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

Q 11 :
If $α = \lim_{x \to 0^{+}} (\frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}})$ and $β = \lim_{x \to 0} {(1 + \sin x)}^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $a x^{2} + b x - \sqrt{e} = 0,$ then $12 \log_{e} (a + b)$ is equal to _______ . [2024]

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(6)

$α = \lim_{x \to 0^{+}} \frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{(\sqrt{\tan x} - \sqrt{x})}$

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

$β = \lim_{x \to 0} {(1 + \sin x)}^{\frac{1}{2} \cot x} = \lim_{x \to 0} e^{(\sin x) (\frac{1}{2} \cot x)}$

$= \lim_{x \to 0} e^{\frac{1}{2} \cos x} = e^{1 / 2} = \sqrt{e}$

Since, $α, β$ be the roots of the quadratic equation, $a x^{2} + b x - \sqrt{e} = 0$

$∴$ Products of roots $= \frac{- \sqrt{e}}{a} = \sqrt{e} \Rightarrow a = - 1$

and sum of roots $= \frac{- b}{a} = 1 + \sqrt{e} \Rightarrow b = \sqrt{e} + 1$

Hence, $12 \log_{e} (a + b) = 12 \log_{e} (\sqrt{e} + 1 - 1)$

$= 12 \log_{e} (e^{1 / 2}) = 6$

Q 12 :
Let ${x}$ denote the fractional part of $x$ and $f (x) = \frac{\cos^{- 1} (1 - {x}^{2}) \sin^{- 1} (1 - {x})}{{x} - {x}^{3}}, x \neq 0 .$

If $L$ and $R$ respectively denotes the left hand limit and the right hand limit of $f (x)$ at $x = 0,$ then $\frac{32}{π^{2}} (L^{2} + R^{2})$ is equal to _____. [2024]

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(18)

$We have, f (x) = \frac{\cos^{- 1} (1 - {x}^{2}) \sin^{- 1} (1 - {x})}{{x} - {x}^{3}}$

$R = \lim_{h \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h)$

$= \lim_{h \to 0^{+}} f (h) = \lim_{h \to 0} \frac{\cos^{- 1} (1 - h^{2})}{h} (\frac{\sin^{- 1} 1}{1})$

$Put \cos^{- 1} (1 - h^{2}) = θ \Rightarrow \cos θ = 1 - h^{2}$

$= \frac{π}{2} \lim_{θ \to 0} \frac{θ}{\sqrt{1 - \cos θ}} = \frac{π}{2} \lim_{θ \to 0} \frac{1}{\sqrt{\frac{1 - \cos θ}{θ^{2}}}} = \frac{π}{2} \frac{1}{\sqrt{1 / 2}} = \frac{π}{\sqrt{2}}$

Also, $L = \lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (- h)$

$= \lim_{h \to 0} \frac{\cos^{- 1} (1 - {- h}^{2}) \sin^{- 1} (1 - {- h})}{{- h} - {- h}^{3}}$

$= \lim_{h \to 0} \frac{\cos^{- 1} (1 - {(- h + 1)}^{2}) \sin^{- 1} (1 - (- h + 1))}{(- h + 1) - {(- h + 1)}^{3}}$

$= \lim_{h \to 0} \frac{\cos^{- 1} (- h^{2} + 2 h) \sin^{- 1} h}{(1 - h) (1 - {(1 - h)}^{2})}$

$= \lim_{h \to 0} (\frac{π}{2}) \frac{\sin^{- 1} h}{(1 - {(1 - h)}^{2})} = \frac{π}{2} \lim_{h \to 0} (\frac{\sin^{- 1} h}{- h^{2} + 2 h})$

$= \frac{π}{2} \lim_{h \to 0} (\frac{\sin^{- 1} h}{h}) (\frac{1}{- h + 2}) = \frac{π}{4}$

