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

Q 11 :

Let $[t]$ denote the largest integer less than or equal to $t .$ If $\int_{0}^{3} ([x^{2}] + [\frac{x^{2}}{2}]) d x = a + b \sqrt{2} - \sqrt{3} - \sqrt{5} + c \sqrt{6} - \sqrt{7},$ where $a, b, c \in Z,$ then $a + b + c$ is equal to _______. [2024]

(23)

Given, $\int_{0}^{3} ([x^{2}] + [\frac{x^{2}}{2}]) d x = a + b \sqrt{2} - \sqrt{3} - \sqrt{5} + c \sqrt{6} - \sqrt{7}$

Now, $\int_{0}^{3} ([x^{2}] + [\frac{x^{2}}{2}]) d x = \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} 1 d x + \int_{\sqrt{2}}^{\sqrt{3}} (2 + 1) d x$

$+ \int_{\sqrt{3}}^{2} (3 + 1) d x + \int_{2}^{\sqrt{5}} 6 d x + \int_{\sqrt{5}}^{\sqrt{6}} 7 d x + \int_{\sqrt{6}}^{\sqrt{7}} 9 d x + \int_{\sqrt{7}}^{\sqrt{8}} 10 d x + \int_{\sqrt{8}}^{3} 12 d x$

$= (\sqrt{2} - 1) + (3 \sqrt{3} - 3 \sqrt{2}) + (8 - 4 \sqrt{3}) + (6 \sqrt{5} - 12) + (7 \sqrt{6} - 7 \sqrt{5}) + (9 \sqrt{7} - 9 \sqrt{6}) + (10 \sqrt{8} - 10 \sqrt{7}) + (36 - 12 \sqrt{8})$

$= 31 - 6 \sqrt{2} - \sqrt{3} - \sqrt{5} - 2 \sqrt{6} - \sqrt{7}$

$\Rightarrow a = 31, b = - 6, c = - 2$

So, $a + b + c = 31 - 8 = 23$

Q 12 :

Let $\lim_{n \to \infty} (\frac{n}{\sqrt{n^{4} + 1}} - \frac{2 n}{(n^{2} + 1) \sqrt{n^{4} + 1}} + \frac{n}{\sqrt{n^{4} + 16}} - \frac{8 n}{(n^{2} + 4) \sqrt{n^{4} + 16}} + \dots + \frac{n}{\sqrt{n^{4} + n^{4}}} - \frac{2 n \cdot n^{2}}{(n^{2} + n^{2}) \sqrt{n^{4} + n^{4}}})$ be $\frac{π}{k},$ using only the principal values of the inverse trigonometric functions. Then $k^{2}$ is equal to ______ . [2024]

(32)

$\lim_{n \to \infty} (\frac{n}{\sqrt{n^{4} + 1}} + \frac{n}{\sqrt{n^{4} + 16}} + \dots \frac{n}{\sqrt{n^{4} + n^{4}}}) - \lim_{n \to \infty} (\frac{2 n}{(n^{2} + 1) (\sqrt{n^{4} + 1})} + \frac{8 n}{(n^{2} + 4) \sqrt{n^{4} + 16}} + \dots \frac{2 n \cdot n^{2}}{(n^{2} + n^{2}) \sqrt{n^{4} + n^{4}}})$

$= \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n \sqrt{1 + \frac{r^{4}}{n^{4}}}} - \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} \frac{2 \cdot {(\frac{r}{n})}^{2}}{(1 + {(\frac{r}{n})}^{2})} \frac{1}{\sqrt{1 + \frac{r^{4}}{n^{4}}}}$

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

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

$Put x + \frac{1}{x} = t, then (1 - \frac{1}{x^{2}}) d x = d t$

$= - \int_{\infty}^{2} \frac{d t}{t \sqrt{t^{2} - 2}} = \int_{2}^{\infty} \frac{d t}{t \sqrt{t^{2} - 2}}$

$Put t^{2} - 2 = u^{2} \Rightarrow 2 t d t = 2 u d u \Rightarrow d t = \frac{u}{t} d u$

$= \int_{\sqrt{2}}^{\infty} \frac{d u}{(u^{2} + 2)} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{u}{\sqrt{2}}$ $|_{\sqrt{2}}^{\infty}$

$= \frac{1}{\sqrt{2}} (\frac{π}{2} - \frac{π}{4}) = \frac{π}{4 \sqrt{2}} = \frac{π}{k}$

$\Rightarrow k = 2^{5 / 2} \Rightarrow k^{2} = 2^{5} = 32$

Q 13 :

Consider the matrices : $A = [\begin{matrix} 2 & - 5 \\ 3 & m \end{matrix}], B = [\begin{matrix} 20 \\ m \end{matrix}]$ and $X = [\begin{matrix} x \\ y \end{matrix}] .$ Let the set of all $m,$ for which the system of equations $A X = B$ has a negative solution (i.e., $x < 0$ and $y < 0$ ), be the interval $(a, b) .$ Then $8 \int_{a}^{b} | A | d m$ is equal to ______ . [2024]

(450)

$A X = B$ has a negative solution

So, $x, y < 0$

Now, $2 x - 5 y = 20$ ...(i)

$3 x + m y = m$ ...(ii)

On solving (i) and (ii), we get

$y = \frac{2 m - 60}{2 m + 15}, Since y < 0$

$\Rightarrow \frac{2 m - 60}{2 m + 15} < 0 \Rightarrow 2 m - 60 < 0 and 2 m + 15 > 0$

$\Rightarrow m < 30 and m > \frac{- 15}{2}$

$\Rightarrow m \in (\frac{- 15}{2}, 30)$

$Also, x = \frac{25 m}{2 m + 15} < 0$

$\Rightarrow m \in (\frac{- 15}{2}, 0)$

$∴ m \in (\frac{- 15}{2}, 0)$

$Now, 8 \int_{a}^{b} | A | d m = 8 \int_{- 15 / 2}^{0} (2 m + 15) d m$

$= 8 {[m^{2} + 15 m]}_{- 15 / 2}^{0} = 450$

Q 14 :

$If \int_{- π / 2}^{π / 2} \frac{8 \sqrt{2} \cos x d x}{(1 + e^{\sin x}) (1 + \sin^{4} x)} = α π + β \log_{e} (3 + 2 \sqrt{2}),$ $where α, β are integers, then α^{2} + β^{2} equals ____ .$ [2024]

(8)

Let $I = \int_{- π / 2}^{π / 2} \frac{8 \sqrt{2} \cos x d x}{(1 + e^{\sin x}) (1 + \sin^{4} x)}$

Using $\int_{- a}^{a} f (a + b - x) d x = \int_{- a}^{a} f (x) d x$

$\Rightarrow 2 I = \int_{- π / 2}^{π / 2} (\frac{8 \sqrt{2} \cos x}{(1 + e^{\sin x}) (1 + \sin^{4} x)} + \frac{e^{\sin x} 8 \sqrt{2} \cos x}{(e^{\sin x} + 1) (1 + \sin^{4} x)}) d x$

$= \int_{- π / 2}^{π / 2} \frac{8 \sqrt{2} \cos x}{1 + \sin^{4} x} d x = 2 \int_{0}^{π / 2} \frac{8 \sqrt{2} \cos x}{1 + \sin^{4} x} d x$ $[∵ \int_{- a}^{a} f (x) d x = \int_{0}^{a} f (x) d x, if f (- x) = f (x)]$

Put $t = \sin x \Rightarrow d t = \cos x d x$

$∴ I = 8 \sqrt{2} \int_{0}^{1} \frac{d t}{1 + t^{4}} = 4 \sqrt{2} \int_{0}^{1} \frac{2 d t}{1 + t^{4}}$

$= 4 \sqrt{2} \int_{0}^{1} (\frac{1 + 1 / t^{2}}{t^{2} + 1 / t^{2}}) d t - 4 \sqrt{2} \int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} d t$

$= 4 \sqrt{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} d t - 4 \sqrt{2} \frac{1}{2 \sqrt{2}} {[\ln \frac{(1 + \frac{1}{t} - \sqrt{2})}{t + \frac{1}{t} + \sqrt{2}}]}_{0}^{1}$

