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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2

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4

(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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2

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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

(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
1

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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]

0%

(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]

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

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

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

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

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

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

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

(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]

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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 (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]

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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

(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]