$∴$ $\frac{32}{π^{2}} (L^{2} + R^{2}) = \frac{32}{π^{2}} (\frac{π^{2}}{2} + \frac{π^{2}}{16}) = 16 + 2 = 18$

Q 13 :
$If \lim_{x \to 0} \frac{a x^{2} e^{x} - b \log_{e} (1 + x) + c x e^{- x}}{x^{2} \sin x} = 1, then 16 (a^{2} + b^{2} + c^{2}) is equal to_____.$ [2024]

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(81)

Given, $\lim_{x \to 0} \frac{a x^{2} e^{x} - b \log_{e} (1 + x) + c x e^{- x}}{x^{2} \sin x} = 1$

$\Rightarrow \lim_{x \to 0} \frac{a x^{2} (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots) - b (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} \dots) + c x (1 - \frac{x}{1!} + \frac{x^{2}}{2!} \dots)}{x^{2} (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} \dots)} = 1$

$\Rightarrow \lim_{x \to 0} (a x^{2} + a x^{3} + \frac{a x^{4}}{2!} + \dots) - b x + \frac{b x^{2}}{2} - \frac{b x^{3}}{3} \dots + c x - c x^{2} + \frac{c x^{3}}{2!} \dots = \lim_{x \to 0} x^{3} - \frac{x^{5}}{3!} + \frac{x^{7}}{5!} \dots$

Comparing the coefficient of $x^{3},$ we get

$a - \frac{b}{3} + \frac{c}{2} = 1$ ...(i)

Comparing the coefficient of $x^{2},$ we get

$a + \frac{b}{2} - c = 0$ ...(ii)

Comparing the coefficient of $x,$ we get

$- b + c = 0$ ...(iii)

On solving (i), (ii) and (iii), we get

$a = \frac{3}{4}, b = c = \frac{3}{2}$

$∴ 16 (a^{2} + b^{2} + c^{2}) = 16 [\frac{9}{16} + \frac{9}{4} + \frac{9}{4}] = 16 [\frac{9 + 36 + 36}{16}] = 81$

Q 14 :
For $α, β, γ \in R$ , if $\lim_{x \to 0} \frac{x^{2} \sin α x + (γ - 1) e^{x^{2}}}{\sin 2 x - β x} = 3$ , then $β + γ - α$ is equal to: [2025]

7
4
6
–1

1

2

3

4

(1)

As, $x \to 0 \Rightarrow \sin 2 x - β x \to 0$

To make the given limit in $\frac{0}{0}$ form; $(γ - 1) e^{0} + 0 \sin (α 0) = 0$

$\Rightarrow (γ - 1) = 0 \Rightarrow γ = 1$

So, $\lim_{x \to 0} \frac{x^{2} \sin (α x)}{(\sin 2 x - β x)} = 3$

$\Rightarrow \lim_{x \to 0} \frac{x^{2} [α x - \frac{{(α x)}^{3}}{3!} + \frac{{(α x)}^{5}}{5!} - . . . . .]}{[2 x - \frac{{(2 x)}^{3}}{3!} + \frac{{(2 x)}^{5}}{5!} - . . . . .] - β x} = 3$

$\Rightarrow \lim_{x \to 0} \frac{α x^{3} - \frac{α^{3} x^{5}}{3!} + \frac{α^{5} x^{7}}{5!} - . . . . .}{x (2 - β) - \frac{8 x^{3}}{6} + \frac{2^{5} \cdot x^{5}}{5!} - . . . . .} = 3$

$\Rightarrow 2 - β = 0 and \frac{α}{\frac{- 8}{6}} = 3$

$\Rightarrow β = 2 and α = 3 (- \frac{8}{6}) = - 4$

$∴ β + γ - α = 2 + 1 - (- 4) = 7$ .