$= 4 \sqrt{2} \cdot \frac{1}{\sqrt{2}} \tan^{- 1} {(\frac{t - \frac{1}{t}}{\sqrt{2}})}_{0}^{1} - \frac{4 \sqrt{2}}{2 \sqrt{2}} {[\ln \frac{(t + \frac{1}{t} - \sqrt{2})}{(t + \frac{1}{t} + \sqrt{2})}]}_{0}^{1}$

$= 2 π + 2 \ln (3 + 2 \sqrt{2}) = α π + β \log_{e} (3 + 2 \sqrt{2})$

$\Rightarrow α = β = 2$

$Then α^{2} + β^{2} = 4 + 4 = 8$

Q 15 :

$Let f : (0, \infty) \to R and F (x) = \int_{0}^{x} t f (t) d t . If F (x^{2}) = x^{4} + x^{5}, then \sum_{r = 1}^{12} f (r^{2}) is equal to____.$ [2024]

(219)

We have, $F (x) = \int_{0}^{x} t \cdot f (t) d t$

$\Rightarrow F' (x) = x \cdot f (x)$ ...(i)

Also, we have $F (x^{2}) = x^{4} + x^{5}$

Differentiating on both sides, we get

$\Rightarrow 2 x \cdot F' (x^{2}) = 4 x^{3} + 5 x^{4}$

$\Rightarrow F' (x^{2}) = 2 x^{2} + \frac{5}{2} x^{3} \Rightarrow F' (x) = 2 x + \frac{5}{2} x^{3 / 2}$ ...(ii)

From (i) & (ii), we get $\Rightarrow x f (x) = 2 x + \frac{5}{2} x^{3 / 2}$

$\Rightarrow f (x) = 2 + \frac{5}{2} x^{1 / 2} \Rightarrow f (x^{2}) = 2 + \frac{5}{2} x$

$∴ \sum_{r = 1}^{12} f (r^{2}) = \sum_{r = 1}^{12} (2 + \frac{5}{2} r) = 24 + \frac{5}{2} \cdot \frac{12 \cdot 13}{2} = 219$

Q 16 :

$Let f (x) = \int_{0}^{x} g (t) \log_{e} (\frac{1 - t}{1 + t}) d t, where g is a continuous odd function.$

$If \int_{- π / 2}^{π / 2}$ $(f (x) + \frac{x^{2} \cos x}{1 + e^{x}})$ $d x = {(\frac{π}{α})}^{2} - α,$ $then α is equal to_____.$ [2024]

(2)

Given, $f (x) = \int_{0}^{x} g (t) \log_{e} (\frac{1 - t}{1 + t}) d t,$ where $g$ is continuous odd function.

$f' (x) = g (x) \log_{e} (\frac{1 - x}{1 + x}) \Rightarrow f' (- x) = g (- x) \log_{e} (\frac{1 + x}{1 - x})$

$= - g (x) [- \log_{e} (\frac{1 - x}{1 + x})] = f (x)$

$∴$ $f' (x)$ is an even function.

$\Rightarrow f (x)$ is an odd function.

Let $I = \int_{- π / 2}^{π / 2} (f (x) + \frac{x^{2} \cos x}{1 + e^{x}}) d x$

$= \int_{- π / 2}^{π / 2} f (x) d x + \int_{- π / 2}^{π / 2} \frac{x^{2} \cos x}{1 + e^{x}} d x$

$I = \int_{- π / 2}^{π / 2} \frac{x^{2} \cos x}{1 + e^{x}} d x$ ...(i) $[∵ \int_{- a}^{a} f (x) d x = 0, as f (x) is an odd function]$

$I = \int_{- π / 2}^{π / 2} \frac{x^{2} \cos x e^{x}}{1 + e^{x}} d x$ ...(ii)

Adding (i) and (ii), we get

$2 I = \int_{- π / 2}^{π / 2} x^{2} \cos x d x = 2 \int_{0}^{π / 2} x^{2} \cos x d x; I = \int_{0}^{π / 2} x^{2} \cos x d x$

$I = {(x^{2} \sin x)}_{0}^{π / 2} - \int_{0}^{π / 2} 2 x \sin x d x$

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

$= {(\frac{π}{2})}^{2} - 2 (1 - 0) = {(\frac{π}{2})}^{2} - 2$

$∴ α = 2$

Q 17 :

$Let the slope of the line 45 x + 5 y + 3 = 0 be 27 r_{1} + \frac{9 r_{2}}{2} for some r_{1}, r_{2} \in R .$ $Then \lim_{x \to 3}$ $(\int_{3}^{x} \frac{8 t^{2}}{\frac{3 r_{2} x}{2} - r_{2} x^{2} - r_{1} x^{3} - 3 x} d t)$ $is equal to______.$ [2024]

(12)

Slope of line $45 x + 5 y + 3 = 0$ is $\frac{- 45}{5} i . e, - 9$

Now, $- 9 = 27 (- 2) + \frac{9 \times 10}{2} .$ So, $r_{1} = - 2$ and $r_{2} = 10,$ we have $\lim_{x \to 3} \int_{3}^{x} \frac{8 t^{2}}{15 x - 10 x^{2} + 2 x^{3} - 3 x} d t$

$= \lim_{x \to 3} \frac{1}{15 x - 10 x^{2} + 2 x^{3} - 3 x} {[\frac{8 t^{3}}{3}]}_{3}^{x}$

$= \lim_{x \to 3} \frac{1}{15 x - 10 x^{2} + 2 x^{3} - 3 x} [\frac{8 x^{3}}{3} - 72]$

$= \lim_{x \to 3} \frac{8 x^{2}}{15 - 20 x + 6 x^{2} - 3} = \frac{72}{15 - 60 + 54 - 3} = 12$

Q 18 :

$If \int_{π / 6}^{π / 3} \sqrt{1 - \sin 2 x} d x = α + β \sqrt{2} + γ \sqrt{3}, where α, β, and γ are rational numbers, then 3 α + 4 β - γ is equal to \underline{}____.$ [2024]

(6)

Let $I = \int_{π / 6}^{π / 3} \sqrt{1 - \sin 2 x} d x$

$= \int_{π / 6}^{π / 3} \sqrt{{(\sin x - \cos x)}^{2}} d x \Rightarrow I = \int_{π / 6}^{π / 3}$ $| \sin x - \cos x |$ $d x$

$= \int_{π / 6}^{π / 4} (\cos x - \sin x) d x + \int_{π / 4}^{π / 3} (\sin x - \cos x) d x$

$= {[\sin x + \cos x]}_{π / 6}^{π / 4} + {[- \cos x - \sin x]}_{π / 4}^{π / 3}$

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

$= \frac{4}{\sqrt{2}} - 1 - \sqrt{3} = - 1 + 2 \sqrt{2} - \sqrt{3}$

$So, α = - 1, β = 2, γ = - 1$

$∴ 3 α + 4 β - γ = - 3 + 8 + 1 = 6$

Q 19 :

The value of $9 \int_{0}^{9} [\sqrt{\frac{10 x}{x + 1}}] d x,$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is ______ . [2024]

(155)

$I = 9 \int_{0}^{9} [\sqrt{\frac{10 x}{x + 1}}] d x$

Now, at $x = 0,$ $[\sqrt{\frac{10 x}{x + 1}}] = 0$ at $x = 9,$ $[\sqrt{\frac{10 x}{x + 1}}] = 3$

Integer values between 0 and 3 are 1 and 2

So $[\sqrt{\frac{10 x}{x + 1}}] = 1 and [\sqrt{\frac{10 x}{x + 1}}] = 2$

$\Rightarrow \frac{10 x}{x + 1} = 1 and \frac{10 x}{x + 1} = 4$

$\Rightarrow 10 x = x + 1 and 10 x = 4 x + 4$

$\Rightarrow x = \frac{1}{9} and x = \frac{2}{3}$

$I = 9 \int_{0}^{1 / 9} [\sqrt{\frac{10 x}{x + 1}}] d x + \int_{1 / 9}^{2 / 3} [\sqrt{\frac{10 x}{x + 1}}] d x + \int_{2 / 3}^{9} [\sqrt{\frac{10 x}{x + 1}}] d x$

$= 9 [\int_{0}^{1 / 9} 0 d x + \int_{1 / 9}^{2 / 3} 1 d x + \int_{2 / 3}^{9} 2 d x]$ [ $∵$ $[t]$ is a greatest integer function]

