Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int\limits_0^x e^{(x-t)} f(t)\,dt$, $x \in \mathbb{R}$ and let
$g(x) = \int\limits_0^x (f(t) + 2)^{15} (t - 4)^6 (t + 12)^{17}\,dt$, $x \in \mathbb{R}$.
If $p$ and $q$ are respectively the points of local minima and local maxima of $g$, then the value of $|p+q|$ is equal to ________.
Explanation:
The given equation is:
$ f(x)=1-2 x+\int_0^x e^{(x-t)} f(t) d t $
Since $e^{(x-t)}=e^x \cdot e^{-t}$, we can move $e^x$ outside the integral because it is constant relative to $t$
$ f(x)=1-2 x+e^x \int_0^x e^{-t} f(t) d t $
Differentiate both sides with respect to $x$ using the Leibniz Rule and the product rule:
$ f^{\prime}(x)=-2+\left[e^x \int_0^x e^{-t} f(t) d t+e^x\left(e^{-x} f(x)\right)\right] $
Now, we use substitution from our original equation. Notice that $e^x \int_0^x e^{-t} f(t) d t=f(x)- (1-2 x)$. Substituting this back into the derivative expression:
$ f^{\prime}(x)=-2+[f(x)-1+2 x]+f(x) $
$\Rightarrow $ $ f^{\prime}(x)=2 f(x)+2 x-3 $
Combine the terms to get the linear differential equatio in the form of $\frac{d f(x)}{d x}+P(x) f(x)=Q(x)$
$ \frac{d f(x)}{d x}-2 f(x)=2 x-3 $
Solve for $f(x)$
Integrating Factor (IF): $e^{\int-2 d x}=e^{-2 x}$
General Solution: $f(x) \cdot e^{-2 x}=\int(2 x-3) e^{-2 x} d x$
By integrating the right side (Integration by Parts), we get $f(x)=1-x+C e^{2 x}$.
Using $f(0)=1$ (from the original equation), we find $C=0$.
Result: $f(x)=1-x$.
Find the Extrema of $g(x)$
$ g(x)=\int_0^x(f(t)+2)^{15}(t-4)^6(t+12)^{17} d t $
Differentiate both sides with respect to $x$ using the Leibniz Rule
$ g^{\prime}(x)=(f(x)+2)^{15}(x-4)^6(x+12)^{17} $
Substitute $f(x)=1-x$ :
$ \begin{gathered} g^{\prime}(x)=(1-x+2)^{15}(x-4)^6(x+12)^{17} \\ g^{\prime}(x)=(3-x)^{15}(x-4)^6(x+12)^{17} \end{gathered} $
Critical points: $x=3,4,-12$.
Plot on number line to find local maxima and minima.
Local Minima ( $p$ ): At $x=-12, g^{\prime}(x)$ changes sign from negative to positive. Thus, $p=-12$.
Local Maxima ( $q$ ): At $x=3, g^{\prime}(x)$ changes sign from positive to negative. Thus, $q=3$.
$x=4$ : No sign change occurs (even power 6), so it is not a local extremum.
Final Answer
$ |p+q|=|-12+3|=|-9|=9 $
The value of $|p+q|$ is 9 .
$ \text { The value of } \sum\limits_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int\limits_0^r x|\sin \pi x| d x\right)}\right|\right) \text { is } $
Explanation:
Let $J_r=\int\limits_0^r x \cdot|\sin \pi x| d x$
So we need to calculate
$ I=\sum\limits_{r=1}^{20} \sqrt{\pi J_r} $
So,
$ I=\sqrt{\pi}\left(J_1+J_2+J_3+\cdots+J_{20}\right) $
Let $\pi x=t \Rightarrow \pi d x=d t$
$ J_r=\int\limits_0^{\pi r} \frac{t}{\pi}|\sin t| \frac{d t}{\pi}=\frac{1}{\pi^2} \int\limits_0^{\pi r} t|\sin t| d t $
Period of $|\sin t|$ is $\pi$, so break the interval
$\begin{aligned} & \int\limits_0^{\pi r} t|\sin t|=\int\limits_0^\pi t|\sin t| d t+\int\limits_\pi^{2 \pi} t|\sin t| d t+\int\limits_{2 \pi}^{3 \pi} t|\sin t| d t+\cdots \int\limits_{(r-1) \pi}^{r \pi} t|\sin t| d t \\ & \text { Now } \\ & \quad|\sin t|=\sin t \text { for Ist and II Quadrant } \\ & \quad|\sin t|=-\sin t \text { for III and IV Quadrant }\end{aligned}$
$ \int\limits_0^{r \pi} t|\sin t| d t=\int\limits_0^\pi t \sin t d t-\int\limits_\pi^{2 \pi} t \sin t d t+\int\limits_{2 \pi}^{3 \pi} t \sin t d t-\int\limits_{3 \pi}^{4 \pi} t \sin t \ldots(-1)^{r-1} \int\limits_{(r-1) \pi}^{r \pi} t \sin t d t $
Solving $\int t \cdot \sin t$ using by parts
$ \int u \cdot v d x=u \cdot \int v d x-\int\left(u^{\prime} \int v d v\right) d x $
$\Rightarrow $ $ \int t \cdot \sin t=(-t \cos t)-\int 1 \cdot(-\cos t)=-t \cos t+\sin t $
Substituting the limits
$\Rightarrow $ $\begin{aligned} & \int\limits_0^{\pi r} t \cdot|\sin t| d t=[-t \cos t+\sin t]_0^\pi-[-t \cos t+\sin t]_\pi^{2 \pi}+[-t \cos t+ \sin t]_{2 \pi}^{3 \pi}-[-t \cos t+\sin t]_{3 \pi}^{4 \pi} \cdots \int\limits_{(r-1) \pi}^{r \pi} t \sin t(-1)^{r-1} d t\end{aligned}$
$\begin{aligned} & =[\pi+0-0]-[-2 \pi+0-\{\pi+0\}]+[3 \pi+0-\{-2 \pi\}+0]-[-4 \pi- \{-3 \pi+0\}]+\cdots \int\limits_{(r-1) \pi}^{r \pi} t \sin t(-1)^{r-1} d t\end{aligned}$
$\Rightarrow $ $\int\limits_0^{r \pi} t|\sin t| d t=[\pi+0]+[2 \pi+\pi]+[3 \pi+2 \pi]+[4 \pi+3 \pi] \ldots[r \pi+(r-1) \pi]$
$ \begin{aligned} & =\pi+3 \pi+5 \pi+7 \pi+\ldots(2 r-1) \pi \\\\ & =\pi(1+3+5+7+\ldots(2 r-1)) \end{aligned} $
Sum of first $r$ odd numbers is $r^2$
$ \int\limits_0^{r \pi} t|\sin t| d t=\pi r^2 $
$ J_r=\frac{1}{\pi^2} \pi r^2=\frac{r^2}{\pi} $
so,
$ I=\sum\limits_{r=1}^{20} \sqrt{\pi \cdot J_r}=\sum\limits_{r=1}^{20} \sqrt{\pi \cdot \frac{r^2}{\pi}} $
$\Rightarrow $ $I=\sum\limits_{r=1}^{20} r=1+2+3+\ldots 20$
$\Rightarrow $ $I=\frac{20 \times 21}{2}=210$
Explanation:
$ \begin{aligned} & f(x)=e^x+\int_0^1\left(y+x e^x\right) f(y) d y \\ & f(x)=e^x+\int_0^1 y f(y) d y+\int_0^1 x e^x f(y) d y \end{aligned} $
Integration is w.r.t $y$, so variable $x$ is taken as a constant.
$ f(x)=e^x+\int_0^1 y f(y) d y+x e^x \int_0^1 f(y) d y . $
Let $A=\int_0^1 y f(y) d y$ and $B=\int_0^1 f(y) d y$, both are constants.
So $f(x)$ becomes,
$ f(x)=e^x+A+B x e^x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)$
Or, $f(y)=e^y+A+B y e^y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)$
Solve, $B=\int_0^1 f(y) d y$.
$ B=\int_0^1\left[e^y+A+B y e^y\right] d y,\left\{\text { from (2) } f(y)=e^y+A+B y e^y\right\} . $
$ B=\int_0^1\left[e^y+A+B y e^y\right] d y \text {, \{from (2) } f(y)=e^y+A+B y e^y \text { \}. } $
Using integration by parts, $\int y e^y d y=y e^y-\int e^y=e^y(y-1)$.
$ \begin{aligned} & B=\left[e^y+A y+B e^y(y-1)\right]_0^1 . \\ & B=\left[e^1+A \cdot 1+0-e^0-0-B \cdot e^0(-1)\right] . \\ & B=e+A-1+B \Longrightarrow A=1-e . \end{aligned} $
Finding $f(0)$ using equation (1).
$ \begin{aligned} & f(0)=1+A+0 . \\ & f(0)=1+1-e,\{\text { substitute } A=1-e\} . \\ & f(0)+e=2\end{aligned} $
Let a differentiable function $f$ satisfy the equation $\int_0^{36} f\left(\frac{t x}{36}\right) d t=4 \alpha f(x)$. If $y=f(x)$ is a standard parabola passing through the points $(2,1)$ and $(-4, \beta)$, then $\beta^\alpha$ is equal to $\_\_\_\_$ .
Explanation:
Given, $\int_0^{36} f\left(\frac{t x}{36}\right) d t=4 \alpha f(x)$
Let $\frac{t x}{36}=u \Rightarrow \frac{x}{36} d t=d u$
$ \int_0^x f(u) \frac{36}{x} d u=4 \alpha f(x) $
Integration is w.r.t $u$, so treat $x$ as a constant
$ \int_0^x f(u) d u=\frac{4 x}{36} \alpha f(x)=\frac{x}{9} \alpha f(x) $
Now, differentiate both sides w.r.t $x$, use Leibniz theorem in LHS:
$ 1 \cdot f(x)=\frac{\alpha}{9}\left[1 \cdot f(x)+x f^{\prime}(x)\right] $
$\Rightarrow $ $9 f(x)=\alpha f(x)+\alpha x f^{\prime}(x)$
$\Rightarrow $ $\alpha x f^{\prime}(x)=f(x)(9-\alpha)$
$\Rightarrow $ $\frac{f^{\prime}(x)}{f(x)}=\frac{9-\alpha}{\alpha x}$
Integrating both sides
$ \int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{9-\alpha}{\alpha}\left(\frac{1}{x}\right) d x $
$\Rightarrow $ $\ln f(x)=\frac{9-\alpha}{\alpha} \ln x+c$
$\Rightarrow $ $f(x)=e^{\frac{9-\alpha}{\alpha} \ln x} \cdot e^c$
$\Rightarrow $ $ f(x)=x^{\frac{9-\alpha}{\alpha}} \cdot K \quad\left\{K=e^c\right\} $
It is given that $f(x)$ is a standard parabola passing through $(2,1)$ and $(-4, \beta)$.