Q 15 :
If $\lim_{x \to 0} \frac{\cos (2 x) + a \cos (4 x) - b}{x^{4}}$ is finite, then (a + b) is equal to : [2025]

$\frac{3}{4}$
–1
$\frac{1}{2}$
0

1

2

3

4

(3)

$\lim_{x \to 0} \frac{\cos 2 x + a \cos 4 x - b}{x^{4}}$

$= \lim_{x \to 0} \frac{(1 - \frac{{(2 x)}^{2}}{2!} + \frac{{(2 x)}^{4}}{4!} . . .) + a (1 - \frac{{(4 x)}^{2}}{2!} + \frac{{(4 x)}^{4}}{4!} . . .) - b}{x^{4}}$

$= \lim_{x \to 0} \frac{(1 + a - b) + (- 2 - 8 a) x^{2} + (\frac{2}{3} + \frac{32}{3} a) x^{4} + . . . . . Higher power of x .}{x^{4}}$

Since limit is finite so, we have

1 + a – b = 0 and – 2 – 8a = 0

$\Rightarrow a = - \frac{1}{4} and b = 1 - \frac{1}{4} = \frac{3}{4}$

$∴ a + b = \frac{- 1}{4} + \frac{3}{4} = \frac{2}{4} = \frac{1}{2}$

Q 16 :
If $\lim_{x \to 1^{+}} \frac{(x - 1) (6 + λ \cos (x - 1)) + μ \sin (1 - x)}{{(x - 1)}^{3}} = - 1$ , where $λ, μ \in R$ , then $λ + μ$ is equal to [2025]

20
19
17
18

1

2

3

4

(4)

We have,

$\lim_{x \to 1^{+}} \frac{(x - 1) (6 + λ \cos (x - 1)) + μ \sin (1 - x)}{{(x - 1)}^{3}} = - 1$

Put x – 1 = t

$\Rightarrow \lim_{t \to 0^{+}} \frac{t (6 + λ \cos t) - μ \sin t}{t^{3}} = - 1$

$\Rightarrow \lim_{t \to 0^{+}} \frac{t [6 + λ (1 - \frac{t^{2}}{2!} + . . . \infty)] - μ [t - \frac{t^{3}}{3!} . . . \infty]}{t^{3}} = - 1$

Now, $6 + λ - μ = 0$ ...(i)

$\frac{- λ}{2} + \frac{μ}{6} = - 1 \Rightarrow 3 λ - μ = 6$ ...(ii)

From (i) and (ii), we get $λ = 6$ , $μ = 12$

$∴ λ + μ = 18$ .

Q 17 :
Let f be a differentiable function on R such that $f (2) = 1, f' (2) = 4$ . Let $\lim_{x \to 0} (f {(2 + x))}^{3 / x} = e^{α}$ . Then the number of times the curve $y = 4 x^{3} - 4 x^{2} - 4 (α - 7) x - α$ meets x-axis is : [2025]

1
0
2
3

1

2

3

4

(3)

Given, $\lim_{x \to 0} (f {(2 + x))}^{3 / x} = e^{α}$

$\Rightarrow e^{\lim_{x \to 0} \frac{3}{x} (f (2 + x) - 1)} = e^{α}$ ( $1^{\infty}$ form)

$\Rightarrow e^{\lim_{x \to 0} 3 f' (2 + x)} = e^{α}$

$\Rightarrow e^{3 f' (2)} = e^{α} \Rightarrow e^{12} = e^{α}$

So, $α = 12$

Now, $y = 4 x^{3} - 4 x^{2} - 4 (α - 7) x - α$

$= 4 x^{3} - 4 x^{2} - 20 x - 12$

$= 4 {(x + 1)}^{2} (x - 3)$

$∴$ Roots are –1, –1 and 3

So, the curve $y = 4 x^{3} - 4 x^{2} - 20 x - 12$ . The curve meets x-axis at 2 points.

