Differential Calculus jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question

Q 21 :

If the local maximum value of the function $f (x) = {(\frac{\sqrt{3 e}}{2 \sin x})}^{\sin^{2} x}, x \in (0, \frac{π}{2})$ is $\frac{k}{e}$ , then ${(\frac{k}{e})}^{8} + \frac{k^{8}}{e^{5}} + k^{8}$ is equal to [2023]

$e^{3} + e^{5} + e^{11}$
$e^{3} + e^{6} + e^{10}$
$e^{5} + e^{6} + e^{11}$
$e^{3} + e^{6} + e^{11}$

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Q 22 :

$\max_{0 \leq x \leq π} {x - 2 \sin x \cos x + \frac{1}{3} \sin 3 x} =$ [2023]

$π$
$\frac{π + 2 - 3 \sqrt{3}}{6}$
$\frac{5 π + 2 + 3 \sqrt{3}}{6}$
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Let $f (x) = x - 2 \sin x \cos x + \frac{1}{3} \sin 3 x, 0 \leq x \leq π$

$f (x) = x - \sin 2 x + \frac{1}{3} \sin 3 x$

$f' (x) = 1 - 2 \cos 2 x + \cos 3 x = 0$

$\Rightarrow 4 \cos^{3} x - 3 \cos x - 2 (2 \cos^{2} x - 1) + 1 = 0$

$\Rightarrow 4 \cos^{3} x - 4 \cos^{2} x - 3 \cos x + 3 = 0$

$\Rightarrow 4 \cos^{2} x (\cos x - 1) - 3 (\cos x - 1) = 0$

$\Rightarrow (\cos x - 1) (4 \cos^{2} x - 3) = 0$

$\Rightarrow \cos x = 1, \pm \frac{\sqrt{3}}{2} ∴ x = 0, \frac{π}{6}, \frac{5 π}{6}$

Now, $f'' (x) = 4 \sin 2 x - 3 \sin 3 x$

Now, $f'' (0) = 0$

$f'' (\frac{π}{6}) > 0 and f'' (\frac{5 π}{6}) < 0 \Rightarrow (\frac{5 π}{6}) is a point of maxima.$

Hence, $f (\frac{5 π}{6}) = \frac{5 π}{6} - 2 \sin \frac{5 π}{6} \cos \frac{5 π}{6} + \frac{1}{3} \sin \frac{5 π}{2}$

$= \frac{5 π}{6} - \sin 2 (\frac{5 π}{6}) + \frac{1}{3} = \frac{5 π}{2} + \frac{\sqrt{3}}{2} + \frac{1}{3}$ $= \frac{5 π + 3 \sqrt{3} + 2}{6}$

Q 23 :

Let $f (x) = 2 x + \tan^{- 1} x$ and $g (x) = \log_{e} (\sqrt{1 + x^{2}} + x), x \in [0, 3] .$ Then [2023]

$\min f' (x) = 1 + \max g' (x)$
$there exist 0 < x_{1} < x_{2} < 3 such that f (x) < g (x), \forall x \in (x_{1}, x_{2})$
$\max f (x) > \max g (x)$
$there exists \hat{x} \in [0, 3] such that f' (\hat{x}) < g' (\hat{x})$

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We have, $f (x) = 2 x + \tan^{- 1} x and g (x) = \log_{e} (\sqrt{1 + x^{2}} + x)$

$g' (x) = \frac{1}{\sqrt{1 + x^{2}} + x} [\frac{1}{2 \sqrt{1 + x^{2}}} \times 2 x + 1]$

$= \frac{x + \sqrt{1 + x^{2}}}{(\sqrt{1 + x^{2}} + x) (\sqrt{1 + x^{2}})} = \frac{1}{\sqrt{1 + x^{2}}}; f' (x) = 2 + \frac{1}{1 + x^{2}}$

Both $f (x)$ and $g (x)$ are strictly increasing functions in [0, 3].