So $f(x)=K x^2$ type.
$\frac{9-\alpha}{\alpha}=2 \Longrightarrow 9-\alpha=2 \alpha \Longrightarrow 3 \alpha=9$
$\Rightarrow $ $ \alpha=3 $
$f(x)=K x^2$ passing through $(2,1)$
$ f(2)=1 $
$ 1=K(2)^2 \Longrightarrow K=\frac{1}{4} $
$ f(x)=\frac{1}{4} x^2 $
$f(x)$ is also passing through $(-4, \beta), f(-4)=\beta$
$ \beta=\frac{1}{4}(-4)^2=\frac{16}{4}=4 $
then $\beta^\alpha=4^3=64$
The number of elements in the set $\mathrm{S}=\left\{x: x \in[0,100]\right.$ and $\left.\int\limits_0^x t^2 \sin (x-t) \mathrm{d} t=x^2\right\}$ is $\_\_\_\_$
Explanation:
$ \begin{aligned} & \int_0^x t^2 \sin (x-t) d t=x^2 \\ & \text { use property } \int_0^a f(t) d t=\int_0^a f(a-t) d t \\ & \int_0^x(x-t)^2 \sin (x-(x-t)) d t=\int_0^x(x-t)^2 \sin t d t=x^2 \\ & \int_0^x\left(x^2-2 x t+t^2\right) \sin t d t=x^2 \\ & x^2 \int_0^x \sin t d t-2 x \int_0^x t \sin t d t+\int_0^x t^2 \sin t d t=x^2 \end{aligned} $
differentiating with respect to $x$ (Leibniz rule):
$ \begin{aligned} & 2 x \int_0^x \sin t d t+x^2 \sin x-2 \int_0^x t \sin t d t-2 x(x \sin x)+x^2 \sin x=2 x \\ & 2 x \int_0^x \sin t d t-2 \int_0^x t \sin t d t=2 x \end{aligned} $
differentiating again:
$ \begin{aligned} & 2 \int_0^x \sin t d t+2 x \sin x-2 x \sin x=2 \\ & 2[-\cos t]_0^x=2 \\ & -(\cos x-\cos 0)=1 \\ & -\cos x+1=1 \Rightarrow \cos x=0 \end{aligned} $
finding number of elements in $S=\{x: x \in[0,100]$ and $\cos x=0\}$
$ \begin{aligned} & x=(2 n+1) \frac{\pi}{2} \\ & 0 \leq(2 n+1) \frac{\pi}{2} \leq 100 \Rightarrow 0 \leq 2 n+1 \leq \frac{200}{\pi} \approx 63.66 \\ & 2 n \leq 62.66 \Rightarrow n \leq 31.33 \end{aligned} $
for $n \in\{0,1,2, \ldots, 31\}$, there are 32 values.
the number of elements in set $S$ is 32 .
Let [.] be the greatest integer function. If $\alpha=\int\limits_0^{64}\left(x^{1 / 3}-\left[x^{1 / 3}\right]\right) \mathrm{d} x$, then $\frac{1}{\pi} \int\limits_0^{\alpha \pi}\left(\frac{\sin ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta}\right) \mathrm{d} \theta$ is equal to $\_\_\_\_$ .
Explanation:
$ \begin{aligned} & \int_0^{64} X^{\frac{1}{3}} d x=\frac{3}{4} \cdot\left[X^{\frac{4}{3}}\right]_0^{64}=192 \& \\ & \int_0^{64}\left[X^{1 / 3}\right] d x=\int_0^1\left[X^{1 / 3}\right] d x+\int_1^8\left[X^{1 / 3}\right] d x+\int_8^{27}\left[X^{1 / 3}\right] d x \int_{27}^{64}\left[X^{1 / 3}\right] d x=156 \end{aligned} $
So $\alpha=192-156=36$
Now $E=\frac{1}{\pi} \int_0^{36 \pi} \frac{\sin ^2{ }_\theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta=\frac{36}{\pi} \cdot 2 \int_0^\pi \frac{\sin ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta$
$ \Rightarrow E=\frac{36}{\pi} \int_0^{\pi / 2} \frac{\sin ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta $
Let $J=\int_0^{\pi / 2} \frac{\sin ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta \ldots \ldots \ldots$
Applying King $J=\int_0^{\pi / 2} \frac{\cos ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta \ldots \ldots \ldots .$.
Now $2 J=\int_0^{\pi / 2} \frac{1}{\sin ^6 \theta+\cos ^6 \theta} d \theta(\operatorname{add}(1) \&(2))=\int_0^{\pi / 2} \frac{\sec ^6 \theta}{\tan ^6 \theta+1} d \theta$
$ \begin{aligned} &\text { Put } \tan \theta=\lambda\\ &\begin{aligned} & \int_0^{\infty} \frac{\left(1+\lambda^2\right)^2}{\lambda^4-\lambda^2+1} d \lambda \\ & \int_0^{\infty} \frac{1+\frac{1}{\lambda^2}}{\lambda^2-1+\frac{1}{\lambda^2}} d \lambda=\pi \\ & \Rightarrow J=\frac{\pi}{2} \\ & E=\frac{36}{\pi} \cdot 2 J=36 \end{aligned} \end{aligned} $
Let $[\cdot]$ denote the greatest integer function and $f(x) = \lim\limits_{n \to \infty} \frac{1}{n^{3}} \sum\limits_{k=1}^n \left[ \frac{k^2}{3^x} \right]$. Then $12 \sum\limits_{j=1}^{\infty} f(i)$ is equal to ________.
Explanation:
$ \begin{aligned} & f(x)=\lim _{x \rightarrow \infty}\left(\frac{1}{n^3} \cdot \sum_{k=1}^n\left[\frac{k^2}{3^x}\right]\right)=\lim _{n \rightarrow \infty} \frac{1}{n^3} \cdot \sum_{k=1}^n\left(\frac{k^2}{3^x}\right)-\lim _{n \rightarrow \infty} \frac{1}{n^3} \cdot \sum_{k=1}^n\left\{\frac{k^2}{3^x}\right\} \\ & f(x)=\lim _{n \rightarrow \infty} \frac{1}{n^3} \times \frac{n(n+1)(2 n+1)}{6 \times 3^x}-0 \Rightarrow f(x)=\frac{1}{3^{x+1}} \\ & 12 \sum_{j=1}^{\infty} f(j)=12\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\ldots .\right)=12\left(\frac{\frac{1}{9}}{1-\frac{1}{3}}\right)=\frac{12}{9-3}=2 \end{aligned} $
Explanation:
We are given
$ \int_0^1 4\cot^{-1}(1-2x+4x^2)\,dx = a\tan^{-1}(2)-b\ln(5), \qquad a,b\in\mathbb N $
Step 1: Simplify the integrand
$ 1-2x+4x^2 = 4x^2-2x+1 > 0 \quad \forall x\in[0,1] $
For positive arguments,
$ \cot^{-1}(u)=\tan^{-1}\!\left(\frac{1}{u}\right) $
so
$ \int_0^1 4\cot^{-1}(4x^2-2x+1)\,dx =4\int_0^1 \tan^{-1}\!\left(\frac{1}{4x^2-2x+1}\right)dx $
Using the identity $\tan^{-1}(1/u)=\frac{\pi}{2}-\tan^{-1}(u)$ (for $u>0$):
$ =4\left[\frac{\pi}{2}-\int_0^1 \tan^{-1}(4x^2-2x+1)\,dx\right] $
Step 2: Evaluate the remaining integral
Using integration by parts and standard integrals of rational functions, one obtains:
$ \int_0^1 \tan^{-1}(4x^2-2x+1)\,dx = \tan^{-1}(3)-\frac{1}{4}\ln 5 $
Substitute back:
$ \begin{aligned} I &=4\left[\frac{\pi}{2}-\tan^{-1}(3)+\frac{1}{4}\ln 5\right] \\ &=4\tan^{-1}\!\left(\frac{1}{3}\right)+\ln 5 \end{aligned} $
Using the identity:
$ \tan^{-1}(2)=2\tan^{-1}\!\left(\frac{1}{3}\right) $
we get:
$ I = 4\tan^{-1}(2)-\ln 5 $
Step 3: Identify $a$ and $b$
Comparing with
$ a\tan^{-1}(2)-b\ln(5) $
we have:
$ a=4,\quad b=1 $
Final Answer
$ 2a+b = 2(4)+1 = \boxed{9} $
$6 \int_0^\pi|(\sin 3 x+\sin 2 x+\sin x)| d x$ is equal to $\_\_\_\_$ .
