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

Q 11 :

Let the function $f (x) = \frac{x}{3} + \frac{3}{x} + 3, x \neq 0$ be strictly increasing in $(- \infty, α_{1}) \cup (α_{2}, \infty)$ and strictly decreasing in $(α_{3}, α_{4}) \cup (α_{4}, α_{5})$ . Then $\sum_{i = 1}^{5} α_{i}^{2}$ is equal to [2025]

40
28
36
48

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4

(3)

We have,

$f (x) = \frac{x}{3} + \frac{3}{x} + 3$

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

For critical points,

$f' (x) = 0 \Rightarrow \frac{1}{3} - \frac{3}{x^{2}} = 0$

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

Now, $f' (x) > 0 \forall x \in (- \infty, - 3) \cup (3, \infty)$

and $f' (x) < 0 \forall x \in (- 3, 0) \cup (0, 3)$

$∴$ f(x) is increasing in $(- \infty, - 3) \cup (3, \infty)$ and decreasing in $(- 3, 0) \cup (0, 3)$

$∴ α_{1} = - 3, α_{2} = 3, α_{3} = - 3, α_{4} = 0, α_{5} = 3$

$∴ \sum_{i = 1}^{5} α_{i}^{2} = 9 + 9 + 9 + 0 + 9 = 36$ .

Q 12 :

Let (2, 3) be the largest open interval in which the function $f (x) = 2 \log_{e} (x - 2) - x^{2} + a x + 1$ is strictly increasing and (b, c) be the largest open interval, in which the function $g (x) = {(x - 1)}^{3} {(x + 2 - a)}^{2}$ is strictly decreasing. Then 100 (a + b – c) is equal to: [2025]

360
160
280
420

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3

4

(1)

Given : $f (x) = 2 \log_{e} (x - 2) - x^{2} + a x + 1$ is strictly increasing on (2, 3)

$g (x) = {(x - 1)}^{3} {(x + 2 - a)}^{2}$ is strictly decreasing on (b, c).

Using f(x),

$f' (x) = \frac{2}{x - 2} - 2 x + a$

$\Rightarrow f'' (x) = \frac{- 2}{{(x - 2)}^{2}} - 2 < 0$

SInce f(x) is strictly increasing, so $f' (x) > 0$ .

But we have given that (2, 3) is the largest open interval where f(x) is strictly increasing.

$\Rightarrow f' (3) = 0 \Rightarrow 2 - 6 + a = 0 \Rightarrow a = 4$

Taking, $g (x) = {(x - 1)}^{3} {(x + 2 - a)}^{2}$

$= {(x - 1)}^{3} {(x - 2)}^{2}$ [ $∵$ a = 4]

$\Rightarrow g' (x) = {(x - 1)}^{2} (x - 2) (2 x - 2 + 3 x - 6)$

$= {(x - 1)}^{2} (x - 2) (5 x - 8)$

Here, $g' (x) < 0$ [ $∵$ g(x) is strictly decreasing]

$\Rightarrow {(x - 1)}^{2} (x - 2) (5 x - 8) < 0$

$\Rightarrow x \in (\frac{8}{5}, 2)$

$\Rightarrow b = \frac{8}{5} and c = 2$

Finally, we get 100 (a + b – c) = $100 (4 + \frac{8}{5} - 2) = 360$ .

Q 13 :

Let $g (x) = f (x) + f (1 - x)$ and $f'' (x) > 0, x \in (0, 1)$ . If $g$ is decreasing in the interval $(0, α)$ and increasing in the interval $(α, 1)$ , then $\tan^{- 1} (2 α) + \tan^{- 1} (\frac{1}{α}) + \tan^{- 1} (\frac{α + 1}{α})$ is equal to [2023]

$\frac{3 π}{2}$
$\frac{3 π}{4}$
$π$
$\frac{5 π}{4}$

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2

3

4

(3)

Given, $g (x) = f (x) + f (1 - x)$ and $f'' (x) > 0, x \in (0, 1)$

$∴$ $g' (x) = f' (x) + f' (1 - x) (- 1) = f' (x) - f' (1 - x)$

$and g'' (x) = f'' (x) + f'' (1 - x)$

At $x = \frac{1}{2}$ , $g' (\frac{1}{2}) = f' (\frac{1}{2}) - f' (1 - \frac{1}{2}) = 0 and g'' (x) > 0$

