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

Q 1 :

For the function $f (x) = \sin x + 3 x - \frac{2}{π} (x^{2} + x), where x \in [0, \frac{π}{2}],$ consider the following two statements :

(I) $f$ is increasing in $(0, \frac{π}{2}) .$

(II) $f'$ is decreasing in $(0, \frac{π}{2})$ .

Between the above two statements, [2024]

neither (I) nor (II) is true.
only (II) is true.
both (I) and (II) are true.
only (I) is true.

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(3)

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

$f' (x) = \cos x + 3 - \frac{2}{π} (2 x + 1)$

$f' (x) > 0 for x \in [0, \frac{π}{4}]$

For $x \in [π / 4, π / 2],$ we have $π / 4 \leq x < π / 2$

$\Rightarrow \frac{π}{2} + 1 \leq 2 x + 1 < π + 1$

$\Rightarrow 1 + \frac{2}{π} \leq \frac{2}{π} (2 x + 1) < 2 + \frac{2}{π}$

Since, $\frac{2}{π} (2 x + 1) < 3$

So, $f' (x) > 0, x \in (0, π / 2)$

So $f (x)$ is increasing in $(0, π / 2)$

Now, $f'' (x) = - \sin x - \frac{4}{π} < 0, x \in (0, π / 2)$

$\Rightarrow f' (x) is decreasing in (0, π / 2) .$

Hence, both statements are true.

Q 2 :

The interval in which the function $f (x) = x^{x}, x > 0,$ is strictly increasing is [2024]

$[\frac{1}{e}, \infty)$
$[\frac{1}{e^{2}}, 1)$
$(0, \frac{1}{e}]$
$(0, \infty)$

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(1)

$f (x) = x^{x}, x > 0$

So, $y = x^{x}$

$\Rightarrow \ln y = x \ln x$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = 1 + \ln x \Rightarrow \frac{d y}{d x} = x^{x} (1 + \ln x)$

For $y$ to be strictly increasing, we have $\frac{d y}{d x} \geq 0$

$\Rightarrow x^{x} (1 + \ln x) \geq 0$

$\Rightarrow 1 + \ln x \geq 0 [∵ x^{x} > 0, as x > 0]$

$\Rightarrow \ln x \geq - 1 \Rightarrow x \geq e^{- 1}$

$\Rightarrow x \in [\frac{1}{e}, \infty)$

Q 3 :

The number of critical points of the function $f (x) = {(x - 2)}^{2 / 3} (2 x + 1)$ is [2024]

1
2
3
0

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4

(2)

Given, $f (x) = {(x - 2)}^{\frac{2}{3}} (2 x + 1)$

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

$= 2 {(x - 2)}^{- \frac{1}{3}} [\frac{2 x + 1}{3} + (x - 2)]$

$= \frac{2}{{(x - 2)}^{\frac{1}{3}}} [\frac{2 x + 3 x + 1 - 6}{3}] = \frac{2}{{(x - 2)}^{\frac{1}{3}}} \frac{(5 x - 5)}{3}$

$= \frac{10}{3} \frac{(x - 1)}{{(x - 2)}^{\frac{1}{3}}}$

For critical points, $f' (x) = 0$ or $f' (x)$ is non-existence.

$∴$ Critical points are $x = 1, 2$ i.e., 2 in number.

Q 4 :

For the function $f (x) = (\cos x) - x + 1, x \in R,$ between the following two statements [2024]

(S1) $f (x) = 0$ for only one value of $x$ in $[0, π] .$

(S2) $f (x)$ is decreasing in $[0, \frac{π}{2}]$ and increasing in $[\frac{π}{2}, π .]$

Only (S2) is correct.
Both (S1) and (S2) are incorrect.
Only (S1) is correct.
Both (S1) and (S2) are correct.

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(3)

We have, $f (x) = \cos x - x + 1, x \in R$

$\Rightarrow f' (x) = - \sin x - 1 < 0 \forall x \in R$

$\Rightarrow f (x) is decreasing \forall x \in R$

For $f (x) = 0$

Now, $f (0) = 2$

$f (π) = - π$

$f (0) f (π) < 0$

So, $f$ is strictly decreasing in $[0, π]$ .