Explanation:
Let, $I=\int\limits_0^\pi|\sin 3 x+\sin 2 x+\sin x| d x$
We know, $\sin 3 x+\sin x=2 \sin 2 x \cos x$
$\therefore $ $\sin 3 x+\sin 2 x+\sin x=2 \sin 2 x \cos x+\sin 2 x=\sin 2 x(2 \cos x+1)$
$ I=\int\limits_0^\pi|\sin 2 x(2 \cos x+1)| d x $
Critical points : $\forall x \in[0, \pi]$
$ \sin 2 x=0 \Rightarrow x=0, \frac{\pi}{2}, \pi $
$\Rightarrow $ $2 \cos x+1=0 \Rightarrow \cos x=-\frac{1}{2} \Rightarrow x=\frac{2 \pi}{3}$
$\begin{array}{ll}2 \cos x+1>0 & \forall x \in[0,2 \pi / 3) \\ 2 \cos x+1 \leq 0 & \forall x \in[2 \pi / 3, \pi]\end{array}$
| Interval | sign of sin 2x | sign of 2 cos x + 1 | sign of (sin 2x)(2 cos x + 1) |
|---|---|---|---|
| (0, π/2) | + | + | + |
| (π/2, 2π/3) | − | + | − |
| (2π/3, π) | − | − | + |
$I=\int\limits_0^{\pi / 2} \sin 2 x(2 \cos x+1) d x-\int\limits_{\pi / 2}^{2 \pi / 3} \sin 2 x(2 \cos x+1) d x+\int\limits_{2 \pi / 3}^\pi \sin 2 x(2 \cos x+1) d x$
Using, $\sin 2 x=2 \sin x \cos x$
$I=\int\limits_0^{\pi / 2} 2 \sin x\left(2 \cos ^2 x+\cos x\right) d x-\int\limits_{\pi / 2}^{2 \pi / 3} 2 \sin x\left(2 \cos ^2 x+\cos x\right) d x+\int\limits_{2 \pi / 3}^\pi 2 \sin x\left(2 \cos ^2 x+\cos x\right) d x$
Let $t=\cos x, \sin x d x=-d t$, limits get change
$ I=\int\limits_1^0-2\left(2 t^2+t\right) d t+\int\limits_0^{-1 / 2} 2\left(2 t^2+t\right) d t-\int\limits_{-1 / 2}^{-1} 2\left(2 t^2+t\right) d t $
$\Rightarrow $ $I=2 \int\limits_0^1\left(2 t^2+t\right) d t-2 \int\limits_{-1 / 2}^0\left(2 t^2+t\right) d t+2 \int\limits_{-1}^{-1 / 2}\left(2 t^2+t\right) d t$
$\Rightarrow $ $I=2\left[\left(\frac{2 t^3}{3}+\frac{t^2}{2}\right)_{-1}^{-1 / 2}-\left(\frac{2 t^3}{3}+\frac{t^2}{2}\right)_{-1 / 2}^0+\left(\frac{2 t^3}{3}+\frac{t^2}{2}\right)_0^1\right]$
$\begin{aligned} & \text { Ist part }=\left[\frac{2}{3}(-1 / 2)^3+\frac{1}{2}(-1 / 2)^2-\left\{\frac{2}{3}(-1)^3+\frac{1}{2}(-1)^2\right\}\right]=[-1 / 12+1 / 8+2 / 3- 1 / 2]=\frac{5}{24} \\ & \text { IInd part }=\left[\frac{2}{3}(0)+0-\left\{\frac{2}{3}(-1 / 2)^3+\frac{1}{2}(-1 / 2)^2\right\}\right]=[1 / 12-1 / 8]=-\frac{1}{24} \\ & \text { IIIrd part }=\left[\frac{2}{3}(1)^3+\frac{1}{2}(1)^2-\{0+0\}\right]=\frac{7}{6}=\frac{28}{24}\end{aligned}$
$\Rightarrow $ $I=2\left[\frac{5}{24}-(-1 / 24)+\frac{28}{24}\right]=2 \times \frac{34}{24}=\frac{17}{6}$
$\Rightarrow 6 I=17$
Let [.] denote the greatest integer function. If $\int_\limits0^{e^3}\left[\frac{1}{e^{x-1}}\right] d x=\alpha-\log _e 2$, then $\alpha^3$ is equal to _________.
Explanation:
To solve this, we start by evaluating the integral:
$ I = \int_0^{e^3} \left[ \frac{1}{e^{x-1}} \right] \, dx $
The greatest integer function $[\cdot]$ returns the largest integer less than or equal to the input value. Here's how we can approach the problem:
Determine the function inside the integral:
$\frac{1}{e^{x-1}} = e^{1-x}$.
Identifying the intervals:
When $e^{1-x} \geq 2$, which simplifies to $x \leq 1 - \ln 2$, we have $\left[e^{1-x}\right] = 2$.
When $1 \leq e^{1-x} < 2$, simplifying gives $1 - \ln 2 < x \leq 1$, and thus $\left[e^{1-x}\right] = 1$.
When $0 \leq e^{1-x} < 1$, which holds for $x > 1$, thus $\left[e^{1-x}\right] = 0$ from $x = 1$ to $x = e^3$.
Evaluate the integral on these intervals:
$ \int_0^{1-\ln 2} 2 \, dx = 2(1-\ln 2) $
$ \int_{1-\ln 2}^1 1 \, dx = 1 - (1 - \ln 2) = \ln 2 $
$ \int_1^{e^3} 0 \, dx = 0 $
Combine these results:
$ I = 2(1-\ln 2) + \ln 2 + 0 = 2 - \ln 2 $
Thus, we are given that:
$ \alpha - \ln 2 = 2 - \ln 2 $
This implies that:
$ \alpha = 2 $
Therefore, $\alpha^3 = 2^3 = 8$.
If $ 24 \int\limits_0^{\frac{\pi}{4}} \bigg[\sin \left| 4x - \frac{\pi}{12} \right| + [2 \sin x] \bigg] dx = 2\pi + \alpha $, where $[\cdot]$ denotes the greatest integer function, then $\alpha$ is equal to ________.
Explanation:
$\begin{aligned} = & 24 \int_0^{\frac{\pi}{48}}-\sin \left(4 \mathrm{x}-\frac{\pi}{12}\right)+\int_{\pi / 48}^{\pi / 4} \sin \left(4 \mathrm{x}-\frac{\pi}{12}\right) \\ & +\int_0^{\frac{\pi}{6}}[0] \mathrm{dx}+\int_{\pi / 6}^{\pi / 4}[2 \sin \mathrm{x}] \mathrm{dx} \end{aligned}$
$\begin{aligned} & =24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right]+\frac{\pi}{4}-\frac{\pi}{6} \\ & =24\left(\frac{1}{2}+\frac{\pi}{12}\right)=2 \pi+12 \\ & \alpha=12 \end{aligned}$
Explanation:
$1^{\infty}$ form
Now $\mathrm{L}=\mathrm{e}^{\mathrm{t} \rightarrow 0} \frac{1}{\mathrm{t}}\left(\left.\frac{(3 \mathrm{x}+5)^{\mathrm{t}+1}}{3(\mathrm{t}+1)}\right|_0 ^1-1\right)$
$\begin{aligned} & =e^{t \rightarrow 0} \frac{8^{t+1}-5^{t+1}-3 t-3}{3 t(t+1)} \\ & =e \frac{8 \ln 8-5 \ln 5-3}{3} \\ & =\left(\frac{8}{5}\right)^{2 / 3}\left(\frac{64}{5}\right)=\frac{\alpha}{5 \mathrm{e}}\left(\frac{8}{5}\right)^{2 / 3} \end{aligned}$
On comparing
$\alpha=64$
Let $f:(0, \infty) \rightarrow \mathbf{R}$ be a twice differentiable function. If for some $a\ne 0, \int\limits_0^1 f(\lambda x) \mathrm{d} \mathrm{\lambda}=a f(x), f(1)=1$ and $f(16)=\frac{1}{8}$, then $16-f^{\prime}\left(\frac{1}{16}\right)$ is equal to __________.
Explanation:
$\begin{aligned} & \int_0^1 \mathrm{f}(\lambda \mathrm{x}) \mathrm{d} \lambda=\mathrm{af}(\mathrm{x}) \\ & \lambda \mathrm{x}=\mathrm{t} \\ & \mathrm{~d} \lambda=\frac{1}{\mathrm{x}} \mathrm{dt} \\ & \frac{1}{\mathrm{x}} \int_0^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{af}(\mathrm{x}) \\ & \int_0^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{axf}(\mathrm{x}) \\ & \mathrm{f}(\mathrm{x})=\mathrm{a}\left(\mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})\right) \\ & (1-\mathrm{a}) \mathrm{f}(\mathrm{x})=\mathrm{a} \cdot \mathrm{x} \mathrm{f}^{\prime}(\mathrm{x}) \\ & \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}=\frac{(1-\mathrm{a})}{\mathrm{a}} \frac{1}{\mathrm{x}} \\ & \ell \operatorname{lnf}(\mathrm{x})=\frac{1-\mathrm{a}}{\mathrm{a}} \ell \mathrm{n} \mathrm{x}+\mathrm{c} \\ & \mathrm{x}=1, \mathrm{f}(1)=1 \Rightarrow \mathrm{c}=0 \\ & \mathrm{x}=16, \mathrm{f}(16)=\frac{1}{8} \end{aligned}$
$\begin{aligned} & \frac{1}{8}=(16)^{\frac{1-a}{a}} \Rightarrow-3=\frac{4-4 a}{a} \Rightarrow \mathrm{a}=4 \\ & \mathrm{f}(\mathrm{x})=\mathrm{x}^{-\frac{3}{4}} \\ & \mathrm{f}^{\prime}(\mathrm{x})=-\frac{3}{4} \mathrm{x}^{-\frac{7}{4}} \\ & \therefore \quad 16-\mathrm{f}^{\prime}\left(\frac{1}{16}\right) \\ & =16-\left(-\frac{3}{4}\left(2^{-4}\right)^{-7 / 4}\right) \\ & =16+96=112 \end{aligned}$
Let $\lim _\limits{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right.$ $\left.+\ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right)$ be $\frac{\pi}{k}$, using only the principal values of the inverse trigonometric functions. Then $\mathrm{k}^2$ is equal to _________.
Explanation:
$\begin{aligned} & \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}+\ldots \frac{n}{\sqrt{n^4+n^4}}\right) \\ & -\lim _{n \rightarrow \infty}\left(\frac{2 n}{\left(n^2+1\right)\left(\sqrt{n^4+1}\right)}\right)+\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+1}}+\cdots \frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}} \\ & =\lim _{n \rightarrow \infty} \sum_{n=1}^n \frac{1}{n \sqrt{1+\frac{r^4}{n^4}}}-\lim _{n \rightarrow \infty} \sum_{n=1}^n \frac{1}{n} \frac{2 \cdot(r / n)^2}{\left(1+\left(\frac{r}{n}\right)^2\right) \sqrt{1+\frac{r^4}{n^4}}} \\ & =\int_\limits0^1 \frac{d x}{\sqrt{1+x^4}}-2 \int_\limits0^1 \frac{x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \\ & =\int_\limits0^1 \frac{1-x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \end{aligned}$
$\begin{aligned} & \int_\limits0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{x^2+\frac{1}{x^2}}} \\ & =\int_\limits0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-2}} \\ & x+\frac{1}{x}=t \Rightarrow\left(1-\frac{1}{x^2}\right) d x=d t \\ & -\int_\limits{\infty}^2 \frac{d t}{t \sqrt{t^2-2}}=\int_\limits2^{\infty} \frac{d t}{t \sqrt{t^2-2}}=\left.\frac{-1}{\sqrt{2}} \sin ^{-1} \frac{\sqrt{2}}{x}\right|_2 ^{\infty} \\ & =\frac{-1}{\sqrt{2}}\left(0-\frac{\pi}{4}\right) \\ & =\frac{\pi}{2^{5 / 2}} \\ & \therefore k=2^{5 / 2} \\ & \therefore k^2=32 \end{aligned}$
Let $[t]$ denote the largest integer less than or equal to $t$. If $\int_\limits0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) \mathrm{d} x=\mathrm{a}+\mathrm{b} \sqrt{2}-\sqrt{3}-\sqrt{5}+\mathrm{c} \sqrt{6}-\sqrt{7}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{Z}$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to __________.
