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

Q 1 :
$\lim_{n \to \infty} \frac{(1^{2} - 1) (n - 1) + (2^{2} - 2) (n - 2) + \dots + ({(n - 1)}^{2} - (n - 1)) \cdot 1}{(1^{3} + 2^{3} + \dots + n^{3}) - (1^{2} + 2^{2} + \dots + n^{2})} is equal to:$ [2024]

$\frac{3}{4}$
$\frac{1}{3}$
$\frac{1}{2}$
$\frac{2}{3}$

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(B)

Let $L = \lim_{n \to \infty} (1^{2} - 1) (n - 1) + (2^{2} - 2) (n - 2) + \dots \frac{. . . + ({(n - 1)}^{2} - (n - 1)) \cdot 1}{(1^{3} + 2^{3} + . . . + n^{3}) - (1^{2} + 2^{2} + . . . + n^{2})}$

$Numerator = \sum_{r = 1}^{n - 1} [(r^{2} - r) (n - r)] = \sum_{r = 1}^{n - 1} (- r^{3} + r^{2} (n + 1) - n r)$

$= - {(\frac{(n - 1) n}{2})}^{2} + \frac{(n + 1) (n - 1) n (2 n - 1)}{6} - \frac{n^{2} (n - 1)}{2}$

So, $L = \lim_{n \to \infty} \frac{\frac{n (n - 1)}{2} [\frac{- n (n - 1)}{2} + \frac{(n + 1) (2 n - 1)}{3} - n]}{{(\frac{n (n + 1)}{2})}^{2} - \frac{n (n + 1) (2 n + 1)}{6}}$

$= \lim_{n \to \infty} \frac{(n - 1) (- 3 n^{2} + 3 n + 2 (2 n^{2} + n - 1) - 6 n)}{(n + 1) (3 n^{2} + 3 n - 4 n - 2)}$

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

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

Q 2 :
$\lim_{x \to 0} \frac{e - {(1 + 2 x)}^{\frac{1}{2 x}}}{x} is equal to$ [2024]

$e$
$0$
$\frac{- 2}{e}$
$e - e^{2}$

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(A)

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

$= \lim_{x \to 0} \frac{e - e^{\frac{1}{2 x} \log (1 + 2 x)}}{x}$

$= \lim_{x \to 0} \frac{- e (e^{\log \frac{(1 + 2 x)}{2 x} - 1} - 1)}{x}$

$= \lim_{x \to 0} \frac{- e [e^{\frac{\log (1 + 2 x) - 2 x}{2 x} - 1}]}{x \times \frac{\log (1 + 2 x) - 2 x}{2 x}} \times \frac{\log (1 + 2 x) - 2 x}{2 x}$

$= \lim_{x \to 0} \frac{- e}{x} \frac{\log (1 + 2 x) - 2 x}{2 x}$ $[∵ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1]$

$= \lim_{x \to 0} - e [\frac{\frac{1}{1 + 2 x} \cdot 2 - 2}{4 x}] [Using L'Hospital Rule]$

$= \lim_{x \to 0} \frac{- e}{4} \frac{- 1}{{(1 + 2 x)}^{2}} \cdot 4 [Again by L'Hospital]$

$= e$

Q 3 :
$If a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}} and b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}, then the value of a b^{3} is: [2024]$

25
36
32
30

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(C)

$a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}}$

Rationalising numerator, we get

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

$= \lim_{x \to 0} \frac{\sqrt{1 + x^{4}} - 1}{x^{4} (\sqrt{1 + \sqrt{1 + x^{4}}} + \sqrt{2})}$

Again rationalising numerator, we get

$= \lim_{x \to 0} \frac{1}{(\sqrt{1 + x^{4}} + 1) (\sqrt{1 + \sqrt{1 + x^{4}}} + \sqrt{2})} = \frac{1}{2 \times 2 \sqrt{2}} = \frac{1}{4 \sqrt{2}}$

Now, $b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}$

$= \lim_{x \to 0} \frac{(1 - \cos^{2} x) (\sqrt{2} + \sqrt{1 + \cos x})}{[2 - (1 + \cos x)]}$ [By rationalising denominator]

