Definite Integration

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 .

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

$ \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} $

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 $\beta^\alpha=4^3=64$

$ \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 .

$ \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} $

$ \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} $

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} $

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 $ $2 \cos x+1=0 \Rightarrow \cos x=-\frac{1}{2} \Rightarrow x=\frac{2 \pi}{3}$

$\begin{array}{ll}2 \cos x+1>0 & \forall x \in[0,2 \pi / 3) \\ 2 \cos x+1 \leq 0 & \forall x \in[2 \pi / 3, \pi]\end{array}$

$I=\int\limits_0^{\pi / 2} \sin 2 x(2 \cos x+1) d x-\int\limits_{\pi / 2}^{2 \pi / 3} \sin 2 x(2 \cos x+1) d x+\int\limits_{2 \pi / 3}^\pi \sin 2 x(2 \cos x+1) d x$

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 $ $I=2\left[\frac{5}{24}-(-1 / 24)+\frac{28}{24}\right]=2 \times \frac{34}{24}=\frac{17}{6}$

$\Rightarrow 6 I=17$

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, $\alpha^3 = 2^3 = 8$.

$\begin{aligned} = & 24 \int_0^{\frac{\pi}{48}}-\sin \left(4 \mathrm{x}-\frac{\pi}{12}\right)+\int_{\pi / 48}^{\pi / 4} \sin \left(4 \mathrm{x}-\frac{\pi}{12}\right) \\ & +\int_0^{\frac{\pi}{6}}[0] \mathrm{dx}+\int_{\pi / 6}^{\pi / 4}[2 \sin \mathrm{x}] \mathrm{dx} \end{aligned}$

$\begin{aligned} & =24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right]+\frac{\pi}{4}-\frac{\pi}{6} \\ & =24\left(\frac{1}{2}+\frac{\pi}{12}\right)=2 \pi+12 \\ & \alpha=12 \end{aligned}$

$1^{\infty}$ 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

$\alpha=64$

$\begin{aligned} & \int_0^1 \mathrm{f}(\lambda \mathrm{x}) \mathrm{d} \lambda=\mathrm{af}(\mathrm{x}) \\ & \lambda \mathrm{x}=\mathrm{t} \\ & \mathrm{~d} \lambda=\frac{1}{\mathrm{x}} \mathrm{dt} \\ & \frac{1}{\mathrm{x}} \int_0^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{af}(\mathrm{x}) \\ & \int_0^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{axf}(\mathrm{x}) \\ & \mathrm{f}(\mathrm{x})=\mathrm{a}\left(\mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})\right) \\ & (1-\mathrm{a}) \mathrm{f}(\mathrm{x})=\mathrm{a} \cdot \mathrm{x} \mathrm{f}^{\prime}(\mathrm{x}) \\ & \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}=\frac{(1-\mathrm{a})}{\mathrm{a}} \frac{1}{\mathrm{x}} \\ & \ell \operatorname{lnf}(\mathrm{x})=\frac{1-\mathrm{a}}{\mathrm{a}} \ell \mathrm{n} \mathrm{x}+\mathrm{c} \\ & \mathrm{x}=1, \mathrm{f}(1)=1 \Rightarrow \mathrm{c}=0 \\ & \mathrm{x}=16, \mathrm{f}(16)=\frac{1}{8} \end{aligned}$

$\begin{aligned} & \frac{1}{8}=(16)^{\frac{1-a}{a}} \Rightarrow-3=\frac{4-4 a}{a} \Rightarrow \mathrm{a}=4 \\ & \mathrm{f}(\mathrm{x})=\mathrm{x}^{-\frac{3}{4}} \\ & \mathrm{f}^{\prime}(\mathrm{x})=-\frac{3}{4} \mathrm{x}^{-\frac{7}{4}} \\ & \therefore \quad 16-\mathrm{f}^{\prime}\left(\frac{1}{16}\right) \\ & =16-\left(-\frac{3}{4}\left(2^{-4}\right)^{-7 / 4}\right) \\ & =16+96=112 \end{aligned}$

$\begin{aligned} & \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}+\ldots \frac{n}{\sqrt{n^4+n^4}}\right) \\ & -\lim _{n \rightarrow \infty}\left(\frac{2 n}{\left(n^2+1\right)\left(\sqrt{n^4+1}\right)}\right)+\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+1}}+\cdots \frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}} \\ & =\lim _{n \rightarrow \infty} \sum_{n=1}^n \frac{1}{n \sqrt{1+\frac{r^4}{n^4}}}-\lim _{n \rightarrow \infty} \sum_{n=1}^n \frac{1}{n} \frac{2 \cdot(r / n)^2}{\left(1+\left(\frac{r}{n}\right)^2\right) \sqrt{1+\frac{r^4}{n^4}}} \\ & =\int_\limits0^1 \frac{d x}{\sqrt{1+x^4}}-2 \int_\limits0^1 \frac{x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \\ & =\int_\limits0^1 \frac{1-x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \end{aligned}$

$\begin{aligned} & \int_\limits0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{x^2+\frac{1}{x^2}}} \\ & =\int_\limits0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-2}} \\ & x+\frac{1}{x}=t \Rightarrow\left(1-\frac{1}{x^2}\right) d x=d t \\ & -\int_\limits{\infty}^2 \frac{d t}{t \sqrt{t^2-2}}=\int_\limits2^{\infty} \frac{d t}{t \sqrt{t^2-2}}=\left.\frac{-1}{\sqrt{2}} \sin ^{-1} \frac{\sqrt{2}}{x}\right|_2 ^{\infty} \\ & =\frac{-1}{\sqrt{2}}\left(0-\frac{\pi}{4}\right) \\ & =\frac{\pi}{2^{5 / 2}} \\ & \therefore k=2^{5 / 2} \\ & \therefore k^2=32 \end{aligned}$