$= 9 [0 + \frac{2}{3} - \frac{1}{9} + 18 - \frac{4}{3}] = 9 [- \frac{2}{3} - \frac{1}{9} + 18]$

$= 162 - 6 - 1 = 155$

Q 20 :

Let $f : R \to R$ be a function defined by $f (x) = \frac{4^{x}}{4^{x} + 2} and M = \int_{f (a)}^{f (1 - a)} x \sin^{4} (x (1 - x)) d x, N = \int_{f (a)}^{f (1 - a)} \sin^{4} (x (1 - x)) d x, a \neq \frac{1}{2} .$ If $α M = β N, α, β \in N,$ then the least value of $α^{2} + β^{2}$ is equal to ______ . [2024]

(5)

We have, $f (a) = \frac{4^{a}}{4^{a} + 2}$

and $f (1 - a) = \frac{4^{(1 - a)}}{4^{1 - a} + 2} = \frac{4^{1} \cdot 4^{- a}}{4^{1} \cdot 4^{- a} + 2} = \frac{4^{1} \cdot 4^{- a}}{4^{- a} (4 + 2 \cdot 4^{a})} = \frac{4}{(4 + 2 \cdot 4^{a})} = \frac{2}{2 + 4^{a}}$

$\Rightarrow f (a) + f (1 - a) = \frac{4^{a}}{4^{a} + 2} + \frac{2}{2 + 4^{a}} = \frac{4^{a} + 2}{4^{a} + 2} = 1$

$M = \int_{f (a)}^{f (1 - a)} x \sin^{4} [x (1 - x)] d x,$

$N = \int_{f (a)}^{f (1 - a)} \sin^{4} [x (1 - x)] d x$

$\Rightarrow M = \int_{f (a)}^{f (1 - a)} (1 - x) \sin^{4} [(1 - x) (1 - 1 + x)] d x$

$\Rightarrow M = \int_{f (a)}^{f (1 - a)} (1 - x) \sin^{4} [x (1 - x)] d x$

$\Rightarrow \int_{f (a)}^{f (1 - a)} \sin^{4} [x (1 - x)] d x - \int_{f (a)}^{f (1 - a)} x \sin^{4} [x (1 - x)] d x$

$\Rightarrow M = N - M \Rightarrow 2 M = N \Rightarrow \frac{M}{N} = \frac{1}{2}$

$\Rightarrow \frac{M}{N} = \frac{β}{α} = \frac{1}{2} \Rightarrow β = 1, α = 2$

Then, the least value of $α^{2} + β^{2} = 4 + 1 = 5$

Q 21 :

If the integral $525 \int_{0}^{\frac{π}{2}} \sin 2 x \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2} - 64),$ then $n$ is equal to ______ . [2024]

(176)

Let $I = 525 \int_{0}^{\frac{π}{2}} \sin 2 x \cdot \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$

Put $\cos x = t^{2}$

$- \sin x d x = 2 t d t$

$I = - 525 \int_{1}^{0} 2 \sin x \cdot t^{2} \cdot t^{11} {(1 + t^{5})}^{\frac{1}{2}} \cdot \frac{2 t d t}{\sin x}$

$= 525 \times 4 \int_{0}^{1} t^{3} \cdot t^{11} {(1 + t^{5})}^{\frac{1}{2}} d t$

$I = 2100 \int_{0}^{1} t^{14} \sqrt{(1 + t^{5})} d t$

Let $1 + t^{5} = k^{2} \Rightarrow 5 t^{4} d t = 2 k d k \Rightarrow t^{4} d t = \frac{2 k d k}{5}$

$∴ I = 2100 \int_{1}^{\sqrt{2}} {(k^{2} - 1)}^{2} \cdot k \cdot \frac{2 k d k}{5}$

$= \frac{2100 \times 2}{5} \int_{1}^{\sqrt{2}} k^{2} {(k^{2} - 1)}^{2} d k$

$= 420 \times 2 \int_{1}^{\sqrt{2}} (k^{6} - 2 k^{4} + k^{2}) d k$

$= 840 {[\frac{k^{7}}{7} - \frac{2 k^{5}}{5} + \frac{k^{3}}{3}]}_{1}^{\sqrt{2}}$

$= 840 [\frac{8 \sqrt{2}}{7} - \frac{8 \sqrt{2}}{5} + \frac{2 \sqrt{2}}{3} - (\frac{1}{7} - \frac{2}{5} + \frac{1}{3})]$

$= 840 [\frac{22 \sqrt{2} - 8}{105}] = 8 [22 \sqrt{2} - 8] = 176 \sqrt{2} - 64$

So, $n = 176$

Q 22 :

$| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x |$ is equal to ________ . [2024]

(15)

$Let I = \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$

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

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

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

$= 2 π \cdot \frac{π}{4} \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} - π^{2} \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$

$= \frac{- π^{2}}{2} \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} = \frac{- π^{2}}{2} \int_{0}^{π / 2} \frac{\sin x \cos x}{1 - 2 \sin^{2} x + \cos^{2} x} d x$

$= \frac{- π^{2}}{2} \int_{0}^{π / 2} \frac{\sin 2 x}{1 + \cos^{2} 2 x} d x$

$Put \cos 2 x = t ∴ I = \frac{- π^{2}}{2} \int_{1}^{- 1} \frac{- d t}{2 (1 + t^{2})}$

$= \frac{- π^{2}}{2} \int_{0}^{1} \frac{d t}{1 + t^{2}} = \frac{- π^{2}}{2} {(\tan^{- 1} t)}_{0}^{1} = \frac{- π^{2}}{2} (\frac{π}{4}) = \frac{- π^{3}}{8}$

$∴$ $| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} |$ $= \frac{120}{π^{3}} \times \frac{π^{3}}{8} = 15$

Q 23 :

If $(a, b)$ be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1) and $I_{1} = \int_{a}^{b} x \sin (4 x - x^{2}) d x, I_{2} = \int_{a}^{b} \sin (4 x - x^{2}) d x,$ then $36 \frac{I_{1}}{I_{2}}$ is equal to : [2024]

66
88
72
80

1

2

3

4

(3)

Let the vertices of the triangle are $A (3, 1), B (1, 2)$ and $C (2, 3)$ .

$Slope of B C = \frac{3 - 2}{2 - 1} = 1$

Since, AD is perpendicular to BC.

Slope of AD = $- 1$

Equation of line $A D$ is

$y - 1 = - 1 (x - 3)$

Since, $(a, b)$ lies on AD.

$∴$ $b - 1 = - 1 (a - 3) \Rightarrow a + b = 4$

Now, $I_{1} = \int_{a}^{b} x \sin [(4 - x) x] d x$

$I_{1} = \int_{a}^{b} (a + b - x) \sin [(4 - (a + b - x)) (a + b - x)] d x$

$I_{1} = 4 \int_{a}^{b} \sin [(4 - x) x] d x - \int_{a}^{b} x \sin [(4 - x) x] d x$

$I_{1} = 4 I_{2} - I_{1}$ $[∵ \int_{a}^{b} \sin [(4 - x) x] d x = I_{2}]$

$\Rightarrow 2 I_{1} = 4 I_{2} \Rightarrow \frac{I_{1}}{I_{2}} = 2$ $∴$ $\frac{36 I_{1}}{I_{2}} = 36 \times 2 = 72$

Q 24 :

Let $S = (- 1, \infty)$ and $f : S \to R$ be defined as $f (x) = \int_{- 1}^{x} {(e^{t} - 1)}^{11} {(2 t - 1)}^{5} {(t - 2)}^{7} {(t - 3)}^{12} {(2 t - 10)}^{61} d t .$ Let $p =$ Sum of squares of the values of $x$ , where $f (x)$ attains local maxima on $S,$ and $q =$ Sum of the values of $x,$ where $f (x)$ attains local minima on $S$ . Then, the value of $p^{2} + 2 q$ is ____________ . [2024]

(27)

Given,

$f (x) = \int_{- 1}^{x} {(e^{t} - 1)}^{11} {(2 t - 1)}^{5} {(t - 2)}^{7} {(t - 3)}^{12} {(2 t - 10)}^{61} d t$

