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

Q 1 :

Let $f (x) = 3 \sqrt{x - 2} + \sqrt{4 - x}$ be a real valued function. If $α$ and $β$ are respectively the minimum and the maximum values of $f,$ then $α^{2} + 2 β^{2}$ is equal to [2024]

42
44
24
38

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(1)

$f (x) = 3 \sqrt{x - 2} + \sqrt{4 - x}$

$f' (x) = \frac{3}{2 \sqrt{x - 2}} - \frac{1}{2 \sqrt{4 - x}}$

$∴$ $f' (x) = 0 \Rightarrow 9 (4 - x) = (x - 2) \Rightarrow x = \frac{19}{5} \in [2, 4]$

Now, $f (2) = \sqrt{2}, f (\frac{19}{5}) = 2 \sqrt{5}, f (4) = 3 \sqrt{2}$

$∴$ $α = \sqrt{2}, β = 2 \sqrt{5} \Rightarrow α^{2} + 2 β^{2} = 2 + 40 = 42$

Q 2 :

If the function $f (x) = {(\frac{1}{x})}^{2 x}; x > 0$ attains the maximum value at $x = \frac{1}{e}$ then [2024]

${(2 e)}^{π} > π^{(2 e)}$
$e^{2 π} < {(2 π)}^{e}$
$e^{π} > π^{e}$
$e^{π} < π^{e}$

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(3)

Q 3 :

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1, a > 0$ has a local maximum at $x = α$ and a local minimum at $x = α^{2},$ then $α$ and $α^{2}$ are the roots of the equation : [2024]

$x^{2} + 6 x + 8 = 0$
$8 x^{2} - 6 x + 1 = 0$
$8 x^{2} + 6 x - 1 = 0$
$x^{2} - 6 x + 8 = 0$

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(4)

$We have, f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$

$\Rightarrow f' (x) = 6 x^{2} - 18 a x + 12 a^{2} = 6 (x^{2} - 3 a x + 2 a^{2})$

$\Rightarrow 6 (x - a) (x - 2 a) = 0$

$f' (x) = 0 \Rightarrow x = a, 2 a$

$Now, f'' (x) = 12 x - 18 a$

$f'' (a) = 12 a - 18 a = - 6 a < 0 (maximum)$

$f'' (2 a) = 24 a - 18 a = 6 a > 0 (minimum)$

$∴ f (x) has a local maximum at x = a and minimum at x = 2 a .$

$∴ α = a and α^{2} = 2 a$

$\Rightarrow a^{2} = 2 a \Rightarrow a = 0, 2 \Rightarrow a = 2 = α (∵ a > 0)$

$\Rightarrow α^{2} = 4$

$∴ Equation having roots α and α^{2} is$

$x^{2} - (α + α^{2}) x + α \cdot α^{2} = 0$

$\Rightarrow x^{2} - (2 + 4) x + 2 \times 4 = 0 \Rightarrow x^{2} - 6 x + 8 = 0$

Q 4 :

A variable line $L$ passes through the point (3, 5) and intersects the positive coordinate axes at the points $A$ and $B$ . The minimum area of the triangle $O A B,$ where $O$ is the origin, is: [2024]

35
40
25
30

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(4)

Equation of line $A B : \frac{x}{a} + \frac{y}{b} = 1$

Since, it passes through (3, 5)

$∴$ $\frac{3}{a} + \frac{5}{b} = 1$

$\Rightarrow 3 b + 5 a = a b \Rightarrow 5 a - a b = - 3 b \Rightarrow a (5 - b) = - 3 b$

$\Rightarrow a = \frac{3 b}{b - 5}$

Area of triangle $=$ $| \frac{1}{2} \times a \times b |$ $= \frac{1}{2} \times \frac{3 b}{b - 5} \times b$

$\Rightarrow f (b) = \frac{3 b^{2}}{2 b - 10}$

$\Rightarrow f' (b) = 0 \Rightarrow (2 b - 10) 6 b - 2 (3 b^{2}) = 0$

$\Rightarrow 12 b^{2} - 60 b - 6 b^{2} = 0 \Rightarrow 6 b^{2} - 60 b = 0$

$\Rightarrow b^{2} - 10 b = 0 \Rightarrow b (b - 10) = 0$

$b = 0 or b = 10$

So for minimum area, $b = 10$

$Hence, minimum area = \frac{3 b^{2}}{2 b - 10} = \frac{3 \times {(10)}^{2}}{2 \times 10 - 10}$

$= \frac{300}{10} = 30 sq. units$

Q 5 :

The maximum area of a triangle whose one vertex is at (0, 0) and the other two vertices lie on the curve $y = - 2 x^{2} + 54$ at points $(x, y)$ and $(- x, y),$ where $y > 0,$ is [2024]

108
122
88
92

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(1)

Let ABC be the required triangle whose area is A.

