Integral Calculus jee mains questions | JEE Main Maths Integral Calculus Previous Year Question

Q 31 :

Let $f (x) = \frac{x}{{(1 + x^{n})}^{\frac{1}{n}}}, x \in ℝ - {- 1}, n \in ℕ, n > 2 .$ If $f^{n} (x) = (f o f o f . . . . . upto n times)$ $(x),$ then $\lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f^{n} (x)) d x$ is equal to _____ . [2023]

(0)

We have, $f (x) = \frac{x}{{(1 + x^{n})}^{1 / n}}$

$\Rightarrow f (f (x)) = \frac{x}{{(1 + 2 x^{n})}^{1 / n}}$

$f (f (f (x))) = \frac{x}{{(1 + 3 x^{n})}^{1 / n}}$

$\Rightarrow f^{n} (x) = \frac{x}{{(1 + n x^{n})}^{1 / n}}$

Now, $\lim_{n \to \infty} \int_{0}^{1} x^{n - 2} \cdot (f^{n} (x)) d x = \lim_{n \to \infty} \int_{0}^{1} \frac{x^{n - 1}}{{(1 + n x^{n})}^{1 / n}} d x$

Put $1 + n x^{n} = t \Rightarrow n^{2} x^{n - 1} d x = d t$

When $x = 0, t = 1$ and $x = 1, t = 1 + n$

$∴$ $\lim_{n \to \infty} \cdot \frac{1}{n^{2}} \int_{1}^{1 + n} \frac{d t}{t^{1 / n}} = \lim_{n \to \infty} \frac{1}{n^{2}} {[\frac{t^{1 - 1 / n}}{1 - 1 / n}]}_{1}^{1 + n}$

$= \lim_{n \to \infty} \frac{1}{n (n - 1)} [{(1 + n)}^{1 - 1 / n} - 1]$

$= \lim_{h \to 0} \frac{h^{2}}{1 - h} [{(1 + \frac{1}{h})}^{1 - h} - 1] (Put n = \frac{1}{h}, n \to \infty, h \to 0)$

$= 0$

Q 32 :

Let $[t]$ denote the greatest integer $\leq f$ . Then $\frac{2}{π} \int_{π / 6}^{5 π / 6} (8 [c o s e c x] - 5 [\cot x]) d x$ is equal to ______ . [2023]

(14)

$Let I = \frac{2}{π} \int_{π / 6}^{5 π / 6} (8 [\cos e c x] - 5 [\cot x]) d x$

$= \frac{2}{π} [8 \int_{π / 6}^{5 π / 6} [\cos e c x] d x - 5 \int_{π / 6}^{5 π / 6} [\cot x] d x]$

$= \frac{2}{π} [8 {\int_{π / 6}^{5 π / 6} 1 \cdot d x} - 5 {\int_{π / 6}^{π / 4} 1 \cdot d x + \int_{π / 4}^{π / 2} 0 \cdot d x + \int_{π / 2}^{3 π / 4} (- 1) d x + \int_{3 π / 4}^{5 π / 6} (- 2) d x}]$

$= \frac{2}{π} [8 \times \frac{2 π}{3} - 5 {\frac{π}{12} - \frac{π}{4} - 2 \times \frac{π}{12}}] = \frac{2}{π} [\frac{16 π}{3} + \frac{5 π}{3}] = 14$

Q 33 :

Let $[t]$ denote the greatest integer function. If $\int_{0}^{2.4} [x^{2}] d x = α + β \sqrt{2} + γ \sqrt{3} + δ \sqrt{5},$ then $α + β + γ + δ$ is equal to _________ . [2023]

(6)

We have,

$\int_{0}^{2.4} [x^{2}] d x = \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} 1 d x + \int_{\sqrt{2}}^{\sqrt{3}} 2 d x + \int_{\sqrt{3}}^{2} 3 d x + \int_{2}^{\sqrt{5}} 4 d x + \int_{\sqrt{5}}^{2.4} 5 d x$

$= \sqrt{2} - 1 + 2 (\sqrt{3} - \sqrt{2}) + 3 (2 - \sqrt{3}) + 4 (\sqrt{5} - 2) + 5 (2.4 - \sqrt{5})$

$= 9 - \sqrt{2} - \sqrt{3} - \sqrt{5}$

$∴$ $α + β + γ + δ = 9 - 1 - 1 - 1 = 6$

Q 34 :

For $m, n > 0$ , let $α (m, n) = \int_{0}^{2} t^{m} {(1 + 3 t)}^{n} d t$ . If $11 α (10, 6) + 18 α (11, 5) = p {(14)}^{6}$ , then $p$ is equal to ______ . [2023]

(32)

We have, $α (m, n) = \int_{0}^{2} t^{m} {(1 + 3 t)}^{n} d t$

Now, $11 α (10, 6) + 18 α (11, 5)$

$= 11 \int_{0}^{2} t^{10} {(1 + 3 t)}^{6} d t + 18 \int_{0}^{2} t^{11} {(1 + 3 t)}^{5} d t$

$= 11 {[{(1 + 3 t)}^{6} \cdot \frac{t^{11}}{11}]}_{0}^{2} - 11 \int_{0}^{2} 6 {(1 + 3 t)}^{5} \cdot$ $3 \frac{t^{11}}{11} d t + 18 \int_{0}^{2} t^{11} {(1 + 3 t)}^{5} d t$

$= {[t^{11} {(1 + 3 t)}^{6}]}_{0}^{2}$ $= 2^{11} {(7)}^{6} = 2^{5} {(14)}^{6}$

$= 32 {(14)}^{6} = p {(14)}^{6}$ $[∵ Given]$

$\Rightarrow p = 32$

Q 35 :

If $\int_{- 0.15}^{0.15}$ $| 100 x^{2} - 1 |$ $d x = \frac{k}{3000}$ , then $k$ is equal to ___________. [2023]

(575)

Let $I = \int_{- 0.15}^{0.15}$ $| 100 x^{2} - 1 |$ $d x$

$= 2 \int_{0}^{0.15}$ $| 100 x^{2} - 1 |$ $d x$

$(∵ \int_{- a}^{a} f (x) d x = {\begin{matrix} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) i.e. f (x) is even \\ 0, & if f (- x) = - f (x) i.e. f (x) is odd \end{matrix})$

$= 2 {\int_{0}^{0.1} - (100 x^{2} - 1) d x + \int_{0.1}^{0.15} (100 x^{2} - 1) d x}$

$= 2 {{[x - \frac{100 x^{3}}{3}]}_{0}^{0.1} + {[\frac{100 x^{3}}{3} - x]}_{0.1}^{0.15}}$ $= \frac{575}{3000}$

$∴$ $k = 575$

Q 36 :

Let for $x \in ℝ, S_{0} (x) = x, S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t,$ where $C_{0} = 1, C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x, k = 1, 2, 3, \dots$ Then $S_{2} (3) + 6 C_{3}$ is equal to _______ . [2023]

(18)

Given, $S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t ...(i)$

Put $k$ = 2 and $x$ = 3 in (i),

$S_{2} (3) = C_{2} (3) + 2 \int_{0}^{3} S_{1} (t) d t ...(ii)$

Put $k$ = 1 in (i),

$S_{1} (x) = C_{1} x + \int_{0}^{x} S_{0} (t) d t = C_{1} x + \int_{0}^{x} t d t [∵ S_{0} (x) = x]$