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

$= \lim_{x \to 0} (1 + \cos x) (\sqrt{2} + \sqrt{1 + \cos x}) = 4 \sqrt{2}$

$∴$ $a b^{3} = \frac{1}{4 \sqrt{2}} \times {(4 \sqrt{2})}^{3} = 32$

Q 4 :
$\lim_{x \to 0} \frac{e^{2 | \sin x |} - 2 | \sin x | - 1}{x^{2}}$ [2024]

does not exist
is equal to -1
is equal to 1
is equal to 2

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(D)

We have, $\lim_{x \to 0} = \frac{e^{2 | \sin x |} - 2 | \sin x | - 1}{x^{2}}$

$R.H.L. \lim_{x \to 0^{+}} = \frac{e^{2 \sin x} - 2 \sin x - 1}{x^{2}}$

$\lim_{x \to 0^{+}} = \frac{(1 + 2 \sin x + \frac{{(2 \sin x)}^{2}}{2!} + \dots) - 2 \sin x - 1}{x^{2}}$

$\lim_{x \to 0^{+}} = \frac{4 \sin^{2} x}{x^{2}} (\frac{1}{2!} + \frac{2 \sin x}{3!} + \dots) = 2$

$L.H.L. = \lim_{x \to 0^{-}} \frac{e^{- 2 \sin x} + 2 \sin x - 1}{x^{2}}$

$\lim_{x \to 0^{-}} \frac{(1 - 2 \sin x + \frac{{(2 \sin x)}^{2}}{2!} - \frac{{(2 \sin x)}^{3}}{3!} + \dots) + 2 \sin x - 1}{x^{2}}$

$\lim_{x \to 0^{-}} \frac{4 \sin^{2} x}{x^{2}} (\frac{1}{2!} - \frac{2 \sin x}{3!} + \dots) = 2$

$L.H.L. = R.H.L. = 2$

Q 5 :
If $\lim_{x \to 0} \frac{3 + α \sin x + β \cos x + \log_{e} (1 - x)}{3 \tan^{2} x} = \frac{1}{3},$ then $2 α - β$ is equal to [2024]

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(A)

Given, $\lim_{x \to 0} \frac{3 + α \sin x + β \cos x + \log_{e} (1 - x)}{3 \tan^{2} x} = \frac{1}{3}$

$\Rightarrow \lim_{x \to 0} \frac{3 + α (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots) + β (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots) + (- x - \frac{x^{2}}{2!} - \frac{2 x^{3}}{3!} - \dots)}{3 {(\frac{\tan x}{x})}^{2} \cdot x^{2}} = \frac{1}{3}$

$\Rightarrow \lim_{x \to 0} \frac{3 + α (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots) + β (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots) + (- x - \frac{x^{2}}{2!} - \frac{2 x^{3}}{3!} - \dots)}{3 x^{2}} = \frac{1}{3}$

$Comparing the coefficient of x^{0}, we get$

$3 + β = 0 \Rightarrow β = - 3$

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

$α - 1 = 0 \Rightarrow α = 1$

$∴$ $2 α - β = 2 (1) - (- 3) = 2 + 3 = 5$

Q 6 :
If $\lim_{x \to 1} \frac{{(5 x + 1)}^{1 / 3} - {(x + 5)}^{1 / 3}}{{(2 x + 3)}^{1 / 2} - {(x + 4)}^{1 / 2}} = \frac{m \sqrt{5}}{n {(2 n)}^{2 / 3}},$ where $\gcd (m, n) = 1,$ then $8 m + 12 n$ is equal to _______ . [2024]

0%

(100)

Let $L = \lim_{x \to 1} \frac{{(5 x + 1)}^{1 / 3} - {(x + 5)}^{1 / 3}}{{(2 x + 3)}^{1 / 2} - {(x + 4)}^{1 / 2}} (\frac{0}{0} form)$

Using L-Hospital's rule

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

$= \frac{(\frac{5}{3} - \frac{1}{3}) 6^{- 2 / 3}}{(\frac{1}{2}) 5^{- 1 / 2}} = \frac{8}{3} \times \frac{5^{1 / 2}}{6^{2 / 3}} = \frac{m \sqrt{5}}{n {(2 n)}^{2 / 3}}$

