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

1

2

3

4

(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
$\sqrt{2}$

1

2

3

4

(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]

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

1

2

3

4

(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]

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

1

2

3

4

(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

1

2

3

4

(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

1

2

3

4

(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

1

2

3

4

(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]

(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]

(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]

(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 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]

(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]

(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]

(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 $= \frac{\frac{e}{1 - e}}{1 + \frac{e}{1 - e}} = e$ .

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]

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

1

2

3

4

(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
$\frac{1}{2 \sqrt{5}}$

1

2

3

4

(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

1

2

3

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

1

2

3

4

(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]

(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]

(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]

(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]

(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

1

2

3

4

(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]

$2 β$
$α$
$β$
$2 α$

1

2

3

4

(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 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]

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]

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]

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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]

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]

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

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]

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]

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

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]

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]

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]

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]

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]

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 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]

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]

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]