Let α be the angle between the lines whose direction cosines satisfy the equations I + m – n = 0 and I² + m² – n² = 0. Then the value of sin⁴α + cos⁴α is:

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

$C=+\frac{\pi }{2}$

now at x = $\mathcal{l}n\sqrt[]{3}$

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$

$x_1=\underset{x\to \infty }{lim}\frac{2^{\frac{x^n}{e^x}}-3^{\frac{x^n}{e^x}}}{\frac{x^n}{e^x}},put\frac{x^n}{e^x}=t$

$x_2=\underset{x\to \infty }{lim}\frac{co{t}^{-1}\left(\sqrt[]{x+1}-\sqrt[]{x}\right)}{se{c}^{-1}\left({\left(\frac{2x+1}{x-1}\right)}^x\right)}$

$x_2=\frac{2}{\pi }\underset{x\to \infty }{lim}ta{n}^{-1}\left(\sqrt[]{x+1}+\sqrt[]{x}\right)$

$x_2=\frac{2}{\pi }.\frac{\pi }{2}=1$

Differentiating

y.  $\frac{-2x}{2\sqrt[]{1-x^2}}+\sqrt[]{1-x^2}\cdot y^{\mathrm{\text{'}}}=\frac{x\cdot 2y{y}^{\mathrm{\text{'}}}}{2\sqrt[]{1-y^2}}-\sqrt[]{1-y^2}$

Put $x=\frac{1}{2},y=-\frac{1}{4}$ and $x,y=-\frac{1}{8}$

$y^{\mathrm{\text{'}}}=\frac{-\sqrt[]{5}}{2}$

dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.

$af\left(x\right)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta }{x}............\left(i\right)$  

Replace x by   $\frac{1}{x}$

$af\left(\frac{1}{x}\right)+\alpha f\left(x\right)=\frac{b}{x}+\beta x..............\left(ii\right)$  

$a\left(f\left(x\right)+f\left(\frac{1}{x}\right)\right)+\alpha \left(f\left(x\right)+f\left(\frac{1}{x}\right)\right)=b\left(x+\frac{1}{x}\right)+\beta \left(x+\frac{1}{x}\right)$           

$\therefore \frac{f\left(x\right)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}=\frac{b+\beta }{a+\alpha }=\frac{2}{1}=2$