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

Q 11 :

Among

$(S1) : \lim_{n \to \infty} \frac{1}{n^{2}} (2 + 4 + 6 + \dots + 2 n) = 1$

$(S2) : \lim_{n \to \infty} \frac{1}{n^{16}} (1^{15} + 2^{15} + 3^{15} + \dots + n^{15}) = \frac{1}{16}$ [2023]

Only (S1) is true
Both (S1) and (S2) are false
Only (S2) is true
Both (S1) and (S2) are true

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$(S1):$ $\lim_{n \to \infty} \frac{1}{n^{2}} (2 + 4 + 6 + \dots + 2 n)$

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

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

$∴$ $(S 1) is true.$

$(S2):$ $\lim_{n \to \infty} \frac{1}{n^{16}} (1^{15} + 2^{15} + 3^{15} + \dots + n^{15}) = \lim_{n \to \infty} \frac{1}{n^{16}} \sum_{r = 1}^{n} r^{15}$

$= \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{15} = \int_{0}^{1} x^{15} d x = | \frac{x^{16}}{16} |_{0}^{1} = \frac{1}{16}$

$∴$ $(S 2) is also true.$

Q 12 :

$If \lim_{x \to 0} \frac{e^{a x} - \cos (b x) - \frac{c x e^{- c x}}{2}}{1 - \cos (2 x)} = 17, then 5 a^{2} + b^{2} is equal to$ [2023]

72
64
68
76

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$\lim_{x \to 0} \frac{a e^{a x} + b \sin b x - \frac{c}{2} e^{- c x} + \frac{c^{2} x}{2} e^{- c x}}{2 \sin 2 x} = 17$

$\Rightarrow a - \frac{c}{2} = 0$ $∴$ $a = \frac{c}{2}$

Again, applying (L'Hospital's rule)

$\lim_{x \to 0} \frac{a^{2} e^{a x} + b^{2} \cos b x + \frac{c^{2}}{2} e^{- c x} - \frac{c^{3} x}{2} e^{- c x} + \frac{c^{2}}{2} e^{- c x}}{4 \cos 2 x} = 17$

$\Rightarrow \frac{a^{2} + b^{2} + \frac{c^{2}}{2} + \frac{c^{2}}{2}}{4} = 17$

$\Rightarrow a^{2} + b^{2} + c^{2} = 68 \Rightarrow a^{2} + b^{2} + 4 a^{2} = 68$

$\Rightarrow 5 a^{2} + b^{2} = 68$

Q 13 :

$\lim_{t \to 0} {(1^{\frac{1}{^{\sin^{2} t}}} + 2^{\frac{1}{^{\sin^{2} t}}} + \dots + n^{\frac{1}{^{\sin^{2} t}}})}^{\sin^{2} t}$ is equal to [2023]

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

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$Let L = \lim_{t \to 0} {(1^{\frac{1}{\sin^{2} t}} + 2^{\frac{1}{\sin^{2} t}} + \dots + n^{\frac{1}{\sin^{2} t}})}^{\sin^{2} t}$

$\Rightarrow L = \lim_{t \to 0} n {[{(\frac{1}{n})}^{c o s e c^{2} t} + {(\frac{2}{n})}^{c o s e c^{2} t} + \dots + {(\frac{n - 1}{n})}^{c o s e c^{2} t} + 1]}^{\sin^{2} t}$

$\Rightarrow L = n (1^{\sin (0)}) = n (1) = n [0 < \frac{k}{n} < 1, {(\frac{k}{n})}^{\infty} = 0 for 1 \leq k < n]$

Q 14 :

$The set of all values of a for which \lim_{x \to a} ([x - 5] - [2 x + 2]) = 0,$ where $[α]$ denotes the greatest integer less than or equal to $α$ is equal to [2023]

$[- 7.5, - 6.5)$
$(- 7.5, - 6.5)$
$(- 7.5, - 6.5]$
$[- 7.5, - 6.5]$

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$\lim_{x \to a} ([x - 5] - [2 x + 2]) = 0$

$\Rightarrow \lim_{x \to a} ([x] - 5 - [2 x] - 2) = 0 \Rightarrow \lim_{x \to a} ([x] - [2 x]) = 7$

$Let a \in [x_{1}, x + \frac{1}{2}), then x - 2 x = 7 \Rightarrow x = - 7$

$∴$ $a \in [- 7, - 6.5)$

$Let a \in (x + \frac{1}{2}, x + 1), then x - (2 x + 1) = 7$

$\Rightarrow - x = 8 \Rightarrow x = - 8;$ $a \in [- 7.5, - 7)$

But limit does not exist at $a = - 7.5$

$Hence, a \in (- 7.5, - 6.5)$

Q 15 :

$The value of \lim_{x \to \infty} \frac{1 + 2 - 3 + 4 + 5 - 6 + \dots + (3 n - 2) + (3 n - 1) - 3 n}{\sqrt{2 n^{4} + 4 n + 3} - \sqrt{n^{4} + 5 n + 4}} is$ [2023]

$3 (\sqrt{2} + 1)$
$\frac{3}{2} (\sqrt{2} + 1)$
$\frac{\sqrt{2} + 1}{2}$
$\frac{3}{2 \sqrt{2}}$

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(2)

$\lim_{n \to \infty} \frac{1 + 2 - 3 + 4 + 5 - 6 + \dots + (3 n - 2) + (3 n - 1) - 3 n}{\sqrt{2 n^{4} + 4 n + 3} - \sqrt{n^{4} + 5 n + 4}}$

$= \lim_{n \to \infty} \frac{(1 + 2 + 3 + 4 + \dots + 3 n) - 2 (3 + 6 + 9 + \dots + 3 n)}{\sqrt{2 n^{4} + 4 n + 3} - \sqrt{n^{4} + 5 n + 4}}$