So, $f (x) = 0$ has one solution in $[0, π]$

(S1) is correct but (S2) is incorrect.

Q 5 :

$Let g (x) = 3 f (\frac{x}{3}) + f (3 - x) and f'' (x) > 0 for all x \in (0, 3) .$

$If g is decreasing in (0, α) and increasing in (α, 3), then 8 α is:$ [2024]

20
18
0
24

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(2)

$Given, g (x) = 3 f (\frac{x}{3}) + f (3 - x)$ ...(i)

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

$g' (x) = 3 f' (\frac{x}{3}) \cdot \frac{1}{3} + f' (3 - x) (- 1)$

$\Rightarrow g' (x) = f' (\frac{x}{3}) - f' (3 - x)$

When $g (x)$ is decreasing, $g' (x) < 0,$

$\Rightarrow f' (\frac{x}{3}) < f' (3 - x) \Rightarrow \frac{x}{3} < (3 - x) \Rightarrow x < \frac{9}{4}$

Therefore, $α = \frac{9}{4};$ So, $8 α = 8 \times \frac{9}{4} = 18$

Q 6 :

Consider the function $f : [\frac{1}{2}, 1] \to R$ defined by $f (x) = 4 \sqrt{2} x^{3} - 3 \sqrt{2} x - 1 .$

Consider the statements

(I) The curve $y = f (x)$ intersects the $x$ -axis exactly at one point.

(II) The curve $y = f (x)$ intersects the $x$ -axis at $x = \cos \frac{π}{12}$

Then

Both (I) and (II) are incorrect.
Only (I) is correct.
Both (I) and (II) are correct.
Only (II) is correct.

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(3)

$We have, f (x) = 4 \sqrt{2} x^{3} - 3 \sqrt{2} x - 1$

$So, f' (x) = 12 \sqrt{2} x^{2} - 3 \sqrt{2} \geq 0 for x \in [\frac{1}{2}, 1]$

$So, f (x) is increasing.$

$Now, f (\frac{1}{2}) < 0 and f (1) > 0$

$∴ f (x) intersects x -axis at exactly one point.$

$So, statement-I is correct.$

$Let \cos α = x$

$∴ \sqrt{2} (4 \cos^{3} α - 3 \cos α) - 1 = 0$

$\Rightarrow \sqrt{2} \cos 3 α = 1 \Rightarrow \cos 3 α = \frac{1}{\sqrt{2}} \Rightarrow 3 α = \frac{π}{4}$

$\Rightarrow α = \frac{π}{12}$

$So, statement-II is correct.$

Q 7 :

The function $f (x) = \frac{x}{x^{2} - 6 x - 16}, x \in R - {- 2, 8}$ [2024]

decreases in $(- 2, 8)$ and increases in $(- \infty, - 2) \cup (8, \infty)$
decreases in $(- \infty, - 2)$ and increases in $(8, \infty)$
increases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$
decreases in $(- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$

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(4)

$f (x) = \frac{x}{x^{2} - 6 x - 16}$

$\Rightarrow f' (x) = \frac{(x^{2} - 6 x - 16) (1) - x (2 x - 6)}{{(x^{2} - 6 x - 16)}^{2}}$

$= \frac{- x^{2} - 16}{{(x^{2} - 6 x - 16)}^{2}} < 0 \forall x$

$∴ f (x) is a decreasing function in its domain.$

$\Rightarrow f (x) decreases in (- \infty, - 2) \cup (- 2, 8) \cup (8, \infty)$

Q 8 :

Let $f : R \to (0, \infty)$ be strictly increasing function such that $\lim_{x \to \infty} \frac{f (7 x)}{f (x)} = 1 .$ Then, the value of $\lim_{x \to \infty} [\frac{f (5 x)}{f (x)} - 1]$ is equal to [2024]

4
0
7/5
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(2)

$For x > 0,$

$x \leq 5 x \leq 7 x \Rightarrow f (x) \leq f (5 x) \leq f (7 x) (∵ f is strictly increasing function)$