$f' (x) = {(e^{x} - 1)}^{11} {(2 x - 1)}^{15} {(x - 2)}^{7} {(x - 3)}^{12} {(2 x - 10)}^{61}$

$So, x = 0, x = \frac{1}{2}, x = 2, x = 3, x = 5$

$∴ p = 0 + 4 = 4$

$q = \frac{1}{2} + 5 = \frac{11}{2}$

$∴ p^{2} + 2 q = 16 + 11 = 27$

Q 25 :

$4 \int_{0}^{1} (\frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}) d x - 3 \log_{e} (\sqrt{3})$ is equal to : [2025]

$2 + \sqrt{2} - \log_{e} (1 + \sqrt{2})$
$2 + \sqrt{2} + \log_{e} (1 + \sqrt{2})$
$2 - \sqrt{2} - \log_{e} (1 + \sqrt{2})$
$2 - \sqrt{2} + \log_{e} (1 + \sqrt{2})$

1

2

3

4

(3)

Let $f (x) = \frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}$

$= \frac{\sqrt{3 + x^{2}} - \sqrt{1 + x^{2}}}{3 + x^{2} - 1 - x^{2}} = \frac{\sqrt{3 + x^{2}} - \sqrt{1 + x^{2}}}{2}$

$∴ 4 \int_{0}^{1} f (x) d x = 4 \int_{0}^{1} \frac{\sqrt{3 + x^{2}} - \sqrt{1 + x^{2}}}{2} d x$

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

$= 2 {[\frac{x}{2} \sqrt{3 + x^{2}} + \frac{3}{2} \log_{e} | x + \sqrt{3 + x^{2}} | - \frac{x}{2} \sqrt{1 + x^{2}} - \frac{1}{2} \log_{e} | x + \sqrt{1 + x^{2}} |]}_{0}^{1}$

$= 2 + 3 \log_{e} 3 - \sqrt{2} - \log_{e} (1 + \sqrt{2}) - 3 \log_{e} \sqrt{3}$

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

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

$= 2 - \sqrt{2} - \log_{e} (1 + \sqrt{2}) + 3 \log_{e} \sqrt{3}$

$∴ 4 \int_{0}^{1} f (x) d x - 3 \log_{e} \sqrt{3} = 2 - \sqrt{2} - \log_{e} (1 + \sqrt{2})$ .

Q 26 :

Let (a, b) be the point of intersection of the curve $x^{2} = 2 y$ and the straight line y – 2x – 6 = 0 in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9 x^{2}}{1 + 5^{x}} d x$ is equal to : [2025]

21
18
24
27

1

2

3

4

(3)

We have, $x^{2} = 2 y$ and y – 2x – 6 = 0

Substitute y = 2x + 6 in $x^{2} = 2 y$ , we get

$x^{2} = (4 x + 12)$

$\Rightarrow x^{2} - 4 x - 12 = 0 \Rightarrow (x - 6) (x + 2) = 0$

$\Rightarrow x = 6 or - 2 \Rightarrow y = 18 or 2$

$∴$ Intersection points are (6, 18) and (–2, 2)

Since, (a, b) is a point in second quadrant

$∴$ (a, b) = (–2, 2)

Now, $I = \int_{- 2}^{2} \frac{9 x^{2}}{1 + 5^{x}} d x$ ... (i)

$∴ I = \int_{- 2}^{2} \frac{9 x^{2}}{1 + 5^{- x}} d x = \int_{- 2}^{2} \frac{9 x^{2} 5^{x}}{5^{x} + 1} d x$ ... (ii)

On adding (i) and (ii), we get

$2 I = \int_{- 2}^{2} \frac{9 x^{2} (5^{x} + 1)}{(5^{x} + 1)} d x = \int_{- 2}^{2} 9 x^{2} d x$

$= {[\frac{9 x^{3}}{3}]}_{- 2}^{2} = 3 [8 - (- 8)] = 48$

$\Rightarrow I = \frac{48}{2} = 24$ .

Q 27 :

Let the domain of the function $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$ be (a, b). If $\int_{0}^{b - a} [x^{2}] d x = p - \sqrt{q} - \sqrt{r}$ , $p, q, r \in ℕ$ , gcd (p, q, r) = 1, where $[\cdot]$ is the greatest integer function, then p + q + r is equal to [2025]

8
11
9
10

1

2

3

4

(4)

We have, $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$

$∴$ f(x) is define when $\log_{4} \log_{6} (3 + 4 x - x^{2}) > 0$

$\Rightarrow \log_{6} (3 + 4 x - x^{2}) > 1$

$\Rightarrow 3 + 4 x - x^{2} > 6$

$\Rightarrow - x^{2} + 4 x - 3 > 0$

$\Rightarrow x^{2} - 4 x + 3 < 0$

$\Rightarrow (x - 1) (x - 3) < 0$

$\Rightarrow x \in (1, 3)$

$∴$ a = 1 and b = 3

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

$= 0 + | x |_{1}^{\sqrt{2}} + 2 | x |_{\sqrt{2}}^{\sqrt{3}} + 3 | x |_{\sqrt{3}}^{\sqrt{4}}$

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

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

$\Rightarrow$ p = 5, q = 2, r = 3

$∴$ p + q + r = 5 + 2 + 3 = 10.

Q 28 :

The integral $\int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ is equal to [2025]

$π^{2}$
$4 π^{2}$
$\frac{3 π^{2}}{2}$
$2 π^{2}$

1

2

3

4

(4)

Let $I = \int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ ... (i)

$\Rightarrow I = \int_{0}^{π} \frac{8 (π - x) d x}{4 \cos^{2} (π - x) + \sin^{2} (π - x)}$

$\Rightarrow I = \int_{0}^{π} \frac{(8 π - 8 x) d x}{4 \cos^{2} x + \sin^{2} x} d x$ ... (ii)

Adding (i) and (ii), we get

$2 I = 8 π \int_{0}^{π} \frac{d x}{4 \cos^{2} x + \sin^{2} x}$

$\Rightarrow 2 I = 8 π \times 2 \int_{0}^{π} \frac{\sec^{2} x}{4 + \tan^{2 x}} d x$

Put $\tan x = t \Rightarrow \sec^{2} xdx = dt$

$∴ I = 8 π \int_{0}^{\infty} \frac{d t}{4 + t^{2}} = 8 π \times \frac{1}{2} {[\tan^{- 1} \frac{t}{2}]}_{0}^{\infty}$

$= 4 π \times \frac{π}{2} = 2 π^{2}$ .

Q 29 :

The value of $\int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}} d x$ is equal to [2025]

$2 + \frac{2 \sqrt{2}}{3}$
$3 - \frac{2 \sqrt{2}}{3}$
$1 - \frac{2 \sqrt{2}}{3}$
$1 + \frac{2 \sqrt{2}}{3}$

1

2

3

4

(4)

Consider, $I = \int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}}$

$= \int_{- 1}^{1} \frac{e^{x} + e^{x} \sqrt{| x | - x} + e^{- x} \sqrt{| x - x |}}{e^{x} + e^{- x}} d x$

$= \int_{- 1}^{1} [\frac{e^{x}}{e^{x} + e^{- x}} + \frac{\sqrt{| x | - x} (e^{x} + e^{- x})}{e^{x} + e^{- x}}] d x$

$= \int_{- 1}^{1} \frac{e^{x}}{e^{x} + e^{- x}} d x + \int_{- 1}^{1} \sqrt{| x | - x} d x = I_{1} + I_{2}$

Now, $I_{1} = \int_{- 1}^{1} \frac{e^{x}}{e^{x} + e^{- x}} d x$ ... (i)

$\Rightarrow I_{1} = \int_{- 1}^{1} \frac{e^{- x}}{e^{- x} + e^{x}} d x$ ... (ii)

On adding (i) and (ii), we get

$\Rightarrow 2 I_{1} = \int_{- 1}^{1} \frac{e^{x} + e^{- x}}{e^{x} + e^{- x}} d x = \int_{- 1}^{1} 1 d x = 2 \Rightarrow I_{1} = 1$

and $I_{2} = \int_{- 1}^{1} \sqrt{| x | - x} d x = \int_{- 1}^{0} \sqrt{| x | - x} d x + \int_{0}^{1} \sqrt{| x | - x} d x$

$\Rightarrow I_{2} = \int_{- 1}^{0} \sqrt{- 2 x} d x + \int_{0}^{1} \sqrt{x - x} d x = \int_{- 1}^{0} \sqrt{- 2 x} d x$

Put $- 2 x = u \Rightarrow d x = - \frac{1}{2} d u$

$∴ I_{2} = \int_{2}^{0} \sqrt{u} (- \frac{1}{2} d u) = \frac{1}{2} \int_{0}^{2} \sqrt{u} d u = \frac{1}{2} {[\frac{u^{3 / 2}}{\frac{3}{2}}]}_{0}^{2} = \frac{2 \sqrt{2}}{3}$

$∴ I = I_{1} + I_{2} = 1 + \frac{2 \sqrt{2}}{3}$ .

