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

Q 21 :
$\lim_{x \to \infty} \frac{(2 x^{2} - 3 x + 5) {(3 x - 1)}^{\frac{x}{2}}}{(3 x^{2} + 5 x + 4) \sqrt{{(3 x + 2)}^{x}}}$ is equal to : [2025]

$\frac{2 e}{\sqrt{3}}$
$\frac{2}{3 \sqrt{e}}$
$\frac{2}{\sqrt{3 e}}$
$\frac{2 e}{3}$

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(2)

$\lim_{x \to \infty} \frac{(2 - \frac{3}{x} + \frac{5}{x^{2}}) {(1 - \frac{1}{3 x})}^{x / 2}}{(3 + \frac{5}{x} + \frac{4}{x^{2}}) {(1 + \frac{2}{3 x})}^{x / 2}}$

$= \lim_{x \to \infty} \frac{2}{3} \cdot \frac{e^{\frac{x}{2} (1 - \frac{1}{3 x} - 1)}}{e^{\frac{x}{2} (1 + \frac{2}{3 x} - 1)}}$

$= \frac{2}{3} \cdot \frac{e^{- \frac{1}{6}}}{e^{\frac{1}{3}}} = \frac{2}{3 \sqrt{e}}$ .

Q 22 :
$\lim_{x \to 0} \cos e c x (\sqrt{2 \cos^{2} x + 3 \cos x} - \sqrt{\cos^{2} x + \sin x + 4})$ is : [2025]

$- \frac{1}{2 \sqrt{5}}$
$\frac{1}{\sqrt{15}}$
0
$\frac{1}{2 \sqrt{5}}$

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(1)

$\lim_{x \to 0} \cos e c x (\sqrt{2 \cos^{2} x + 3 \cos x} - \sqrt{\cos^{2} x + \sin x + 4})$

$= \lim_{x \to 0} \frac{1}{\sin x} [\frac{2 \cos^{2} x + 3 \cos x - \cos^{2} x - \sin x - 4}{\sqrt{2 \cos^{2} x + 3 \cos x} + \sqrt{\cos^{2} x + \sin x + 4}}]$

$= \lim_{x \to 0} \frac{1}{\sin x} [\frac{\cos^{2} x + 3 \cos x - \sin x - 4}{\sqrt{2 \cos^{2} x + 3 \cos x} + \sqrt{\cos^{2} x + \sin x + 4}}]$

$= \lim_{x \to 0} [\frac{(\cos x + 4) (\cos x - 1) - \sin x}{\sin x}] \times \lim_{x \to 0} [\frac{1}{\sqrt{2 \cos^{2} x + 3 \cos x} + \sqrt{\cos^{2} x + \sin x + 4}}]$

$= \frac{1}{2 \sqrt{5}} \lim_{x \to 0} [\frac{(\cos x + 4) (- 2 \sin^{2} \frac{x}{2}) - 2 \sin \frac{x}{2} \cos \frac{x}{2})}{2 \sin \frac{x}{2} \cos \frac{x}{2}}]$

$= \frac{1}{2 \sqrt{5}} \lim_{x \to 0} [\frac{(- \sin \frac{x}{2}) (\cos x + 4) - \cos \frac{x}{2}}{\cos \frac{x}{2}}]$

$= \frac{1}{2 \sqrt{5}} [- 1] = \frac{- 1}{2 \sqrt{5}}$ .

Q 23 :
Let $f : R - {0} \to R$ be a function such that $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2}$ . If the $\lim_{x \to 0} (\frac{1}{α x} + f (x)) = β; α, β \in R$ , than $α + 2 β$ is equal to [2025]

4
3
5
6

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4

(1)

We have, $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2}$ ... (i)

Apply $x \to \frac{1}{x}$ , then

$f (\frac{1}{x}) - 6 f (x) = \frac{35 x}{3} - \frac{5}{2}$ ... (ii)

Using (i) and (ii), we get $f (x) = - 2 x - \frac{1}{3 x} + \frac{1}{2}$

Now, $β = \lim_{x \to 0} (\frac{1}{α x} + f (x))$

$= \lim_{x \to 0} (\frac{1}{α x} - 2 x - \frac{1}{3 x} + \frac{1}{2})$

$= [\lim_{x \to 0} \frac{1}{x} (\frac{1}{α} - \frac{1}{3})] + \frac{1}{2} \Rightarrow α = 3, β = \frac{1}{2}$