$= C_{1} x + \frac{x^{2}}{2}$

Put $S_{1} (t) = C_{1} t + \frac{t^{2}}{2}$ in (ii),

$S_{2} (3) = C_{2} (3) + 2 \int_{0}^{3} (C_{1} t + \frac{t^{2}}{2}) d t$

$= 3 C_{2} + 2 {[C_{1} \frac{t^{2}}{2} + \frac{t^{3}}{6}]}_{0}^{3} = 3 C_{2} + 2 [C_{1} (\frac{9}{2}) + (\frac{27}{6})]$

$= 3 C_{2} + 9 C_{1} + 9$

Given, $C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x$ $...(iii)$

Put $k$ = 1 in (iii),

$C_{1} = 1 - \int_{0}^{1} S_{0} (x) d x = 1 - \int_{0}^{1} x d x = 1 - {(\frac{x^{2}}{2})}_{0}^{1} = \frac{1}{2}$

Put $k$ = 2 in (iii),

$C_{2} = 1 - \int_{0}^{1} S_{1} (x) d x = 1 - \int_{0}^{1} (C_{1} x + \frac{x^{2}}{2}) d x$

$= 1 - {[C_{1} \frac{x^{2}}{2} + \frac{x^{3}}{6}]}_{0}^{1} = 1 - [\frac{1}{2} \times \frac{1}{2} + \frac{1}{6}] = 1 - [\frac{1}{4} + \frac{1}{6}] = 1 - \frac{5}{12} = \frac{7}{12}$

Put $k$ = 3 in (iii), $C_{3} = 1 - \int_{0}^{1} S_{2} (x) d x$

Put $k$ = 2 in (i), $S_{2} (x) = C_{2} x + 2 \int_{0}^{x} S_{1} (t) d t = C_{2} x + C_{1} x^{2} + \frac{x^{3}}{3}$

So, $C_{3} = 1 - \int_{0}^{1} (C_{2} x + C_{1} x^{2} + \frac{x^{3}}{3}) d x$

$= 1 - {[C_{2} \frac{x^{2}}{2} + C_{1} \frac{x^{3}}{3} + \frac{x^{4}}{12}]}_{0}^{1}$

$= 1 - \frac{7}{12} \times \frac{1}{2} - \frac{1}{2} \times \frac{1}{3} - \frac{1}{12} = 1 - \frac{7}{24} - \frac{1}{6} - \frac{1}{12}$

$= \frac{24 - 7 - 4 - 2}{24} = \frac{11}{24}$

$S_{3} (3) + 6 C_{3} = 3 C_{2} + 9 C_{1} + 9 + 6 C_{3}$

$= 3 \times \frac{7}{12} + 9 \times \frac{1}{2} + 9 + 6 \times \frac{11}{24} = \frac{7}{4} + \frac{9}{2} + 9 + \frac{11}{4}$

$= \frac{7 + 18 + 36 + 11}{4} = \frac{72}{4} = 18$

Q 37 :

Let $f_{n} = \int_{0}^{\frac{π}{2}} (\sum_{k = 1}^{n} \sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) \sin^{k - 1} x) \cos x d x, n \in ℕ .$ Then $f_{21} - f_{20}$ is equal to ________. [2023]

(41)

Let $f_{n} = \int_{0}^{π / 2} (\sum_{k = 1}^{n} \sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) \sin^{k - 1} x) \cos x d x, x \in ℕ$

$f_{n} (x) = \int_{0}^{π / 2} (1 + \sin x + \sin^{2} x + \dots + \sin^{n - 1} x) (1 + 3 \sin x + 5 \sin^{2} x + \dots + (2 n - 1) \sin^{n - 1} x) \cdot \cos x d x$

Multiply and divide by $\sqrt{\sin x}$ , we get:

$\int_{0}^{π / 2} [{(\sin x)}^{1 / 2} + {(\sin x)}^{3 / 2} + {(\sin x)}^{5 / 2} + \dots + {(\sin x)}^{\frac{2 n - 1}{2}}] (1 + 3 \sin x + 5 \sin^{2} x + \dots + (2 n - 1) \sin^{n - 1} x) \frac{\cos x}{\sqrt{\sin x}} d x$

Put ${(\sin x)}^{1 / 2} + {(\sin x)}^{3 / 2} + {(\sin x)}^{5 / 2} + \dots + {(\sin x)}^{n - \frac{1}{2}} = t$

$\Rightarrow \frac{1}{2} \frac{(1 + 3 \sin x + 5 \sin^{2} x + \dots + (2 n - 1) \sin^{n - 1} x)}{\sqrt{\sin x}} \cos x d x = d t$

$\Rightarrow f_{n} = 2 \int_{0}^{n} t d t \Rightarrow f_{n} = n^{2}$

$∴$ $f_{21} - f_{20} = {(21)}^{2} - {(20)}^{2} = 441 - 400 = 41$

Q 38 :

$If \int_{0}^{1} (x^{21} + x^{14} + x^{7}) {(2 x^{14} + 3 x^{7} + 6)}^{1 / 7} d x = \frac{1}{l} {(11)}^{m}$ $where l, m, n \in ℕ, m and n are coprime, then l + m + n is equal to$ _______ . [2023]

(63)

$\int_{0}^{1} (x^{21} + x^{14} + x^{7}) {(2 x^{14} + 3 x^{7} + 6)}^{1 / 7} d x = \frac{1}{l} {(11)}^{m / n}, l, m, n \in ℕ$

L.H.S. $= \int_{0}^{1} x (x^{20} + x^{13} + x^{6}) {(2 x^{14} + 3 x^{7} + 6)}^{1 / 7} d x$

$= \int_{0}^{1} (x^{20} + x^{13} + x^{6}) {(2 x^{21} + 3 x^{14} + 6 x^{7})}^{1 / 7} d x$

Let $2 x^{21} + 3 x^{14} + 6 x^{7} = t^{7} \Rightarrow 42 (x^{20} + x^{13} + x^{6}) d x = 7 t^{6} d t$

$\Rightarrow (x^{20} + x^{13} + x^{6}) d x = \frac{t^{6}}{6} d t,$ When $x$ = 0, $t$ = 0 and when $x$ = 1,

$t = 11^{1 / 7}$ $\Rightarrow L.H.S. = \int_{0}^{11^{1 / 7}} (\frac{t^{6}}{6} \cdot t) d t$

$= \frac{1}{6} \int_{0}^{11^{1 / 7}} t^{7} d t$ $= \frac{1}{6} {[\frac{t^{8}}{8}]}_{0}^{{(11)}^{1 / 7}} = \frac{11^{8 / 7}}{48}$

Comparing with R.H.S., we get $l = 48, m = 8, n = 7$

$∴$ $l + m + n = 48 + 8 + 7 = 63$

Q 39 :

$The value of \frac{8}{π} \int_{0}^{π / 2} \frac{{(\cos x)}^{2023}}{{(\sin x)}^{2023} + {(\cos x)}^{2023}} d x is$ _______ . [2023]

(2)