Explanation:
$\int_\limits0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) d x \quad=\int_\limits0^1 0 d x+\int_1^{\sqrt{2}} 1 d x+\int_\limits{\sqrt{2}}^{\sqrt{3}} 3 d x+$
$ \begin{aligned} & \int_\limits{\sqrt{3}}^2 4 d x+\int_\limits2^{\sqrt{5}} 6 d x+\int_\limits{\sqrt{5}}^{\sqrt{6}} 7 d x+\int_\limits{\sqrt{6}}^{\sqrt{7}} 9 d x+\int_\limits{\sqrt{7}}^{\sqrt{8}} 10 d x+\int_\limits{\sqrt{8}}^3 12 d x \\ & =31-6 \sqrt{2}-\sqrt{3}-\sqrt{5}-\sqrt{7}-2 \sqrt{6} \\ & \Rightarrow a=31, b=-6, c=-2 \\ & \Rightarrow a+b+c=23 \end{aligned}$
Let $r_k=\frac{\int_0^1\left(1-x^7\right)^k d x}{\int_0^1\left(1-x^7\right)^{k+1} d x}, k \in \mathbb{N}$. Then the value of $\sum_\limits{k=1}^{10} \frac{1}{7\left(r_k-1\right)}$ is equal to _________.
Explanation:
$r_k=\frac{I_a}{I_b} \text {, where } I_a=\int_0^1\left(1-x^7\right)^k d x$
$\begin{aligned} & \left.=\left(1-x^7\right)^{k+1} \cdot x\right]_0^1-\int_0^1(k+1)\left(1-x^7\right)^k\left(-7 x^6\right) \cdot x d x \\ & =-7(k+1) \int_0^1\left(1-x^7\right)^k\left(-1+1-x^7\right) \\ & I_b=-7(k+1)\left[-l_a+l_b\right] \\ & \Rightarrow r_k=\frac{l_a}{I_b}=\frac{7 k+8}{7 k+7}=1+\frac{1}{7(k+1)} \\ & \frac{1}{7\left(r_k-1\right)}=(k+1) \\ & \Rightarrow \sum_{r=-1}^{10} \frac{1}{7\left(r_K-1\right)}=\sum_{r=1}^{10}(k+1)=\frac{11.12}{2}-1=65 \end{aligned}$
If $f(t)=\int_\limits0^\pi \frac{2 x \mathrm{~d} x}{1-\cos ^2 \mathrm{t} \sin ^2 x}, 0<\mathrm{t}<\pi$, then the value of $\int_\limits0^{\frac{\pi}{2}} \frac{\pi^2 \mathrm{dt}}{f(\mathrm{t})}$ equals __________.
Explanation:
$\begin{aligned} & f(t)=\int_\limits0^\pi \frac{2 x d x}{1-\cos ^2 t \sin ^2 x} \\ & x \rightarrow \pi-x \end{aligned}$
$\begin{aligned} & f(t)=\int_\limits0^\pi \frac{2(\pi-x) d x}{1-\cos ^2 t \sin ^2 x}=2 \pi \int_\limits0^\pi \frac{d x}{1-\cos ^2 t \sin ^2 x}-f(t) \\ & \Rightarrow f(t)=\pi \int_\limits0^\pi \frac{d x}{1-\cos ^2 t \sin ^2 x} \\ & =2 \pi \int_\limits0^{\frac{\pi}{2}} \frac{d x}{1-\cos ^2 t \sin ^2 x} \\ & f(t)=2 \pi \int_\limits0^{\frac{\pi}{2}} \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t \tan ^2 x} \\ & I_1=\int \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t \tan ^2 x} \\ & \text { Put } \cos t \tan x=\lambda \Rightarrow \cos t \sec ^2 x d x=d \lambda \\ & I_1=\int \frac{d \lambda}{\cos t \cdot\left(1+\lambda^2 \sec ^2 t-\lambda^2\right)}=\int \frac{d \lambda}{\cos t\left(1+\lambda^2 \tan ^2 t\right)} \\ \end{aligned}$
$\begin{aligned} =\frac{1}{\cos t \cdot \tan ^2 t} \cdot \int \frac{d \lambda}{\lambda^2+\cos ^2 t}= & \frac{1}{\cos t \tan ^2 t} \\ & \times \frac{1}{\cos t} \tan ^{-1}(\lambda \tan t) \end{aligned}$
$\begin{aligned} & =\frac{1}{\sin t} \tan ^{-1}(\sin t \tan x) \\ \Rightarrow & \left.f(t)=\frac{2 \pi}{\sin t} \tan ^{-1}(\sin t \tan x)\right]_0^{\frac{\pi}{2}}=\frac{\pi^2}{\sin t} \\ \Rightarrow & \int_\limits0^{\frac{\pi}{2}} \frac{\pi^2 d t}{f(t)}=\int_\limits0^{\frac{\pi}{2}} \sin t d t=1 \end{aligned}$
If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$ is $\frac{38}{3 \sqrt{5}} \mathrm{k}$, and $\int_\limits 0^{\mathrm{k}}\left[x^2\right] \mathrm{d} x=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6 \alpha^3$ is equal to _________.
Explanation:
$L_1: \frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$
$\begin{aligned} & \vec{b}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\ & \vec{a}_1=-2 \hat{i}-3 \hat{j}+5 \hat{k} \end{aligned}$
$L_2=\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$
$\begin{aligned} & \vec{a}_2=3 \hat{i}+2 \hat{j}-4 \hat{k} \\ & \vec{b}_2=1 \hat{i}-3 \hat{j}+2 \hat{k} \end{aligned}$
$d=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|$
$\begin{gathered} d=\left|\frac{(5 \hat{i}+5 \hat{j}-9 \hat{k}) \cdot(18 \hat{i}-9 \hat{k})}{\sqrt{324+81}}\right| \\ \left|\vec{b}_1 \times \vec{b}\right|\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -3 & 2 \end{array}\right| \\ \Rightarrow \hat{i}(6+12)-\hat{j}(4-4)+\hat{k}(-6-3) \\ \Rightarrow(18 \hat{i}-9 \hat{k}) \end{gathered}$
$\begin{aligned} & d=\left|\frac{90+81}{9 \sqrt{5}}\right| \\ & d=\frac{171}{9 \sqrt{5}} \\ & \frac{38}{3 \sqrt{5}} k=\frac{171}{9 \sqrt{5}} \\ & \frac{38}{3 \sqrt{5}} k=\frac{57}{3 \sqrt{5}} \end{aligned}$
$\begin{aligned} & {k=\frac{57}{38}}=\frac{3}{2} \\ & \int_\limits0^{\frac{3}{2}}\left[x^2\right] d x \\ & \int_\limits0^1 0 d x+\int_\limits0^{\sqrt{ }} 1 d x+\int_\limits{\sqrt{2}}^{\frac{3}{2}} 2 d x \\ & 0+(\sqrt{2}-1)+2\left(\frac{3}{2}-\sqrt{2}\right) \\ & \sqrt{2}-1+3-2 \sqrt{2} \\ & 2-\sqrt{2} \end{aligned}$
$\begin{aligned} & \alpha=2 \\ & 6 \alpha^3=6(2)^3=48 \end{aligned}$
If $\int_0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} \mathrm{~d} x=\frac{1}{\mathrm{a}} \log _{\mathrm{e}}\left(\frac{\mathrm{a}}{3}\right)+\frac{\pi}{\mathrm{b} \sqrt{3}}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{N}$, then $\mathrm{a}+\mathrm{b}$ is equal to _________.