$\Rightarrow m = 8, n = 3 \Rightarrow 8 m + 12 n = 100$

Q 7 :
$Let a > 0 be a root of the equation 2 x^{2} + x - 2 = 0 . If \lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos (2 + x - 2 x^{2}))}{{(1 - a x)}^{2}} = α + β \sqrt{17}, where α, β \in Z,$ $then α + β is equal to___.$ [2024]

0%

(170)

$We have, 2 x^{2} + x - 2 = 0$

$Given one root is a . Let another root be b .$

$Here, a = \frac{- 1 + \sqrt{17}}{4}, b = \frac{- 1 - \sqrt{17}}{4}$

$∴ 2 + x - 2 x^{2} = 0 has roots \frac{1}{a} and \frac{1}{b}$

$Now, \lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos 2 (x - \frac{1}{a}) (x - \frac{1}{b}))}{a^{2} {(x - \frac{1}{a})}^{2}}$

$\lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos 2 (x - \frac{1}{a}) (x - \frac{1}{b}))}{a^{2} {(x - \frac{1}{a})}^{2}} \times \frac{{(x - \frac{1}{b})}^{2}}{{(x - \frac{1}{b})}^{2}}$

$= \lim_{x \to \frac{1}{a}} \frac{16 \cdot 2 \sin^{2} (x - \frac{1}{a}) (x - \frac{1}{b})}{a^{2} {(x - \frac{1}{a})}^{2}} \times \frac{{(x - \frac{1}{b})}^{2}}{{(x - \frac{1}{b})}^{2}}$

$= 16 \times \frac{2}{a^{2}} {(\frac{1}{a} - \frac{1}{b})}^{2}$

$= \frac{32}{a^{2}} (\frac{17}{4}) = \frac{17 \times 8}{a^{2}} = \frac{17 \times 8 \times 16}{(- 1 + \sqrt{17})^{2}} = 153 + 17 \sqrt{17}$

$= α + β \sqrt{17}$

$\Rightarrow α = 153, β = 17$

$\Rightarrow α + β = 170$

Q 8 :
The value of $\lim_{x \to 0} 2 (\frac{1 - \cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \dots \sqrt[10]{\cos 10 x}}{x^{2}})$ is _______ . [2024]

0%

(55)

$\lim_{x \to 0} 2 (\frac{1 - \cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \dots \sqrt[10]{\cos 10 x}}{x^{2}})$

Let $f = \cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \dots \sqrt[10]{\cos 10 x}$

$f = \cos x {(\cos 2 x)}^{1 / 2} {(\cos 3 x)}^{1 / 3} \dots {(\cos 10 x)}^{1 / 10}$

Taking log on both sides, we get

$\log f = \log \cos x + \frac{1}{2} \cos 2 x + \frac{1}{3} \cos 3 x + \dots + \frac{1}{10} \cos 10 x$

Differentiating w.r.t. $x$

$\frac{1}{f} \frac{d f}{d x} = - \tan x - \tan 2 x \dots - \tan 10 x$

$\frac{d f}{d x} = - f (\tan x + \tan 2 x \dots + \tan 10 x)$

Using L'Hospital's Rule

$\lim_{x \to 0} 2 \frac{(f (\tan x + \tan 2 x \dots + \tan 10 x))}{2 x}$ $(∵ f = 1)$

$= 1 + 2 + \dots + 10 = 55$

Q 9 :
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%

(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 10 :
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]

0%

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

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

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

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

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

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

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

Topic Question Set

Q 1 :
$\lim_{n \to \infty} \frac{(1^{2} - 1) (n - 1) + (2^{2} - 2) (n - 2) + \dots + ({(n - 1)}^{2} - (n - 1)) \cdot 1}{(1^{3} + 2^{3} + \dots + n^{3}) - (1^{2} + 2^{2} + \dots + n^{2})} is equal to:$ [2024]

Q 2 :
$\lim_{x \to 0} \frac{e - {(1 + 2 x)}^{\frac{1}{2 x}}}{x} is equal to$ [2024]

Q 3 :
$If a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}} and b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}, then the value of a b^{3} is: [2024]$

Q 4 :
$\lim_{x \to 0} \frac{e^{2 | \sin x |} - 2 | \sin x | - 1}{x^{2}}$ [2024]

Q 5 :
If $\lim_{x \to 0} \frac{3 + α \sin x + β \cos x + \log_{e} (1 - x)}{3 \tan^{2} x} = \frac{1}{3},$ then $2 α - β$ is equal to [2024]