Hence, $α + 2 β = 3 + 2 \times \frac{1}{2} = 4$

Q 24 :
The value of $\lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 5}{(k + 3)!})$ is : [2025]

5/3
4/3
7/3
2

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(1)

$\lim_{n \to \infty} [\sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 5}{(k + 3)!}]$

$= \lim_{n \to \infty} [\sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 6 - 1}{(k + 3)!}]$

$= \lim_{n \to \infty} [\sum_{k = 1}^{n} \frac{(k + 3) (k + 2) (k + 1) - 1}{(k + 3)!}]$

$= \lim_{n \to \infty} [\sum_{k = 1}^{n} (\frac{1}{k!} - \frac{1}{(k + 3)!})]$

$= \lim_{n \to \infty} [\frac{1}{1!} - \frac{1}{4!} + \frac{1}{2!} - \frac{1}{5!} + \frac{1}{3!} - \frac{1}{6!} + \frac{1}{4!} - \frac{1}{7!} + . . . + \frac{1}{n!} - \frac{1}{(n + 3)!}]$

$= 1 + \frac{1}{2!} + \frac{1}{3!} = \frac{5}{3}$ .

Q 25 :
If $\lim_{x \to 0} {(\frac{\tan x}{x})}^{\frac{1}{x^{2}}} = p$ , then $96 \log_{e} p$ is equal to __________. [2025]

0%

(32)

Given, $p = \lim_{x \to 0} {(\frac{\tan x}{x})}^{\frac{1}{x^{2}}}$

$\Rightarrow p = e^{\lim_{x \to 0} (\frac{\tan x - x}{x^{3}})}$

$= e^{\lim_{x \to 0} (\frac{x + \frac{x^{3}}{3} + \frac{2 x^{5}}{5} + . . . - x}{x^{3}})}$

$[∵ For 1^{\infty} form limits, \lim_{x \to a} f {(x)}^{g (x)} = e^{\lim_{x \to a} (f (x) - 1) g (x)}]$

$= e^{1 / 3}$

$∴ 96 \log_{e} p = 96 \times \frac{1}{3} = 32$

Q 26 :
For t > –1, let $α_{t}$ and $β_{t}$ be the roots of the equation $({(t + 2)}^{1 / 7} - 1) x^{2} + ({(t + 2)}^{1 / 6} - 1) x + ({(t + 2)}^{1 / 21} - 1) = 0$ . If $\lim_{t \to - 1^{+}} α_{t} = a and \lim_{t \to - 1^{+}} β_{t} = b$ , then $72 {(a + b)}^{2}$ is equal to __________. [2025]

0%

(98)

$a + b = \lim_{t \to - 1^{+}} (α_{t} + β_{t})$

$= \lim_{t \to - 1^{+}} \frac{- [{(t + 2)}^{\frac{1}{6}} - 1]}{{(t + 2)}^{\frac{1}{7}} - 1}$ [Sum of roots]

Let t + 2 = y, we get

$a + b = \lim_{y \to 1^{+}} - \frac{y^{1 / 6} - 1}{y^{1 / 7} - 1} = - \frac{7}{6}$

So, $72 {(a + b)}^{2} = 72 (\frac{49}{36}) = 98$ .

Q 27 :
Let $f (x) = \lim_{n \to \infty} \sum_{r = 0}^{n} (\frac{\tan (x / 2^{r + 1}) + \tan^{3} (x / 2^{r + 1})}{1 - \tan^{2} (x / 2^{r + 1})})$ . Then $\lim_{x \to 0} \frac{e^{x} - e^{f (x)}}{(x - f (x))}$ is equal to __________. [2025]

0%

(1)

We have,

$= \lim_{n \to \infty} \sum_{r = 0}^{n} [\frac{2 \tan (\frac{x}{2^{r + 1}}) - \tan (\frac{x}{2^{r + 1}}) + \tan^{3} (\frac{x}{2^{r + 1}})}{1 - \tan^{2} (\frac{x}{2^{r + 1}})}]$