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

$= \lim_{n \to \infty} \frac{(3 n^{2} - 3 n) \times n^{2} \times (\sqrt{2 + \frac{4}{n^{3}} + \frac{3}{n^{4}}} + \sqrt{1 + \frac{5}{n^{3}} + \frac{4}{n^{4}}})}{2 (n^{4} - n - 1)}$

$= \lim_{n \to \infty} \frac{(3 - \frac{3}{n}) (\sqrt{2 + \frac{4}{n^{3}} + \frac{3}{n^{4}}} + \sqrt{1 + \frac{5}{n^{3}} + \frac{4}{n^{4}}})}{2 (1 - \frac{1}{n^{3}} - \frac{1}{n^{4}})}$

$= \frac{3 (\sqrt{2} + 1)}{2}$

Q 16 :

$Let x = 2 be a root of the equation x^{2} + p x + q = 0 and$ $f (x) =$ ${\begin{matrix} \frac{1 - \cos (x^{2} - 4 p x + q^{2} + 8 q + 16)}{{(x - 2 p)}^{4}}, & x \neq 2 p \\ 0, & x = 2 p \end{matrix}$

Then $\lim_{x \to 2 p^{+}} [f (x)],$ where $[\cdot]$ denotes the greatest integer function, is [2023]

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$Put x = 2 in x^{2} + p x + q = 0 \Rightarrow 4 + 2 p + q = 0 \Rightarrow - 2 p = (q + 4)$

Now, $\lim_{x \to 2 p^{+}} \frac{1 - \cos (x^{2} - 4 p x + q^{2} + 8 q + 16)}{{(x - 2 p)}^{4}}$

$= \lim_{x \to 2 p^{+}} \frac{1 - \cos [x^{2} - 2 x \cdot 2 p + q^{2} + 2 \cdot q \cdot 4 + 4^{2}]}{{(x - 2 p)}^{4}}$

$= \lim_{x \to 2 p^{+}} \frac{1 - \cos {(x - 2 p)}^{2}}{{(x - 2 p)}^{4}} = \lim_{x \to 2 p^{+}} \frac{2 \sin^{2} (\frac{{(x - 2 p)}^{2}}{2})}{\frac{{(x - 2 p)}^{4}}{4}} \cdot \frac{1}{4}$

$= \lim_{x \to 2 p^{+}} 2 \times 1 \times \frac{1}{4} = \frac{1}{2}$ $(∵ \lim_{θ \to 0} \frac{\sin θ}{θ} = 1)$

Now, $[1 / 2] = 0$

Q 17 :

$Let f, g and h be real-valued functions defined on R$ $f (x) =$ ${\begin{matrix} \frac{x}{| x |}, & x \neq 0 \\ 1, & x = 0 \end{matrix},$ $g (x) =$ ${\begin{matrix} \frac{\sin (x + 1)}{(x + 1)}, & x \neq - 1 \\ 1, & x = - 1 \end{matrix}$ and $h (x) = 2 [x] - f (x),$ where $[x]$ is the greatest integer $\leq x$ . Then the value of $\lim_{x \to 1} g (h (x - 1))$ is [2023]

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0
sin (1)
- 1

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(1)

$f (x) =$ ${\begin{matrix} \frac{x}{| x |}, & x \neq 0 \\ 1, & x = 0 \end{matrix}$

and $g (x) =$ ${\begin{matrix} \frac{\sin (x + 1)}{x + 1}, & x \neq - 1 \\ 1, & x = - 1 \end{matrix}$

$h (x) = 2 [x] - f (x), [x]$ is the greatest integer $\leq x$

$LHL$ $= \lim_{x \to 1^{-}} g (h (x - 1)) = \lim_{k \to 0} g (h (1 - k - 1))$

$= \lim_{k \to 0} g [(2 [- k] - \frac{- k}{| - k |})] = g [2 (- 1) + 1] = \lim_{x \to 1^{-}} g (- 1) = 1$

$RHL$ $= \lim_{x \to 1^{+}} g (h (x - 1)) = \lim_{h \to 0} g (h (1 + k - 1))$

$= \lim_{k \to 0} g (2 [k] - f (k)) = \lim_{h \to 0} g (0 - 1) = \lim_{x \to 1^{+}} g (- 1) = 1$

$∴$ $\lim_{x \to 1} g (h (x - 1)) = 1$

Q 18 :

$\lim_{x \to \infty} \frac{{(\sqrt{3 x + 1} + \sqrt{3 x - 1})}^{6} + {(\sqrt{3 x + 1} - \sqrt{3 x - 1})}^{6}}{{(x + \sqrt{x^{2} - 1})}^{6} + {(x - \sqrt{x^{2} - 1})}^{6}} x^{3}$ [2023]

is equal to $\frac{27}{2}$
is equal to 9
does not exist
is equal to 27

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$\lim_{x \to \infty} \frac{{(\sqrt{3 x + 1} + \sqrt{3 x - 1})}^{6} + {(\sqrt{3 x + 1} - \sqrt{3 x - 1})}^{6}}{{(x + \sqrt{x^{2} - 1})}^{6} + {(x - \sqrt{x^{2} - 1})}^{6}} \times x^{3}$

$\lim_{x \to \infty} x^{3} \times {\frac{x^{3} [{(\sqrt{3 + \frac{1}{x}} + \sqrt{3 - \frac{1}{x}})}^{6} + {(\sqrt{3 + \frac{1}{x}} - \sqrt{3 - \frac{1}{x}})}^{6}]}{x^{6} [{(1 + \sqrt{1 - \frac{1}{x^{2}}})}^{6} + {(1 - \sqrt{1 - \frac{1}{x^{2}}})}^{6}]}}$

$= \frac{{(\sqrt{3} + \sqrt{3})}^{6} + {(\sqrt{3} - \sqrt{3})}^{6}}{{(1 + \sqrt{1})}^{6} + {(1 - \sqrt{1})}^{6}} = \frac{{(2 \sqrt{3})}^{6} + 0}{2^{6} + 0} = \frac{2^{6} \cdot 3^{3}}{2^{6}} = 3^{3} = 27$