Q 30 :

Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ , x > 0. If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is : [2025]

0
11
10
1

1

2

3

4

(2)

We have, $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ ... (i)

$\Rightarrow f (\frac{1}{x}) + 2 f (x) = \frac{1}{x^{2}} + 5$ ... (ii)

On solving (i) and (ii), we get

$f (x) = \frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}$

So, $α = \int_{1}^{2} f (x) d x = \int_{1}^{2} (\frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}) d x$

$= {(- \frac{2}{3 x} - \frac{x^{3}}{9} + \frac{5 x}{3})}_{1}^{2} = \frac{6}{3} - \frac{7}{9} = \frac{11}{9}$

Also, we have $2 g (x) - 3 g (\frac{1}{2}) = x$

$\Rightarrow 2 g (\frac{1}{2}) - 3 g (\frac{1}{2}) = \frac{1}{2} \Rightarrow g (\frac{1}{2}) = - \frac{1}{2}$

so, $g (x) = \frac{x}{2} - \frac{3}{4}$

$β = \int_{1}^{2} g (x) d x = \int_{1}^{2} (\frac{x}{2} - \frac{3}{4}) d x$

$= {(\frac{x^{2}}{4} - \frac{3 x}{4})}_{1}^{2} = \frac{3}{4} - \frac{3}{4} = 0$

$∴ 9 α + β = 9 (\frac{11}{9}) + 0 = 11$ .

Q 31 :

The integral $\int_{0}^{π} \frac{(x + 3) \sin x}{1 + 3 \cos^{2} x} d x$ is equal to [2025]

$\frac{π}{2 \sqrt{3}} (π + 4)$
$\frac{π}{\sqrt{3}} (π + 1)$
$\frac{π}{\sqrt{3}} (π + 2)$
$\frac{π}{3 \sqrt{3}} (π + 6)$

1

2

3

4

(4)

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

Replace x to $π$ – x,

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

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

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

$2 I = \int_{0}^{π} \frac{(x + 6) \sin x}{1 + 3 \cos^{2} x} d x$

Let cos x = t $\Rightarrow$ – sin x dx = dt

$2 I = (π + 6) \int_{1}^{- 1} \frac{- d t}{(1 + 3 t^{2})} = (\frac{π + 6}{3}) \int_{- 1}^{1} \frac{d t}{t^{2} + {(\frac{1}{\sqrt{3}})}^{2}}$

$\Rightarrow 2 I = (\frac{π + 6}{3}) \cdot \frac{1}{(\frac{1}{\sqrt{3}})} \cdot \tan^{- 1} (\frac{t}{1 / \sqrt{3}}) |_{- 1}^{1}$

$\Rightarrow 2 I = (\frac{π + 6}{\sqrt{3}}) [\tan^{- 1} (\sqrt{3}) - \tan^{- 1} (- \sqrt{3})]$

$\Rightarrow 2 I = (\frac{π + 6}{\sqrt{3}}) \cdot (\frac{π}{3} - (- \frac{π}{3})) = (\frac{π + 6}{\sqrt{3}} \cdot \frac{2 π}{3})$

$\Rightarrow I = \frac{π (π + 6)}{3 \sqrt{3}}$ .

Q 32 :

The integral $\int_{- 1}^{3 / 2} (| π^{2} x \sin (π x) |) d x$ is equal to : [2025]

$3 + 2 π$
$4 + π$
$1 + 3 π$
$2 + 3 π$

1

2

3

4

(3)

Let $I = \int_{- 1}^{3 / 2}$ $| π^{2} x \sin π x |$ $d x$

$= π^{2} [\int_{- 1}^{1} x \sin π x d x - \int_{1}^{3 / 2} x \sin π x d x]$

$= π^{2} [2 \int_{0}^{1} x \sin π x d x - \int_{1}^{3 / 2} x \sin π x d x]$

$= π^{2} [2 {[\frac{- x \cos π x}{π} + \frac{\sin π x}{π^{2}}]}_{0}^{1} - {[\frac{- x \cos π x}{π} + \frac{\sin π x}{π^{2}}]}_{1}^{3 / 2}]$

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

$= 3 π + 1$ .

Q 33 :

Let f(x) be a positive function and $I_{1} = \int_{- 1 / 2}^{1} 2 x f (2 x (1 - 2 x)) d x and I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I_{2}}{I_{1}}$ is equal to _______ [2025]

12
6
4
9

1

2

3

4

(3)

$I_{1} = \int_{- 1 / 2}^{1} 2 x f (2 x (1 - 2 x)) d x$

Let $2 x = t \Rightarrow 2 d x = d t$

when $x = \frac{- 1}{2}, t = - 1$ and

When for x = 1, t = 2

$∴ I_{1} = \frac{1}{2} \int_{- 1}^{2} t f (t (1 - t)) d t$

$\Rightarrow 2 I_{1} = \int_{- 1}^{2} (1 - t) f ((1 - t) (1 - (1 - t))) d t$

$\Rightarrow 2 I_{1} = \int_{- 1}^{2} f (t (1 - t)) d t - \int_{- 1}^{2} t f (t (1 - t)) d t$

$\Rightarrow 2 I_{1} = I_{2} - 2 I_{1} \Rightarrow 4 I_{1} = I_{2}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{4} i . e ., \frac{I_{2}}{I_{1}} = 4$

Q 34 :

Let for $f (x) = 7 \tan^{8} x + 7 \tan^{6} x - 3 \tan^{4} x - 3 \tan^{2} x$ , $I_{1} = \int_{0}^{π / 4} f (x) d x$ and $I_{2} = \int_{0}^{π / 4} x f (x) d x$ . Then $7 I_{1} + 12 I_{2}$ is equal to : [2025]

2
$π$
1
$2 π$

1

2

3

4

(3)

We have, $f (x) = 7 \tan^{8} x + 7 \tan^{6} x - 3 \tan^{4} x - 3 \tan^{2} x$

$= 7 \tan^{6} x (\tan^{2} x + 1) - 3 \tan^{2} x (\tan^{2} x + 1)$

$= (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x$

$I_{1} = \int_{0}^{π / 4} (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x d x$

Putting tan x = t $\Rightarrow \sec^{2} x d x = d t$

$I_{1} = \int_{0}^{1} (7 t^{6} - 3 t^{2}) d t = {[t^{7} - t^{3}]}_{0}^{1} = 0$

$I_{2} = \int_{0}^{π / 4} x (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x d x$

$= \int_{0}^{1} (7 t^{6} - 3 t^{2}) \tan^{- 1} t d t$

$= {[\tan^{- 1} t (t^{7} - t^{3})]}_{0}^{1} - \int_{0}^{1} \frac{t^{7} - t^{3}}{1 + t^{2}} d t = 0 - \int_{0}^{1} \frac{t^{3} (t^{4} - 1)}{1 + t^{2}} d t$

$= - \int_{0}^{1} t^{3} (t^{2} - 1) d t = - \int_{0}^{1} (t^{5} - t^{3}) d t = - {[\frac{t^{6}}{6} - \frac{t^{4}}{4}]}_{0}^{1} = \frac{1}{12}$

$∴ 7 I_{1} + 12 I_{2} = 0 + 12 (\frac{1}{12}) = 1$ .