$\Rightarrow \frac{f (x)}{f (x)} \leq \frac{f (5 x)}{f (x)} \leq \frac{f (7 x)}{f (x)} \Rightarrow 1 \leq \frac{f (5 x)}{f (x)} \leq \frac{f (7 x)}{f (x)}$

$\Rightarrow 1 \leq \lim_{x \to \infty} \frac{f (5 x)}{f (x)} \leq 1 [∵ \lim_{x \to \infty} \frac{f (7 x)}{f (x)} = 1]$

$∴ \lim_{x \to \infty} \frac{f (5 x)}{f (x)} - 1 = 0$

Q 9 :

Let the set of all values of $p,$ for which $f (x) = (p^{2} - 6 p + 8) (\sin^{2} 2 x - \cos^{2} 2 x) + 2 (2 - p) x + 7$ does not have any critical point, be the interval $(a, b) .$ Then $16 a b$ is equal to ______ . [2024]

(252)

$f (x) = (p^{2} - 6 p + 8) (\sin^{2} 2 x - \cos^{2} 2 x) + 2 (2 - p) x + 7$

$= - \cos 4 x (p^{2} - 6 p + 8) + 2 (2 - p) x + 7$

$f' (x) = 4 \sin 4 x (p^{2} - 6 p + 8) + 2 (2 - p) = 0$ [ $∵$ $f (x)$ has no critical points]

$\Rightarrow \sin 4 x \neq \frac{2 p - 4}{4 (p^{2} - 6 p + 8)}$

$\Rightarrow \sin 4 x \neq \frac{2 (p - 2)}{4 (p - 2) (p - 4)}, p \neq 2 \Rightarrow \sin 4 x \neq \frac{1}{2 (p - 4)}$

$\Rightarrow$ $| \frac{1}{2 (p - 4)} |$ $> 1 \Rightarrow | 2 (p - 4) | < 1$

$\Rightarrow$ $| p - 4 |$ $< \frac{1}{2} \Rightarrow \frac{- 1}{2} < p - 4 < \frac{1}{2}$

$\Rightarrow \frac{7}{2} < p < \frac{9}{2} \Rightarrow a = \frac{7}{2}, b = \frac{9}{2}$

$So, 16 a b = 16 \times \frac{7}{2} \times \frac{9}{2} = 4 \times 62 = 252$

Q 10 :

If $5 f (x) + 4 f (\frac{1}{x}) = x^{2} - 2, \forall x \neq 0$ and $y = 9 x^{2} f (x),$ then $y$ is strictly increasing in [2024]

$(- \infty, \frac{1}{\sqrt{5}}) \cup (0, \frac{1}{\sqrt{5}})$
$(- \frac{1}{\sqrt{5}}, 0) \cup (0, \frac{1}{\sqrt{5}})$
$(0, \frac{1}{\sqrt{5}}) \cup (\frac{1}{\sqrt{5}}, \infty)$
$(- \frac{1}{\sqrt{5}}, 0) \cup (\frac{1}{\sqrt{5}}, \infty)$

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(4)

$5 f (x) + 4 f (\frac{1}{x}) = x^{2} - 2$ ...(i)

Replace x by $\frac{1}{x}$ , we get:

$5 f (\frac{1}{x}) + 4 f (x) = \frac{1}{x^{2}} - 2$ ...(ii)

$5 \times (i)$ and $4 \times (i i)$ , we get

$\Rightarrow$ $f (x) = \frac{5}{9} x^{2} - \frac{4}{9 x^{2}} - \frac{2}{9}$ $∴$ $y = 9 x^{2} (\frac{5}{9} x^{2} - \frac{4}{9 x^{2}} - \frac{2}{9})$

$y = 5 x^{4} - 4 - 2 x^{2}$

As $y$ is strictly increasing when $y^{'} > 0,$ so

$y' = 20 x^{3} - 4 x > 0$

$\Rightarrow$ $4 x (5 x^{2} - 1) > 0$

$x \in (\frac{- 1}{\sqrt{5}}, 0) \cup (\frac{1}{\sqrt{5}}, \infty)$

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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(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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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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(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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(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 $x \in R and$ $f' (a - 1) = 0$ 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

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4

(3)