$∴$ $A = \frac{1}{2}$ $| 0 (y - y) + x (y - 0) - x (0 - y) |$

$= \frac{1}{2}$ $| x y + x y |$ $= \frac{1}{2} (2 x y) = x y$

$= x (- 2 x^{2} + 54) = - 2 (x^{3} - 27 x)$

$\Rightarrow A = - 2 (x^{3} - 27 x) \Rightarrow \frac{d A}{d x} = - 6 x^{2} + 54$

For critical point, $\frac{d A}{d x} = 0$

$\Rightarrow 3 x^{2} - 27 = 0 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3$

Now, $\frac{d^{2} A}{d x^{2}} = - 12 x,$ $\frac{d^{2} A}{d x^{2}} |_{x = 3}$ $< 0$

$∴$ Maximum area, $A = - 2 (27 - (27 \times 3))$

$= - 2 \times (- 54) = 108 sq. units.$

Q 6 :

$Let f (x) = {(x + 3)}^{2} {(x - 2)}^{3}, x \in [- 4, 4] . If M and m are the maximum and minimum values of f, respectively, in [- 4, 4],$ $then the value of M - m is:$ [2024]

600
392
108
608

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(4)

Given, $f (x) = {(x + 3)}^{2} {(x - 2)}^{3}$ ...(i)

Differentiate (i), w.r.t. $x,$ we get

$f' (x) = 2 (x + 3) {(x - 2)}^{3} + 3 {(x + 3)}^{2} {(x - 2)}^{2}$

$= (x + 3) {(x - 2)}^{2} [2 (x - 2) + (x + 3) \times 3]$

$= (x + 3) {(x - 2)}^{2} (5 x + 5)$

For maxima / minima, $f' (x) = 0$

$\Rightarrow (x + 3) {(x - 2)}^{2} (5 x + 5) = 0 \Rightarrow x = - 3, - 1, 2$

Now find the value of (i) at $x = - 4, - 3, - 1, 2, 4$

$f (- 4) = {(- 4 + 3)}^{2} {(- 4 - 2)}^{3} = 1 \times (- 216) = - 216$

$f (- 3) = 0$

$f (- 1) = {(- 1 + 3)}^{2} {(- 1 - 2)}^{3} = 4 \times (- 27) = - 108$

$f (2) = 0$

$f (4) = {(4 + 3)}^{2} {(4 - 2)}^{3} = 49 \times 8 = 392$

$∴$ Maximum value, $M = 392$

Minimum value, $m = - 216$

Hence, $M - m = 392 + 216 = 608$

Q 7 :

Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let $a$ and $b$ be the sides of the rectangle PQRS when its area is maximum. Then ${(a + b)}^{2}$ is equal to: [2024]

80
60
64
72

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(4)

In $∆ C D S$ , we have

$\cos θ = \frac{C S}{C D}$

$\Rightarrow \cos θ = \frac{C S}{4}$

$\Rightarrow C S = 4 \cos θ$

Similarly, in $∆ B C R$ we have

$\sin θ = \frac{C R}{2}$

$\Rightarrow C R = 2 \sin θ$

$∴$ $S R = C S + C R = 4 \cos θ + 2 \sin θ$

Similarly, $P S = 4 \sin θ + 2 \cos θ$

$∴$ Area of $P Q R S = P S \times S R$

$= 16 \cos θ \sin θ + 8 \cos^{2} θ + 8 \sin^{2} θ + 4 \sin θ \cos θ$

$= 8 (\cos^{2} θ + \sin^{2} θ) + 10 (2 \sin θ \cos θ)$

$= 8 + 10 \sin (2 θ)$

Area is maximum, when $\sin 2 θ = 1 \Rightarrow θ = 45 °$

$∴$ $Side a = 4 \sin θ + 2 \cos θ$ $[∵ θ = 45^{o}]$

$= \frac{4}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3 \sqrt{2}$

and $b = 4 \cos θ + 2 \sin θ = 3 \sqrt{2}$

Now, ${(a + b)}^{2} = {(6 \sqrt{2})}^{2} = 72$

Q 8 :

Let $f (x) = 4 \cos^{3} x + 3 \sqrt{3} \cos^{2} x - 10$ . The number of points of local maxima of $f$ in the interval $(0, 2 π)$ is: [2024]

2
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(1)

