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

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]

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(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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2

3

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
4
5
2

1

2

3

4

(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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2

3

4

(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

4

(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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4

(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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2

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4

(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]

0%

(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

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

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

$∴ x \in (- 10, 10)$

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

Topic Question Set

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]

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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]

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Topic Question Set

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]

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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]

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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]