$Use \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x to solve.$

Q 40 :

$The value of 12 \int_{0}^{3}$ $| x^{2} - 3 x + 2 |$ $d x is$ ________ . [2023]

(22)

$Let I = 12 \int_{0}^{3}$ $| x^{2} - 3 x + 2 |$ $d x$

$= 12 \int_{0}^{3}$ $| (x - 2) (x - 1) |$ $d x$

$= 12 \times [\int_{0}^{1} (x^{2} - 3 x + 2) d x - \int_{1}^{2} (x^{2} - 3 x + 2) d x + \int_{2}^{3} (x^{2} - 3 x + 2) d x]$

$= 12 \times [{(\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x)}_{0}^{1} + {(\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x)}_{2}^{3} - {(\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x)}_{1}^{2}] = 22$

Q 41 :

If $\int_{0}^{π} \frac{5^{\cos x} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x) d x}{1 + 5^{\cos x}} = \frac{k π}{16},$ then $k$ is equal to ______. [2023]

(13)

$Let$

$I = \int_{0}^{π} \frac{5^{\cos x} (1 + \cos x \cdot \cos 3 x + \cos^{2} x + \cos^{3} x \cdot \cos 3 x)}{1 + 5^{\cos x}} d x$ ...(i)

$I = \int_{0}^{π} \frac{5^{\cos (π - x)} (1 + \cos (π - x) \cdot \cos 3 (π - x) + \cos^{2} (π - x) + \cos^{3} (π - x) \cdot \cos 3 (π - x))}{1 + 5^{\cos (π - x)}} d x$

$\Rightarrow I = \int_{0}^{π} \frac{5^{- \cos x} (1 + \cos x \cdot \cos 3 x + \cos^{2} x + \cos^{3} x \cdot \cos 3 x) d x}{1 + 5^{- \cos x}}$ ...(ii)

$Adding (i) and (ii), we get$

$2 I = \int_{0}^{π} (1 + \cos x \cdot \cos 3 x + \cos^{2} x + \cos^{3} x \cdot \cos 3 x) d x$

$\Rightarrow 2 I = 2 \int_{0}^{π / 2} (1 + \cos x \cdot \cos 3 x + \cos^{2} x + \cos^{3} x \cdot \cos 3 x) d x$ ...(iii)

$\Rightarrow I = \int_{0}^{π / 2} (1 + \sin x (- \sin 3 x) + \sin^{2} x - \sin^{3} x \cdot \sin 3 x) d x$ ...(iv)

$Adding (iii) and (iv), we get$

$\Rightarrow 2 I = \int_{0}^{π / 2} (3 + \cos 4 x + \cos^{3} x \cdot \cos 3 x - \sin^{3} x \cdot \sin 3 x) d x$

$2 I = \int_{0}^{π / 2} 3 + \cos 4 x + (\frac{\cos 3 x + 3 \cos x}{4}) \cos 3 x - \sin 3 x (\frac{3 \sin x - \sin 3 x}{4}) d x$

$\Rightarrow 2 I = \int_{0}^{π / 2} (3 + \cos 4 x + \frac{1}{4} + \frac{3}{4} \cos 4 x) d x$

$\Rightarrow 2 I = \frac{13}{4} \times \frac{π}{2} + \frac{7}{4} {(\frac{\sin 4 x}{4})}_{0}^{π / 2} \Rightarrow I = \frac{13 π}{16}$

$Hence, k = 13$

Q 42 :

$If \int_{1 / 3}^{3}$ $| \log_{e} x |$ $d x = \frac{m}{n} \log_{e} (\frac{n^{2}}{e}), where m and n are coprime natural numbers, then m^{2} + n^{2} - 5 is equal to$ ______ . [2023]

(20)

$Let I = \int_{1 / 3}^{3}$ $| \log_{e} x |$ $d x = \int_{1 / 3}^{1} (- \log_{e} x) d x + \int_{1}^{3} (\log_{e} x) d x$

$= - {[x \log_{e} x - x]}_{1 / 3}^{1} + {[x \log_{e} x - x]}_{1}^{3}$

$= - [- 1 - (\frac{1}{3} \log_{e} \frac{1}{3} - \frac{1}{3})] + [3 \log_{e} 3 - 3 - (- 1)]$

$= - \frac{4}{3} + \frac{8}{3} \log_{e} 3 = \frac{4}{3} (2 \log_{e} 3 - 1) = \frac{4}{3} (\log_{e} \frac{9}{e})$

$Comparing with the given condition, we get m = 4, n = 3 .$

$Now, m^{2} + n^{2} - 5 = 16 + 9 - 5 = 20$

Q 43 :

$\lim_{x \to 0} \frac{48}{x^{4}} \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t is equal to$ ______. [2023]

(12)

Let $I = \lim_{x \to 0} \frac{48 \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t}{x^{4}}$ $(\frac{0}{0} form)$

$I = \lim_{x \to 0} \frac{48 (\frac{x^{3}}{x^{6} + 1})}{4 x^{3}} = \lim_{x \to 0} \frac{48 \cdot x^{3}}{x^{6} + 1} \times \frac{1}{4 x^{3}} = \lim_{x \to 0} \frac{12}{x^{6} + 1}$

$I = 12$

Q 44 :

The value of the integral $\int_{\frac{π}{24}}^{\frac{5 π}{24}} \frac{d x}{1 + \sqrt[3]{\tan 2 x}}$ is: [2026]

$\frac{π}{3}$
$\frac{π}{18}$
$\frac{π}{12}$
$\frac{π}{6}$

1

2

3

4

(3)

$I = \int_{\frac{π}{24}}^{\frac{5 π}{24}} \frac{d x}{1 + \sqrt[3]{\tan 2 x}} . . . (1)$

Apply King

$I = \int_{\frac{π}{24}}^{\frac{5 π}{24}} \frac{d x}{1 + \sqrt[3]{\tan 2 (\frac{π}{4} - x)}}$

$I = \int_{\frac{π}{24}}^{\frac{5 π}{24}} \frac{d x}{1 + \sqrt[3]{\cot 2 x}} . . . (2)$

Add (1) + (2) $\Rightarrow 2 I = \int_{\frac{π}{24}}^{\frac{5 π}{24}} (1) d x$

$I = \frac{1}{2} (\frac{π}{6}) = \frac{π}{12}$

Q 45 :

Let $f$ be a twice differentiable non-negative function such that $(f {(x))}^{2} = 25 + \int_{0}^{x} ((f {(t))}^{2} + (f' {(t))}^{2}) d t .$ Then the mean of $f (\log_{e} (1)), f (\log_{e} (2)), \dots, f (\log_{e} (625))$ is equal to _________ . [2026]

(1565)

$2 f (x) f' (x) = f^{2} (x) + (f' {(x))}^{2}$

$\Rightarrow (f (x) - f' {(x))}^{2} = 0$

$\Rightarrow f (x) = f' (x)$

$\Rightarrow l n (f (x)) = x + c \Rightarrow f (x) = c' e^{x}$

$f (0) = 5 \Rightarrow f (x) = 5 e^{x}$

$Mean = \frac{f (l n 1) + f (l n 2) + \dots + f (l n 625)}{625}$

$= \frac{5 (1 + 2 + \dots + 625)}{625} = 1565$

Q 46 :