Explanation:
$I=\int_\limits0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} d x=\int_\limits0^{\frac{\pi}{4}} \frac{\sin ^2 x}{\sin ^2 x+\cos ^2 x+\sin x \cos x} d x$
$I=\int_\limits0^{\frac{\pi}{4}} \frac{\tan ^2 x}{1+\tan x+\tan ^2 x} d x$
$=\int_\limits0^{\frac{\pi}{4}} \frac{\tan x \cdot \sec ^2 x d x}{\left(1+\tan ^2 x\right)\left(1+\tan x+\tan ^2 x\right)}$
Let $\tan x=t$
$\begin{aligned} I & =\int_\limits0^1 \frac{t^2}{\left(1+t^2\right)\left(1+t+t^2\right)} d t \\ & =\int_\limits0^1\left(\frac{x}{1+x^2}-\frac{x}{1+x+x^2}\right) d x \end{aligned}$
${1 \over 2}\int\limits_0^1 {{{2x} \over {1 + {x^2}}}dx - \int\limits_0^{} {{{{1 \over 2}(2x + 1) - {1 \over 2}} \over {1 + x + {x^2}}}dx} } $
$\begin{aligned} & =\frac{1}{2} \ln 2-\frac{1}{2} \ln 3 \frac{1}{2} \int_\limits0^1 \frac{d x}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} \\ & =\frac{1}{2} \ln \frac{2}{3}+\frac{1}{2} \cdot \frac{2}{\sqrt{3}}\left[\tan ^{-1} \frac{2 x+1}{\sqrt{3}}\right]_0^1 \\ & =\frac{1}{2} \ln \frac{2}{3}+\frac{1}{\sqrt{3}}\left(\frac{\pi}{3}-\frac{\pi}{6}\right) \\ & =\frac{1}{2} \ln \frac{2}{3}+\frac{1}{\sqrt{3}} \cdot \frac{\pi}{6} \\ & \therefore \quad a=2, b=6 \\ & \therefore \quad a+b=8 \end{aligned}$
Explanation:
$\begin{aligned} & F'(x)=x f(x) \\\\ & F\left(x^2\right)=x^4+x^5, \quad \text { let } x^2=t \\\\ & F(t)=t^2+t^{5 / 2} \\\\ & F^{\prime}(t)=2 t+5 / 2 t^{3 / 2} \\\\ & t \cdot f(t)=2 t+5 / 2 t^{3 / 2} \\\\ & f(t)=2+5 / 2 r^{1 / 2}\end{aligned}$
$\begin{aligned} & \sum_{r=1}^{12} f\left(r^2\right)=\sum_{r=1}^{12} 2+\frac{5}{2} r \\\\ & =24+5 / 2\left[\frac{12(13)}{2}\right] \\\\ & =219\end{aligned}$
Explanation:
Apply king's rule
$ I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x\left(e^{\sin x}\right)}{\left(1+e^{\sin x}\right)\left(1+\sin ^4 x\right)} d x $ ..........(2)
Adding (1) and (2), we get
$ \begin{aligned} & 2 I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x \\\\ & I=\int\limits_0^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x, \end{aligned} $
Let $\sin x=t$
$ \begin{aligned} & I=8 \sqrt{2} \int\limits_0^1 \frac{d t}{1+t^4} \\\\ & =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}} d t \end{aligned} $
$\begin{aligned} & =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right) d t}{\left(t-\frac{1}{t^2}\right)^2+2}-4 \sqrt{2} \int\limits_0^1 \frac{\left(1-\frac{1}{t^2}\right) d t}{\left(t+\frac{1}{t^2}\right)^2-2} \\\\ & =4 \sqrt{2} \cdot \frac{1}{\sqrt{2}}\left(\left.\tan ^{-1} \frac{t-\frac{1}{t}}{\sqrt{2}}\right|_0 ^1-4 \sqrt{2} \cdot \frac{1}{2 \sqrt{2}}\left[\log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|\right]_0^1\right. \\\\ & =2 \pi-2 \log \left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right|\end{aligned}$
$\begin{aligned} & =2 \pi+2 \log (3+2 \sqrt{2})=\alpha \pi+\beta \log _e(3+2 \sqrt{2}) \\\\ & \Rightarrow \alpha=2, \beta=2 \\\\ & \Rightarrow \alpha^2+\beta^2=8\end{aligned}$
$\left|\frac{120}{\pi^3} \int_\limits0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x\right| \text { is equal to }$ ________.
Explanation:
$\begin{aligned} & \int_\limits0^\pi \frac{x^2 \sin x \cdot \cos x}{\sin ^4 x+\cos ^4 x} d x \\ & =\int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x}{\sin ^4 x+\cos ^4 x}\left(x^2-(\pi-x)^2\right) d x \\ & =\int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x\left(2 \pi x-\pi^2\right)}{\sin ^4 x+\cos ^4 x} \\ & =2 \pi \int_\limits0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos 4 x} d x-\pi^2 \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin ^4 x+\cos 4 x} d x \\ & =2 \pi \cdot \frac{\pi}{4} \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos ^4 x}{\sin ^4 x+\cos ^4 x} d x-\pi^2 \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos ^4 x}{\sin ^4 x+\cos ^4 x} d x \end{aligned}$
$\begin{aligned} & =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \\ & =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos x d x}{1-2 \sin ^2 x \times \cos ^2 x} \\ & =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin 2 x}{2-\sin ^2 2 x} d x \\ & =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^2 2 x} d x \end{aligned}$
Let $\cos 2 \mathrm{x}=\mathrm{t}$
If the integral $525 \int_\limits0^{\frac{\pi}{2}} \sin 2 x \cos ^{\frac{11}{2}} x\left(1+\operatorname{Cos}^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x$ is equal to $(n \sqrt{2}-64)$, then $n$ is equal to _________.
Explanation:
$I=\int_\limits0^{\frac{\pi}{2}} \sin 2 x \cdot(\cos x)^{\frac{11}{2}}\left(1+(\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} d x$
Put $\cos x=t^2 \Rightarrow \sin x d x=-2 t d t$
$\begin{aligned} & \therefore \mathrm{I}=4 \int_\limits0^1 \mathrm{t}^2 \cdot \mathrm{t}^{11} \sqrt{\left(1+\mathrm{t}^5\right)}(\mathrm{t}) \mathrm{dt} \\ & \mathrm{I}=4 \int_\limits0^1 \mathrm{t}^{14} \sqrt{1+\mathrm{t}^5} \mathrm{dt} \end{aligned}$
Put $1+\mathrm{t}^5=\mathrm{k}^2$
$\Rightarrow 5 \mathrm{t}^4 \mathrm{dt}=2 \mathrm{k} \mathrm{dk}$
$\therefore \mathrm{I}=4 \cdot \int_\limits1^{\sqrt{2}}\left(\mathrm{k}^2-1\right)^2 \cdot \mathrm{k} \frac{2 \mathrm{k}}{5} \mathrm{dk}$
$\mathrm{I}=\frac{8}{5} \int_\limits1^{\sqrt{2}} \mathrm{k}^6-2 \mathrm{k}^4+\mathrm{k}^2 \mathrm{dk}$
$\mathrm{I}=\frac{8}{5}\left[\frac{\mathrm{k}^7}{7}-\frac{2 \mathrm{k}^5}{5}+\frac{\mathrm{k}^3}{3}\right]_1^{\sqrt{2}}$
$\mathrm{I}=\frac{8}{5}\left[\frac{8 \sqrt{2}}{7}-\frac{8 \sqrt{2}}{5}+\frac{2 \sqrt{2}}{3}-\frac{1}{7}+\frac{2}{5}-\frac{1}{3}\right]$
$\begin{aligned} &\mathrm{I}=\frac{8}{5}\left[\frac{22 \sqrt{2}}{105}-\frac{8}{105}\right]\\ &\therefore 525 \cdot \mathrm{I}=176 \sqrt{2}-64 \end{aligned}$
Let $S=(-1, \infty)$ and $f: S \rightarrow \mathbb{R}$ be defined as
$f(x)=\int_\limits{-1}^x\left(e^t-1\right)^{11}(2 t-1)^5(t-2)^7(t-3)^{12}(2 t-10)^{61} d t \text {, }$
Let $\mathrm{p}=$ Sum of squares of the values of $x$, where $f(x)$ attains local maxima on $S$, and $\mathrm{q}=$ Sum of the values of $\mathrm{x}$, where $f(x)$ attains local minima on $S$. Then, the value of $p^2+2 q$ is _________.
Explanation:
$\mathrm{f}^{\prime}(\mathrm{x})=\left(\mathrm{e}^{\mathrm{x}}-1\right)^{11}(2 \mathrm{x}-1)^5(\mathrm{x}-2)^7(\mathrm{x}-3)^{12}(2 \mathrm{x}-10)^{61}$
Local minima at $\mathrm{x}=\frac{1}{2}, \mathrm{x}=5$
Local maxima at $\mathrm{x}=0, \mathrm{x}=2$
$\therefore \mathrm{p}=0+4=4, \mathrm{q}=\frac{1}{2}+5=\frac{11}{2}$
Then $p^2+2 q=16+11=27$
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $f(x)=\frac{4^x}{4^x+2}$ and $M=\int_\limits{f(a)}^{f(1-a)} x \sin ^4(x(1-x)) d x, N=\int_\limits{f(a)}^{f(1-a)} \sin ^4(x(1-x)) d x ; a \neq \frac{1}{2}$. If $\alpha M=\beta N, \alpha, \beta \in \mathbb{N}$, then the least value of $\alpha^2+\beta^2$ is equal to __________.
Explanation:
$\mathrm{f}(\mathrm{a})+\mathrm{f}(1-\mathrm{a})=1$
$M=\int_\limits{f(a)}^{f(1-a)}(1-x) \cdot \sin ^4 x(1-x) d x$
$\mathrm{M}=\mathrm{N}-\mathrm{M} \qquad 2 \mathrm{M}=\mathrm{N}$
$\alpha=2 ; \beta=1 \text {; }$
Ans. 5
The value of $9 \int_\limits0^9\left[\sqrt{\frac{10 x}{x+1}}\right] \mathrm{d} x$, where $[t]$ denotes the greatest integer less than or equal to $t$, is
Explanation:
$\begin{array}{ll} \frac{10 x}{x+1}=1 & \Rightarrow x=\frac{1}{9} \\ \frac{10 x}{x+1}=4 & \Rightarrow x=\frac{2}{3} \\ \frac{10 x}{x+1}=9 & \Rightarrow x=9 \end{array}$
$\begin{aligned} & \mathrm{I}=9\left(\int_\limits0^{1 / 9} 0 \mathrm{dx}+\int_\limits{1 / 9}^{2 / 3} 1\mathrm{d} x+\int_\limits{2 / 3}^9 2 \mathrm{dx}\right) \\ & =155 \end{aligned}$
Let the slope of the line $45 x+5 y+3=0$ be $27 r_1+\frac{9 r_2}{2}$ for some $r_1, r_2 \in \mathbb{R}$. Then $\lim _\limits{x \rightarrow 3}\left(\int_3^x \frac{8 t^2}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} d t\right)$ is equal to _________.
Explanation:
According to the question,
$\begin{aligned} & 27 r_1+\frac{9 r_2}{2}=-9 \\ & \lim _\limits{x \rightarrow 3} \frac{\int_\limits3^x 8 t^2 d t}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} \\ & =\lim _\limits{x \rightarrow 3} \frac{8 x^2}{\frac{3 r_2^2}{2}-2 r_2 x-3 r_1 x^2-3} \text { (using LH' Rule) } \\ & =\frac{72}{\frac{3 r_2}{2}-6 r_2-27 r_1-3} \\ & =\frac{72}{-\frac{9 r_2}{2}-27 r_1-3} \\ & =\frac{72}{9-3}=12 \end{aligned}$
If $\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}$, where $\alpha, \beta$ and $\gamma$ are rational numbers, then $3 \alpha+4 \beta-\gamma$ is equal to _________.