Q 19 :

Let $f : ℝ \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f' (3) = 0$ and $f'' (3) = 4$ . Then $\lim_{x \to 1} (l {og}_{e} {(\frac{f (2 + x)}{f (3)})}^{\frac{18}{{(x - 1)}^{2}}})$ is equal to [2026]

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18

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$Let T = \lim_{x \to 1} {(\frac{f (x + 2)}{f (3)})}^{\frac{18}{{(x - 1)}^{2}}}; 1^{\infty} form$

$\Rightarrow T = e^{\lim_{x \to 1} \frac{18}{{(x - 1)}^{2}} (\frac{f (x + 2) - f (3)}{f (3)})}$

$\Rightarrow T = e^{\lim_{x \to 1} \frac{18}{{(x - 1)}^{2}} (\frac{f (x + 2) - f (3)}{18})}$

$\Rightarrow T = e^{\lim_{x \to 1} (\frac{f (x + 2) - f (3)}{{(x - 1)}^{2}}) \frac{0}{0} form}$ apply L' opital

$\Rightarrow T = e^{\lim_{x \to 1} \frac{f' (x + 2)}{2 (x - 1)}; \frac{0}{0} form}$ apply L' opital

$\Rightarrow T = e^{\lim_{x \to 1} \frac{f'' (x + 2)}{2}}$ ; $= e^{\frac{4}{2}} = e^{2}$

$\Rightarrow \log_{e} (T) = 2$

Q 20 :

Let $f (x) = \lim_{θ \to 0} (\frac{\cos (π x) - x^{(\frac{2}{θ})} \sin (x - 1)}{1 + x^{(\frac{2}{θ})} (x - 1)}), x \in ℝ .$

Consider the following two statements:

I. f(x) is discontinuous at x = 1

II. f(x) is continuous at x = -1

Then, [2026]

Only (I) is True
Only (II) is True
Neither (I) nor (II) is True
Both (I) and (II) are True

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(3)

$f (x)$ $=$ ${\begin{matrix} \cos (π x) & x \to 1^{-} \\ \frac{- \sin (x - 1)}{x - 1} & x \to 1^{+} \end{matrix}$

$RHL = \lim_{x \to 1} \frac{- \sin (x - 1)}{x - 1} = - 1$

$LHL = \lim_{x \to 1} \cos (π x) = - 1, f (1) = - 1$

$f(x) is continuous at x = 1$

$f (x)$ $=$ ${\begin{matrix} \frac{- \sin (x - 1)}{- (x - 1)} & x \to - 1^{-} \\ \cos (π x) & x \to - 1^{+} \end{matrix}$

$RHL = \lim_{x \to - 1} \cos (π x) = - 1$

$LHL = \lim_{x \to - 1} \frac{- \sin (x - 1)}{x - 1} = \frac{\sin 2}{- 2}$

$f(x) is discontinuous at x = - 1$

Q 21 :

If $\lim_{x \to 0} \frac{e^{(a - 1) x} + 2 \cos b x + (c - 2) e^{- x}}{x \cos x - \log_{e} (1 + x)} = 2,$ then $a^{2} + b^{2} + c^{2}$ is equal to: [2026]

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

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

$For which$

$∵$ $c + 1 = 0 \Rightarrow c = - 1$

$∵$ $a - c = - 1 \Rightarrow a = - 2$

$∵$ $\frac{{(a - 1)}^{2}}{2} - b^{2} + (\frac{c - 2}{2}) = 1$

$\frac{9}{2} - b^{2} - \frac{3}{2} = 1 \Rightarrow b^{2} = 2$

$a^{2} + b^{2} + c^{2} = 4 + 2 + 1 = 7$

Q 22 :

The value of $\lim_{x \to 0} \frac{\log_{e} ((\sec (e x) \cdot \sec ((e^{2} x)) \dots \sec ((e^{10} x))))}{e^{2} - e^{2 \cos x}}$ is equal to [2026]

$\frac{(e^{10} - 1)}{2 e^{2} (e^{2} - 1)}$
$\frac{(e^{20} - 1)}{2 e^{2} (e^{2} - 1)}$
$\frac{(e^{10} - 1)}{2 (e^{2} - 1)}$
$\frac{(e^{20} - 1)}{2 (e^{2} - 1)}$

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$\Rightarrow \lim_{x \to 0} \frac{\ln (\sec (e x)) + \ln (\sec (e^{2} x)) + \dots + l n (\sec (e^{10} x))}{e^{2 \cos x} (\frac{e^{2 - 2 \cos x} - 1}{2 - 2 \cos x}) (\frac{2 - 2 \cos x}{x^{2}}) x^{2}}$

$\Rightarrow \lim_{x \to 0} \frac{l n (\sec (e x)) + l n (\sec (e^{2} x)) + \dots + l n (\sec (e^{10} x))}{e^{2} x^{2}}$

$Using L'H rule$

$\Rightarrow \lim_{x \to 0} \frac{e \tan (e x) + e^{2} \tan (e^{2} x) + \dots + e^{10} \tan (e^{10} x)}{2 e^{2} x}$

$= \frac{1}{2 e^{2}} [e^{2} + e^{4} + e^{6} + \dots + e^{20}]$

$= \frac{1}{2} \cdot \frac{e^{2} ({(e^{2})}^{10} - 1)}{e^{2} (e^{2} - 1)}$

$= \frac{1}{2} \cdot \frac{(e^{20} - 1)}{(e^{2} - 1)}$

Q 23 :

$\lim_{x \to \infty} \frac{3}{x} [\frac{x}{4}] = \frac{p}{q}$ (where $[.]$ denotes greatest integer function), then $p + q$ (where $p, q$ are relative prime) is:

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$\lim_{x \to \infty} \frac{3}{x} (\frac{x}{4} - {\frac{x}{4}}) = \frac{3}{4} - 0 = \frac{3}{4} \Rightarrow p + q = 7$

Q 24 :

If $f (x) = \lim_{n \to \infty} (\prod_{i = 1}^{n} \cos (\frac{x}{2^{i}}))$ then $f' (x)$ is equal to:

$\frac{\sin x}{x}$
$\frac{x}{\sin x}$
$\frac{x \cos x - \sin x}{x^{2}}$
$\frac{\sin x - x \cos x}{\sin^{2} x}$

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(3)

Using the identity

$\sin x = 2^{n} \sin (\frac{x}{2^{n}}) \prod_{i = 1}^{n} \cos (\frac{x}{2^{i}})$

we get

$\prod_{i = 1}^{n} \cos (\frac{x}{2^{i}}) = \frac{\sin x}{2^{n} \sin (\frac{x}{2^{n}})}$

As $n \to \infty$ ,

$2^{n} \sin (\frac{x}{2^{n}}) \to x$

Hence,

$f (x) = \frac{\sin x}{x}$

Differentiating,

$f' (x) = \frac{x \cos x - \sin x}{x^{2}}$

Q 25 :

$\lim_{n \to \infty} \frac{\sum_{k = 1}^{n - 1} (k - 1) (n k - k^{2})}{2 \sum_{r = 1}^{n} r^{3} - \sum_{s = 0}^{n} (s^{2} + {(n - s)}^{2})} = t then$ $[43 t]$ $=____$ $([.] denotes GIF)$

(7)

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

Q 26 :

The value of $\lim_{n \to \infty} \frac{(\lim_{x \to 0^{-}} \sum_{r = 1}^{2 n + 1} [x^{r}]) + n + 3}{ℓ n {(1 + n)}^{n} - ℓ n {(n)}^{n}}$ , where $[\cdot]$ denotes greatest integer function

(2)

$\lim_{x \to 0^{-}} \sum_{r = 1}^{2 n + 1} [x^{r}] = \lim_{x \to 0^{-}} [x] + [x^{2}] + [x^{3}] + \dots + [x^{2 n + 1}]$

$= (- 1) + 0 + (- 1) + \dots + 0 + (- 1) = (- 1) (n + 1) = - n - 1$

$\Rightarrow L = \lim_{n \to \infty} \frac{- n - 1 + n + 3}{n \ln (\frac{1 + n}{n})}$

$= \lim_{n \to \infty} \frac{2 (\frac{1}{n})}{\ln (1 + \frac{1}{n})}$ $\Rightarrow \frac{1}{n} = t$

$\Rightarrow L = \lim_{t \to 0} \frac{2 t}{\ln (1 + t)} = 2$

Q 27 :

Consider the following Statements

Statement-I: If both functions $f (x)$ and $g (x)$ have limits at $x = a$ , then $\lim_{x \to a} [f (g (x))] = [\lim_{x \to a} f (g (x))]$ [ ] denotes G.I.F

Statement-2: A function may fail to have a limit at a point in its domain but functional value at $x = a$ exists (i.e. $f (a)$ exists).

Only Statement I is true
Only Statement II is true
Both Statement I and II are True
Both Statement I and II are false

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Apply Limits & Properties

Q 28 :

If $f (x) =$ $| \begin{matrix} \cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix} |$ , then $\lim_{x \to 0} \frac{f' (x)}{x}$ is equal to

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$Applying R_{1} \to R_{1} - R_{2}$

$f (x) =$ $| \begin{matrix} \cos x - \tan x & 0 & 0 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix} |$ $= (\cos x - \tan x) (x^{2} - 2 x^{2})$

$= - x^{2} (\cos x - \tan x)$

$∴$ $f' (x) = - 2 x (\cos x - \tan x) - x^{2} (- \sin x - \sec^{2} x)$

$∴$ $\lim_{x \to \infty} \frac{f' (x)}{x} = \lim_{x \to \infty} [- 2 (\cos x - \tan x)] + \lim_{x \to \infty} [x (\sin x + \sec^{2} x)] = - 2 \times 1 = - 2$

Q 29 :

$\lim_{x \to 0^{+}} \frac{\tan x \sqrt{\tan x} - \sin x \sqrt{\sin x}}{x^{3} \sqrt{x}} Equals$

0.75
0.85
0.95
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(1)

$\lim_{x \to 0} \frac{{(\tan x)}^{3 / 2} [1 - {(\cos x)}^{3 / 2}]}{x^{3 / 2} \cdot x^{2}}$

$= 1 \times \lim_{x \to 0} \frac{1 - \cos^{3} x}{x^{2}} \cdot \frac{1}{1 + {(\cos x)}^{3 / 2}}$

$= \frac{1}{2} \cdot \frac{1}{2} (1 + \cos x + \cos^{2} x) = \frac{3}{4}$

Q 30 :

$\lim_{x \to 0} \frac{e^{x^{2}} - \cos x}{\sin^{2} x} is equal to$

$3$
$\frac{3}{2}$
$\frac{5}{4}$
$2$

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(2)

$\lim_{x \to 0} \frac{e^{x^{2}} - \cos x}{\sin^{2} x} = \lim_{x \to 0} \frac{2 x e^{x^{2}} + \sin x}{2 \sin x \cos x}$

$(Using L'Hospital Rule)$

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

Q 31 :

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 $24 \log_{e} (a + b)$ is equal to _______

(12)

$α = \lim_{x \to 0^{+}} \frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}} = 1, β = \lim_{x \to 0} {(1 + \sin x)}^{\frac{1}{2} \cot x} = e^{1 / 2}$

The equation having roots $1, \sqrt{e}$ is $x^{2} - (1 + \sqrt{e}) x + \sqrt{e} = 0$