$f (x) = 4 \cos^{3} x + 3 \sqrt{3} \cos^{2} x - 10$

$f' (x) = - 12 \cos^{2} x \sin x - 6 \sqrt{3} \cos x \sin x$

$= - 6 \sin x \cos x [2 \cos x + \sqrt{3}]$

Critical points are $x = 0, \frac{π}{2}, π - \frac{π}{6}, π + \frac{π}{6}, π + \frac{π}{2}, π, 2 π$

Sign of $f (x)$

Points of local maxima $= \frac{5 π}{6}, \frac{7 π}{6}$

Q 9 :

The function $f (x) = 2 x + 3 {(x)}^{2 / 3}, x \in R,$ has [2024]

exactly one point of local minima and no point of local maxima
exactly two points of local maxima and exactly one point of local minima
exactly one point of local maxima and no point of local minima
exactly one point of local maxima and exactly one point of local minima

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(4)

We have, $f (x) = 2 x + 3 x^{2 / 3}$

$\Rightarrow f' (x) = 2 + 3 \cdot \frac{2}{3} x^{- 1 / 3} = 2 (1 + x^{- 1 / 3})$

For critical points, $f' (x) = 0$

$\Rightarrow 1 + x^{- 1 / 3} = 0 \Rightarrow x^{1 / 3} = - 1 \Rightarrow x = - 1$

The sign scheme of $f' (x)$ is

$∴$ $x = - 1 is the point of local maxima and x = 0 is the point of local minima.$

Q 10 :

Let A be the region enclosed by the parabola $y^{2} = 2 x$ and the line $x = 24$ . Then the maximum area of the rectangle inscribed in the region A is _____. [2024]

(128)

$y^{2} = 2 x and x = 24$

$∴$ Area of the rectangle, $A = (24 - x) 2 y$

$= (24 - \frac{y^{2}}{2}) 2 y$

$= 48 y - y^{3}$

$\frac{d A}{d y} = 48 - 3 y^{2}$

$\frac{d A}{d y} = 0 \Rightarrow 48 - 3 y^{2} = 0 \Rightarrow y^{2} = 16 \Rightarrow y = \pm 4$

Now, $\frac{d^{2} A}{d y^{2}} = - 6 y$

$\frac{d^{2} A}{d y^{2}} |_{y = 4} = - 24 < 0 (Maximum)$

$\frac{d^{2} A}{d y^{2}} |_{y = - 4} = 24 > 0 (Minimum)$

Hence, for maximum area, $y = 4 \Rightarrow x = 8$

Hence, maximum area $= (24 - 8) \times 2 \times 4 = 128$

Q 11 :

Let the set of all positive values of $λ$ , for which the point of local minimum of the function $(1 + x (λ^{2} - x^{2}))$ satisfies $\frac{x^{2} + x + 2}{x^{2} + 5 x + 6} < 0, be (α, β) .$ Then $α^{2} + β^{2}$ is equal to _____________ . [2024]

(39)

Let $f (x) = 1 + x (λ^{2} - x^{2})$

$f (x) = - x^{3} + (λ^{2} x + 1)$

$\Rightarrow f' (x) = - 3 x^{2} + λ^{2}$

Put $f' (x) = 0$

$\Rightarrow λ^{2} - 3 x^{2} = 0 \Rightarrow (λ - \sqrt{3} x) (λ + \sqrt{3} x) = 0$

$x = \pm \frac{λ}{\sqrt{3}}, f'' (x) = - 6 x$

So, $x = \frac{- λ}{\sqrt{3}}$ is point of minima.

Now, $- \frac{λ}{\sqrt{3}}$ should satisfy the given condition $\frac{x^{2} + x + 2}{x^{2} + 5 x + 6} < 0$

$i.e., \frac{1}{(x + 2) (x + 3)} < 0$ $[∵ x^{2} + x + 2 > 0]$

$\Rightarrow x \in (- 3, - 2)$

$\Rightarrow - 3 < - \frac{λ}{\sqrt{3}} < - 2 \Rightarrow - 3 \sqrt{3} < - λ < - 2 \sqrt{3}$

$2 \sqrt{3} < λ < 3 \sqrt{3} \Rightarrow λ \in (2 \sqrt{3}, 3 \sqrt{3}) \equiv (α, β)$ $(∵ Given)$

$∴$ ${(2 \sqrt{3})}^{2} + {(3 \sqrt{3})}^{2} = 12 + 27 = 39$

Q 12 :

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$ , where a > 0, attains its local maximum and local minimum values at p and q, respectively, such that $p^{2} = q$ , then f(3) is equal to : [2025]

55
10
37
23

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4

(3)