Q 18 :
$\lim_{x \to 0^{+}} \frac{\tan (5 {(x)}^{\frac{1}{3}}) \log_{e} (1 + 3 x^{2})}{{(\tan^{- 1} 3 \sqrt{x})}^{2} (e^{5 {(x)}^{\frac{4}{3}}} - 1)}$ is equal to [2025]

1
$\frac{5}{3}$
$\frac{1}{3}$
$\frac{1}{15}$

1

2

3

4

(3)

We have, $\lim_{x \to 0^{+}} \frac{\tan (5 {(x)}^{\frac{1}{3}}) \log_{e} (1 + 3 x^{2})}{{(\tan^{- 1} (3 \sqrt{x}))}^{2} (e^{5 {(x)}^{\frac{4}{3}}} - 1)}$

$\Rightarrow \lim_{x \to 0^{+}} \frac{\frac{\tan (5 {(x)}^{\frac{1}{3}}) \log_{e} (1 + 3 x^{2})}{3 x^{2} \times 5 {(x)}^{\frac{1}{3}}} (3 x^{2}) 5 {(x)}^{\frac{1}{3}}}{\frac{{(\tan^{- 1} (3 \sqrt{x}))}^{2}}{{(3 \sqrt{x})}^{2}} \frac{(e^{5 x^{\frac{4}{3}}} - 1)}{5 x^{\frac{4}{3}}} \times 9 x \times 5 x^{\frac{4}{3}}}$

$\Rightarrow \lim_{x \to 0^{+}} \frac{\frac{\tan (5 {(x)}^{\frac{1}{3}})}{5 {(x)}^{\frac{1}{3}}} \frac{\log_{e} (1 + 3 x^{2})}{3 x^{2}} \times 15 x^{\frac{7}{3}}}{\frac{{(\tan^{- 1} (3 \sqrt{x}))}^{2}}{{(3 \sqrt{x})}^{2}} \frac{(e^{5 {(x)}^{\frac{4}{3}}} - 1)}{5 x^{\frac{4}{3}}} \times 45 x^{\frac{7}{3}}} = \frac{1}{3}$ .

Q 19 :
Given below are two statements :

Statement I : $\lim_{x \to 0} (\frac{\tan^{- 1} x + \log_{e} \sqrt{\frac{1 + x}{1 - x}} - 2 x}{x^{5}}) = \frac{2}{5}$

Statement II : $\lim_{x \to 1} (x^{\frac{2}{1 - x}}) = \frac{1}{e^{2}}$

In the light of the above statements, choose the correct answer from the potions given below. [2025]

Statement I is true but Statement II is false
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is false but Statement II is true

1

2

3

4

(2)

Let $L = \lim_{x \to 0} (\frac{\tan^{- 1} x + \log_{e} \sqrt{\frac{1 + x}{1 - x}} - 2 x}{x^{5}})$

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

$= \lim_{x \to 0} \frac{(x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - . . .) + \frac{1}{2} [x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - . . . . . - (- x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - . . . . .)] - 2 x}{x^{5}}$

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

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

$∴$ Statement I is true.

Let $I = \lim_{x \to 1} x^{(\frac{2}{1 - x})} = \lim_{x \to 1} e^{\ln (x^{2 / 1 - x})} = \lim_{x \to 1} e^{2 \ln x / (1 - x)}$

Let us evaluate $\lim_{x \to 1} \frac{2 \ln x}{1 - x}$ $(\frac{0}{0} form)$

$= \lim_{x \to 1} \frac{2 / x}{- 1}$ [Using L'Hospital rule]

= – 2

$∴ I = \lim_{x \to 1} e^{- 2} = e^{- 2}$

$∴$ Statement II is also true.