$g (x) = f ({(\tan x - 1)}^{2} + a - 1)$

$g' (x) = f' ({(\tan x - 1)}^{2} + a - 1) \cdot 2 (\tan x - 1) \sec^{2} x$

$∵$ $f' (a - 1) = 0 and f'' (x) > 0$

$∴$ $f' ({(\tan x - 1)}^{2} + a - 1) > 0$

$g' (x) > 0 if (\tan x - 1) > 0$

$g is increasing in x \in (\frac{π}{4}, \frac{π}{2})$

$g' (x) < 0 if \tan x - 1 < 0$

$g is decreasing in x \in (0, \frac{π}{4})$

Topic Question Set

Q 1 :

For the function $f (x) = \sin x + 3 x - \frac{2}{π} (x^{2} + x), where x \in [0, \frac{π}{2}],$ consider the following two statements :

(I) $f$ is increasing in $(0, \frac{π}{2}) .$

(II) $f'$ is decreasing in $(0, \frac{π}{2})$ .

Between the above two statements, [2024]

Q 2 :

The interval in which the function $f (x) = x^{x}, x > 0,$ is strictly increasing is [2024]

Q 3 :

The number of critical points of the function $f (x) = {(x - 2)}^{2 / 3} (2 x + 1)$ is [2024]

Q 4 :

For the function $f (x) = (\cos x) - x + 1, x \in R,$ between the following two statements [2024]

(S1) $f (x) = 0$ for only one value of $x$ in $[0, π] .$

(S2) $f (x)$ is decreasing in $[0, \frac{π}{2}]$ and increasing in $[\frac{π}{2}, π .]$

Q 5 :

$Let g (x) = 3 f (\frac{x}{3}) + f (3 - x) and f'' (x) > 0 for all x \in (0, 3) .$

$If g is decreasing in (0, α) and increasing in (α, 3), then 8 α is:$ [2024]

Q 7 :

The function $f (x) = \frac{x}{x^{2} - 6 x - 16}, x \in R - {- 2, 8}$ [2024]

Q 8 :

Let $f : R \to (0, \infty)$ be strictly increasing function such that $\lim_{x \to \infty} \frac{f (7 x)}{f (x)} = 1 .$ Then, the value of $\lim_{x \to \infty} [\frac{f (5 x)}{f (x)} - 1]$ is equal to [2024]

Q 9 :

Let the set of all values of $p,$ for which $f (x) = (p^{2} - 6 p + 8) (\sin^{2} 2 x - \cos^{2} 2 x) + 2 (2 - p) x + 7$ does not have any critical point, be the interval $(a, b) .$ Then $16 a b$ is equal to ______ . [2024]

Q 10 :

If $5 f (x) + 4 f (\frac{1}{x}) = x^{2} - 2, \forall x \neq 0$ and $y = 9 x^{2} f (x),$ then $y$ is strictly increasing in [2024]

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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Topic Question Set

Q 1 : For the function f(x)=sinx+3x-2π(x2+x), where x∈[0,π2], consider the following two statements : (I) f is increasing in (0,π2). (II) f' is decreasing in (0,π2). Between the above two statements, [2024]

Q 2 : The interval in which the function f(x)=xx, x>0, is strictly increasing is [2024]

Q 3 : The number of critical points of the function f(x)=(x-2)2/3(2x+1) is [2024]

Q 4 : For the function f(x)=(cosx)-x+1, x∈R, between the following two statements [2024] (S1) f(x)=0 for only one value of x in [0,π]. (S2) f(x) is decreasing in [0,π2] and increasing in [π2,π.]

Q 5 : Let g(x)=3f(x3)+f(3-x) and f''(x)>0 for all x∈(0,3). If g is decreasing in (0,α) and increasing in (α,3), then 8α is: [2024]

Q 6 : Consider the function f:[12,1]→R defined by f(x)=42x3-32x-1. Consider the statements (I) The curve y=f(x) intersects the x-axis exactly at one point. (II) The curve y=f(x) intersects the x-axis at x=cosπ12 Then

Q 7 : The function f(x)=xx2-6x-16, x∈R-{-2,8} [2024]