Q 35 :

Let $f (x) = \int_{0}^{x^{2}} \frac{t^{2} - 8 t + 15}{e^{t}} d t, x \in R$ . Then the numbers of local maximum and local minimum points of f, respectively, are : [2025]

2 and 3
3 and 2
2 and 2
1 and 3

1

2

3

4

(1)

From Leibnitz theorem,

$f' (x) = (\frac{x^{4} - 8 x^{2} + 15}{e^{x^{2}}}) (2 x)$

$= \frac{(x^{2} - 3) (x^{2} - 5) (2 x)}{e^{x^{2}}}$

$= \frac{(x - \sqrt{3}) (x + \sqrt{3}) (x - \sqrt{5}) (x + \sqrt{5}) 2 x}{e^{x^{2}}}$

$\Rightarrow f' (x) = 0 \Rightarrow x = \pm \sqrt{3}, 0, \pm \sqrt{5}$

Maxima at $x \in {- \sqrt{3}, \sqrt{3}}$

Minima at $x \in {- \sqrt{5}, 0, \sqrt{5}}$

Hence, 2 points of maxima and 3 points of minima.

Q 36 :

The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}}}{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}} + e^{({{(6 - \log_{e} x)}^{2} + 1)}^{- 1}}}) d x$ is [2025]

$\log_{e} 2$
1
$e^{2}$
2

1

2

3

4

(2)

Let $\log_{e}$ $x = t \Rightarrow \frac{1}{x} d x = d t$

$\Rightarrow I = \int_{2}^{4} \frac{e^{\frac{1}{t^{2} + 1}}}{e^{\frac{1}{t^{2} + 1}} + e^{\frac{1}{{(6 - t)}^{2} + 1}}} d t$ ... (i)

$\Rightarrow I = \int_{2}^{4} \frac{e^{\frac{1}{{(6 - t)}^{2} + 1}}}{e^{\frac{1}{{(6 - t)}^{2} + 1}} + e^{\frac{1}{t^{2 + 1}}}} d t$ $[∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (b + a - x) d x]$ ... (ii)

Adding (i) and (ii), we get

$2 I = \int_{2}^{4} d t = {[t]}_{2}^{4} = 4 - 2 = 2 \Rightarrow 2 I = 2 \Rightarrow I = 1$ .

Q 37 :

If $I = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} d x$ , then $\int_{0}^{2 I} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ equals: [2025]

$\frac{π^{2}}{12}$
$\frac{π^{2}}{16}$
$\frac{π^{2}}{8}$
$\frac{π^{2}}{4}$

1

2

3

4

(2)

We have, $I = \int_{0}^{π / 2} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} d x$ ... (i)

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

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

$2 I = \int_{0}^{π / 2} d x = \frac{π}{2} \Rightarrow I = \frac{π}{4}$

Let, $I_{2} = \int_{0}^{2 I} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = \int_{0}^{π / 2} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ ... (iii)

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

Adding (iii) and (iv), we get

$I_{2} = \frac{π}{4} \int_{0}^{π / 2} \frac{\tan x \sec^{2} x d x}{\tan^{4} x + 1}$

Now, put $\tan^{2} x = t$

$I_{2} = \frac{π}{8} \int_{0}^{\infty} \frac{d t}{t^{2} + 1} = \frac{π}{8} \tan^{- 1} (t) |_{0}^{\infty} = \frac{π}{8} \cdot \frac{π}{2} = \frac{π^{2}}{16}$

Q 38 :

If $I (m, n) = \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} d x$ , m, n > 0, then $I$ (9, 14) + $I$ (10, 13) is [2025]

$I (19, 27)$
$I (1, 13)$
$I (9, 13)$
$I (9, 1)$

1

2

3

4

(3)

$I (m, n) = \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} d x, m, n > 0$

Let $x = \sin^{2} θ \Rightarrow d x = 2 sinθ cosθ dθ$

$∴ I (m, n) = \int_{0}^{π / 2} \sin^{2 m - 2} θ \times {(1 - \sin^{2} θ)}^{n - 1} \times 2 sinθcosθ d θ$

$= 2 \int_{0}^{π / 2} \sin^{2 m - 1} θ \times \cos^{2 n - 1} θ d θ$

$∴ I (9, 14) + I (10, 13) = 2 \int_{0}^{π / 2} \sin^{17} θ \cos^{27} θ dθ + 2 \int_{0}^{π / 2} \sin^{19} θ \cos^{25} θ dθ$

$= 2 \int_{0}^{π / 2} \sin^{17} θ \cos^{25} θ (\cos^{2} θ + \sin^{2} θ) d θ$

$= 2 \int_{0}^{π / 2} \sin^{17} θ \cos^{25} θ dθ = I (9, 13)$ .

Q 39 :

If $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x^{2} \cos^{2} x}{(1 + e^{x})} d x = π (α π^{2} + β)$ , $α, β \in Z$ , then ${(α + β)}^{2}$ be equals [2025]

64
144
100
196

1

2

3

4

(3)

Let $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x^{2} \cos^{2} x}{(1 + e^{x})} d x$ ... (i)

$\Rightarrow I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x^{2} \cos^{2} x}{(1 + e^{- x})} d x$ ... (ii)

$[∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x]$

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

$2 I = \int_{- π / 2}^{π / 2} \frac{96 x^{2} \cos^{2} x}{1 + e^{x}} d x + \int_{- π / 2}^{π / 2} \frac{e^{x} (96 x^{2} \cos^{2} x)}{e^{x} + 1} d x$

$= 2 \int_{0}^{π / 2} 96 x^{2} \cos^{2} x d x$

$\Rightarrow I = \int_{0}^{π / 2} 96 x^{2} \cos^{2} x d x$

$= 48 \int_{0}^{π / 2} x^{2} (1 + \cos 2 x) d x$

$= 48 [{[\frac{x^{3}}{3}]}_{0}^{π / 2} + \int_{0}^{π / 2} x^{2} \cos 2 x d x]$

$\Rightarrow I = 48 \cdot \frac{π^{3}}{24} + {[\frac{x^{2} \sin 2 x}{2}]}_{0}^{π / 2} - \int_{0}^{π / 2} x \sin 2 x$

$= 48 [\frac{π^{3}}{24} + {[\frac{x \cos 2 x}{2}]}_{0}^{π / 2} - \frac{1}{4} {[\sin 2 x]}_{0}^{π / 2}]$

$\Rightarrow I = 48 [\frac{π^{3}}{24} - \frac{π}{4}] = π (2 π^{2} - 12)$

So, $α = 2, β = - 12$

$∴ {(α + β)}^{2} = {(2 - 12)}^{2} = {(- 10)}^{2} = 100$ .

Q 40 :

Let f be a real valued continuous function defined on the positive real axis such that $g (x) = \int_{0}^{x} t f (t) d t$ . If $g (x^{3}) = x^{6} + x^{7}$ , then value of $\sum_{r = 1}^{15} f (r^{3})$ is : [2025]

270
340
320
310

1

2

3

4

(4)

We have, $g (x) = \int_{0}^{x} t f (t) d t$

$g (x) = x^{2} + x^{7 / 3} \Rightarrow g' (x) = 2 x + \frac{7}{3} x^{4 / 3}$

As, $f (x) = \frac{g' (x)}{x} = 2 + \frac{7}{3} x^{1 / 3} \Rightarrow f (r^{3}) = 2 + \frac{7 r}{3}$

$∴ \sum_{r = 1}^{15} (2 + \frac{7}{3} r) = 30 + \frac{7}{3} [1 + 2 + . . . + 15] = 310$ .