The value of $\int_{- π / 6}^{π / 6} (\frac{π + 4 x^{11}}{1 - \sin (| x | + π / 6)}) d x$ is equal to: [2026]

$8 π$
$2 π$
$6 π$
$4 π$

1

2

3

4

(4)

$= 2 π \int_{0}^{π / 6} \frac{1}{1 - \sin (x + \frac{π}{6})} d x let x + \frac{π}{6} = t \Rightarrow d x = d t$

$= 2 π \int_{π / 6}^{π / 3} \frac{d t}{1 - \sin t} = 2 π \int_{π / 6}^{π / 3} \frac{1 + \sin t}{\cos^{2} t} d t$

$= 2 π [\int_{π / 6}^{π / 3} \sec^{2} t d t + \int_{π / 6}^{π / 3} \sec t \tan t d t]$

$= 2 π [{(\tan t)}_{π / 6}^{π / 3} + {(\sec t)}_{π / 6}^{π / 3}]$

$= 2 π [(\sqrt{3} - \frac{1}{\sqrt{3}}) + (2 - \frac{2}{\sqrt{3}})]$

$= 2 π [\sqrt{3} + 2 - \sqrt{3}] = 4 π$

Q 47 :

$6 \int_{0}^{π}$ $| (\sin 3 x + \sin 2 x + \sin x) |$ $d x$ is equal to_____. [2026]

(17)

$6 \int_{0}^{π}$ $| 2 \sin 2 x \cos x + \sin 2 x |$ $d x$

$= 6 \int_{0}^{π}$ $| 4 \sin x \cos^{2} x + 2 \sin x \cos x |$ $d x$

$I = 12 \int_{0}^{π}$ $| \sin x (2 \cos^{2} x + \cos x) |$ $d x$

Put $\cos x = t, - \sin x d x = d t$

$I = - 12 \int_{1}^{- 1}$ $| 2 t^{2} + t |$ $d t$

$I = 12 (\int_{- 1}^{- \frac{1}{2}} (2 t^{2} + t) d t + \int_{- \frac{1}{2}}^{0} - (2 t^{2} + t) d t + \int_{0}^{1} (2 t^{2} + t) d t)$

$I = 17$

Q 48 :

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]

(64)

$\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x), Put \frac{t x}{36} = y$

$\frac{d y}{d t} = \frac{x}{36}$

$\int_{0}^{x} \frac{f (y) 36 d y}{x} = 4 α f (x)$

$\int_{0}^{x} f (y) d y = \frac{α f (x) x}{9}$

$f (x) = \frac{α}{9} (f (x) + x f' (x))$

$(1 - \frac{α}{9}) f (x) = \frac{α x}{9} f' (x)$ $\Rightarrow (9 - α) f (x) = α x f' (x)$

$\frac{f' (x)}{f (x)} = (\frac{9}{α} - 1) \frac{1}{x}$

$\log_{e} f (x) = (\frac{9}{α} - 1) \log_{e} x + \log_{e} c$

$f (x) = c x^{(\frac{9}{α} - 1)}$ For standard parabola,

$\frac{9}{α} - 1 = 2$

$α = 3$

$f (x) = c x^{2}$

$Passing through (2, 1)$

$1 = 4 c \Rightarrow c = \frac{1}{4}$

$y = \frac{x^{2}}{4} passing through (- 4, β)$

$β = 4$

$β^{x} = 4^{3} = 64$

Q 49 :

Let $f : [1, \infty) \to ℝ$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is [2026]

$- 4$
$3$
$- 3$
$4$

1

2

3

4

(2)

$6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$

$Differentiate both sides$

$6 f (x) = 3 x f' (x) + 3 f (x) + 3 x^{2}$

$3 f (x) = 3 x f' (x) + 3 x^{2}$

$x \frac{d y}{d x} - y = - x^{2}$

$\frac{x \frac{d y}{d x} - 9}{x^{2}} = - 1$

$\Rightarrow \frac{d}{d x} (\frac{y}{x}) = - 1$

$\frac{y}{x} = - x + C$

$\Rightarrow f (x) = - x^{2} + C x$

At $x = 1, y = 1$ $\Rightarrow C = 2$

$f (x) = - x^{2} + 2 x$

$f (2) - f (3) = 3$

Q 50 :

The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{[x] + 4}) d x,$ where $[\cdot]$ denotes the greatest integer function, is [2026]

$\frac{7}{60} (3 π - 1)$
$\frac{1}{60} (21 π - 1)$
$\frac{7}{60} (π - 3)$
$\frac{1}{60} (π - 7)$

1

2

3

4

(1)

$I = \int_{- π / 2}^{π / 2} \frac{1}{[x] + 4} d x$

$I = \int_{- π / 2}^{- 1} \frac{d x}{2} + \int_{- 1}^{0} \frac{d x}{3} + \int_{0}^{1} \frac{d x}{4} + \int_{1}^{π / 2} \frac{d x}{5}$

$= \frac{1}{2} (- 1 + \frac{π}{2}) + \frac{1}{3} (1) + \frac{1}{4} (1) + (\frac{π}{2} - 1) \frac{1}{5}$

$= \frac{7 π}{20} - \frac{7}{60} = \frac{7}{60} (3 π - 1)$

Q 51 :

Let [.] denote the greatest integer function. Then $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{12 (3 + [x])}{3 + [\sin x] + [\cos x]})$ is equal to: [2026]

$13 π + 1$
$15 π + 4$
$11 π + 2$
$12 π + 5$

1

2

3

4

(3)

$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{12 (3 + [x])}{3 + [\sin x] + [\cos x]} d x$

$I = \int_{- \frac{π}{2}}^{- 1} \frac{12 (1) d x}{2} + \int_{- 1}^{0} \frac{12 (2) d x}{2} + \int_{0}^{1} \frac{12 (3) d x}{3} + \int_{1}^{\frac{π}{2}} \frac{12 (4) d x}{3}$

$I = 6 (\frac{π}{2} - 1) + 12 (0 + 1) + 12 (1 - 0) + 16 (\frac{π}{2} - 1)$

$I = 3 π - 6 + 12 + 12 + 8 π - 16$

$I = 11 π + 2$

Q 52 :

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y,$ then $e + f (0)$ is equal to _______ . [2026]

(2)

$f (x) = e^{x} + \int_{0}^{1} y f (y) d y + x e^{x} \int_{0}^{1} f (y) d y$

$f (x) = e^{x} + A + B x e^{x}$

$A = \int_{0}^{1} y f (y) d y = \int_{0}^{1} y (A + e^{y} + B y e^{y}) d y$

$A = \frac{A}{2} + 0 - (- 1) + B (e - 1)$

$\frac{A}{2} + B (1 - e) = 1$

$B = \int_{0}^{1} f (y) d y$

$B = \int_{0}^{1} (e^{y} + A + B y e^{y}) d y$

$B = (e - 1) + A + B (0 - (- 1))$

$B = e - 1 + A + B \Rightarrow A = 1 - e$

$f (x) = e^{x} + A + B x e^{x}$

$f (0) = 1 + A = 1 - e + 1 = 2 - e$

$e + f (0) = 2$

Q 53 :