It is given as $a x^{2} + b x - \sqrt{e} = 0$

By comparing we get $\frac{a}{1} = \frac{b}{- (1 + \sqrt{e})} = \frac{- 1}{1} \Rightarrow a = - 1, b = 1 + \sqrt{e}$

$∴$ $12 \log_{e} (a + b) = 12 \log_{e} (- 1 + 1 + \sqrt{e}) = 6$

Q 32 :

$\lim_{x \to 0} \frac{x \tan 4 x}{\sin^{2} x} is$ _______

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$\lim_{x \to 0} \frac{\frac{x \tan 4 x}{4 x} 4 x}{{(\frac{\sin x}{x})}^{2} x^{2}}$

Q 33 :

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

$\log_{e}^{2}$
$\log_{e}^{3}$
$\log_{e}^{4}$
$\log_{e}^{5}$

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$\lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{r + n}, \int_{0}^{1} \frac{1}{1 + x} d x$

Q 34 :

If $\lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{(\sqrt{x - 1} - 1) \log_{e} (x - 1)} = m$ , then a + b + m is equal to: [2026]

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4

(2)

We have, $\lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{(\sqrt{x - 1} - 1) \log_{e} (x - 1)} = m$

$\Rightarrow \lim_{x \to 2} \frac{(\sqrt{x - 1} - 1) \sin (x^{3} - 5 x^{2} + a x + b)}{[(x - 1) - 1] \log_{e} (1 + (x - 2))} = m$

$\Rightarrow 2 \lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{{(x - 2)}^{2}} = m$ ... (i)

Let $f (x) = x^{3} - 5 x^{2} + a x + b$

$\Rightarrow f' (x) = 3 x^{2} - 10 x + a$

For limit to be finite, $f (2) = 0, f' (2) = 0$

For $f (2) = 0 \Rightarrow 8 - 20 + 2 a + b, f' (2) = 0$

$\Rightarrow 2 a + b = 12$

For $f' (2) = 0 \Rightarrow 3 (4) - 10 (2) + a = 0$

$\Rightarrow a = 8 ∴ b = - 4$

$∴$ From (i),

$2 \lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + 8 x - 4)}{{(x - 2)}^{2}} = m$

$\Rightarrow 2 \lim_{x \to 2} \frac{\sin [{(x 2)}^{2} (x - 1)]}{{(x - 2)}^{2} (x - 1)} (x - 1) = m$

$\Rightarrow 2 [1] [2 - 1] = m \Rightarrow m = 2$

$∴ a + b + m = 8 - 4 + 2 = 6$

Q 35 :

The product of all possible values of $α$ , for which $\lim_{x \to 0} (\frac{1 - \cos (α x) \cos ((α + 1) x) \cos (α + 2) x)}{\sin^{2} ((α + 1) x)}) = 2$ is: [2026]

–2
1
–1
5/4

1

2

3

4

(3)

We have, $\lim_{x \to 0} \frac{1 - \cos (α x) \cos (α + 1) x \cos (α + 2) x}{(\frac{\sin^{2} (α + 1) x}{{(α + 1)}^{2} x^{2}}) {(α + 1)}^{2} x^{2}} = 2$

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

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

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

$\Rightarrow 3 α^{2} + 6 α + 5 = 4 α^{2} + 8 a + 4$

$\Rightarrow α^{2} + 2 α - 1 = 0$

$∴$ Products of roots = –1

Q 36 :

Let $f (x) = \lim_{y \to 0} \frac{(1 - \cos (x y)) \tan (x y)}{y^{3}}$ . Then the number of solutions of the equation f(x) = sin x, $x \in R$ is: [2026]

0
2
3
1

1

2

3

4

(3)

Given, $f (x) = \lim_{y \to 0} \frac{(1 - \cos (x y)) \tan (x y)}{y^{3}}$

$= \lim_{y \to 0} x^{2} \frac{(1 - \cos (x y))}{{(x y)}^{2}} \times \frac{\tan (x y)}{(x y)} \times x$

$\Rightarrow f (x) = \frac{x^{3}}{2}$

Now for solution f(x) = sin x

$∴$ Number of solutions = 3

Q 37 :

The value of $\lim_{x \to 0} (\frac{x^{2} \sin^{2} x}{x^{2} - \sin^{2} x})$ is: [2026]

2
3
4
6

1

2

3

4

(2)

We have, $\lim_{x \to 0} (\frac{x^{2} \sin^{2} x}{x^{2} - \sin^{2} x})$

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

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

$= \lim_{x \to 0} \frac{x^{3} {(1)}^{2}}{(\frac{x + \sin x}{x}) x^{3} (\frac{1}{6})}$

$= \lim_{x \to 0} \frac{1}{(2 \times \frac{1}{6})} = \frac{6}{2} = 3$

Q 38 :

Let $\lim_{x \to 2} \frac{(\tan (x - 2)) (r x^{2} + (p - 2) x - 2 p)}{{(x - 2)}^{2}} = 5$ for some $r, p \in R$ . If the set of all possible values of q, such that the roots of the equation in (0, 2), be the interval ( $α, β$ ], then 4( $α + β$ ) equals: [2026]

11
13
17
21

1

2

3

4

(3)

$\lim_{x \to 2} [\frac{\tan (x - 2)}{(x - 2)}] [\frac{r x^{2} + (p - 2) x - 2 p}{x - 2}] = 5$

$\Rightarrow \lim_{x \to 2} \frac{r x^{2} + (p - 2) x - 2 p}{x - 2} = 5$ $[∵ \lim_{x \to 2} [\frac{\tan (x - 2)}{(x - 2)} = 1]$

