 - 6 \sqrt{6} + \frac{9}{2} - 18 + 18 \sqrt{6} - 27 - 8 + 4 \sqrt{6} + 12$
$M = 12 \sqrt{6} - \frac{33}{2}$

JEE Main-2023-25.01.2023

Application of Derivatives

85692 Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p$ -9) $x - 6$ have a maxima for some value of $x <$ 0 and a minima for some value of $x > 0$ . Then, the set of all values of $p$ is

1

(\frac{9}{2}, \infty)

2

(0, \frac{9}{2})

3

(- \frac{9}{2}, \frac{9}{2})

4

(- \infty, \frac{9}{2})

Explanation:

(D) : Given,
$f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$
Then, $f^{'} (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)$
From equation, $f^{'} (0) < 0$
$3 (2 p - 9) < 0$
$p < \frac{9}{2}$
So,
$p \in (- \infty, \frac{9}{2})$

JEE Main-2023-25.01.2023

Application of Derivatives

85693 If the function $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$ and
$g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b$ have
common extreme point, then $a + 2 b + 7$ is equal to :

1

\frac{3}{2}

2 3

3 6

4 4

Explanation:

(C) : Given, $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$
and, $g (x) = \frac{x^{3}}{3} + a x + b x^{2}$
Then, $f^{'} (x) = x^{2} + 2 b + ax$
And,
$g^{'} (x) = x^{2} + a + 2 b x$
Then, $f^{'} (x) - g^{'} (x) = 0$ (for find $x)$
$x^{2} + 2 b + a x - x^{2} - a - 2 b x = 0$
$(2 b - a) + a x - 2 b x = 0$
$(2 b - a) - x (2 b - a) = 0$
$(2 b - a) (1 - x) = 0$
$2 b = a$ or $x = 1$
But given $a \neq 2 b$ then $x = 1$
Put, $x = 1$ in $f^{'} (x) = 0$ or $g^{'} (x) = 0$
$1 + 2 b + a = 0$
$7 + 2 b + a = 6$
So, $a + 2 b + 7 = 6$

JEE Main-2023-30.01.2023