The number of elements in the set

$S = {x : x \in [0, 100] and \int_{0}^{x} t^{2} \sin (x - t) d t = x^{2}}$ is_____. [2026]

(16)

$\int_{0}^{x} t^{2} \sin (x - t) d t = x^{2}$

$Use by parts$

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

$- x^{2} \cos x + 2 x \sin x + 2 \cos x - 2 = x$

$\cos x = 1$

$x = 0, 2 π, 4 π, \dots, 30 π$

$Total Elements = 16$

Q 54 :

If $\int_{0}^{1} 4 \cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a \tan^{- 1} (2) - b \log_{e} (5),$ where $a, b \in ℕ, then (2 a + b)$ is equal to _______. [2026]

(9)

$Let I = \int_{0}^{1} \cot^{- 1} (1 - 2 x + 4 x^{2}) d x$

$I = \int_{0}^{1} (\cot^{- 1} (2 x - 1) - \cot^{- 1} (2 x)) d x . . . (1)$

Applying king

$I = \int_{0}^{1} (- \cot^{- 1} (2 x - 1) + \cot^{- 1} (2 x - 2)) d x . . . (2)$

From (1) & (2)

$2 I = \int_{0}^{1} (\cot^{- 1} (2 x - 2) - \cot^{- 1} (2 x)) d x$

$= \int_{0}^{1} \cot^{- 1} (2 x - 2) d x - \int_{0}^{1} \cot^{- 1} (2 x) d x$

Applying King

$= \int_{0}^{1} \cot^{- 1} (- 2 x) d x - \int_{0}^{1} \cot^{- 1} (2 x) d x$

$= \int_{0}^{1} (π - \cot^{- 1} (2 x)) d x - \int_{0}^{1} \cot^{- 1} (2 x) d x$

$= \int_{0}^{1} (π - 2 \cot^{- 1} (2 x)) d x$

$= π - 2 \int_{0}^{1} \cot^{- 1} (2 x) d x$

By parts

$= π - 2 [(x \cot^{- 1} {(2 x))}_{0}^{1} + \int_{0}^{1} \frac{2 x}{1 + 4 x^{2}} d x]$

Let $1 + 4 x^{2} = t$

$8 x d x = d t$

$= π - 2 [\cot^{- 1} 2 + \frac{1}{4} \int_{1}^{5} \frac{d t}{t}]$

$= π - 2 \cot^{- 1} 2 - \frac{1}{2} l n 5$

$2 I = 2 \tan^{- 1} 2 - \frac{1}{2} l n 5$

$\Rightarrow 4 I = 4 \tan^{- 1} 2 - l n 5$

$∴$ $2 a + b = 8 + 1 = 9$

Q 55 :

Let $[\cdot]$ denote the greatest integer function and $f (x) = \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [[\frac{k^{2}}{3^{x}}]] .$ Then $12 \sum_{j = 1}^{\infty} f (j)$ is equal to ________. [2026]

(2)

$\sum_{k = 1}^{n} (\frac{k^{2}}{3^{x}} - 1) < \sum_{k = 1}^{n} [\frac{k^{2}}{3^{x}}] \leq \sum_{k = 1}^{n} \frac{k^{2}}{3^{x}}$

$\frac{n (n + 1) (2 n + 1)}{6 \cdot 3^{x}} < \sum_{k = 1}^{n} [\frac{k^{2}}{3^{x}}] \leq \frac{n (n + 1) (2 n + 1)}{6 \cdot 3^{x}}$

$\lim_{n \to \infty} \frac{n (n + 1) (2 n + 1)}{6 n^{3} \cdot 3^{x}} < \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [\frac{k^{2}}{3^{x}}] \leq \lim_{n \to \infty} \frac{n (n + 1) (2 n + 1)}{6 \cdot 3^{x} \cdot n^{3}}$

$\frac{1}{3^{x + 1}} < \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [\frac{k^{2}}{3^{x}}] \leq \frac{1}{3^{x + 1}}$

$\Rightarrow f (x) = \frac{1}{3^{x + 1}}$

$\Rightarrow 12 \sum_{j = 1}^{\infty} f (j) = 12 \sum_{j = 1}^{\infty} \frac{1}{3^{j + 1}} = 12 [\frac{1}{9} + \frac{1}{27} + \dots \infty]$

$= 12 (\frac{\frac{1}{9}}{1 - \frac{1}{3}}) = 2$

Q 56 :

Let $[\cdot]$ be the greatest integer function. If $α = \int_{0}^{64} (x^{1 / 3} - [x^{1 / 3}]) d x,$ then $\frac{1}{π} \int_{0}^{α π} (\frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ}) d θ$ is equal to_______. [2026]

(36)

$∵$ $\int_{0}^{64} x^{\frac{1}{3}} d x = \frac{3}{4} {[x^{\frac{4}{3}}]}_{0}^{64} = 192$

and

$\int_{0}^{64} [x^{1 / 3}] d x = \int_{0}^{1} [x^{1 / 3}] d x + \int_{1}^{8} [x^{1 / 3}] d x + \int_{8}^{27} [x^{1 / 3}] d x + \dots + \int_{27}^{64} [x^{1 / 3}] d x = 156$

$So α = 192 - 156 = 36$

Now

$E = \frac{1}{π} \int_{0}^{36 π} \frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ} d θ$

$= \frac{36}{π} \int_{0}^{π} \frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ} d θ$

$\Rightarrow E = \frac{36 \cdot 2}{π} \int_{0}^{π / 2} \frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ} d θ$

Let $J = \int_{0}^{π / 2} \frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ} d θ . . . (1)$

Applying King property,

$J = \int_{0}^{π / 2} \frac{\cos^{2} θ}{\sin^{6} θ + \cos^{6} θ} d θ . . . (2)$

Now

$2 J = \int_{0}^{π / 2} \frac{1}{\sin^{6} θ + \cos^{6} θ} d θ (add (1) & (2))$

$= \int_{0}^{π / 2} \frac{\sec^{6} θ}{\tan^{6} θ + 1} d θ$

$= \int_{0}^{\infty} \frac{1 + λ^{2}}{λ^{4} - λ^{2} + 1} d λ$

$= \int_{0}^{\infty} \frac{1 + \frac{1}{λ^{2}}}{λ^{2} - 1 + \frac{1}{λ^{2}}} d λ$

$= π$

$\Rightarrow J = \frac{π}{2}$

$\Rightarrow E = \frac{36 \cdot 2}{π} \times J = 36$

Q 57 :

Let f be a polynomial function such that $f (x^{2} + 1) = x^{4} + 5 x^{2} + 2, for all x \in ℝ .$ Then $\int_{0}^{3} f (x) d x$ is equal to: [2026]

$\frac{27}{2}$
$\frac{33}{2}$
$\frac{41}{3}$
$\frac{5}{3}$

1

2

3

4

(2)

$∵$ $f (x^{2} + 1) = x^{4} + 5 x^{2} + 2$

$Put x^{2} + 1 = t$

$\Rightarrow f (t) = {(t - 1)}^{2} + 5 (t - 1) + 2$

$\Rightarrow f (t) = t^{2} + 3 t - 2$

$Now, \int_{0}^{3} f (t) d t = \int_{0}^{3} (t^{2} + 3 t - 2) d t$

$= {[\frac{t^{3}}{3} + \frac{3 t^{2}}{2} - 2 t]}_{0}^{3}$

$= [\frac{27}{3} + \frac{27}{2} - 6]$

$= \frac{33}{2}$

Q 58 :