We have, $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$

$\Rightarrow f' (x) = 6 x^{2} - 18 a x + 12 a^{2}$

$\Rightarrow f (x) = 6 (x^{2} - 3 a x + 2 a^{2}) = 6 (x - a) (x - 2 a)$

$\Rightarrow f' (x) = 0 \Rightarrow 6 (x - a) (x - 2 a) = 0 \Rightarrow x = a, 2 a$

Now, $f'' (x) = 6 (2 x - 3 a)$

$f'' (a) = 6 (2 a - 3 a) = - 6 a < 0$

$∴$ x = a is point of maxima { $∵$ a > 0}

$f'' (2 a) = 6 (4 a - 3 a) = 6 a > 0$

$∴$ x = 2a is a point of minima

So, p = a and q = 2a.

Given, $p^{2} = q \Rightarrow a^{2} = 2 a \Rightarrow a = 2$

Now, $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$

f(3) = 54 – 162 + 144 + 1 = 37.

Q 13 :

Let $f : R \to R$ be a function defined by $f (x) = | | x + 2 | - 2 | x | |$ . If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is [2025]

3
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5
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(1)

We have, $f (x) = | | x + 2 | - 2 | x | |$

$\Rightarrow$ $f (x)$ $= {\begin{matrix} (- x + 2), & x < - 2 \\ (- 3 x - 2), & - 2 \leq x < \frac{- 2}{3} \\ (3 x + 2), & \frac{- 2}{3} \leq x < 0 \\ (- x + 2), & 0 \leq x < 2 \\ (x - 2), & x > 2 \end{matrix}$

Critical points are $x = - 2, \frac{- 2}{3}, 0, 2$

Number of local maxima = 1 = n

Number of local minima = 2 = m

$∴$ m + n = 2 + 1 = 3.

Q 14 :

Let a > 0. If the function $f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1$ attains its local maximum and minimum values at the points $x_{1}$ and $x_{2}$ respectively such that $x_{1} x_{2} = 54$ , then $a + x_{1} + x_{2}$ is equal to : [2025]

18
24
13
15

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(1)

We have, $f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1$ .

Since local max. and min. values occur when $f' (x) = 0$

$f' (x) = 18 x^{2} - 90 a x + 108 a^{2} = 0 \Rightarrow x = 2 a and 3 a$

i.e., $x_{1} = 2 a, x_{2} = 3 a$

Also, we have $x_{1} x_{2} = 54 \Rightarrow 6 a^{2} = 54 \Rightarrow a = 3$

$∴ a + x_{1} + x_{2} = 3 + 6 + 9 = 18$ .

Q 15 :

Let x = –1 and x = 2 be the critical points of the function $f (x) = x^{3} + a x^{2} + b \log_{e} | x | + 1, x \neq 0$ . Let m and M respectively be the absolute minimum and the absolute maximum values of f in the interval $[- 2, - \frac{1}{2}]$ . Then |M + m| is equal to

(Take $\log_{e} 2 = 0.7$ ): [2025]

19.8
22.1
21.1
20.9

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(3)

We have, $f (x) = x^{3} + a x^{2} + b \log_{e} | x | + 1, x \neq 0$

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

$f' (- 1) = 3 - 2 a - b = 0$

$f' (2) = 12 + 4 a + \frac{b}{2} = 0 \Rightarrow a = \frac{- 9}{2}, b = 12$

$∴ f (x) = x^{3} - \frac{9}{2} x^{2} + 12 \log_{e} | x | + 1$

$f (- 1) = - 1 - \frac{9}{2} + 1 = - \frac{9}{2}$

M = – 4.5

Min. value at x = – 2

$f (- 2) = - 8 - 18 + 12 \log_{e} 2 + 1$

m = – 25 + 12(0.7) = – 16.6

$∴$ |M + m| = 21.1

Q 16 :

Let the length of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be 10. If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}, t \in R$ , then $a^{2} + b^{2}$ is equal to: [2025]

126
120
115
125

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(1)

Length of latus rectum $= \frac{2 b^{2}}{a} = 10$ [Given]

$\Rightarrow 5 a = b^{2}$ ... (i)

Now, eccentricity is minimum value of

$f (t) = t^{2} + t + \frac{11}{12}$

$\Rightarrow f' (t) = 2 t + 1$

For critical point $f' (t) = 0$

$\Rightarrow 2 t + 1 = 0 \Rightarrow t = \frac{- 1}{2}$

Since, $f'' (t) = 2 > 0$ , so at $t = \frac{- 1}{2}$ , f(t) will give the minimum value.