Topic Question Set

Q 11 :

Let $[t]$ denote the largest integer less than or equal to $t .$ If $\int_{0}^{3} ([x^{2}] + [\frac{x^{2}}{2}]) d x = a + b \sqrt{2} - \sqrt{3} - \sqrt{5} + c \sqrt{6} - \sqrt{7},$ where $a, b, c \in Z,$ then $a + b + c$ is equal to _______. [2024]

Q 14 :

$If \int_{- π / 2}^{π / 2} \frac{8 \sqrt{2} \cos x d x}{(1 + e^{\sin x}) (1 + \sin^{4} x)} = α π + β \log_{e} (3 + 2 \sqrt{2}),$ $where α, β are integers, then α^{2} + β^{2} equals ____ .$ [2024]

Q 15 :

$Let f : (0, \infty) \to R and F (x) = \int_{0}^{x} t f (t) d t . If F (x^{2}) = x^{4} + x^{5}, then \sum_{r = 1}^{12} f (r^{2}) is equal to____.$ [2024]

Q 16 :

$Let f (x) = \int_{0}^{x} g (t) \log_{e} (\frac{1 - t}{1 + t}) d t, where g is a continuous odd function.$

$If \int_{- π / 2}^{π / 2}$ $(f (x) + \frac{x^{2} \cos x}{1 + e^{x}})$ $d x = {(\frac{π}{α})}^{2} - α,$ $then α is equal to_____.$ [2024]

Q 17 :

$Let the slope of the line 45 x + 5 y + 3 = 0 be 27 r_{1} + \frac{9 r_{2}}{2} for some r_{1}, r_{2} \in R .$ $Then \lim_{x \to 3}$ $(\int_{3}^{x} \frac{8 t^{2}}{\frac{3 r_{2} x}{2} - r_{2} x^{2} - r_{1} x^{3} - 3 x} d t)$ $is equal to______.$ [2024]

Q 18 :

$If \int_{π / 6}^{π / 3} \sqrt{1 - \sin 2 x} d x = α + β \sqrt{2} + γ \sqrt{3}, where α, β, and γ are rational numbers, then 3 α + 4 β - γ is equal to \underline{}____.$ [2024]

Q 19 :

The value of $9 \int_{0}^{9} [\sqrt{\frac{10 x}{x + 1}}] d x,$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is ______ . [2024]

Q 21 :

If the integral $525 \int_{0}^{\frac{π}{2}} \sin 2 x \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2} - 64),$ then $n$ is equal to ______ . [2024]

Q 22 :

$| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x |$ is equal to ________ . [2024]

Q 23 :

If $(a, b)$ be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1) and $I_{1} = \int_{a}^{b} x \sin (4 x - x^{2}) d x, I_{2} = \int_{a}^{b} \sin (4 x - x^{2}) d x,$ then $36 \frac{I_{1}}{I_{2}}$ is equal to : [2024]

Q 25 :

$4 \int_{0}^{1} (\frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}) d x - 3 \log_{e} (\sqrt{3})$ is equal to : [2025]

Q 26 :

Let (a, b) be the point of intersection of the curve $x^{2} = 2 y$ and the straight line y – 2x – 6 = 0 in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9 x^{2}}{1 + 5^{x}} d x$ is equal to : [2025]

Q 27 :

Let the domain of the function $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$ be (a, b). If $\int_{0}^{b - a} [x^{2}] d x = p - \sqrt{q} - \sqrt{r}$ , $p, q, r \in ℕ$ , gcd (p, q, r) = 1, where $[\cdot]$ is the greatest integer function, then p + q + r is equal to [2025]

Q 28 :

The integral $\int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ is equal to [2025]

Q 29 :

The value of $\int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}} d x$ is equal to [2025]

Q 30 :

Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ , x > 0. If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is : [2025]

Q 31 :

The integral $\int_{0}^{π} \frac{(x + 3) \sin x}{1 + 3 \cos^{2} x} d x$ is equal to [2025]

Q 32 :

The integral $\int_{- 1}^{3 / 2} (| π^{2} x \sin (π x) |) d x$ is equal to : [2025]

Q 33 :

Let f(x) be a positive function and $I_{1} = \int_{- 1 / 2}^{1} 2 x f (2 x (1 - 2 x)) d x and I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I_{2}}{I_{1}}$ is equal to _______ [2025]

Q 34 :

Let for $f (x) = 7 \tan^{8} x + 7 \tan^{6} x - 3 \tan^{4} x - 3 \tan^{2} x$ , $I_{1} = \int_{0}^{π / 4} f (x) d x$ and $I_{2} = \int_{0}^{π / 4} x f (x) d x$ . Then $7 I_{1} + 12 I_{2}$ is equal to : [2025]

Q 35 :

Let $f (x) = \int_{0}^{x^{2}} \frac{t^{2} - 8 t + 15}{e^{t}} d t, x \in R$ . Then the numbers of local maximum and local minimum points of f, respectively, are : [2025]

Q 36 :

The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}}}{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}} + e^{({{(6 - \log_{e} x)}^{2} + 1)}^{- 1}}}) d x$ is [2025]

Q 37 :

If $I = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} d x$ , then $\int_{0}^{2 I} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ equals: [2025]

Q 38 :

If $I (m, n) = \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} d x$ , m, n > 0, then $I$ (9, 14) + $I$ (10, 13) is [2025]

Q 39 :

If $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x^{2} \cos^{2} x}{(1 + e^{x})} d x = π (α π^{2} + β)$ , $α, β \in Z$ , then ${(α + β)}^{2}$ be equals [2025]

Q 40 :

Let f be a real valued continuous function defined on the positive real axis such that $g (x) = \int_{0}^{x} t f (t) d t$ . If $g (x^{3}) = x^{6} + x^{7}$ , then value of $\sum_{r = 1}^{15} f (r^{3})$ is : [2025]

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Topic Question Set

Q 11 : Let [t] denote the largest integer less than or equal to t. If ∫03([x2]+[x22])dx=a+b2-3-5+c6-7, where a,b,c∈Z, then a+b+c is equal to _______. [2024]

Q 12 : Let limn→∞(nn4+1-2n(n2+1)n4+1+nn4+16-8n(n2+4)n4+16+⋯+nn4+n4-2n·n2(n2+n2)n4+n4) be πk, using only the principal values of the inverse trigonometric functions. Then k2 is equal to ______ . [2024]

Q 13 : Consider the matrices : A=[2-53m], B=[20m] and X=[xy]. Let the set of all m, for which the system of equations AX=B has a negative solution (i.e., x<0 and y<0), be the interval (a,b). Then 8∫ab|A| dm is equal to ______ . [2024]

Q 14 : If ∫-π/2π/282cosx dx(1+esinx)(1+sin4x)=απ+βloge(3+22), where α,β are integers, then α2+β2 equals ____ . [2024]

Q 15 : Let f:(0,∞)→R and F(x)=∫0xtf(t) dt. If F(x2)=x4+x5, then ∑r=112f(r2) is equal to ____. [2024]

Q 16 : Let f(x)=∫0xg(t)loge(1-t1+t)dt, where g is a continuous odd function. If ∫-π/2π/2(f(x)+x2cosx1+ex)dx=(πα)2-α, then α is equal to _____. [2024]

Q 17 : Let the slope of the line 45x+5y+3=0 be 27r1+9r22 for some r1,r2∈R. Then limx→3 (∫3x8t23r2x2-r2x2-r1x3-3x dt) is equal to ______ . [2024]

Q 18 : If∫π/6π/31-sin2x dx=α+β2+γ3, where α,β, and γ are rational numbers, then 3α+4β-γ is equal to ¯____. [2024]

Q 19 : The value of 9∫09[10xx+1]dx, where [t] denotes the greatest integer less than or equal to t, is ______ . [2024]

Q 20 : Let f:R→R be a function defined by f(x)=4x4x+2 and M=∫f(a)f(1-a)xsin4(x(1-x))dx, N=∫f(a)f(1-a)sin4(x(1-x))dx, a≠12. If αM=βN, α,β∈N, then the least value of α2+β2 is equal to ______ . [2024]

Q 21 : If the integral 525∫0π2sin2xcos112x(1+cos52x)12dx is equal to (n2-64), then n is equal to ______ . [2024]

Q 22 : |120π3∫0πx2sinxcosxsin4x+cos4x dx| is equal to ________ . [2024]

Q 23 : If (a,b) be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1) and I1=∫abxsin(4x-x2) dx, I2=∫absin(4x-x2) dx, then 36I1I2 is equal to : [2024]

Q 24 : Let S=(-1,∞) and f:S→R be defined as f(x)=∫-1x(et-1)11(2t-1)5(t-2)7(t-3)12(2t-10)61 dt. Let p=Sum of squares of the values of x, where f(x) attains local maxima on S, and q= Sum of the values of x, where f(x) attains local minima on S. Then, the value of p2+2q is ____________ . [2024]

Q 25 : 4∫01(13+x2+1+x2)dx–3 loge (3) is equal to : [2025]

Q 26 : Let (a, b) be the point of intersection of the curve x2=2y and the straight line y – 2x – 6 = 0 in the second quadrant. Then the integral I=∫ab9x21+5xdx is equal to : [2025]