The value of $\sum_{r = 1}^{20} (| \sqrt[]{π (\int_{0}^{r} x | \sin π x | d x)} |)$ is _______. [2026]

(210)

$Let I_{r} = \int_{0}^{r} x | \sin π x | d x . . . (1)$

$Apply King Property$

$= \int_{0}^{r} (r - x) | \sin π x | d x . . . (2)$

$By (1) + (2)$

$2 I_{r} = \int_{0}^{r} r | \sin π x | d x \Rightarrow I_{r} = \frac{r}{2} \int_{0}^{r} | \sin π x | d x$

$I_{1} = \frac{1}{2} \int_{0}^{1} | \sin π x | d x = \frac{1}{2 π} \int_{0}^{π} | \sin t | d t = \frac{1}{2 π} (2)$

$I_{2} = \frac{2}{2} \int_{0}^{2} | \sin π x | d x = \frac{2}{2 π} \int_{0}^{2 π} | \sin t | d t = \frac{2}{2 π} (4)$

$S = \sqrt{π \cdot \frac{1}{2 π} \cdot 2} + \sqrt{π \cdot \frac{2}{2 π} \cdot 4} + \sqrt{π \cdot \frac{3}{2 π} \cdot 6} + \dots + \sqrt{π \cdot \frac{20}{2 π} \cdot (2 \cdot 20)}$

$= 1 + 2 + 3 + \dots + 20$

$= \frac{20 \times 21}{2} = 210$

Q 59 :

The value of the definite integral $\int_{- 2}^{2} x^{3} \ln (1^{x} + 3^{x} + 5^{x} + 15^{x}) d x =$

$\frac{\ln 15}{4}$
$\frac{64}{5} \ln 15$
$\frac{32}{5} \ln 15$
$\frac{64}{5} \ln 30$

1

2

3

4

(3)

$I = \int_{- 2}^{2} {(- x)}^{3} \ln (\frac{15^{x} + 5^{x} + 3^{x} + 1^{x}}{15^{x}}) d x$

$2 I = \int_{- 2}^{2} x^{4} \ln 15 d x$

Q 60 :

Let $I_{1} = \int_{0}^{x} e^{t x} . e^{- t^{2}} d t$ and $I_{2} = \int_{0}^{x} e^{- t^{2} / 4} d t$ where $x > 0,$ then the value of $\frac{I_{1}}{I_{2}}$ is

$e^{- x^{2} / 2}$
$e^{x^{2} / 4}$
$e^{- x^{2} / 4}$
$e^{x^{2} / 2}$

1

2

3

4

(2)

$I_{1} = \int_{0}^{x} e^{\frac{x^{2}}{4} - {(t - \frac{x}{2})}^{2}} d t = e^{\frac{x^{2}}{4}} \int_{- \frac{x}{2}}^{\frac{x}{2}} e^{- z^{2}} d z t - \frac{x}{2} = z$

$= 2 e^{\frac{x^{2}}{4}} \int_{0}^{\frac{x}{2}} e^{- z^{2}} d z$

$I_{2} = \int_{0}^{x} e^{- \frac{t^{2}}{4} d t}$

$Let \frac{t}{2} = z = \int_{0}^{\frac{x}{2}} e^{- z^{2}} \times 2 d z$

Topic Question Set

Q 31 :

Let $f (x) = \frac{x}{{(1 + x^{n})}^{\frac{1}{n}}}, x \in ℝ - {- 1}, n \in ℕ, n > 2 .$ If $f^{n} (x) = (f o f o f . . . . . upto n times)$ $(x),$ then $\lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f^{n} (x)) d x$ is equal to _____ . [2023]

Q 32 :

Let $[t]$ denote the greatest integer $\leq f$ . Then $\frac{2}{π} \int_{π / 6}^{5 π / 6} (8 [c o s e c x] - 5 [\cot x]) d x$ is equal to ______ . [2023]

Q 33 :

Let $[t]$ denote the greatest integer function. If $\int_{0}^{2.4} [x^{2}] d x = α + β \sqrt{2} + γ \sqrt{3} + δ \sqrt{5},$ then $α + β + γ + δ$ is equal to _________ . [2023]

Q 34 :

For $m, n > 0$ , let $α (m, n) = \int_{0}^{2} t^{m} {(1 + 3 t)}^{n} d t$ . If $11 α (10, 6) + 18 α (11, 5) = p {(14)}^{6}$ , then $p$ is equal to ______ . [2023]

Q 35 :

If $\int_{- 0.15}^{0.15}$ $| 100 x^{2} - 1 |$ $d x = \frac{k}{3000}$ , then $k$ is equal to ___________. [2023]

Q 36 :

Let for $x \in ℝ, S_{0} (x) = x, S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t,$ where $C_{0} = 1, C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x, k = 1, 2, 3, \dots$ Then $S_{2} (3) + 6 C_{3}$ is equal to _______ . [2023]

Q 37 :

Let $f_{n} = \int_{0}^{\frac{π}{2}} (\sum_{k = 1}^{n} \sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) \sin^{k - 1} x) \cos x d x, n \in ℕ .$ Then $f_{21} - f_{20}$ is equal to ________. [2023]

Q 38 :

$If \int_{0}^{1} (x^{21} + x^{14} + x^{7}) {(2 x^{14} + 3 x^{7} + 6)}^{1 / 7} d x = \frac{1}{l} {(11)}^{m}$ $where l, m, n \in ℕ, m and n are coprime, then l + m + n is equal to$ _______ . [2023]

Q 39 :

$The value of \frac{8}{π} \int_{0}^{π / 2} \frac{{(\cos x)}^{2023}}{{(\sin x)}^{2023} + {(\cos x)}^{2023}} d x is$ _______ . [2023]

Q 40 :

$The value of 12 \int_{0}^{3}$ $| x^{2} - 3 x + 2 |$ $d x is$ ________ . [2023]

Q 41 :

If $\int_{0}^{π} \frac{5^{\cos x} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x) d x}{1 + 5^{\cos x}} = \frac{k π}{16},$ then $k$ is equal to ______. [2023]

Q 42 :

$If \int_{1 / 3}^{3}$ $| \log_{e} x |$ $d x = \frac{m}{n} \log_{e} (\frac{n^{2}}{e}), where m and n are coprime natural numbers, then m^{2} + n^{2} - 5 is equal to$ ______ . [2023]

Q 43 :

$\lim_{x \to 0} \frac{48}{x^{4}} \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t is equal to$ ______. [2023]

Q 44 :

The value of the integral $\int_{\frac{π}{24}}^{\frac{5 π}{24}} \frac{d x}{1 + \sqrt[3]{\tan 2 x}}$ is: [2026]

Q 45 :

Let $f$ be a twice differentiable non-negative function such that $(f {(x))}^{2} = 25 + \int_{0}^{x} ((f {(t))}^{2} + (f' {(t))}^{2}) d t .$ Then the mean of $f (\log_{e} (1)), f (\log_{e} (2)), \dots, f (\log_{e} (625))$ is equal to _________ . [2026]