$∴ f (\frac{- 1}{2}) = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{2}{3}$

$\Rightarrow e = \frac{2}{3} \Rightarrow \frac{2}{3} = \sqrt{1 - \frac{b^{2}}{a^{2}}}$

$\Rightarrow \frac{4}{9} = 1 - \frac{5 a}{a^{2}}$ [Using (i)]

$\Rightarrow \frac{4}{9} = 1 - \frac{5}{a} \Rightarrow \frac{5}{a} = 1 - \frac{4}{9} = \frac{5}{9} \Rightarrow a = 9$

$\Rightarrow b = \sqrt{45} = 3 \sqrt{5}$

$∴ a^{2} + b^{2} = {(9)}^{2} + {(3 \sqrt{5})}^{2} = 81 + 45 = 126$ .

Q 17 :

Let $f : R \to R$ be a polynomial function of degree four having extreme values at x = 4 and x = 5. If $\lim_{x \to 0} \frac{f (x)}{x^{2}} = 5$ , then f(2) is equal to : [2025]

12
14
8
10

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(4)

We have, $\lim_{x \to 0} \frac{f (x)}{x^{2}} = 5$

Let $f (x) = a x^{4} + b x^{3} + c x^{2} + d x + e$

$\Rightarrow \lim_{x \to 0} \frac{(a x^{4} + b x^{3} + c x^{2} + d x + e)}{x^{2}} = 5$

$\Rightarrow c = 5 and d = e = 0$

$∴ f (x) = a x^{4} + b x^{3} + 5 x^{2}$

$f' (x) = 4 a x^{3} + 3 b x^{2} + 10 x$

$= x (4 a x^{2} + 3 b x + 10)$

$∵$ f(x) has extreme values at x = 4 and x = 5, so f(4) = 0 and f(5) = 0.

Using derivative and its values, we get

$a = \frac{1}{8} and b = \frac{- 3}{2}$

Now, $f (2) = \frac{1}{8} \times 2^{4} - \frac{3}{2} \times 2^{3} + 5 \times 2^{2}$

= 2 – 12 + 20 = 10.

Q 18 :

Consider the region $R = {(x, y) : x \leq y \leq 9 - \frac{11}{3} x^{2}, x \geq 0}$ .

The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is : [2025]

$\frac{821}{123}$
$\frac{625}{111}$
$\frac{567}{121}$
$\frac{730}{119}$

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(3)

The given region R is shown below:

Here, x = t and $y = 9 - \frac{11 t^{3}}{3} - t$

Area of required rectangle,

$A = x y = t (9 - \frac{11 t^{2}}{3} - t) = 9 t - t^{2} - \frac{11}{3} t^{3}$

$\frac{d A}{d t} = 9 - 2 t - 11 t^{2}$

For critical points, $\frac{d A}{d t} = 0$

$\Rightarrow 11 t^{2} + 2 t - 9 = 0$

$\Rightarrow 11 t^{2} + 11 t - 9 t - 9 = 0$

$\Rightarrow t = - 1 or t = \frac{9}{11}$

$\frac{d^{2} A}{d t^{2}} = - 2 - 22 t$

$\frac{d^{2} A}{d t^{2}} > 0 at t = - 1$ i.e., minima and

$\frac{d^{2} A}{d t^{2}} < 0 at t = \frac{9}{11}$ i.e., maxima

$∴$ Maxima at $t = \frac{9}{11}$

$∴$ Largest area $= \frac{9}{11} (9 - \frac{11}{3} \times \frac{81}{121} - \frac{9}{11}) = \frac{9}{11} \times \frac{63}{11} = \frac{567}{121}$ .

Q 19 :

If the set of all values of a, for which the equation $5 x^{3} - 15 x - a = 0$ has three distinct real roots, is the interval $(α, β)$ , then $β - 2 α$ is equal to __________. [2025]

(30)

Given, $5 x^{3} - 15 x - a = 0 \Rightarrow a = 5 x^{3} - 15 x$

Let $P (x) = 5 x^{3} - 15 x$

Differentiating w.r.t. x, we get

$P' (x) = 15 x^{2} - 15 = 15$

$(x - 1) (x + 1)$

$∴ x \in (- 10, 10)$

$∴ β - 2 α = 10 - 2 (- 10) = 10 + 20 = 30$ .