$= \lim_{n \to \infty} \sum_{r = 0}^{n} [\frac{2 \tan (\frac{x}{2^{r + 1}})}{1 - \tan^{2} (\frac{x}{2^{r + 1}})} - \frac{\tan (\frac{x}{2^{r + 1}}) {1 - \tan^{2} (\frac{x}{2^{r + 1}})}}{{1 - \tan^{2} (\frac{x}{2^{r + 1}})}}]$

$= \lim_{n \to \infty} \sum_{r = 0}^{n} (\tan \frac{x}{2^{r}} - \tan (\frac{x}{2^{r + 1}})) = \tan x$

Now, $\lim_{x \to 0} (\frac{e^{x} - e^{\tan x}}{x - \tan x})$

$= \lim_{x \to 0} e^{\tan x} \frac{(e^{x - \tan x} - 1)}{(x - \tan x)} = 1$ $[∵ \lim_{y \to 0} \frac{e^{y} - 1}{y} = 1]$

Q 28 :
Let [t] be the greatest integer less than or equal to t. Then the least value of $p \in N$ for which $\lim_{x \to 0^{+}} (x ([\frac{1}{x}] + [\frac{2}{x}] + . . . + [\frac{p}{x}]) - x^{2} ([\frac{1}{x^{2}}] + [\frac{2^{2}}{x^{2}}] + . . . + [\frac{9^{2}}{x^{2}}])) \geq 1$ is equal to _________. [2025]

0%

(24)

We have,

$\lim_{x \to 0^{+}} (x ([\frac{1}{x}] + [\frac{2}{x}] + . . . + [\frac{p}{x}]) - x^{2} ([\frac{1}{x^{2}}] + [\frac{2^{2}}{x^{2}}] + . . . + [\frac{9^{2}}{x^{2}}])) \geq 1$

$\Rightarrow (1 + 2 + . . . + p) - (1^{2} + 2^{2} + . . . + 9^{2}) \geq 1$

$\Rightarrow \frac{p (p + 1)}{2} - \frac{9.10.19}{6} \geq 1$

$\Rightarrow p (p + 1) \geq 572 \Rightarrow p \geq 24$ .

Q 29 :
$\lim_{x \to 0} ((\frac{1 - \cos^{2} (3 x)}{\cos^{3} (4 x)}) (\frac{\sin^{3} (4 x)}{(\log_{e} {(2 x + 1))}^{5}}))$ is equal to [2023]

24
15
9
18

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(4)

$\lim_{x \to 0} [(\frac{(1 - \cos^{2} 3 x)}{\cos^{3} (4 x)}) (\frac{\sin^{3} (4 x)}{(\log_{e} {(2 x + 1))}^{5}})]$

$= \lim_{x \to 0} [\frac{\sin^{2} (3 x)}{\cos^{3} (4 x)} \times \frac{\sin^{3} (4 x)}{[\log_{e} (2 x + 1)]^{5}}]$

$= \lim_{x \to 0} [\frac{\frac{\sin^{2} (3 x)}{{(3 x)}^{2}} \times \frac{\sin^{3} (4 x)}{{(4 x)}^{3}} \times {(3 x)}^{2} \times {(4 x)}^{3}}{\cos^{3} (4 x) \cdot {[\frac{\log_{e} (2 x + 1)}{2 x}]}^{5} \times {(2 x)}^{5}}]$

$= \frac{9 \times 64}{32} = 18$

Q 30 :
$If α > β > 0 are the roots of the equation a x^{2} + b x + 1 = 0,$ and $\lim_{x \to \frac{1}{α}} {(\frac{1 - \cos (x^{2} + b x + a)}{2 {(1 - α x)}^{2}})}^{\frac{1}{2}} = \frac{1}{k} (\frac{1}{β} - \frac{1}{α}),$ then $k$ is equal to [2023]

$2 β$
$α$
$β$
$2 α$

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(4)

Given, $a x^{2} + b x + 1 = 0$ has roots $α, β$ , then $x^{2} + b x + a = 0$ has roots $\frac{1}{α}, \frac{1}{β}$ .