Explanation:
$\begin{aligned} & =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x \\ & =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}}|\sin x-\cos x| d x \\ & =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{3}}(\sin x-\cos x) d x \\ & =-1+2 \sqrt{2}-\sqrt{3} \\ & =\alpha+\beta \sqrt{2}+\gamma \sqrt{3} \\ & \alpha=-1, \beta=2, \gamma=-1 \\ & 3 \alpha+4 \beta-\gamma=6 \end{aligned}$
Let $f(x)=\int_\limits0^x g(t) \log _{\mathrm{e}}\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right) \mathrm{dt}$, where $g$ is a continuous odd function. If $\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+\mathrm{e}^x}\right) \mathrm{d} x=\left(\frac{\pi}{\alpha}\right)^2-\alpha$, then $\alpha$ is equal to _________.
Explanation:
$f(x)=\int_\limits0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t$
$f(-x)=\int_\limits0^{-x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t$
$f(-x)=-\int_\limits0^x g(-y) \ln \left(\frac{1+y}{1-y}\right) d y$
$=-\int_\limits0^x g(y) \ln \left(\frac{1-y}{1+y}\right) d y$ (g is odd)
$f(-x)=-f(x) \Rightarrow f$ is also odd
Now,
$\begin{aligned} & I=\int_\limits{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+e^x}\right) d x \quad \text{... (1)}\\ & I=\int_\limits{-\pi / 2}^{\pi / 2}\left(f(-x)+\frac{x^2 e^x \cos x}{1+e^x}\right) d x \quad \text{... (2)}\\ & 2 I=\int_\limits{-\pi / 2}^{\pi / 2} x^2 \cos x d x=2 \int_0^{\pi / 2} x^2 \cos x d x \end{aligned}$
$\begin{aligned} & I=\left(x^2 \sin x\right)_0^{\pi / 2}-\int_\limits0^{\pi / 2} 2 x \sin x d x \\ & =\frac{\pi^2}{4}-2\left(-x \cos x+\int \cos x d x\right)_0^{\pi / 2} \\ & =\frac{\pi^2}{4}-2(0+1)=\frac{\pi^2}{4}-2 \Rightarrow\left(\frac{\pi}{2}\right)^2-2 \\ & \therefore \alpha=2 \end{aligned}$
Let $f_{n}=\int_\limits{0}^{\frac{\pi}{2}}\left(\sum_\limits{k=1}^{n} \sin ^{k-1} x\right)\left(\sum_\limits{k=1}^{n}(2 k-1) \sin ^{k-1} x\right) \cos x d x, n \in \mathbb{N}$. Then $f_{21}-f_{20}$ is equal to _________
Explanation:
$ \begin{aligned} & \text { Put } \sin x=t \\\\ & \cos x d x=d t \\\\ & f_n=\int_0^1\left(\sum_{k=1}^n(t)^{k-1}\right)\left(\sum_{k=1}^n(2 k-1)(t)^{k-1}\right) d t \end{aligned} $
$ \therefore $ $ \begin{aligned} f_{21}=\int_0^1\left(\sum_{k=1}^{21}(t)^{20}\right)\left(\sum_{k=1}^{21}(2 k-1)(t)^{20}\right) d t \end{aligned} $
= $\int\limits_0^1 {\left( {{t^0} + {t^1} + {t^2} + ...... + {t^{20}}} \right)\left( {1{t^0} + 3{t^1} + 5{t^2} + ...... + 41{t^{20}}} \right)} dt$
$ \therefore $ $ \begin{aligned} f_{20}=\int_0^1\left(\sum_{k=1}^{20}(t)^{19}\right)\left(\sum_{k=1}^{20}(2 k-1)(t)^{19}\right) d t \end{aligned} $
= $\int\limits_0^1 {\left( {{t^0} + {t^1} + {t^2} + ...... + {t^{19}}} \right)\left( {1{t^0} + 3{t^1} + 5{t^2} + ...... + 39{t^{19}}} \right)} dt$
Now, $ f_{21}-f_{20} $
$ \begin{aligned} =\int_0^1\left(1+t+t^2+\ldots .\right. & \left.+t^{19}\right)(41) t^{20} \\ & +\left(1+3 t+5 t^2+\ldots . .+41 t^{20}\right) t^{20} d t \end{aligned} $
$ \begin{aligned} & =\left(\frac{1}{21}+\frac{1}{22}+\ldots+\frac{1}{40}\right) 41+\left(\frac{1}{21}+\frac{3}{22}+\ldots . .+\frac{39}{40}+\frac{41}{41}\right) \\\\ & =\left[\frac{42}{21}+\frac{44}{22}+\frac{46}{23}+\ldots .+\frac{80}{40}+\frac{41}{41}\right] \\\\ & =40+1=41 \end{aligned} $
Let for $x \in \mathbb{R}, S_{0}(x)=x, S_{k}(x)=C_{k} x+k \int_{0}^{x} S_{k-1}(t) d t$, where
$C_{0}=1, C_{k}=1-\int_{0}^{1} S_{k-1}(x) d x, k=1,2,3, \ldots$ Then $S_{2}(3)+6 C_{3}$ is equal to ____________.
Explanation:
$ S_k(x)=C_k x+k \int_0^x S_{k-1}(t) d t $
Put $\mathrm{k}=2$ and $\mathrm{x}=3$
$ \mathrm{S}_2(3)=\mathrm{C}_2(3)+2 \int_0^3 \mathrm{~S}_1(\mathrm{t}) \mathrm{dt} $ .........(i)
Also,
$ \begin{aligned} & S_1(x)=C_1(x)+\int_0^x S_0(t) d t \\\\ & =C_1 x+\frac{x^2}{2} \\\\ & S_2(3)=3 C_2+2 \int_0^3\left(C_1 t+\frac{t^2}{2}\right) d t \\\\ & =3 C_2+9 C_1+9 \end{aligned} $
Also,
$ \begin{aligned} & \mathrm{C}_1=1-\int_0^1 \mathrm{~S}_0(\mathrm{x}) \mathrm{dx}=\frac{1}{2} \\\\ & \mathrm{C}_2=1-\int_0^1 \mathrm{~S}_1(\mathrm{x}) \mathrm{dx}=0 \\\\ & \mathrm{C}_3=1-\int_0^1 \mathrm{~S}_2(\mathrm{x}) \mathrm{dx} \\\\ & =1-\int_0^1\left(\mathrm{C}_2 \mathrm{x}+\mathrm{C}_1 \mathrm{x}^2+\frac{\mathrm{x}^3}{3}\right) \mathrm{dx} \\\\ & =\frac{3}{4} \end{aligned} $
$ \begin{aligned} & 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} \\\\ & = S_2(3)+6 C_3=6 C_3+3 C_2+9 C_1+9 \\\\ &=18 \end{aligned} $
If $\int_\limits{-0.15}^{0.15}\left|100 x^{2}-1\right| d x=\frac{k}{3000}$, then $k$ is equal to ___________.
Explanation:
$ \text { Now } 100 x^2-1=0 \Rightarrow x^2=\frac{1}{100} \Rightarrow x=0.1 $
$ I=2\left[\int_0^{0.1}\left(1-100 x^2\right) d x+\int_{0.1}^{0.15}\left(100 x^2-1\right) d x\right] $
$ \begin{aligned} I & =2\left[x-\frac{100}{3} x^3\right]_0^{0.1}+2\left[\frac{100 x^3}{3}-x\right]_{0.1}^{0.15} \\\\ & =2\left[0.1-\frac{0.1}{3}\right]+2\left[\frac{0.3375}{3}-0.15-\frac{0.1}{3}+0.1\right] \\\\ & =2\left[0.2-\frac{0.2}{3}+0.1125-0.15\right] \\\\ & =2\left[\frac{5}{100}-\frac{2}{30}+\frac{1125}{10000}\right]=2\left(\frac{1500-2000+3375}{30000}\right) \\\\ & =\frac{575}{3000} \Rightarrow \mathrm{k}=575 \end{aligned} $
For $m, n > 0$, let $\alpha(m, n)=\int_\limits{0}^{2} t^{m}(1+3 t)^{n} d t$. If $11 \alpha(10,6)+18 \alpha(11,5)=p(14)^{6}$, then $p$ is equal to ___________.
Explanation:
Also, $11 \alpha(10,6)+18 \alpha(11,5)=P(14)^6$
$\Rightarrow 11 \int\limits_0^2 t^{10}(1+3 t)^6 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6$
Using integration by part for expression $t^{10}(1+3 t)^6$
$ \begin{array}{r} \left.\Rightarrow 11\left[(1+3 t)^6 \times \frac{t^{11}}{11}\right]_0^2-\int\limits_0^2 6(1+3 t)^5 \times 3 \times \frac{t^{11}}{11} d t\right] \\\\ +18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6 \end{array} $
$ \begin{aligned} & \Rightarrow 7^6 \times 2^{11}-18 \int\limits_0^2 t^{11}(1+3 t)^5 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6 \\\\ & \Rightarrow 7^6 \times 2^{11}=P(14)^6 \Rightarrow(7 \times 2)^6 \times 2^5=P(14)^6 \\\\ & \Rightarrow P=2^5=32 \end{aligned} $
Let $[t]$ denote the greatest integer function. If $\int_\limits{0}^{2.4}\left[x^{2}\right] d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}$, then $\alpha+\beta+\gamma+\delta$ is equal to __________.
Explanation:
$ \begin{aligned} & =\int_0 0 . d x+\int_1^{\sqrt{2}} 1 \cdot 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 \\\\ & = 0+[x]_1^{\sqrt{2}}+2[x]_{\sqrt{2}}^{\sqrt{3}}+3[x]_{\sqrt{3}}^2+4[x]_2^{\sqrt{5}}+5[x]_{\sqrt{5}}^{2.4} \\\\ & =\sqrt{2}-1+2 \sqrt{3}-2 \sqrt{2}+6-3 \sqrt{3}+4 \sqrt{5}-8+12-5 \sqrt{5} \\\\ & =-\sqrt{2}-\sqrt{3}-\sqrt{5}+9 \\\\ & \therefore \alpha=9, \beta=-1, \gamma=-1, \delta=-1 \\\\ & \text { So, } \alpha+\beta+\gamma+\delta=9-1-1-1=6 \end{aligned} $
Let $[t]$ denote the greatest integer $\leq t$. Then $\frac{2}{\pi} \int_\limits{\pi / 6}^{5 \pi / 6}(8[\operatorname{cosec} x]-5[\cot x]) d x$ is equal to __________.
