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

Q 1 :

Let $a_{1}, a_{2}, a_{3}, . ., a_{n}$ be $n$ positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference, then

$\lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}) i s$ [2023]

0
$\sqrt{d}$
1
$\frac{1}{\sqrt{d}}$

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(3)

$W e h a v e, a_{1}, a_{2}, . . . ., a_{n} a r e i n A . P .$

$∴ a_{2} - a_{1} = a_{3} - a_{2} = . . . . . . = a_{n} - a_{n - 1} = d$

$N o w, \lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}})$

$= \lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{a_{2} - a_{1}} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{a_{3} - a_{2}} + . . . + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{a_{n} - a_{n - 1}})$

$= \lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{d} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{d} + . . . + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{d})$

$= \lim_{n \to \infty} \frac{1}{\sqrt{n d}} (\sqrt{a_{n}} - \sqrt{a_{1}})$

$= \lim_{n \to \infty} \frac{1}{\sqrt{n d}} (\frac{a_{n} - a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}}) = \lim_{n \to \infty} \frac{1}{\sqrt{n d}} \frac{(n - 1) d}{\sqrt{a_{n}} + \sqrt{a_{1}}}$

$= \lim_{n \to \infty} \frac{1}{\sqrt{n d}} \frac{n (1 - \frac{1}{n}) d}{\sqrt{n} (\sqrt{\frac{a_{1}}{n} + (1 - \frac{1}{n})} d + \sqrt{\frac{a_{1}}{n}})}$

$= \lim_{n \to \infty} \frac{(1 - \frac{1}{n}) d}{\sqrt{d} (\sqrt{\frac{a_{1}}{n} + d - \frac{d}{n}}) + \sqrt{\frac{a_{1}}{n}}} = \frac{d}{\sqrt{d} \cdot \sqrt{d}} = 1$

Q 2 :
$\lim_{n \to \infty} {(2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{3}}}) (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{5}}}) \dots (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{2 n + 1}}})}$ is equal to [2023]

$\frac{1}{\sqrt{2}}$
1
0
$\sqrt{2}$

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(3)

Let $L = \lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / 3}) (2^{1 / 2} - 2^{1 / 5}) \dots (2^{1 / 2} - 2^{1 / (2 n + 1)})}$

By Sandwich Theorem,

${(2^{1 / 2} - 2^{1 / 3})}^{n} < (2^{1 / 2} - 2^{1 / 3}) (2^{1 / 2} - 2^{1 / 5}) (2^{1 / 2} - 2^{1 / 7}) \dots (2^{1 / 2} - 2^{1 / (2 n + 1)}) < {(2^{1 / 2} - 2^{1 / (2 n + 1)})}^{n}$

$\Rightarrow \lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / 3})}^{n} < L < \lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / (2 n + 1)})}^{n}$

As, $\lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / 3})}^{n} = 0 and \lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / (2 n + 1)})}^{n} = 0$

$∴$ $L = 0$

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

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

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(2)

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 4 :
$\lim_{x \to 0} \frac{e - {(1 + 2 x)}^{\frac{1}{2 x}}}{x} is equal to$ [2024]

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

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(1)

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 5 :
$If a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}} and b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}, then the value of a b^{3} is: [2024]$

25
36
32
30

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(3)

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

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 6 :
$\lim_{x \to 0} \frac{e^{2 | \sin x |} - 2 | \sin x | - 1}{x^{2}}$ [2024]

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

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(4)

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 7 :
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]

5
1
7
2

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(1)

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 8 :
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 9 :
$Let a > 0 be a root of the equation 2 x^{2} + x - 2 = 0 . If \lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos (2 + x - 2 x^{2}))}{{(1 - a x)}^{2}} = α + β \sqrt{17}, where α, β \in Z,$ $then α + β is equal to___.$ [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 10 :
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

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

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

Topic Question Set

Q 2 :
$\lim_{n \to \infty} {(2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{3}}}) (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{5}}}) \dots (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{2 n + 1}}})}$ is equal to [2023]

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

Q 4 :
$\lim_{x \to 0} \frac{e - {(1 + 2 x)}^{\frac{1}{2 x}}}{x} is equal to$ [2024]

Q 5 :
$If a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}} and b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}, then the value of a b^{3} is: [2024]$

Q 6 :
$\lim_{x \to 0} \frac{e^{2 | \sin x |} - 2 | \sin x | - 1}{x^{2}}$ [2024]

Q 7 :
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 8 :
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 9 :
$Let a > 0 be a root of the equation 2 x^{2} + x - 2 = 0 . If \lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos (2 + x - 2 x^{2}))}{{(1 - a x)}^{2}} = α + β \sqrt{17}, where α, β \in Z,$ $then α + β is equal to___.$ [2024]

0%

Q 10 :
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%

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Topic Question Set

Q 1 : Let a1, a2, a3, .., an be n positive consecutive terms of an arithmetic progression. If d>0 is its common difference, then limn→∞dn(1a1+a2+1a2+a3+...+1an-1+an) is [2023]

Q 2 : limn→∞{(212-213)(212-215)⋯(212-212n+1)} is equal to [2023]

Q 3 : 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 4 : limx→0e-(1+2x)12xx is equal to [2024]

Q 5 : If a=limx→01+1+x4-2x4 and b=limx→0sin2x2-1+cosx,then the value of ab3 is: [2024]

Q 6 : limx→0e2|sinx|-2|sinx|-1x2 [2024]

Q 7 : If limx→03+αsinx+βcosx+loge(1-x)3tan2x=13, then 2α-β is equal to [2024]

Q 8 : 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]

0%

Q 9 : 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]

0%

Q 10 : The value of limx→02(1-cosxcos2xcos3x3⋯cos10x10x2) is _______ . [2024]

0%

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Q 2 :
$\lim_{n \to \infty} {(2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{3}}}) (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{5}}}) \dots (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{2 n + 1}}})}$ is equal to [2023]

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

Q 4 :
$\lim_{x \to 0} \frac{e - {(1 + 2 x)}^{\frac{1}{2 x}}}{x} is equal to$ [2024]

Q 5 :
$If a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}} and b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}, then the value of a b^{3} is: [2024]$

Q 6 :
$\lim_{x \to 0} \frac{e^{2 | \sin x |} - 2 | \sin x | - 1}{x^{2}}$ [2024]

Q 7 :
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 8 :
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 9 :
$Let a > 0 be a root of the equation 2 x^{2} + x - 2 = 0 . If \lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos (2 + x - 2 x^{2}))}{{(1 - a x)}^{2}} = α + β \sqrt{17}, where α, β \in Z,$ $then α + β is equal to___.$ [2024]

Q 10 :
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]