$Max f (x) = f (3) = 6 + \tan^{- 1} 3$

$Max g (x) = g (3) = \ln (\sqrt{10} + 3)$

$Max f (x) > Max g (x), x \in [0, 3]$

Q 24 :

Let

$f (x)$ $=$ $| \begin{matrix} 1 + \sin^{2} x & \cos^{2} x & \sin 2 x \\ \sin^{2} x & 1 + \cos^{2} x & \sin 2 x \\ \sin^{2} x & \cos^{2} x & 1 + \sin 2 x \end{matrix} |$ , $x \in [\frac{π}{6}, \frac{π}{3}] .$

If $α$ and $β$ respectively are the maximum and the minimum values of $f$ , then [2023]

$β^{2} + 2 \sqrt{α} = \frac{19}{4}$
$α^{2} + β^{2} = \frac{9}{2}$
$α^{2} - β^{2} = 4 \sqrt{3}$
$β^{2} - 2 \sqrt{α} = \frac{19}{4}$

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Q 25 :

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 [2023]

12
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24
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$f (x) =$ $| x^{2} - 5 x + 6 |$ $- 3 x + 2$

$f (x)$ $=$ ${\begin{matrix} x^{2} - 8 x + 8, & x \in [- 1, 2] \\ - x^{2} + 2 x - 4, & x \in [2, 3] \end{matrix}$

From the graph, Maximum value = 17

Minimum value = - 7 as $f$ (-1) = 17 and $f (3) = - 7$

$∴$ Sum of the absolute maximum and minimum values = $17 - 7 = 10$

Q 26 :

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 the local maximum value of the function $f$ in (- 4, 4), then $M =$ [2023]

$18 \sqrt{6} - \frac{33}{2}$
$12 \sqrt{6} - \frac{33}{2}$
$12 \sqrt{6} - \frac{31}{2}$
$18 \sqrt{6} - \frac{31}{2}$

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$f (x) = 2 x^{4} - 18 x^{2} + 8 x + 12$

$f' (x) = 8 x^{3} - 36 x + 8 = 4 (x - 2) (2 x^{2} + 4 x - 1)$

Sign of $f' (x)$ is given below

Point of maxima = $\frac{\sqrt{6} - 2}{2}$

$M = 12 \sqrt{6} - \frac{33}{2}$

Q 27 :

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 [2023]

$(\frac{9}{2}, \infty)$
$(0, \frac{9}{2})$
$(- \frac{9}{2}, \frac{9}{2})$
$(- \infty, \frac{9}{2})$

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We have, $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$

$Differentiating w.r.t. x, we get$

$f' (x) = 6 x^{2} + 2 (2 p - 7) x + 3 (2 p - 9)$

$∵$ $f' (0) < 0 \Rightarrow 3 (2 p - 9) < 0$

$\Rightarrow p < \frac{9}{2}$

$So, p \in (- \infty, \frac{9}{2})$

Q 28 :

If the functions $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$ and $g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b,$ have a common extreme point, then $a + 2 b + 7$ is equal to [2023]

$\frac{3}{2}$
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We have, $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$

and $g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b$

For critical points,

$f' (x) = x^{2} + 2 b + a x = 0 ...(i)$

$g' (x) = x^{2} + 2 b x + a = 0 ...(ii)$

Since $f (x)$ and $g (x)$ have a common extreme point,

$∴$ condition for a common root is

$α = \frac{a_{2} c_{1} - a_{1} c_{2}}{a_{1} b_{2} - a_{2} b_{1}} = \frac{b_{1} c_{2} - b_{2} c_{1}}{a_{2} c_{1} - a_{1} c_{2}}, α \neq 0$ $= \frac{2 b - a}{2 b - a} = \frac{a^{2} - 4 b^{2}}{2 b - a}$

$\Rightarrow \frac{(a + 2 b) (a - 2 b)}{- (a - 2 b)} = 1 \Rightarrow a + 2 b = - 1$

$∴$ $a + 2 b + 7 = - 1 + 7 = 6$

Q 29 :

A wire of length 20 m is to be cut into two pieces. A piece of length $l_{1}$ is bent to make a square of area $A_{1}$ and the other piece of length $l_{2}$ is made into a circle of area $A_{2}$ . If $2 A_{1} + 3 A_{2}$ is minimum, then $(π l_{1}) : l_{2}$ is equal to: [2023]

6 : 1
3 : 1
4 : 1
1 : 6

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Let $x$ be the side of the square and $r =$ radius of the circle.