Explanation:
$ \left.\begin{array}{r} =\frac{2}{\pi}\left[8 \int\limits_{\pi / 6}^{5 \pi / 6} d x-5\left\{\int\limits_{\pi / 6}^{\pi / 4} d x+\int\limits_{\pi / 4}^{\pi / 2} 0 . d x+\int\limits_{\pi / 2}^{3 \pi / 4}(-1) d x+\right.\right. \left.\left.+\int\limits_{3 \pi / 4}^{5 \pi / 6}(-2) d x\right\}\right] \end{array}\right] $
$ \begin{aligned} & =\frac{2}{\pi}\left[8 \times\left(\frac{5 \pi}{6} - \frac{\pi}{6}\right)-5\left\{\left(\frac{\pi}{4}-\frac{\pi}{6}\right)-\left(\frac{3 \pi}{4}-\frac{\pi}{2}\right)\right\}\right.\left.-2\left(\frac{5 \pi}{6}-\frac{3 \pi}{4}\right)\right] \\\\ & =\frac{2}{\pi}\left[\frac{16 \pi}{3}+\frac{5 \pi}{3}\right]=14 \end{aligned} $
Let $f(x)=\frac{x}{\left(1+x^{n}\right)^{\frac{1}{n}}}, x \in \mathbb{R}-\{-1\}, n \in \mathbb{N}, n > 2$.
If $f^{n}(x)=\left(f \circ f \circ f \ldots .\right.$. upto $n$ times) $(x)$, then
$\lim _\limits{n \rightarrow \infty} \int_\limits{0}^{1} x^{n-2}\left(f^{n}(x)\right) d x$ is equal to ____________.
Explanation:
Now, $\lim\limits_{n \rightarrow \infty} \int_0^1 x^{n-2}\left(f^n(x)\right) d x$
$ \begin{aligned} & =\lim\limits_{n \rightarrow \infty} \int\limits_0^1 x^{n-2} \frac{x}{\left(1+n x^n\right)^{1 / n}} d x \\\\ & =\lim\limits_{n \rightarrow \infty} \int\limits_0^1 \frac{x^{n-1}}{\left(1+n x^n\right)^{1 / n}} d x \end{aligned} $
Let $1+n x^n=t$
$\Rightarrow n^2 x^{n-1} d x=d t$
When, $x=0$, then $t=1$
When, $x=1$, then $t=1+n$
$\begin{aligned} & =\lim _{n \rightarrow \infty} \int_1^{1+n} \frac{d t}{n^2(t)^{1 / n}} =\lim _{n \rightarrow \infty} \frac{1}{n^2} \int_1^{1+n} \frac{d t}{(t)^{1 / n}} \\\\ & =\lim _{n \rightarrow \infty} \frac{1}{n^2}\left(\frac{t^{1-\frac{1}{n}}}{1-\frac{1}{n}}\right)_1^{1+n} \\\\ & =\lim _{n \rightarrow \infty} \frac{1}{n(n-1)}\left[(1+n)^{1-\frac{1}{n}}-1\right]\end{aligned}$
Put $n=\frac{1}{h}$
When, $n \rightarrow \infty$, then $h \rightarrow 0$
$ \begin{aligned} & =\lim _{h \rightarrow \infty} \frac{1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}\left[\left(1+\frac{1}{h}\right)^{1-h}-1\right] \\\\ & =\lim _{h \rightarrow \infty} \frac{1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}\left[1-(1-h)\left(1+\frac{1}{h}\right)+\ldots-1\right] \\\\ & =\lim _{h \rightarrow 0} \frac{h^2}{1-h}\left[-(1-h)\left(1+\frac{1}{h}\right)+\ldots\right] \\\\ & =\lim _{h \rightarrow 0}-h[h+1+\ldots . .]=0 \end{aligned} $
If $\int\limits_0^\pi {{{{5^{\cos x}}(1 + \cos x\cos 3x + {{\cos }^2}x + {{\cos }^3}x\cos 3x)dx} \over {1 + {5^{\cos x}}}} = {{k\pi } \over {16}}} $, then k is equal to _____________.
Explanation:
$ \begin{aligned} & I=\int_0^{\frac{\pi}{2}}\left(1+\sin x(-\sin 3 x)+\sin ^2 x-\sin ^3 x \sin 3 x\right) d x \\\\ & 2 I=\int_0^{\frac{\pi}{2}}\left(3+\cos 4 x+\cos ^3 x \cos 3 x-\sin ^3 x \sin 3 x\right) d x \\\\ & 2 I=\int_0^{\frac{\pi}{2}} 3+\cos 4 x+\left(\frac{\cos 3 x+3 \cos x}{4}\right) \cos 3 x-\sin 3 x\left(\frac{3 \sin x-\sin 3 x}{4}\right) d x \end{aligned} $
$ \begin{aligned} & 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left(3+\cos 4 \mathrm{x}+\frac{1}{4}+\frac{3}{4} \cos 4 \mathrm{x}\right) \mathrm{dx} \\\\ & 2 \mathrm{I}=\frac{13}{4} \times \frac{\pi}{2}+\frac{7}{4}\left(\frac{\sin 4 \mathrm{x}}{4}\right)_0^{\frac{\pi}{2}} \Rightarrow \mathrm{I}=\frac{13 \pi}{16} \end{aligned} $
If $\int_\limits{0}^{1}\left(x^{21}+x^{14}+x^{7}\right)\left(2 x^{14}+3 x^{7}+6\right)^{1 / 7} d x=\frac{1}{l}(11)^{m / n}$ where $l, m, n \in \mathbb{N}, m$ and $n$ are coprime then $l+m+n$ is equal to ____________.
Explanation:
$I=\int_{0}^{1}\left(x^{20}+x^{13}+x^{6}\right)\left(2 x^{21}+3 x^{14}+6 x^{7}\right)^{1 / 7} d x$
Let $2 x^{21}+3 x^{14}+6 x^{7}=t$
$\Rightarrow 42\left(x^{20}+x^{13}+x^{6}\right) d x=d t$
$I=\frac{1}{42} \int_{0}^{11} t^{1 / 7} d t=\frac{1}{42} \frac{7}{8}\left[t^{8 / 7}\right]_{0}^{11}$
$=\frac{1}{48} (11)^{8/7}$
$\therefore \quad I=48, m=8, n=7$
$\therefore \quad l+m+n=63$
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f^{\prime}(x)+f(x)=\int_\limits{0}^{2} f(t) d t$. If $f(0)=e^{-2}$, then $2 f(0)-f(2)$ is equal to ____________.
Explanation:
$ \begin{aligned} & \Rightarrow e^{x} f(x)=k e^{x}+c \\\\ & f(x)=k+c e^{-x} \\\\ & k=\int_{0}^{2}\left(k+c e^{-t}\right) d t \\\\ & k=2 k+\left.c \cdot \frac{e^{-t}}{-1}\right|_{0} ^{2} \\\\ & k=2 k+c\left(\frac{e^{-2}}{-1}+1\right) \\\\ & -k=c\left(1-\frac{1}{e^{2}}\right) \\\\ & f(x)=c e^{-x}-c\left(1-\frac{1}{e^{2}}\right) \\\\ & f(0)=c-c+\frac{c}{e^{2}}=\frac{1}{e^{2}} \Rightarrow c=1 \\\\ & f(2)=e^{-2}-r\left(1-e^{-2}\right) \\\\ & =2 e^{-2}-1 \\\\ & 2f(0)-f(2)=1 \end{aligned} $
$\lim_\limits{x \rightarrow 0} \frac{48}{x^{4}} \int_\limits{0}^{x} \frac{t^{3}}{t^{6}+1} \mathrm{~d} t$ is equal to ___________.
Explanation:
Applying L' Hospitals Rule
$ \begin{aligned} 48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}\\\\ \end{aligned} $
= ${{48} \over 4}$$\mathop {\lim }\limits_{x \to 0} {{1} \over {{x^6} + 1}} = 12$
If $\int\limits_{{1 \over 3}}^3 {|{{\log }_e}x|dx = {m \over n}{{\log }_e}\left( {{{{n^2}} \over e}} \right)} $, where m and n are coprime natural numbers, then ${m^2} + {n^2} - 5$ is equal to _____________.
Explanation:
$ \begin{aligned} & \left.=-[x \ln x-x]_{\frac{1}{3}}^{1}+x \ln x-x\right]_{1}^{3} \\\\ & =-\left[(0-1)-\left(\frac{1}{3} \ln 3-\frac{1}{3}\right)\right]+[(3 \ln 3-3)-(0-1)] \\\\ & =\frac{2}{3}-\frac{1}{3} \ln 3+3 \ln 3-2 \end{aligned} $
$ \begin{aligned} & =\frac{8}{3} \ln 3-\frac{4}{3} \\\\ & =\frac{4}{3}(2 \ln 3-\ln e) \\\\ & =\frac{4}{3} \ln \left(\frac{3^{2}}{e}\right) \\\\ & m=4, m^{m}=3 \\\\ & m^{2}+n^{2}-5=20 \end{aligned} $
Let $f$ be $a$ differentiable function defined on $\left[ {0,{\pi \over 2}} \right]$ such that $f(x) > 0$ and $f(x) + \int_0^x {f(t)\sqrt {1 - {{({{\log }_e}f(t))}^2}} dt = e,\forall x \in \left[ {0,{\pi \over 2}} \right]}$. Then $\left( {6{{\log }_e}f\left( {{\pi \over 6}} \right)} \right)^2$ is equal to __________.