Q 20 :
If $\lim_{x \to \infty} ((\frac{e}{1 - e}) {(\frac{1}{e} - \frac{x}{1 + x}))}^{x} = α$ , then the value of $\frac{\log_{e} α}{1 + \log_{e} α}$ equals : [2025]

$e^{- 2}$
$e^{- 1}$
e
$e^{2}$

1

2

3

4

(3)

$α = \lim_{x \to \infty} ((\frac{e}{1 - e}) {(\frac{1}{e} - \frac{x}{1 + x}))}^{x}$ ( $1^{\infty}$ form)

$∴ α = e^{L}, where L = \lim_{x \to \infty} x ((\frac{e}{1 - e}) \times (\frac{1}{e} - \frac{x}{1 + x}) - 1)$

$\Rightarrow L = \lim_{x \to \infty} (\frac{e}{1 - e}) x (\frac{1}{e} - \frac{x}{1 + x} - (\frac{1 - e}{e}))$

$\Rightarrow L = \frac{e}{1 - e} \lim_{x \to \infty} x (1 - \frac{x}{1 + x}) \Rightarrow L = \frac{e}{1 - e} \lim_{x \to \infty} \frac{x}{1 + x}$

$\Rightarrow L = \frac{e}{1 - e} \cdot 1 \Rightarrow L = \frac{e}{1 - e}$

$∴ α = e^{\frac{e}{1 - e}} \Rightarrow \log_{e} α = \frac{e}{1 - e}$

$∴$ Required value $= \frac{\frac{e}{1 - e}}{1 + \frac{e}{1 - e}} = e$ .

Topic Question Set

Q 11 :
If $α = \lim_{x \to 0^{+}} (\frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}})$ and $β = \lim_{x \to 0} {(1 + \sin x)}^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $a x^{2} + b x - \sqrt{e} = 0,$ then $12 \log_{e} (a + b)$ is equal to _______ . [2024]

0%

0%

Q 13 :
$If \lim_{x \to 0} \frac{a x^{2} e^{x} - b \log_{e} (1 + x) + c x e^{- x}}{x^{2} \sin x} = 1, then 16 (a^{2} + b^{2} + c^{2}) is equal to_____.$ [2024]

0%

Q 14 :
For $α, β, γ \in R$ , if $\lim_{x \to 0} \frac{x^{2} \sin α x + (γ - 1) e^{x^{2}}}{\sin 2 x - β x} = 3$ , then $β + γ - α$ is equal to: [2025]

Q 15 :
If $\lim_{x \to 0} \frac{\cos (2 x) + a \cos (4 x) - b}{x^{4}}$ is finite, then (a + b) is equal to : [2025]

Q 16 :
If $\lim_{x \to 1^{+}} \frac{(x - 1) (6 + λ \cos (x - 1)) + μ \sin (1 - x)}{{(x - 1)}^{3}} = - 1$ , where $λ, μ \in R$ , then $λ + μ$ is equal to [2025]

Q 17 :
Let f be a differentiable function on R such that $f (2) = 1, f' (2) = 4$ . Let $\lim_{x \to 0} (f {(2 + x))}^{3 / x} = e^{α}$ . Then the number of times the curve $y = 4 x^{3} - 4 x^{2} - 4 (α - 7) x - α$ meets x-axis is : [2025]

Q 18 :
$\lim_{x \to 0^{+}} \frac{\tan (5 {(x)}^{\frac{1}{3}}) \log_{e} (1 + 3 x^{2})}{{(\tan^{- 1} 3 \sqrt{x})}^{2} (e^{5 {(x)}^{\frac{4}{3}}} - 1)}$ is equal to [2025]

Q 20 :
If $\lim_{x \to \infty} ((\frac{e}{1 - e}) {(\frac{1}{e} - \frac{x}{1 + x}))}^{x} = α$ , then the value of $\frac{\log_{e} α}{1 + \log_{e} α}$ equals : [2025]

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Topic Question Set

Q 11 : If α=limx→0+(etanx-extanx-x) and β=limx→0(1+sinx)12cotx are the roots of the quadratic equation ax2+bx-e=0, then 12loge(a+b) is equal to _______ . [2024]