Q 6 :
If $\lim_{x \to 1} \frac{{(5 x + 1)}^{1 / 3} - {(x + 5)}^{1 / 3}}{{(2 x + 3)}^{1 / 2} - {(x + 4)}^{1 / 2}} = \frac{m \sqrt{5}}{n {(2 n)}^{2 / 3}},$ where $\gcd (m, n) = 1,$ then $8 m + 12 n$ is equal to _______ . [2024]

0%

Q 7 :
$Let a > 0 be a root of the equation 2 x^{2} + x - 2 = 0 . If \lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos (2 + x - 2 x^{2}))}{{(1 - a x)}^{2}} = α + β \sqrt{17}, where α, β \in Z,$ $then α + β is equal to___.$ [2024]

0%

Q 8 :
The value of $\lim_{x \to 0} 2 (\frac{1 - \cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \dots \sqrt[10]{\cos 10 x}}{x^{2}})$ is _______ . [2024]

0%

Q 9 :
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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Topic Question Set

Q 1 : limn→∞(12-1)(n-1)+(22-2)(n-2)+…+((n-1)2-(n-1))·1(13+23+…+n3)-(12+22+…+n2) is equal to: [2024]

Q 2 : limx→0e-(1+2x)12xx is equal to [2024]

Q 3 : If a=limx→01+1+x4-2x4 and b=limx→0sin2x2-1+cosx,then the value of ab3 is: [2024]

Q 4 : limx→0e2|sinx|-2|sinx|-1x2 [2024]

Q 5 : If limx→03+αsinx+βcosx+loge(1-x)3tan2x=13, then 2α-β is equal to [2024]

Q 6 : If limx→1(5x+1)1/3-(x+5)1/3(2x+3)1/2-(x+4)1/2=m5n(2n)2/3, where gcd(m,n)=1, then 8m+12n is equal to _______ . [2024]

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Q 7 : Let a>0 be a root of the equation 2x2+x-2=0. If limx→1a16(1-cos(2+x-2x2))(1-ax)2=α+β17, where α,β∈Z,then α+β is equal to ___. [2024]

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Q 8 : The value of limx→02(1-cosxcos2xcos3x3⋯cos10x10x2) is _______ . [2024]

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Q 9 : 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 10 : 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]

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Q 1 :
$\lim_{n \to \infty} \frac{(1^{2} - 1) (n - 1) + (2^{2} - 2) (n - 2) + \dots + ({(n - 1)}^{2} - (n - 1)) \cdot 1}{(1^{3} + 2^{3} + \dots + n^{3}) - (1^{2} + 2^{2} + \dots + n^{2})} is equal to:$ [2024]

Q 2 :
$\lim_{x \to 0} \frac{e - {(1 + 2 x)}^{\frac{1}{2 x}}}{x} is equal to$ [2024]

Q 3 :
$If a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}} and b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}, then the value of a b^{3} is: [2024]$

Q 4 :
$\lim_{x \to 0} \frac{e^{2 | \sin x |} - 2 | \sin x | - 1}{x^{2}}$ [2024]

Q 5 :
If $\lim_{x \to 0} \frac{3 + α \sin x + β \cos x + \log_{e} (1 - x)}{3 \tan^{2} x} = \frac{1}{3},$ then $2 α - β$ is equal to [2024]

Q 6 :
If $\lim_{x \to 1} \frac{{(5 x + 1)}^{1 / 3} - {(x + 5)}^{1 / 3}}{{(2 x + 3)}^{1 / 2} - {(x + 4)}^{1 / 2}} = \frac{m \sqrt{5}}{n {(2 n)}^{2 / 3}},$ where $\gcd (m, n) = 1,$ then $8 m + 12 n$ is equal to _______ . [2024]

Q 7 :
$Let a > 0 be a root of the equation 2 x^{2} + x - 2 = 0 . If \lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos (2 + x - 2 x^{2}))}{{(1 - a x)}^{2}} = α + β \sqrt{17}, where α, β \in Z,$ $then α + β is equal to___.$ [2024]

Q 8 :
The value of $\lim_{x \to 0} 2 (\frac{1 - \cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \dots \sqrt[10]{\cos 10 x}}{x^{2}})$ is _______ . [2024]

Q 9 :
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]