Q 27 : Let the domain of the function f(x)=log2 log4 log6 (3+4x–x2) be (a, b). If ∫0b–a[x2]dx=p–q–r, p, q, r∈ℕ, gcd (p, q, r) = 1, where [·] is the greatest integer function, then p + q + r is equal to [2025]

Q 28 : The integral ∫0π8 xdx4cos2x+sin2x is equal to [2025]

Q 29 : The value of ∫–11(1+|x|–x)ex+(|x|–x)e–xex+e–xdx is equal to [2025]

Q 30 : Let f(x)+2f(1x)=x2+5 and 2g(x)–3g(12)=x, x > 0. If α=∫12f(x)dx, and β=∫12g(x)dx, then the value of 9α+β is : [2025]

Q 31 : The integral ∫0π(x+3) sin x1+3 cos2 xdx is equal to [2025]

Q 32 : The integral ∫–13/2(|π2xsin(πx)|)dx is equal to : [2025]

Q 33 : Let f(x) be a positive function and I1=∫–1/212xf(2x(1–2x))dx and I2=∫–12f(x(1–x))dx. Then the value of I2I1 is equal to _______ [2025]

Q 34 : Let for f(x)=7tan8x+7tan6x–3tan4x–3tan2x, I1=∫0π/4f(x)dx and I2=∫0π/4xf(x)dx. Then 7I1+12I2 is equal to : [2025]

Q 35 : Let f(x)=∫0x2t2–8t+15etdt, x∈R. Then the numbers of local maximum and local minimum points of f, respectively, are : [2025]

Q 36 : The value of ∫e2e41x(e((logex)2+1)–1e((logex)2+1)–1+e((6–logex)2+1)–1)dx is [2025]

Q 37 : If I=∫0π2sin32xsin32x+cos32xdx, then ∫02Ixsinx cosxsin4x+cos4xdx equals: [2025]

Q 38 : If I(m,n)=∫01xm–1(1–x)n–1dx, m, n > 0, then I(9, 14) + I(10, 13) is [2025]

Q 39 : If ∫–π2π296x2cos2x(1+ex)dx=π(απ2+β), α, β∈Z, then(α+β)2 be equals [2025]

Q 40 : Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫0xtf(t)dt. If g(x3)=x6+x7, then value of ∑r=115f(r3) is : [2025]

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Q 11 :

Let $[t]$ denote the largest integer less than or equal to $t .$ If $\int_{0}^{3} ([x^{2}] + [\frac{x^{2}}{2}]) d x = a + b \sqrt{2} - \sqrt{3} - \sqrt{5} + c \sqrt{6} - \sqrt{7},$ where $a, b, c \in Z,$ then $a + b + c$ is equal to _______. [2024]

Q 14 :

$If \int_{- π / 2}^{π / 2} \frac{8 \sqrt{2} \cos x d x}{(1 + e^{\sin x}) (1 + \sin^{4} x)} = α π + β \log_{e} (3 + 2 \sqrt{2}),$ $where α, β are integers, then α^{2} + β^{2} equals ____ .$ [2024]

Q 15 :

$Let f : (0, \infty) \to R and F (x) = \int_{0}^{x} t f (t) d t . If F (x^{2}) = x^{4} + x^{5}, then \sum_{r = 1}^{12} f (r^{2}) is equal to____.$ [2024]

Q 16 :

$Let f (x) = \int_{0}^{x} g (t) \log_{e} (\frac{1 - t}{1 + t}) d t, where g is a continuous odd function.$

$If \int_{- π / 2}^{π / 2}$ $(f (x) + \frac{x^{2} \cos x}{1 + e^{x}})$ $d x = {(\frac{π}{α})}^{2} - α,$ $then α is equal to_____.$ [2024]

Q 17 :

$Let the slope of the line 45 x + 5 y + 3 = 0 be 27 r_{1} + \frac{9 r_{2}}{2} for some r_{1}, r_{2} \in R .$ $Then \lim_{x \to 3}$ $(\int_{3}^{x} \frac{8 t^{2}}{\frac{3 r_{2} x}{2} - r_{2} x^{2} - r_{1} x^{3} - 3 x} d t)$ $is equal to______.$ [2024]

Q 18 :

$If \int_{π / 6}^{π / 3} \sqrt{1 - \sin 2 x} d x = α + β \sqrt{2} + γ \sqrt{3}, where α, β, and γ are rational numbers, then 3 α + 4 β - γ is equal to \underline{}____.$ [2024]

Q 19 :

The value of $9 \int_{0}^{9} [\sqrt{\frac{10 x}{x + 1}}] d x,$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is ______ . [2024]

Q 21 :

If the integral $525 \int_{0}^{\frac{π}{2}} \sin 2 x \cos^{\frac{11}{2}} x {(1 + \cos^{\frac{5}{2}} x)}^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2} - 64),$ then $n$ is equal to ______ . [2024]

Q 22 :

$| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x |$ is equal to ________ . [2024]

Q 23 :

If $(a, b)$ be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1) and $I_{1} = \int_{a}^{b} x \sin (4 x - x^{2}) d x, I_{2} = \int_{a}^{b} \sin (4 x - x^{2}) d x,$ then $36 \frac{I_{1}}{I_{2}}$ is equal to : [2024]

Q 25 :

$4 \int_{0}^{1} (\frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}) d x - 3 \log_{e} (\sqrt{3})$ is equal to : [2025]

Q 26 :

Let (a, b) be the point of intersection of the curve $x^{2} = 2 y$ and the straight line y – 2x – 6 = 0 in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9 x^{2}}{1 + 5^{x}} d x$ is equal to : [2025]

Q 27 :

Let the domain of the function $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$ be (a, b). If $\int_{0}^{b - a} [x^{2}] d x = p - \sqrt{q} - \sqrt{r}$ , $p, q, r \in ℕ$ , gcd (p, q, r) = 1, where $[\cdot]$ is the greatest integer function, then p + q + r is equal to [2025]

Q 28 :

The integral $\int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ is equal to [2025]

Q 29 :

The value of $\int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}} d x$ is equal to [2025]

Q 30 :

Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ , x > 0. If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is : [2025]

Q 31 :

The integral $\int_{0}^{π} \frac{(x + 3) \sin x}{1 + 3 \cos^{2} x} d x$ is equal to [2025]

Q 32 :

The integral $\int_{- 1}^{3 / 2} (| π^{2} x \sin (π x) |) d x$ is equal to : [2025]

Q 33 :

Let f(x) be a positive function and $I_{1} = \int_{- 1 / 2}^{1} 2 x f (2 x (1 - 2 x)) d x and I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I_{2}}{I_{1}}$ is equal to _______ [2025]

Q 34 :

Let for $f (x) = 7 \tan^{8} x + 7 \tan^{6} x - 3 \tan^{4} x - 3 \tan^{2} x$ , $I_{1} = \int_{0}^{π / 4} f (x) d x$ and $I_{2} = \int_{0}^{π / 4} x f (x) d x$ . Then $7 I_{1} + 12 I_{2}$ is equal to : [2025]

Q 35 :

Let $f (x) = \int_{0}^{x^{2}} \frac{t^{2} - 8 t + 15}{e^{t}} d t, x \in R$ . Then the numbers of local maximum and local minimum points of f, respectively, are : [2025]

Q 36 :

The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}}}{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}} + e^{({{(6 - \log_{e} x)}^{2} + 1)}^{- 1}}}) d x$ is [2025]

Q 37 :

If $I = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} d x$ , then $\int_{0}^{2 I} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ equals: [2025]

Q 38 :

If $I (m, n) = \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} d x$ , m, n > 0, then $I$ (9, 14) + $I$ (10, 13) is [2025]

Q 39 :

If $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x^{2} \cos^{2} x}{(1 + e^{x})} d x = π (α π^{2} + β)$ , $α, β \in Z$ , then ${(α + β)}^{2}$ be equals [2025]

Q 40 :

Let f be a real valued continuous function defined on the positive real axis such that $g (x) = \int_{0}^{x} t f (t) d t$ . If $g (x^{3}) = x^{6} + x^{7}$ , then value of $\sum_{r = 1}^{15} f (r^{3})$ is : [2025]