Since $D^{r} \to 0, N^{r} \to 0$

$\Rightarrow r = 1$

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

$\Rightarrow \lim_{x \to 2} \frac{x (x - 2) + p (x - 2)}{(x - 2)} = 5$

$\Rightarrow \lim_{x \to 2} (x + p) = 5$

$\Rightarrow 2 + p = 5 \Rightarrow p = 3$

Now given equation : $r x^{2} - p x + q = 0$

$\Rightarrow x^{2} - 3 x + q = 0$

$f (0) > 0 \Rightarrow q > 0, f (2) > 0 \Rightarrow q - 2 > 0 \Rightarrow q > 2$

And $D \geq 0 \Rightarrow 9 - 4 q \geq 0 \Rightarrow \frac{9}{4} \geq q$

$\Rightarrow q \in (2, \frac{9}{4}]$

Thus, $α = 2, β = \frac{9}{4}$

$∴ 4 (α + β) = 4 (2 + \frac{9}{4}) = 17$

Topic Question Set

Q 11 :

Among

$(S1) : \lim_{n \to \infty} \frac{1}{n^{2}} (2 + 4 + 6 + \dots + 2 n) = 1$

$(S2) : \lim_{n \to \infty} \frac{1}{n^{16}} (1^{15} + 2^{15} + 3^{15} + \dots + n^{15}) = \frac{1}{16}$ [2023]

Q 12 :

$If \lim_{x \to 0} \frac{e^{a x} - \cos (b x) - \frac{c x e^{- c x}}{2}}{1 - \cos (2 x)} = 17, then 5 a^{2} + b^{2} is equal to$ [2023]

Q 13 :

$\lim_{t \to 0} {(1^{\frac{1}{^{\sin^{2} t}}} + 2^{\frac{1}{^{\sin^{2} t}}} + \dots + n^{\frac{1}{^{\sin^{2} t}}})}^{\sin^{2} t}$ is equal to [2023]

Q 14 :

$The set of all values of a for which \lim_{x \to a} ([x - 5] - [2 x + 2]) = 0,$ where $[α]$ denotes the greatest integer less than or equal to $α$ is equal to [2023]

Q 15 :

$The value of \lim_{x \to \infty} \frac{1 + 2 - 3 + 4 + 5 - 6 + \dots + (3 n - 2) + (3 n - 1) - 3 n}{\sqrt{2 n^{4} + 4 n + 3} - \sqrt{n^{4} + 5 n + 4}} is$ [2023]

Q 18 :

$\lim_{x \to \infty} \frac{{(\sqrt{3 x + 1} + \sqrt{3 x - 1})}^{6} + {(\sqrt{3 x + 1} - \sqrt{3 x - 1})}^{6}}{{(x + \sqrt{x^{2} - 1})}^{6} + {(x - \sqrt{x^{2} - 1})}^{6}} x^{3}$ [2023]

Q 19 :

Let $f : ℝ \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f' (3) = 0$ and $f'' (3) = 4$ . Then $\lim_{x \to 1} (l {og}_{e} {(\frac{f (2 + x)}{f (3)})}^{\frac{18}{{(x - 1)}^{2}}})$ is equal to [2026]

Q 20 :

Let $f (x) = \lim_{θ \to 0} (\frac{\cos (π x) - x^{(\frac{2}{θ})} \sin (x - 1)}{1 + x^{(\frac{2}{θ})} (x - 1)}), x \in ℝ .$

Consider the following two statements:

I. f(x) is discontinuous at x = 1

II. f(x) is continuous at x = -1

Then, [2026]

Q 21 :

If $\lim_{x \to 0} \frac{e^{(a - 1) x} + 2 \cos b x + (c - 2) e^{- x}}{x \cos x - \log_{e} (1 + x)} = 2,$ then $a^{2} + b^{2} + c^{2}$ is equal to: [2026]

Q 22 :

The value of $\lim_{x \to 0} \frac{\log_{e} ((\sec (e x) \cdot \sec ((e^{2} x)) \dots \sec ((e^{10} x))))}{e^{2} - e^{2 \cos x}}$ is equal to [2026]

Q 23 :

$\lim_{x \to \infty} \frac{3}{x} [\frac{x}{4}] = \frac{p}{q}$ (where $[.]$ denotes greatest integer function), then $p + q$ (where $p, q$ are relative prime) is:

Q 24 :

If $f (x) = \lim_{n \to \infty} (\prod_{i = 1}^{n} \cos (\frac{x}{2^{i}}))$ then $f' (x)$ is equal to:

Q 25 :

$\lim_{n \to \infty} \frac{\sum_{k = 1}^{n - 1} (k - 1) (n k - k^{2})}{2 \sum_{r = 1}^{n} r^{3} - \sum_{s = 0}^{n} (s^{2} + {(n - s)}^{2})} = t then$ $[43 t]$ $=____$ $([.] denotes GIF)$

Q 26 :

The value of $\lim_{n \to \infty} \frac{(\lim_{x \to 0^{-}} \sum_{r = 1}^{2 n + 1} [x^{r}]) + n + 3}{ℓ n {(1 + n)}^{n} - ℓ n {(n)}^{n}}$ , where $[\cdot]$ denotes greatest integer function

Q 28 :

If $f (x) =$ $| \begin{matrix} \cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix} |$ , then $\lim_{x \to 0} \frac{f' (x)}{x}$ is equal to

Q 29 :

$\lim_{x \to 0^{+}} \frac{\tan x \sqrt{\tan x} - \sin x \sqrt{\sin x}}{x^{3} \sqrt{x}} Equals$

Q 30 :

$\lim_{x \to 0} \frac{e^{x^{2}} - \cos x}{\sin^{2} x} is equal to$

Q 31 :

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 $24 \log_{e} (a + b)$ is equal to _______

Q 32 :

$\lim_{x \to 0} \frac{x \tan 4 x}{\sin^{2} x} is$ _______

Q 33 :

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

Q 34 :

If $\lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{(\sqrt{x - 1} - 1) \log_{e} (x - 1)} = m$ , then a + b + m is equal to: [2026]

Q 35 :