$Now, \lim_{x \to \frac{1}{α}} {(\frac{1 - \cos (x^{2} + b x + a)}{2 {(1 - α x)}^{2}})}^{1 / 2}$

$= \lim_{x \to \frac{1}{α}} {(\frac{2 \sin^{2} (\frac{x^{2} + b x + a}{2})}{2 {(1 - α x)}^{2}})}^{1 / 2}$

$= \lim_{x \to \frac{1}{α}} {(\frac{2 \sin^{2} (\frac{(x - \frac{1}{α}) (x - \frac{1}{β})}{2})}{2 {(1 - α x)}^{2}})}^{1 / 2}$

$\lim_{x \to \frac{1}{α}} {(\frac{\frac{\sin^{2} (\frac{(1 - α x) (1 - β x)}{2 α β})}{{(\frac{(1 - α x) (1 - β x)}{2 α β})}^{2}} \times {(\frac{(1 - α x) (1 - β x)}{2 α β})}^{2}}{{(1 - α x)}^{2}})}^{\frac{1}{2}}$

$= \lim_{x \to \frac{1}{α}} (\frac{1 - β x}{2 α β}) = (\frac{α - β}{2 α^{2} β}) = \frac{1}{2 α} (\frac{1}{β} - \frac{1}{α})$

$= \frac{1}{k} (\frac{1}{β} - \frac{1}{α}) (\Rightarrow Given)$

$∴$ $k = 2 α$

Topic Question Set

Q 21 :
$\lim_{x \to \infty} \frac{(2 x^{2} - 3 x + 5) {(3 x - 1)}^{\frac{x}{2}}}{(3 x^{2} + 5 x + 4) \sqrt{{(3 x + 2)}^{x}}}$ is equal to : [2025]

Q 22 :
$\lim_{x \to 0} \cos e c x (\sqrt{2 \cos^{2} x + 3 \cos x} - \sqrt{\cos^{2} x + \sin x + 4})$ is : [2025]

Q 23 :
Let $f : R - {0} \to R$ be a function such that $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2}$ . If the $\lim_{x \to 0} (\frac{1}{α x} + f (x)) = β; α, β \in R$ , than $α + 2 β$ is equal to [2025]

Q 24 :
The value of $\lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 5}{(k + 3)!})$ is : [2025]

Q 25 :
If $\lim_{x \to 0} {(\frac{\tan x}{x})}^{\frac{1}{x^{2}}} = p$ , then $96 \log_{e} p$ is equal to __________. [2025]

0%

Q 26 :
For t > –1, let $α_{t}$ and $β_{t}$ be the roots of the equation $({(t + 2)}^{1 / 7} - 1) x^{2} + ({(t + 2)}^{1 / 6} - 1) x + ({(t + 2)}^{1 / 21} - 1) = 0$ . If $\lim_{t \to - 1^{+}} α_{t} = a and \lim_{t \to - 1^{+}} β_{t} = b$ , then $72 {(a + b)}^{2}$ is equal to __________. [2025]

0%

Q 27 :
Let $f (x) = \lim_{n \to \infty} \sum_{r = 0}^{n} (\frac{\tan (x / 2^{r + 1}) + \tan^{3} (x / 2^{r + 1})}{1 - \tan^{2} (x / 2^{r + 1})})$ . Then $\lim_{x \to 0} \frac{e^{x} - e^{f (x)}}{(x - f (x))}$ is equal to __________. [2025]

0%

0%

Q 29 :
$\lim_{x \to 0} ((\frac{1 - \cos^{2} (3 x)}{\cos^{3} (4 x)}) (\frac{\sin^{3} (4 x)}{(\log_{e} {(2 x + 1))}^{5}}))$ is equal to [2023]

Q 30 :
$If α > β > 0 are the roots of the equation a x^{2} + b x + 1 = 0,$ and $\lim_{x \to \frac{1}{α}} {(\frac{1 - \cos (x^{2} + b x + a)}{2 {(1 - α x)}^{2}})}^{\frac{1}{2}} = \frac{1}{k} (\frac{1}{β} - \frac{1}{α}),$ then $k$ is equal to [2023]

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Topic Question Set

Q 21 : limx→∞(2x2–3x+5)(3x–1)x2(3x2+5x+4)(3x+2)x is equal to : [2025]

Q 22 : limx→0cosec x(2 cos2x+3 cos x–cos2x+sin x+4) is : [2025]

