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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4

(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]

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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]

0%

(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 $= (24 - 8) \times 2 \times 4 = 128$

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]

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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]

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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]