The product of all possible values of $α$ , for which $\lim_{x \to 0} (\frac{1 - \cos (α x) \cos ((α + 1) x) \cos (α + 2) x)}{\sin^{2} ((α + 1) x)}) = 2$ is: [2026]

Q 36 :

Let $f (x) = \lim_{y \to 0} \frac{(1 - \cos (x y)) \tan (x y)}{y^{3}}$ . Then the number of solutions of the equation f(x) = sin x, $x \in R$ is: [2026]

Q 37 :

The value of $\lim_{x \to 0} (\frac{x^{2} \sin^{2} x}{x^{2} - \sin^{2} x})$ is: [2026]

Q 38 :

Let $\lim_{x \to 2} \frac{(\tan (x - 2)) (r x^{2} + (p - 2) x - 2 p)}{{(x - 2)}^{2}} = 5$ for some $r, p \in R$ . If the set of all possible values of q, such that the roots of the equation in (0, 2), be the interval ( $α, β$ ], then 4( $α + β$ ) equals: [2026]

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Topic Question Set

Q 11 : Among (S1):limn→∞1n2(2+4+6+…+2n)=1 (S2):limn→∞1n16(115+215+315+…+n15)=116 [2023]

Q 12 : If limx→0 eax-cos(bx)-cxe-cx21-cos(2x) =17, then 5a2+b2 is equal to [2023]

Q 13 : limt→0(11sin2t+21sin2t+⋯+n1sin2t)sin2t is equal to [2023]

Q 14 : The set of all values of a for which limx→a[x-5]-[2x+2]=0, where [α] denotes the greatest integer less than or equal to α is equal to [2023]

Q 15 : The value of limx→∞1+2-3+4+5-6+…+(3n-2)+(3n-1)-3n2n4+4n+3-n4+5n+4 is [2023]

Q 16 : Let x=2 be a root of the equation x2+px+q=0 and f(x)={1-cos(x2-4px+q2+8q+16)(x-2p)4,x≠2p0,x=2p Then limx→2p+[f(x)], where [·] denotes the greatest integer function, is [2023]

Q 17 : Let f,g and h be real-valued functions defined on R f(x)={x|x|,x≠01,x=0, g(x)={sin(x+1)(x+1),x≠-11,x=-1 and h(x)=2[x]-f(x), where [x] is the greatest integer ≤x. Then the value of limx→1g(h(x-1)) is [2023]

Q 18 : limx→∞(3x+1+3x-1)6+(3x+1-3x-1)6(x+x2-1)6+(x-x2-1)6x3 [2023]

Q 19 : Let f:ℝ→(0,∞) be a twice differentiable function such that f(3)=18, f'(3)=0 and f''(3)=4. Then limx→1(loge(f(2+x)f(3))18(x-1)2) is equal to [2026]

Q 20 : Let f(x)=limθ→0(cos(πx)-x(2θ)sin(x-1)1+x(2θ)(x-1)), x∈ℝ. Consider the following two statements: I. f(x) is discontinuous at x = 1 II. f(x) is continuous at x = -1 Then, [2026]

Q 21 : If limx→0e(a-1)x+2cosbx+(c-2)e-xxcosx-loge(1+x)=2, then a2+b2+c2 is equal to: [2026]

Q 22 : The value of limx→0loge(sec(ex)·sec (e2x)⋯sec (e10x))e2-e2cosx is equal to [2026]

Q 23 : limx→∞3x[x4]=pq (where [.] denotes greatest integer function), then p+q (where p,q are relative prime) is:

Q 24 : If f(x)=limn→∞(∏i=1ncos(x2i)) then f'(x) is equal to:

Q 25 : limn→∞∑k=1n-1(k-1)(nk-k2)2∑r=1nr3-∑s=0n(s2+(n-s)2)=t then [43t]=____ ([.] denotes GIF)

Q 26 : The value of limn→∞(limx→0-∑r=12n+1[xr])+n+3ℓn(1+n)n-ℓn(n)n, where [·] denotes greatest integer function

Q 27 : Consider the following Statements Statement-I: If both functions f(x) and g(x) have limits at x=a, then limx→a[f(g(x))]=[limx→af(g(x))] [ ] denotes G.I.F Statement-2: A function may fail to have a limit at a point in its domain but functional value at x=a exists (i.e. f(a) exists).

Q 28 : If f(x)=|cosxx12sinxx22xtanxx1|, then limx→0f'(x)x is equal to

Q 29 : limx→0+tanxtanx-sinxsinxx3x Equals

Q 30 : limx→0ex2-cosxsin2x is equal to

Q 31 : 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 24loge(a+b) is equal to _______

Q 32 : limx→0x tan4xsin2x is_______

Q 33 : limn→∞(11+n+12+n+13+n+⋯+1n+n)=

Q 34 : If limx→2sin(x3–5x2+ax+b)(x–1–1)loge(x–1)=m, then a + b + m is equal to: [2026]

Q 35 : The product of all possible values of α, for which limx→0(1–cos(αx)cos((α+1)x)cos(α+2)x)sin2((α+1)x))=2 is: [2026]

Q 36 : Let f(x)=limy→0(1–cos(xy))tan(xy)y3. Then the number of solutions of the equation f(x) = sin x, x∈R is: [2026]

Q 37 : The value of limx→0(x2sin2xx2–sin2x) is: [2026]

Q 38 : Let limx→2(tan(x–2))(rx2+(p–2)x–2p)(x–2)2=5 for some r, p∈R. If the set of all possible values of q, such that the roots of the equation in (0, 2), be the interval (α,β], then 4(α+β) equals: [2026]

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Q 11 :

Among

$(S1) : \lim_{n \to \infty} \frac{1}{n^{2}} (2 + 4 + 6 + \dots + 2 n) = 1$

$(S2) : \lim_{n \to \infty} \frac{1}{n^{16}} (1^{15} + 2^{15} + 3^{15} + \dots + n^{15}) = \frac{1}{16}$ [2023]