Explanation:
So, $f(0)=e$
Now differentiate w.r. to $x$
$ \begin{gathered} f^{\prime}(x)+f(x) \sqrt{1-\left(\log _{e} f(x)^{2}\right.}=0 \\\\ \frac{f^{\prime}(x)}{f(x) \sqrt{1-\left(\log _{e} f(x)\right)^{2}}}=-1 \end{gathered} $
Let $\log _{e} f(x)=t$
$\therefore \int \frac{d t}{\sqrt{1-t^{2}}}=-x+c$
$\Rightarrow \sin ^{-1} t=-x+c$
Now $f(0)=e \Rightarrow t=1$ So, $c=\frac{\pi}{2}$
$\therefore t=\sin \left(\frac{\pi}{2}-x\right)=\cos x \quad\left(\because x \in\left[0, \frac{\pi}{2}\right]\right)$
$\therefore\left(6 \log _{e} f\left(\frac{\pi}{6}\right)\right)^{2}=27$
The value of $12\int\limits_0^3 {\left| {{x^2} - 3x + 2} \right|dx} $ is ____________
Explanation:
Let I = $\int\limits_0^3 {|{x^2} - 3x + 2|dx} $
$ = \int\limits_0^3 {|(x - 1)(x - 2)|dx} $
$ = \int\limits_0^1 { + ({x^2} - 3x + 2)dx + \int\limits_1^2 { - ({x^2} - 3x + 2)dx + \int\limits_2^3 {({x^2} - 3x + 2)dx} } } $
$ = \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_0^1 - \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_1^2 + \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_2^3$
$ = \left( {{1 \over 3} - {3 \over 2} + 2} \right) - \left[ {\left( {{8 \over 3} - 6 + 4} \right) - \left( {{1 \over 3} - {3 \over 2} + 2} \right)} \right] + \left[ {\left( {{{27} \over 3} - {{27} \over 2} + 6} \right) - \left( {{8 \over 3} - 6 + 4} \right)} \right]$
$ = {5 \over 6} - \left( {{2 \over 3} - {5 \over 6}} \right) + \left( {{3 \over 2} - {2 \over 3}} \right)$
$ = {5 \over 6} + {5 \over 6} - {2 \over 3} - {2 \over 3} + {3 \over 2}$
$ = {{10} \over 6} - {4 \over 3} + {3 \over 2}$
$ = {{10 - 8 + 9} \over 6} = {{11} \over 6}$
$ \therefore $ 12I = $12 \times {{11} \over 6}$ = 22
The value of ${8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}}}dx} $ is ___________
Explanation:
Let, $I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}}}dx} $ ..... (1)
Using formula,
$\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } $
$I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{2023}}} \over {{{\left[ {\sin \left( {{\pi \over 2} - x} \right)} \right]}^{2023}} + {{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{2023}}}}dx} $
$ = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{2023}}} \over {{{(\cos x)}^{2023}} + {{(\sin x)}^{2023}}}}dx} $ ..... (2)
Adding equation (1) and (2), we get
$2I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{203}}}}dx} $
$ \Rightarrow 2I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {dx} $
$ \Rightarrow 2I = {8 \over \pi }\left[ x \right]_0^{{\pi \over 2}}$
$ \Rightarrow 2I = {8 \over \pi } \times {\pi \over 2}$
$ \Rightarrow 2I = 4$
$ \Rightarrow I = 2$
The value of the integral $\int\limits_{0}^{\frac{\pi}{2}} 60 \frac{\sin (6 x)}{\sin x} d x$ is equal to _________.
Explanation:
$I = \int\limits_0^{{\pi \over 2}} {60\,.\,{{\sin 6x} \over {\sin x}}dx} $
$ = 60\,.\,2\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(4{{\cos }^2}x - 3)\cos x\,dx} $
$ = 120\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(1 - 4{{\sin }^2}x)\cos x\,dx} $
Let $\sin x = t \Rightarrow \cos xdx = dt$
$ = 120\int\limits_0^1 {(3 - 4{t^2})(1 - 4{t^2})dt} $
$ = 120\int\limits_0^1 {(3 - 16{t^2} + 16{t^4})dt} $
$ = 120\left[ {3t - {{16{t^3}} \over 3} + {{16{t^5}} \over 5}} \right]_0^1$
$ = 104$
If $\int\limits_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{\left(1+x^{2}\right)^{3}}}} \mathrm{~d} x=\alpha \sqrt{2}+\beta \sqrt{3}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ is equal to __________.
Explanation:
Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta $
$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^3}\theta \,.\,{{\sec }^2}\theta \,d\theta } \over {\sqrt {1 + {{\tan }^2}\theta + \sqrt {{{\sec }^6}\theta } } }}} $
$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^2}\theta {{\sec }^2}\theta \,d\theta } \over {\sec \theta \sqrt {1 + \sec \theta } }}} $
$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15({{\sec }^2}\theta - 1)\sec \theta \tan \theta \,d\theta } \over {\left( {\sqrt {1 + \sec \theta } } \right)}}} $
Now put $1 + \sec \theta = {t^2}$
$ \Rightarrow \sec \theta \tan \theta \,d\theta = 2tdt$
$ \Rightarrow I = \int\limits_{\sqrt 2 }^{\sqrt 3 } {{{15\left( {{{({t^2} - 1)}^2} - 1} \right)2t\,dt} \over t}} $
$ \Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2} + 1 - 1)\,dt} $
$ \Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2})\,dt} $
$ \Rightarrow I = \left. {30\left( {{{{t^5}} \over 5} - {{2{t^3}} \over 3}} \right)} \right|_{\sqrt 2 }^{\sqrt 3 }$
$ = 30\left[ {\left( {{9 \over 5}\sqrt 3 - 2\sqrt 3 } \right) - \left( {{{4\sqrt 2 } \over 5} - {{4\sqrt 2 } \over 3}} \right)} \right]$
$ = \left( {54\sqrt 3 - 60\sqrt 3 } \right) - \left( {24\sqrt 2 - 40\sqrt 2 } \right)$
$ = 16\sqrt 2 - 6\sqrt 3 $
$\therefore$ $\alpha = 16$ and $\beta = - 6$
$\alpha + \beta = 10.$
Let $f(x)=\min \{[x-1],[x-2], \ldots,[x-10]\}$ where [t] denotes the greatest integer $\leq \mathrm{t}$. Then $\int\limits_{0}^{10} f(x) \mathrm{d} x+\int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x+\int\limits_{0}^{10}|f(x)| \mathrm{d} x$ is equal to ________________.
Explanation:
$\because$ $f(x) = \min \,\{ [x - 1],[x - 2],\,......,\,[x - 10]\} = [x - 10]$
Also $|f(x)| = \left\{ {\matrix{ { - f(x),} & {if\,x \le 10} \cr {f(x),} & {if\,x \ge 10} \cr } } \right.$
$\therefore$ $\int\limits_0^{10} {f(x)dx + \int\limits_0^{10} {{{(f(x))}^2}dx + \int\limits_0^{10} {( - f(x))dx} } } $
$ = \int\limits_0^{10} {{{(f(x))}^2}dx} $
$ = {10^2} + {9^2} + {8^2}\, + \,.....\, + \,{1^2}$
$ = {{10 \times 11 \times 21} \over 6} = 385$
Let f be a differentiable function satisfying $f(x)=\frac{2}{\sqrt{3}} \int\limits_{0}^{\sqrt{3}} f\left(\frac{\lambda^{2} x}{3}\right) \mathrm{d} \lambda, x>0$ and $f(1)=\sqrt{3}$. If $y=f(x)$ passes through the point $(\alpha, 6)$, then $\alpha$ is equal to _____________.
Explanation:
$\because$ $f(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {f\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda ,\,x > 0} $
On differentiating both sides w.r.t., x, we get
$f'(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {{{{\lambda ^2}} \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $
$f'(x) = {1 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {\lambda \,.\,{{2\lambda } \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $
$\therefore$ $\sqrt 3 f'(x) = \left[ {{\lambda \over x}\,.\,f\left( {{{{\lambda ^2}x} \over 3}} \right)} \right]_0^{\sqrt 3 } - \int\limits_0^{\sqrt 3 } {{1 \over x}f\left( {{{{\lambda ^2}x} \over 3}} \right)dx} $
$\sqrt 3 x\,f'(x) = \sqrt 3 f(x) - {{\sqrt 3 } \over 2}f(x)$
$xf'(x) = {{f(x)} \over 2}$
On integrating we get : $\ln y = {1 \over 2}\ln x + \ln c$
$\because$ $f(1) = \sqrt 3 $ then $c = \sqrt 3 $
$\therefore$ ($\alpha$, 6) lies on
$\therefore$ $y = \sqrt {3x} $
$\therefore$ $6 = \sqrt {3\alpha } \Rightarrow \alpha = 12$.
If $\mathrm{n}(2 \mathrm{n}+1) \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}} \mathrm{d} x=1177 \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}+1} \mathrm{~d} x$, then $\mathrm{n} \in \mathbf{N}$ is equal to ______________.
Explanation:
$\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = \int_0^1 {1\,.\,{{(1 - {x^n})}^{2n + 1}}dx} } $
$ = \left[ {{{(1 - {x^n})}^{2n + 1}}\,.\,x} \right]_0^1 - \int_0^1 {x\,.\,(2n + 1){{(1 - {x^n})}^{2n}}\,.\, - n{x^{n - 1}}dx} $
$ = n(2n + 1)\int_0^1 {(1 - (1 - {x^n})){{(1 - {x^n})}^{2n}}dx} $
$ = n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx - n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} } $
$(1 + n(2n + 1))\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx} } $
$(2{n^2} + n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = 1177\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} } $
$\therefore$ $2{n^2} + n + 1 = 1177$
$2{n^2} + n - 1176 = 0$
$\therefore$ $n = 24$ or $ - {{49} \over 2}$
$\therefore$ $n = 24$
Let $f$ be a twice differentiable function on $\mathbb{R}$. If $f^{\prime}(0)=4$ and $f(x) + \int\limits_0^x {(x - t)f'(t)dt = \left( {{e^{2x}} + {e^{ - 2x}}} \right)\cos 2x + {2 \over a}x} $, then $(2 a+1)^{5}\, a^{2}$ is equal to _______________.
Explanation:
Here $f(0)=2 \hspace{0.5cm} ...(ii)$
On differentiating equation (i) w.r.t. $x$ we get :
$ \begin{aligned} & f^{\prime}(x)+\int_0^x f^{\prime}(t) d t+x f^{\prime}(x)-x f^{\prime}(x) \\\\ & = 2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\ & \Rightarrow \quad f(x)+f(x)-f(0)\\\\ &=2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\ & \quad \text { Replace } x \text { by } 0 \text { we get }: \\\\ & \Rightarrow \quad 4=\frac{2}{a} \Rightarrow a=\frac{1}{2} \cdot \\\\ & \therefore \quad(2 a+1)^5 \cdot a^2=2^5 \cdot \frac{1}{2^2}=2^3=8 \end{aligned} $