Q 46 :

The value of $\int_{- π / 6}^{π / 6} (\frac{π + 4 x^{11}}{1 - \sin (| x | + π / 6)}) d x$ is equal to: [2026]

Q 47 :

$6 \int_{0}^{π}$ $| (\sin 3 x + \sin 2 x + \sin x) |$ $d x$ is equal to_____. [2026]

Q 48 :

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]

Q 49 :

Let $f : [1, \infty) \to ℝ$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is [2026]

Q 50 :

The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{[x] + 4}) d x,$ where $[\cdot]$ denotes the greatest integer function, is [2026]

Q 51 :

Let [.] denote the greatest integer function. Then $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{12 (3 + [x])}{3 + [\sin x] + [\cos x]})$ is equal to: [2026]

Q 52 :

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y,$ then $e + f (0)$ is equal to _______ . [2026]

Q 53 :

The number of elements in the set

$S = {x : x \in [0, 100] and \int_{0}^{x} t^{2} \sin (x - t) d t = x^{2}}$ is_____. [2026]

Q 54 :

If $\int_{0}^{1} 4 \cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a \tan^{- 1} (2) - b \log_{e} (5),$ where $a, b \in ℕ, then (2 a + b)$ is equal to _______. [2026]

Q 55 :

Let $[\cdot]$ denote the greatest integer function and $f (x) = \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [[\frac{k^{2}}{3^{x}}]] .$ Then $12 \sum_{j = 1}^{\infty} f (j)$ is equal to ________. [2026]

Q 56 :

Let $[\cdot]$ be the greatest integer function. If $α = \int_{0}^{64} (x^{1 / 3} - [x^{1 / 3}]) d x,$ then $\frac{1}{π} \int_{0}^{α π} (\frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ}) d θ$ is equal to_______. [2026]

Q 57 :

Let f be a polynomial function such that $f (x^{2} + 1) = x^{4} + 5 x^{2} + 2, for all x \in ℝ .$ Then $\int_{0}^{3} f (x) d x$ is equal to: [2026]

Q 58 :

The value of $\sum_{r = 1}^{20} (| \sqrt[]{π (\int_{0}^{r} x | \sin π x | d x)} |)$ is _______. [2026]

Q 59 :

The value of the definite integral $\int_{- 2}^{2} x^{3} \ln (1^{x} + 3^{x} + 5^{x} + 15^{x}) d x =$

Q 60 :

Let $I_{1} = \int_{0}^{x} e^{t x} . e^{- t^{2}} d t$ and $I_{2} = \int_{0}^{x} e^{- t^{2} / 4} d t$ where $x > 0,$ then the value of $\frac{I_{1}}{I_{2}}$ is

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Topic Question Set

Q 31 : Let f(x)=x(1+xn)1n, x∈ℝ-{-1}, n∈ℕ, n>2. If fn(x)=(fofof ..... upto n times) (x), then limn→∞∫01xn-2(fn(x)) dx is equal to _____ . [2023]

Q 32 : Let [t] denote the greatest integer ≤f. Then 2π∫π/65π/6(8[cosecx]-5[cotx])dx is equal to ______ . [2023]

Q 33 : Let [t] denote the greatest integer function. If ∫02.4[x2] dx=α+β2+γ3+δ5, then α+β+γ+δ is equal to _________ . [2023]

Q 34 : For m,n>0, let α(m,n)=∫02tm(1+3t)ndt. If 11α(10,6)+18α(11,5)=p(14)6, then p is equal to ______ . [2023]

Q 35 : If ∫-0.150.15|100x2-1|dx=k3000, then k is equal to ___________. [2023]

Q 36 : Let for x∈ℝ, S0(x)=x,Sk(x)=Ckx+k∫0xSk-1(t) dt, where C0=1, Ck=1-∫01Sk-1(x) dx, k=1,2,3,… Then S2(3)+6C3 is equal to _______ . [2023]

Q 37 : Let fn=∫0π2(∑k=1nsink-1x)(∑k=1n(2k-1)sink-1x)cosx dx, n∈ℕ. Then f21-f20 is equal to ________. [2023]

Q 38 : If ∫01(x21+x14+x7)(2x14+3x7+6)1/7dx=1l(11)m where l,m,n∈ℕ, m and n are coprime, then l+m+n is equal to _______ . [2023]

Q 39 : The value of 8π∫0π/2(cosx)2023(sinx)2023+(cosx)2023dx is _______ . [2023]

Q 40 : The value of 12∫03|x2-3x+2|dx is ________ . [2023]

Q 41 : If ∫0π5cosx(1+cosxcos3x+cos2x+cos3xcos3x)dx1+5cosx=kπ16, then k is equal to ______. [2023]

Q 42 : If ∫1/33|logex|dx=mnloge(n2e), where m and n are coprime natural numbers, then m2+n2-5 is equal to ______ . [2023]

Q 43 : limx→048x4∫0xt3t6+1dt is equal to ______. [2023]

Q 44 : The value of the integral ∫π245π24dx1+tan2x3 is: [2026]

Q 45 : Let f be a twice differentiable non-negative function such that (f(x))2=25+∫0x((f(t))2+(f'(t))2) dt. Then the mean of f(loge(1)), f(loge(2)), …, f(loge(625)) is equal to _________ . [2026]

Q 46 : The value of ∫-π/6π/6(π+4x111-sin(|x|+π/6))dx is equal to: [2026]

Q 47 : 6∫0π|(sin3x+sin2x+sinx)|dx is equal to_____. [2026]

Q 48 : Let a differentiable function f satisfy the equation ∫036f(tx36)dt=4αf(x). If y=f(x) is a standard parabola passing through the points (2, 1) and (-4,β), then βα is equal to _____. [2026]

Q 49 : Let f:[1,∞)→ℝ be a differentiable function. If 6∫1xf(t)dt=3xf(x)+x3-4 for all x≥1, then the value of f(2)-f(3) is [2026]

Q 50 : The value of ∫-π2π2(1[x]+4)dx, where [·] denotes the greatest integer function, is [2026]

Q 51 : Let [.] denote the greatest integer function. Then ∫-π2π2(12(3+[x])3+[sinx]+[cosx]) is equal to: [2026]

Q 52 : If f(x) satisfies the relation f(x)=ex+∫01(y+xex) f(y) dy, then e+f(0) is equal to _______ . [2026]

Q 53 : The number of elements in the set S={x:x∈[0,100] and ∫0xt2sin(x-t)dt=x2} is_____. [2026]

Q 54 : If ∫014cot-1(1-2x+4x2)dx=a tan-1(2)-bloge(5), where a,b∈ℕ, then (2a+b) is equal to _______. [2026]

Q 55 : Let [⋅] denote the greatest integer function and f(x)=limn→∞1n3∑k=1n[k23x]. Then 12∑j=1∞f(j) is equal to ________. [2026]

Q 56 : Let [·] be the greatest integer function. If α=∫064(x1/3-[x1/3]) dx, then 1π∫0απ(sin2θsin6θ+cos6θ) dθ is equal to_______. [2026]