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Q 12 : Let {x} denote the fractional part of x and f(x)=cos-1(1-{x}2)sin-1(1-{x}){x}-{x}3, x≠0. If L and R respectively denotes the left hand limit and the right hand limit of f(x) at x=0, then 32π2(L2+R2) is equal to _____. [2024]

0%

Q 13 : If limx→0ax2ex-bloge(1+x)+cxe-xx2sinx=1, then 16(a2+b2+c2) is equal to _____. [2024]

0%

Q 14 : For α, β, γ∈R, if limx→0x2 sin αx+(γ–1)ex2sin 2x–βx=3, then β+γ–α is equal to: [2025]

Q 15 : If limx→0cos(2x)+acos(4x)–bx4 is finite, then (a + b) is equal to : [2025]

Q 16 : If limx→1+(x–1)(6+λcos(x–1))+μsin(1–x)(x–1)3=–1, where λ, μ∈R, then λ+μ is equal to [2025]

Q 17 : Let f be a differentiable function on R such that f(2)=1, f'(2)=4. Let limx→0(f(2+x))3/x=eα. Then the number of times the curve y=4x3–4x2–4(α–7)x–α meets x-axis is : [2025]

Q 18 : limx→0+tan (5(x)13)loge(1+3x2)(tan–13x)2(e5(x)43–1) is equal to [2025]

Q 19 : Given below are two statements : Statement I : limx→0(tan–1x+loge1+x1–x–2xx5)=25 Statement II : limx→1(x21-x)=1e2 In the light of the above statements, choose the correct answer from the potions given below. [2025]

Q 20 : If limx→∞((e1–e)(1e–x1+x))x=α, then the value of loge α1+loge α equals : [2025]

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Q 11 :
If $α = \lim_{x \to 0^{+}} (\frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}})$ and $β = \lim_{x \to 0} {(1 + \sin x)}^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $a x^{2} + b x - \sqrt{e} = 0,$ then $12 \log_{e} (a + b)$ is equal to _______ . [2024]

Q 13 :
$If \lim_{x \to 0} \frac{a x^{2} e^{x} - b \log_{e} (1 + x) + c x e^{- x}}{x^{2} \sin x} = 1, then 16 (a^{2} + b^{2} + c^{2}) is equal to_____.$ [2024]

Q 14 :
For $α, β, γ \in R$ , if $\lim_{x \to 0} \frac{x^{2} \sin α x + (γ - 1) e^{x^{2}}}{\sin 2 x - β x} = 3$ , then $β + γ - α$ is equal to: [2025]

Q 15 :
If $\lim_{x \to 0} \frac{\cos (2 x) + a \cos (4 x) - b}{x^{4}}$ is finite, then (a + b) is equal to : [2025]

Q 16 :
If $\lim_{x \to 1^{+}} \frac{(x - 1) (6 + λ \cos (x - 1)) + μ \sin (1 - x)}{{(x - 1)}^{3}} = - 1$ , where $λ, μ \in R$ , then $λ + μ$ is equal to [2025]

Q 17 :
Let f be a differentiable function on R such that $f (2) = 1, f' (2) = 4$ . Let $\lim_{x \to 0} (f {(2 + x))}^{3 / x} = e^{α}$ . Then the number of times the curve $y = 4 x^{3} - 4 x^{2} - 4 (α - 7) x - α$ meets x-axis is : [2025]

Q 18 :
$\lim_{x \to 0^{+}} \frac{\tan (5 {(x)}^{\frac{1}{3}}) \log_{e} (1 + 3 x^{2})}{{(\tan^{- 1} 3 \sqrt{x})}^{2} (e^{5 {(x)}^{\frac{4}{3}}} - 1)}$ is equal to [2025]

Q 20 :
If $\lim_{x \to \infty} ((\frac{e}{1 - e}) {(\frac{1}{e} - \frac{x}{1 + x}))}^{x} = α$ , then the value of $\frac{\log_{e} α}{1 + \log_{e} α}$ equals : [2025]