Now, $l_{1} = 4 x and l_{2} = 2 π r .$ Then $4 x + 2 π r = l (let) ...(i)$

Also, $A_{1} = x^{2} and A_{2} = π r^{2}$

Let $A = 2 A_{1} + 3 A_{2} = 2 x^{2} + 3 π r^{2}$

$= 2 x^{2} + 3 π {(\frac{l - 4 x}{2 π})}^{2} = 2 x^{2} + \frac{3}{4 π} {(l - 4 x)}^{2} [From (i)]$

Now, $\frac{d A}{d x} = 4 x + \frac{3}{2 π} (l - 4 x) (- 4)$

For minima, $\frac{d A}{d x} = 0;$ $4 x + \frac{3}{2 π} (l - 4 x) \times (- 4) = 0$

$\Rightarrow x = \frac{6 l}{4 π + 24}$

Now, $l_{1} = 4 x = \frac{6 l}{π + 6} and l_{2} = 2 π r = l - 4 x$

$\Rightarrow l_{2} = l - \frac{6 l}{π + 6} = \frac{π l}{π + 6}$

Now,
$\frac{π l_{1}}{l_{2}} = \frac{π (\frac{6 l}{π + 6})}{(\frac{π l}{π + 6})} = 6 : 1$

Q 30 :

The number of points, where the curve $y = x^{5} - 20 x^{3} + 50 x + 2$ crosses the $x$ -axis, is __________ . [2023]

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$Let f (x) = y = x^{5} - 20 x^{3} + 50 x + 2$

$\Rightarrow \frac{d y}{d x} = 5 x^{4} - 60 x^{2} + 50 = 5 (x^{4} - 12 x^{2} + 10)$

$Put \frac{d y}{d x} = 0 \Rightarrow x^{4} - 12 x^{2} + 10 = 0$

$\Rightarrow x^{2} = \frac{12 \pm \sqrt{144 - 40}}{2} \Rightarrow x^{2} = 6 \pm \sqrt{26} \Rightarrow x^{2} \approx 6 \pm 5.1$

$\Rightarrow x^{2} \approx 11.1, 0.9 \Rightarrow x \approx \pm 3.3, \pm 0.95$

$f (0) = 2, f (1) = + v e, f (2) = - v e; f (- 1) = - v e, f (- 2) = + v e$

$∴$ $y crosses the x -axis at 5 points.$

Topic Question Set

Q 21 :

If the local maximum value of the function $f (x) = {(\frac{\sqrt{3 e}}{2 \sin x})}^{\sin^{2} x}, x \in (0, \frac{π}{2})$ is $\frac{k}{e}$ , then ${(\frac{k}{e})}^{8} + \frac{k^{8}}{e^{5}} + k^{8}$ is equal to [2023]

(4)

Q 22 :

$\max_{0 \leq x \leq π} {x - 2 \sin x \cos x + \frac{1}{3} \sin 3 x} =$ [2023]

Q 23 :

Let $f (x) = 2 x + \tan^{- 1} x$ and $g (x) = \log_{e} (\sqrt{1 + x^{2}} + x), x \in [0, 3] .$ Then [2023]

(4)

Q 25 :

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 [2023]

Q 26 :

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 the local maximum value of the function $f$ in (- 4, 4), then $M =$ [2023]

Q 27 :

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 [2023]

Q 28 :

If the functions $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$ and $g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b,$ have a common extreme point, then $a + 2 b + 7$ is equal to [2023]

Q 29 :

A wire of length 20 m is to be cut into two pieces. A piece of length $l_{1}$ is bent to make a square of area $A_{1}$ and the other piece of length $l_{2}$ is made into a circle of area $A_{2}$ . If $2 A_{1} + 3 A_{2}$ is minimum, then $(π l_{1}) : l_{2}$ is equal to: [2023]

Q 30 :

The number of points, where the curve $y = x^{5} - 20 x^{3} + 50 x + 2$ crosses the $x$ -axis, is __________ . [2023]

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Topic Question Set

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Q 22 : max0≤x≤π{x-2sinxcosx+13sin3x}= [2023] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Q 23 : Let f(x)=2x+tan-1x and g(x)=loge(1+x2+x), x∈[0,3]. Then [2023] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

(4)

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Q 22 :

$\max_{0 \leq x \leq π} {x - 2 \sin x \cos x + \frac{1}{3} \sin 3 x} =$ [2023]

Q 23 :

Let $f (x) = 2 x + \tan^{- 1} x$ and $g (x) = \log_{e} (\sqrt{1 + x^{2}} + x), x \in [0, 3] .$ Then [2023]