Q 23 : Let f:R–{0}→R be a function such that f(x)–6f(1x)=353x–52. If the limx→0(1αx+f(x))=β; α,β∈R, than α+2β is equal to [2025]

Q 24 : The value of limn→∞(∑k=1nk3+6k2+11k+5(k+3)!) is : [2025]

Q 25 : If limx→0(tan xx)1x2=p, then 96logep is equal to __________. [2025]

0%

Q 26 : For t > –1, let αt and βt be the roots of the equation ((t+2)1/7–1)x2+((t+2)1/6–1)x+((t+2)1/21–1)=0. If limt→–1+αt=a and limt→–1+βt=b, then 72(a+b)2 is equal to __________. [2025]

0%

Q 27 : Let f(x)=limn→∞∑r=0n(tan(x/2r+1)+tan3(x/2r+1)1–tan2(x/2r+1)). Then limx→0ex–ef(x)(x–f(x)) is equal to __________. [2025]

0%

Q 28 : Let [t] be the greatest integer less than or equal to t. Then the least value of p∈N for which limx→0+(x([1x]+[2x]+...+[px])–x2([1x2]+[22x2]+...+[92x2]))≥1 is equal to _________. [2025]

0%

Q 29 : limx→0((1-cos2(3x)cos3(4x))(sin3(4x)(loge(2x+1))5)) is equal to [2023]

Q 30 : If α>β>0 are the roots of the equation ax2+bx+1=0, and limx→1α(1-cos(x2+bx+a)2(1-αx)2)12=1k(1β-1α), then k is equal to [2023]

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Q 21 :
$\lim_{x \to \infty} \frac{(2 x^{2} - 3 x + 5) {(3 x - 1)}^{\frac{x}{2}}}{(3 x^{2} + 5 x + 4) \sqrt{{(3 x + 2)}^{x}}}$ is equal to : [2025]

Q 22 :
$\lim_{x \to 0} \cos e c x (\sqrt{2 \cos^{2} x + 3 \cos x} - \sqrt{\cos^{2} x + \sin x + 4})$ is : [2025]

Q 23 :
Let $f : R - {0} \to R$ be a function such that $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2}$ . If the $\lim_{x \to 0} (\frac{1}{α x} + f (x)) = β; α, β \in R$ , than $α + 2 β$ is equal to [2025]

Q 24 :
The value of $\lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 5}{(k + 3)!})$ is : [2025]

Q 25 :
If $\lim_{x \to 0} {(\frac{\tan x}{x})}^{\frac{1}{x^{2}}} = p$ , then $96 \log_{e} p$ is equal to __________. [2025]

Q 26 :
For t > –1, let $α_{t}$ and $β_{t}$ be the roots of the equation $({(t + 2)}^{1 / 7} - 1) x^{2} + ({(t + 2)}^{1 / 6} - 1) x + ({(t + 2)}^{1 / 21} - 1) = 0$ . If $\lim_{t \to - 1^{+}} α_{t} = a and \lim_{t \to - 1^{+}} β_{t} = b$ , then $72 {(a + b)}^{2}$ is equal to __________. [2025]

Q 27 :
Let $f (x) = \lim_{n \to \infty} \sum_{r = 0}^{n} (\frac{\tan (x / 2^{r + 1}) + \tan^{3} (x / 2^{r + 1})}{1 - \tan^{2} (x / 2^{r + 1})})$ . Then $\lim_{x \to 0} \frac{e^{x} - e^{f (x)}}{(x - f (x))}$ is equal to __________. [2025]

Q 29 :
$\lim_{x \to 0} ((\frac{1 - \cos^{2} (3 x)}{\cos^{3} (4 x)}) (\frac{\sin^{3} (4 x)}{(\log_{e} {(2 x + 1))}^{5}}))$ is equal to [2023]

Q 30 :
$If α > β > 0 are the roots of the equation a x^{2} + b x + 1 = 0,$ and $\lim_{x \to \frac{1}{α}} {(\frac{1 - \cos (x^{2} + b x + a)}{2 {(1 - α x)}^{2}})}^{\frac{1}{2}} = \frac{1}{k} (\frac{1}{β} - \frac{1}{α}),$ then $k$ is equal to [2023]