Q 20 :

A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in ${cm}^{2}$ ) is equal to [2023]

675
900
1025
800

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(4)

Volume of cuboid, $v (x) = {(30 - 2 x)}^{2} x$

$∴$ $\frac{d v}{d x} = {(30 - 2 x)}^{2} + 2 x (30 - 2 x) (- 2) = 0$

$\Rightarrow (30 - 2 x) (30 - 2 x - 4 x) = 0$

$\Rightarrow (30 - 2 x) (30 - 6 x) = 0 \Rightarrow x = 5, 15$

$∵$ $x = 15 not possible,$ $∴$ $x = 5$

$\frac{d^{2} v}{d x^{2}} = 24 x - 240$

$At x = 5; \frac{d^{2} v}{d x^{2}} = 24 (5) - 240 = - 120 < 0$

$∴$ $The volume of the box is maximum at x = 5$

$∴$ Surface area $= {(30 - 2 x)}^{2} + 4 x (30 - 2 x)$

$= {(20)}^{2} + 4 \times 5 \times 20 = 400 + 400 = 800 {cm}^{2}$

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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(4)

Q 22 :

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

$π$
$\frac{π + 2 - 3 \sqrt{3}}{6}$
$\frac{5 π + 2 + 3 \sqrt{3}}{6}$
0

1

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(3)

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]

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

1

2

3

4

(3)

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]

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

1

2

3

4

(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]

12
10
24
13

1

2

3

4

(2)

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

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

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]

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

1

2

3

4

(2)

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

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

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]

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

1

2

3

4

(4)

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]

$\frac{3}{2}$
3
6
4

1

2

3

4

(3)

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

1

2

3

4

(1)

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]

(5)

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

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

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

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

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

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

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

Topic Question Set

Q 1 :

Let $f (x) = 3 \sqrt{x - 2} + \sqrt{4 - x}$ be a real valued function. If $α$ and $β$ are respectively the minimum and the maximum values of $f,$ then $α^{2} + 2 β^{2}$ is equal to [2024]

Q 2 :

If the function $f (x) = {(\frac{1}{x})}^{2 x}; x > 0$ attains the maximum value at $x = \frac{1}{e}$ then [2024]

(3)

Q 3 :

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1, a > 0$ has a local maximum at $x = α$ and a local minimum at $x = α^{2},$ then $α$ and $α^{2}$ are the roots of the equation : [2024]

Q 4 :

A variable line $L$ passes through the point (3, 5) and intersects the positive coordinate axes at the points $A$ and $B$ . The minimum area of the triangle $O A B,$ where $O$ is the origin, is: [2024]

Q 5 :

The maximum area of a triangle whose one vertex is at (0, 0) and the other two vertices lie on the curve $y = - 2 x^{2} + 54$ at points $(x, y)$ and $(- x, y),$ where $y > 0,$ is [2024]

Q 6 :

$Let f (x) = {(x + 3)}^{2} {(x - 2)}^{3}, x \in [- 4, 4] . If M and m are the maximum and minimum values of f, respectively, in [- 4, 4],$ $then the value of M - m is:$ [2024]

Q 7 :

Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let $a$ and $b$ be the sides of the rectangle PQRS when its area is maximum. Then ${(a + b)}^{2}$ is equal to: [2024]

Q 8 :

Let $f (x) = 4 \cos^{3} x + 3 \sqrt{3} \cos^{2} x - 10$ . The number of points of local maxima of $f$ in the interval $(0, 2 π)$ is: [2024]

Q 9 :

The function $f (x) = 2 x + 3 {(x)}^{2 / 3}, x \in R,$ has [2024]

Q 10 :

Let A be the region enclosed by the parabola $y^{2} = 2 x$ and the line $x = 24$ . Then the maximum area of the rectangle inscribed in the region A is _____. [2024]

Q 11 :

Let the set of all positive values of $λ$ , for which the point of local minimum of the function $(1 + x (λ^{2} - x^{2}))$ satisfies $\frac{x^{2} + x + 2}{x^{2} + 5 x + 6} < 0, be (α, β) .$ Then $α^{2} + β^{2}$ is equal to _____________ . [2024]

Q 12 :

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$ , where a > 0, attains its local maximum and local minimum values at p and q, respectively, such that $p^{2} = q$ , then f(3) is equal to : [2025]

Q 13 :

Let $f : R \to R$ be a function defined by $f (x) = | | x + 2 | - 2 | x | |$ . If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is [2025]

Q 14 :

Let a > 0. If the function $f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1$ attains its local maximum and minimum values at the points $x_{1}$ and $x_{2}$ respectively such that $x_{1} x_{2} = 54$ , then $a + x_{1} + x_{2}$ is equal to : [2025]

Q 16 :

Let the length of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be 10. If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}, t \in R$ , then $a^{2} + b^{2}$ is equal to: [2025]

Q 17 :

Let $f : R \to R$ be a polynomial function of degree four having extreme values at x = 4 and x = 5. If $\lim_{x \to 0} \frac{f (x)}{x^{2}} = 5$ , then f(2) is equal to : [2025]

Q 18 :

Consider the region $R = {(x, y) : x \leq y \leq 9 - \frac{11}{3} x^{2}, x \geq 0}$ .