Q 8 : Let f:R→(0,∞) be strictly increasing function such that limx→∞f(7x)f(x)=1. Then, the value of limx→∞[f(5x)f(x)-1] is equal to [2024]

Q 9 : Let the set of all values of p, for which f(x)=(p2-6p+8)(sin22x-cos22x)+2(2-p)x+7 does not have any critical point, be the interval (a,b). Then 16ab is equal to ______ . [2024]

Q 10 : If 5f(x)+4f(1x)=x2-2,∀x≠0 and y=9x2 f(x), then y is strictly increasing in [2024]

Q 11 : Let the function f(x)=x3+3x+3, x≠0 be strictly increasing in (–∞,α1)∪(α2,∞) and strictly decreasing in (α3,α4)∪(α4,α5). Then ∑i=15αi2 is equal to [2025]

Q 12 : Let (2, 3) be the largest open interval in which the function f(x)=2 loge (x–2)–x2+ax+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]

Q 13 : Let g(x)=f(x)+f(1-x) and f''(x)>0, x∈(0,1). If g is decreasing in the interval (0,α) and increasing in the interval (α,1), then tan-1(2α)+tan-1(1α)+tan-1(α+1α) is equal to [2023]

Q 14 : Let f:[2,4]→R be a differentiable function such that (xlogex)f'(x)+(logex)f(x)+f(x)≥1, x∈[2,4] with f(2)=12 and f(4)=14. Consider the following two statements: (A) : f(x)≤1, for all x∈[2,4] (B) : f(x)≥18, for all x∈[2,4] Then,

(3)

Q 15 : Let f:(0,1)→ℝ be a function defined by f(x)=11-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]

Q 16 : Let f:R→R be a twice differentiable function such that f''(x)>0 for all x∈R and f'(a-1)=0 where a is a real number. Let g(x)=f(tan2x-2tanx+a), 0<x<π2. Consider the following two statements: (I) g is increasing in (0,π4) (II) g is decreasing in (π4,π2) Then, [2026]

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Q 1 :

For the function $f (x) = \sin x + 3 x - \frac{2}{π} (x^{2} + x), where x \in [0, \frac{π}{2}],$ consider the following two statements :

(I) $f$ is increasing in $(0, \frac{π}{2}) .$

(II) $f'$ is decreasing in $(0, \frac{π}{2})$ .

Between the above two statements, [2024]

Q 2 :

The interval in which the function $f (x) = x^{x}, x > 0,$ is strictly increasing is [2024]

Q 3 :

The number of critical points of the function $f (x) = {(x - 2)}^{2 / 3} (2 x + 1)$ is [2024]

Q 4 :

For the function $f (x) = (\cos x) - x + 1, x \in R,$ between the following two statements [2024]

(S1) $f (x) = 0$ for only one value of $x$ in $[0, π] .$

(S2) $f (x)$ is decreasing in $[0, \frac{π}{2}]$ and increasing in $[\frac{π}{2}, π .]$

Q 5 :

$Let g (x) = 3 f (\frac{x}{3}) + f (3 - x) and f'' (x) > 0 for all x \in (0, 3) .$

$If g is decreasing in (0, α) and increasing in (α, 3), then 8 α is:$ [2024]

Q 7 :

The function $f (x) = \frac{x}{x^{2} - 6 x - 16}, x \in R - {- 2, 8}$ [2024]

Q 8 :

Let $f : R \to (0, \infty)$ be strictly increasing function such that $\lim_{x \to \infty} \frac{f (7 x)}{f (x)} = 1 .$ Then, the value of $\lim_{x \to \infty} [\frac{f (5 x)}{f (x)} - 1]$ is equal to [2024]

Q 9 :

Let the set of all values of $p,$ for which $f (x) = (p^{2} - 6 p + 8) (\sin^{2} 2 x - \cos^{2} 2 x) + 2 (2 - p) x + 7$ does not have any critical point, be the interval $(a, b) .$ Then $16 a b$ is equal to ______ . [2024]

Q 10 :

If $5 f (x) + 4 f (\frac{1}{x}) = x^{2} - 2, \forall x \neq 0$ and $y = 9 x^{2} f (x),$ then $y$ is strictly increasing in [2024]

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]

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]