Q 57 : Let f be a polynomial function such that f(x2+1)=x4+5x2+2, for all x∈ℝ. Then ∫03f(x) dx is equal to: [2026]

Q 58 : The value of ∑r=120(|π(∫0rx|sinπx|dx)|) is _______. [2026]

Q 59 : The value of the definite integral ∫-22x3ln(1x+3x+5x+15x)dx=

Q 60 : Let I1=∫0xetx.e-t2dt and I2=∫0xe-t2/4dt where x>0, then the value of I1I2 is

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Q 31 :

Let $f (x) = \frac{x}{{(1 + x^{n})}^{\frac{1}{n}}}, x \in ℝ - {- 1}, n \in ℕ, n > 2 .$ If $f^{n} (x) = (f o f o f . . . . . upto n times)$ $(x),$ then $\lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f^{n} (x)) d x$ is equal to _____ . [2023]

Q 32 :

Let $[t]$ denote the greatest integer $\leq f$ . Then $\frac{2}{π} \int_{π / 6}^{5 π / 6} (8 [c o s e c x] - 5 [\cot x]) d x$ is equal to ______ . [2023]

Q 33 :

Let $[t]$ denote the greatest integer function. If $\int_{0}^{2.4} [x^{2}] d x = α + β \sqrt{2} + γ \sqrt{3} + δ \sqrt{5},$ then $α + β + γ + δ$ is equal to _________ . [2023]

Q 34 :

For $m, n > 0$ , let $α (m, n) = \int_{0}^{2} t^{m} {(1 + 3 t)}^{n} d t$ . If $11 α (10, 6) + 18 α (11, 5) = p {(14)}^{6}$ , then $p$ is equal to ______ . [2023]

Q 35 :

If $\int_{- 0.15}^{0.15}$ $| 100 x^{2} - 1 |$ $d x = \frac{k}{3000}$ , then $k$ is equal to ___________. [2023]

Q 36 :

Let for $x \in ℝ, S_{0} (x) = x, S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t,$ where $C_{0} = 1, C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x, k = 1, 2, 3, \dots$ Then $S_{2} (3) + 6 C_{3}$ is equal to _______ . [2023]

Q 37 :

Let $f_{n} = \int_{0}^{\frac{π}{2}} (\sum_{k = 1}^{n} \sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) \sin^{k - 1} x) \cos x d x, n \in ℕ .$ Then $f_{21} - f_{20}$ is equal to ________. [2023]

Q 38 :

$If \int_{0}^{1} (x^{21} + x^{14} + x^{7}) {(2 x^{14} + 3 x^{7} + 6)}^{1 / 7} d x = \frac{1}{l} {(11)}^{m}$ $where l, m, n \in ℕ, m and n are coprime, then l + m + n is equal to$ _______ . [2023]

Q 39 :

$The value of \frac{8}{π} \int_{0}^{π / 2} \frac{{(\cos x)}^{2023}}{{(\sin x)}^{2023} + {(\cos x)}^{2023}} d x is$ _______ . [2023]

Q 40 :

$The value of 12 \int_{0}^{3}$ $| x^{2} - 3 x + 2 |$ $d x is$ ________ . [2023]

Q 41 :

If $\int_{0}^{π} \frac{5^{\cos x} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x) d x}{1 + 5^{\cos x}} = \frac{k π}{16},$ then $k$ is equal to ______. [2023]

Q 42 :

$If \int_{1 / 3}^{3}$ $| \log_{e} x |$ $d x = \frac{m}{n} \log_{e} (\frac{n^{2}}{e}), where m and n are coprime natural numbers, then m^{2} + n^{2} - 5 is equal to$ ______ . [2023]

Q 43 :

$\lim_{x \to 0} \frac{48}{x^{4}} \int_{0}^{x} \frac{t^{3}}{t^{6} + 1} d t is equal to$ ______. [2023]

Q 44 :

The value of the integral $\int_{\frac{π}{24}}^{\frac{5 π}{24}} \frac{d x}{1 + \sqrt[3]{\tan 2 x}}$ is: [2026]

Q 45 :

Let $f$ be a twice differentiable non-negative function such that $(f {(x))}^{2} = 25 + \int_{0}^{x} ((f {(t))}^{2} + (f' {(t))}^{2}) d t .$ Then the mean of $f (\log_{e} (1)), f (\log_{e} (2)), \dots, f (\log_{e} (625))$ is equal to _________ . [2026]

Q 46 :

The value of $\int_{- π / 6}^{π / 6} (\frac{π + 4 x^{11}}{1 - \sin (| x | + π / 6)}) d x$ is equal to: [2026]

Q 47 :

$6 \int_{0}^{π}$ $| (\sin 3 x + \sin 2 x + \sin x) |$ $d x$ is equal to_____. [2026]

Q 48 :

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f (\frac{t x}{36}) d t = 4 α f (x) .$ If $y = f (x)$ is a standard parabola passing through the points (2, 1) and $(- 4, β)$ , then $β^{α}$ is equal to _____. [2026]

Q 49 :

Let $f : [1, \infty) \to ℝ$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is [2026]

Q 50 :

The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{[x] + 4}) d x,$ where $[\cdot]$ denotes the greatest integer function, is [2026]

Q 51 :

Let [.] denote the greatest integer function. Then $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{12 (3 + [x])}{3 + [\sin x] + [\cos x]})$ is equal to: [2026]

Q 52 :

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y,$ then $e + f (0)$ is equal to _______ . [2026]

Q 53 :

The number of elements in the set

$S = {x : x \in [0, 100] and \int_{0}^{x} t^{2} \sin (x - t) d t = x^{2}}$ is_____. [2026]

Q 54 :

If $\int_{0}^{1} 4 \cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a \tan^{- 1} (2) - b \log_{e} (5),$ where $a, b \in ℕ, then (2 a + b)$ is equal to _______. [2026]

Q 55 :

Let $[\cdot]$ denote the greatest integer function and $f (x) = \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [[\frac{k^{2}}{3^{x}}]] .$ Then $12 \sum_{j = 1}^{\infty} f (j)$ is equal to ________. [2026]

Q 56 :

Let $[\cdot]$ be the greatest integer function. If $α = \int_{0}^{64} (x^{1 / 3} - [x^{1 / 3}]) d x,$ then $\frac{1}{π} \int_{0}^{α π} (\frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ}) d θ$ is equal to_______. [2026]

Q 57 :

Let f be a polynomial function such that $f (x^{2} + 1) = x^{4} + 5 x^{2} + 2, for all x \in ℝ .$ Then $\int_{0}^{3} f (x) d x$ is equal to: [2026]

Q 58 :

The value of $\sum_{r = 1}^{20} (| \sqrt[]{π (\int_{0}^{r} x | \sin π x | d x)} |)$ is _______. [2026]

Q 59 :

The value of the definite integral $\int_{- 2}^{2} x^{3} \ln (1^{x} + 3^{x} + 5^{x} + 15^{x}) d x =$

Q 60 :

Let $I_{1} = \int_{0}^{x} e^{t x} . e^{- t^{2}} d t$ and $I_{2} = \int_{0}^{x} e^{- t^{2} / 4} d t$ where $x > 0,$ then the value of $\frac{I_{1}}{I_{2}}$ is