$∴$ $g (x) is concave upward for α = \frac{1}{2} .$

Now, $\tan^{- 1} (2 α) + \tan^{- 1} (\frac{1}{α}) + \tan^{- 1} (\frac{α + 1}{α})$

$= \tan^{- 1} (1) + \tan^{- 1} (2) + \tan^{- 1} (3)$

$= \frac{π}{4} + \tan^{- 1} (\frac{2 + 3}{1 - (2 \times 3)}) = \frac{π}{4} + \tan^{- 1} (- 1) = \frac{π}{4} + (π - \frac{π}{4}) = π$

Q 14 :

Let $f : [2, 4] \to R$ be a differentiable function such that $(x \log_{e} x) f' (x) + (\log_{e} x) f (x) + f (x) \geq 1, x \in [2, 4]$ with $f (2) = \frac{1}{2}$ and $f (4) = \frac{1}{4}$ .

Consider the following two statements:

(A) : $f (x) \leq 1$ , for all $x \in [2, 4]$

(B) : $f (x) \geq \frac{1}{8}$ , for all $x \in [2, 4]$

Then,

Only statement (B) is true
Neither statement (A) nor statement (B) is true
Both the statements (A) and (B) are true
Only statement (A) is true

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(3)

Q 15 :

Let $f : (0, 1) \to ℝ$ be a function defined by $f (x) = \frac{1}{1 - e^{- x}} and g (x) = (f (- x) - f (x)) .$ Consider the following two statements:

(I) $g$ is an increasing function in (0, 1)

(II) $g$ is one-one in (0, 1)

Then, [2023]

Only (I) is true
Both (I) and (II) are true
Neither (I) nor (II) is true
Only (II) is true

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4

(2)

$f (x) = \frac{e^{x}}{e^{x} - 1}$

$f (- x) = \frac{e^{- x}}{e^{- x} - 1} = \frac{1}{1 - e^{x}} = \frac{- 1}{e^{x} - 1};$ $g (x) = - \frac{(e^{x} + 1)}{e^{x} - 1}$

$g' (x) = - [\frac{(e^{x} - 1) e^{x} - (e^{x} + 1) e^{x}}{{(e^{x} - 1)}^{2}}] = \frac{2 e^{x}}{{(e^{x} - 1)}^{2}} > 0 \forall x \in (0, 1)$

$Hence, g is an increasing function in (0, 1) and is also one-one.$

Q 16 :

Let $f : ℝ \to ℝ$ be a twice differentiable function such that $f'' (x) > 0$ for all $x \in ℝ a n d$ where a is a real number. Let $g (x) = f (\tan^{2} x - 2 \tan x + a), 0 < x < \frac{π}{2} .$ Consider the following two statements:

(I) g is increasing in $(0, \frac{π}{4})$

(II) g is decreasing in $(\frac{π}{4}, \frac{π}{2})$

Then, [2026]

Only (I) is True
Both (I) and (II) are True
Neither (I) nor (II) is True
Only (II) is True

1

2

3

4

(4)

Topic Question Set

Q 11 :

Let the function $f (x) = \frac{x}{3} + \frac{3}{x} + 3, x \neq 0$ be strictly increasing in $(- \infty, α_{1}) \cup (α_{2}, \infty)$ and strictly decreasing in $(α_{3}, α_{4}) \cup (α_{4}, α_{5})$ . Then $\sum_{i = 1}^{5} α_{i}^{2}$ is equal to [2025]

Q 13 :

Let $g (x) = f (x) + f (1 - x)$ and $f'' (x) > 0, x \in (0, 1)$ . If $g$ is decreasing in the interval $(0, α)$ and increasing in the interval $(α, 1)$ , then $\tan^{- 1} (2 α) + \tan^{- 1} (\frac{1}{α}) + \tan^{- 1} (\frac{α + 1}{α})$ is equal to [2023]

(3)

Q 15 :

Let $f : (0, 1) \to ℝ$ be a function defined by $f (x) = \frac{1}{1 - e^{- x}} and g (x) = (f (- x) - f (x)) .$ Consider the following two statements:

(I) $g$ is an increasing function in (0, 1)

(II) $g$ is one-one in (0, 1)

Then, [2023]

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