The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is : [2025]

Q 19 :

If the set of all values of a, for which the equation $5 x^{3} - 15 x - a = 0$ has three distinct real roots, is the interval $(α, β)$ , then $β - 2 α$ is equal to __________. [2025]

Q 20 :

A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in ${cm}^{2}$ ) is equal to [2023]

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

Q 1 : Let f(x)=3x-2+4-x be a real valued function. If α and β are respectively the minimum and the maximum values of f, then α2+2β2 is equal to [2024]

Q 2 : If the function f(x)=(1x)2x; x>0 attains the maximum value at x=1e then [2024]

(3)

Q 3 : If the function f(x)=2x3-9ax2+12a2x+1, a>0 has a local maximum at x=α and a local minimum at x=α2, then α and α2 are the roots of the equation : [2024]

Q 4 : A variable line L passes through the point (3, 5) and intersects the positive coordinate axes at the points A and B. The minimum area of the triangle OAB, where O is the origin, is: [2024]

Q 5 : The maximum area of a triangle whose one vertex is at (0, 0) and the other two vertices lie on the curve y=-2x2+54 at points (x,y) and (-x,y), where y>0, is [2024]

Q 6 : Let f(x)=(x+3)2(x-2)3, x∈[-4,4]. If M and m are the maximum and minimum values of f, respectively, in [-4,4], then the value of M-m is: [2024]

Q 7 : Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a+b)2 is equal to: [2024]

Q 8 : Let f(x)=4cos3x+33cos2x-10. The number of points of local maxima of f in the interval (0,2π) is: [2024]

Q 9 : The function f(x)=2x+3(x)2/3, x∈R, has [2024]

Q 10 : Let A be the region enclosed by the parabola y2=2x and the line x=24. Then the maximum area of the rectangle inscribed in the region A is _____. [2024]

Q 11 : Let the set of all positive values of λ, for which the point of local minimum of the function (1+x(λ2-x2)) satisfies x2+x+2x2+5x+6<0, be (α,β). Then α2+β2 is equal to _____________ . [2024]

Q 12 : If the function f(x)=2x3–9ax2+12a2x+1, where a > 0, attains its local maximum and local minimum values at p and q, respectively, such that p2=q, then f(3) is equal to : [2025]

Q 13 : Let f:R→R be a function defined by f(x)=||x+2|–2|x||. If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is [2025]

Q 14 : Let a > 0. If the function f(x)=6x3–45ax2+108a2x+1 attains its local maximum and minimum values at the points x1 and x2 respectively such that x1x2=54, then a+x1+x2 is equal to : [2025]

Q 15 : Let x = –1 and x = 2 be the critical points of the function f(x)=x3+ax2+b loge |x|+1, x≠0. Let m and M respectively be the absolute minimum and the absolute maximum values of f in the interval [–2,–12]. Then |M + m| is equal to (Take loge 2=0.7): [2025]

Q 16 : Let the length of a latus rectum of an ellipse x2a2+y2b2=1 be 10. If its eccentricity is the minimum value of the function f(t)=t2+t+1112, t∈R, then a2+b2 is equal to: [2025]

Q 17 : Let f:R→R be a polynomial function of degree four having extreme values at x = 4 and x = 5. If limx→0f(x)x2=5, then f(2) is equal to : [2025]

Q 18 : Consider the region R={(x,y):x≤y≤9–113x2, x≥0}. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is : [2025]

Q 19 : If the set of all values of a, for which the equation 5x3–15x–a=0 has three distinct real roots, is the interval (α, β), then β–2α is equal to __________. [2025]

Q 20 : A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm2) is equal to [2023]

Q 21 : If the local maximum value of the function f(x)=(3e2sinx)sin2x, x∈(0,π2) is ke, then (ke)8+k8e5+k8 is equal to [2023]

(4)

Q 22 : max0≤x≤π{x-2sinxcosx+13sin3x}= [2023]

Q 23 : Let f(x)=2x+tan-1x and g(x)=loge(1+x2+x), x∈[0,3]. Then [2023]

Q 24 : Let f(x)=|1+sin2xcos2xsin2xsin2x1+cos2xsin2xsin2xcos2x1+sin2x|, x∈[π6,π3]. If α and β respectively are the maximum and the minimum values of f, then [2023]

(4)