Q 12 :

$If \lim_{x \to 0} \frac{e^{a x} - \cos (b x) - \frac{c x e^{- c x}}{2}}{1 - \cos (2 x)} = 17, then 5 a^{2} + b^{2} is equal to$ [2023]

Q 13 :

$\lim_{t \to 0} {(1^{\frac{1}{^{\sin^{2} t}}} + 2^{\frac{1}{^{\sin^{2} t}}} + \dots + n^{\frac{1}{^{\sin^{2} t}}})}^{\sin^{2} t}$ is equal to [2023]

Q 14 :

$The set of all values of a for which \lim_{x \to a} ([x - 5] - [2 x + 2]) = 0,$ where $[α]$ denotes the greatest integer less than or equal to $α$ is equal to [2023]

Q 15 :

$The value of \lim_{x \to \infty} \frac{1 + 2 - 3 + 4 + 5 - 6 + \dots + (3 n - 2) + (3 n - 1) - 3 n}{\sqrt{2 n^{4} + 4 n + 3} - \sqrt{n^{4} + 5 n + 4}} is$ [2023]

Q 18 :

$\lim_{x \to \infty} \frac{{(\sqrt{3 x + 1} + \sqrt{3 x - 1})}^{6} + {(\sqrt{3 x + 1} - \sqrt{3 x - 1})}^{6}}{{(x + \sqrt{x^{2} - 1})}^{6} + {(x - \sqrt{x^{2} - 1})}^{6}} x^{3}$ [2023]

Q 19 :

Let $f : ℝ \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f' (3) = 0$ and $f'' (3) = 4$ . Then $\lim_{x \to 1} (l {og}_{e} {(\frac{f (2 + x)}{f (3)})}^{\frac{18}{{(x - 1)}^{2}}})$ is equal to [2026]

Q 20 :

Let $f (x) = \lim_{θ \to 0} (\frac{\cos (π x) - x^{(\frac{2}{θ})} \sin (x - 1)}{1 + x^{(\frac{2}{θ})} (x - 1)}), x \in ℝ .$

Consider the following two statements:

I. f(x) is discontinuous at x = 1

II. f(x) is continuous at x = -1

Then, [2026]

Q 21 :

If $\lim_{x \to 0} \frac{e^{(a - 1) x} + 2 \cos b x + (c - 2) e^{- x}}{x \cos x - \log_{e} (1 + x)} = 2,$ then $a^{2} + b^{2} + c^{2}$ is equal to: [2026]

Q 22 :

The value of $\lim_{x \to 0} \frac{\log_{e} ((\sec (e x) \cdot \sec ((e^{2} x)) \dots \sec ((e^{10} x))))}{e^{2} - e^{2 \cos x}}$ is equal to [2026]

Q 23 :

$\lim_{x \to \infty} \frac{3}{x} [\frac{x}{4}] = \frac{p}{q}$ (where $[.]$ denotes greatest integer function), then $p + q$ (where $p, q$ are relative prime) is:

Q 24 :

If $f (x) = \lim_{n \to \infty} (\prod_{i = 1}^{n} \cos (\frac{x}{2^{i}}))$ then $f' (x)$ is equal to:

Q 25 :

$\lim_{n \to \infty} \frac{\sum_{k = 1}^{n - 1} (k - 1) (n k - k^{2})}{2 \sum_{r = 1}^{n} r^{3} - \sum_{s = 0}^{n} (s^{2} + {(n - s)}^{2})} = t then$ $[43 t]$ $=____$ $([.] denotes GIF)$

Q 26 :

The value of $\lim_{n \to \infty} \frac{(\lim_{x \to 0^{-}} \sum_{r = 1}^{2 n + 1} [x^{r}]) + n + 3}{ℓ n {(1 + n)}^{n} - ℓ n {(n)}^{n}}$ , where $[\cdot]$ denotes greatest integer function

Q 28 :

If $f (x) =$ $| \begin{matrix} \cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{matrix} |$ , then $\lim_{x \to 0} \frac{f' (x)}{x}$ is equal to

Q 29 :

$\lim_{x \to 0^{+}} \frac{\tan x \sqrt{\tan x} - \sin x \sqrt{\sin x}}{x^{3} \sqrt{x}} Equals$

Q 30 :

$\lim_{x \to 0} \frac{e^{x^{2}} - \cos x}{\sin^{2} x} is equal to$

Q 31 :

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 $24 \log_{e} (a + b)$ is equal to _______

Q 32 :

$\lim_{x \to 0} \frac{x \tan 4 x}{\sin^{2} x} is$ _______

Q 33 :

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

Q 34 :

If $\lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{(\sqrt{x - 1} - 1) \log_{e} (x - 1)} = m$ , then a + b + m is equal to: [2026]

Q 35 :

The product of all possible values of $α$ , for which $\lim_{x \to 0} (\frac{1 - \cos (α x) \cos ((α + 1) x) \cos (α + 2) x)}{\sin^{2} ((α + 1) x)}) = 2$ is: [2026]

Q 36 :

Let $f (x) = \lim_{y \to 0} \frac{(1 - \cos (x y)) \tan (x y)}{y^{3}}$ . Then the number of solutions of the equation f(x) = sin x, $x \in R$ is: [2026]

Q 37 :

The value of $\lim_{x \to 0} (\frac{x^{2} \sin^{2} x}{x^{2} - \sin^{2} x})$ is: [2026]

Q 38 :

Let $\lim_{x \to 2} \frac{(\tan (x - 2)) (r x^{2} + (p - 2) x - 2 p)}{{(x - 2)}^{2}} = 5$ for some $r, p \in R$ . If the set of all possible values of q, such that the roots of the equation in (0, 2), be the interval ( $α, β$ ], then 4( $α + β$ ) equals: [2026]