Q 25 : The sum of the absolute maximum and minimum values of the function f(x)=|x2-5x+6|-3x+2 in the interval [-1,3] is equal to [2023]

Q 26 : Let x = 2 be a local minima of the function f(x)=2x4-18x2+8x+12, x∈(-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)=2x3+(2p-7)x2+3(2p-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)=x33+2bx+ax22 and g(x)=x33+ax+bx2, a≠2b, have a common extreme point, then a+2b+7 is equal to [2023]

Q 29 : A wire of length 20 m is to be cut into two pieces. A piece of length l1 is bent to make a square of area A1 and the other piece of length l2 is made into a circle of area A2. If 2A1+3A2 is minimum, then (πl1):l2 is equal to: [2023]

Q 30 : The number of points, where the curve y=x5-20x3+50x+2 crosses the x-axis, is __________ . [2023]

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Q 1 :

Let $f (x) = 3 \sqrt{x - 2} + \sqrt{4 - x}$ be a real valued function. If $α$ and $β$ are respectively the minimum and the maximum values of $f,$ then $α^{2} + 2 β^{2}$ is equal to [2024]

Q 2 :

If the function $f (x) = {(\frac{1}{x})}^{2 x}; x > 0$ attains the maximum value at $x = \frac{1}{e}$ then [2024]

Q 3 :

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1, a > 0$ has a local maximum at $x = α$ and a local minimum at $x = α^{2},$ then $α$ and $α^{2}$ are the roots of the equation : [2024]

Q 4 :

A variable line $L$ passes through the point (3, 5) and intersects the positive coordinate axes at the points $A$ and $B$ . The minimum area of the triangle $O A B,$ where $O$ is the origin, is: [2024]

Q 5 :

The maximum area of a triangle whose one vertex is at (0, 0) and the other two vertices lie on the curve $y = - 2 x^{2} + 54$ at points $(x, y)$ and $(- x, y),$ where $y > 0,$ is [2024]

Q 6 :

$Let f (x) = {(x + 3)}^{2} {(x - 2)}^{3}, x \in [- 4, 4] . If M and m are the maximum and minimum values of f, respectively, in [- 4, 4],$ $then the value of M - m is:$ [2024]

Q 7 :

Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let $a$ and $b$ be the sides of the rectangle PQRS when its area is maximum. Then ${(a + b)}^{2}$ is equal to: [2024]

Q 8 :

Let $f (x) = 4 \cos^{3} x + 3 \sqrt{3} \cos^{2} x - 10$ . The number of points of local maxima of $f$ in the interval $(0, 2 π)$ is: [2024]

Q 9 :

The function $f (x) = 2 x + 3 {(x)}^{2 / 3}, x \in R,$ has [2024]

Q 10 :

Let A be the region enclosed by the parabola $y^{2} = 2 x$ and the line $x = 24$ . Then the maximum area of the rectangle inscribed in the region A is _____. [2024]

Q 11 :

Let the set of all positive values of $λ$ , for which the point of local minimum of the function $(1 + x (λ^{2} - x^{2}))$ satisfies $\frac{x^{2} + x + 2}{x^{2} + 5 x + 6} < 0, be (α, β) .$ Then $α^{2} + β^{2}$ is equal to _____________ . [2024]

Q 12 :

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$ , where a > 0, attains its local maximum and local minimum values at p and q, respectively, such that $p^{2} = q$ , then f(3) is equal to : [2025]

Q 13 :

Let $f : R \to R$ be a function defined by $f (x) = | | x + 2 | - 2 | x | |$ . If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is [2025]

Q 14 :

Let a > 0. If the function $f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1$ attains its local maximum and minimum values at the points $x_{1}$ and $x_{2}$ respectively such that $x_{1} x_{2} = 54$ , then $a + x_{1} + x_{2}$ is equal to : [2025]

Q 16 :

Let the length of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be 10. If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}, t \in R$ , then $a^{2} + b^{2}$ is equal to: [2025]

Q 17 :

Let $f : R \to R$ be a polynomial function of degree four having extreme values at x = 4 and x = 5. If $\lim_{x \to 0} \frac{f (x)}{x^{2}} = 5$ , then f(2) is equal to : [2025]

Q 18 :

Consider the region $R = {(x, y) : x \leq y \leq 9 - \frac{11}{3} x^{2}, x \geq 0}$ .

The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is : [2025]

Q 19 :

If the set of all values of a, for which the equation $5 x^{3} - 15 x - a = 0$ has three distinct real roots, is the interval $(α, β)$ , then $β - 2 α$ is equal to __________. [2025]

Q 20 :

A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in ${cm}^{2}$ ) is equal to [2023]

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]

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]

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]