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E[T]=\int_0^1 \frac{t}{2}dt+e^2\int_1^{\infty}te^{-2t}dt =[\frac{t^2}{4}]_0^1+e^2[-\frac{t}{2}e^{-2t}]_1^{\infty}+e^2\int_1^{\infty}\frac{1}{2}e^{-2t}dt=\frac{1}{4}+\freac{1}{2}+e^2[-\frac{1}{4}e^{-2t}=\frac{1}{4}+ \frac{1}{2}+ \frac{1}{4}=1
cos\theta=\dfrac{d^2-x^2-h^2-x^2-(h+d)^2}{-2\sqrt{(x^2+h^2)(x^2+(d+h)^2)}}[br]\Rightarrow Tan\theta=\frac{\sqrt{1-cos^2\theta}}{cos\theta}[br]=\dfrac{\sqrt{(x^2+h^2)(x^2+(d+h)^2)-(x^2+hd+x^2)^2}}{x^2+hd+h^2}}[br]=\dfrac{\sqrt{x^4+x^2h^2+2x^2hd+x^2d^2+h^2x^2+h^4+2h^3d+h^2d^2-x^4-2x^2hd-2x^2h^2-h^2d^2-2h^3d-h^4}}{x^2+hd+h^2}[br]=\dfrac{xd}{x^2+hd+h^2}
STEP I - Q11
E[T]=\int_0^1 \frac{t}{2}dt+e^2\int_1^{\infty}te^{-2t}dt =[\frac{t^2}{4}]_0^1+e^2[-\frac{t}{2}e^{-2t}]_1^{\infty}+e^2\int_1^{\infty}\frac{1}{2}e^{-2t}dt=\frac{1}{4}+\freac{1}{2}+e^2[-\frac{1}{4}e^{-2t}=\frac{1}{4}+ \frac{1}{2}+ \frac{1}{4}=1
[\impliesA(x+a)(x+b)+Bx(x+b)+Cx(x+a)=3x^2+2(a+b)x+ab
a\neqb,
\frac{z-x}{c-a}=\frac{x-y}{a-b}=\frtac{1}{x+y+z}
\implies (x+y+z)^2=\frac{2(a^2+b^2+c^2)-2(ab+bc+ca)}[2(x^2+y^2+z^2)-2(xy+yz+zx)}=\frac{a^2+b^2+c^2-bc-ca-ab}{(x^2-yz)+(y^2-zx)+(z^2-xy)}=\frac{a^2+b^2+c^2-bc-ca-ab}{a+b+c}=\Delta^2
(\cos\theta+i\sin\theta)^n=\cosn\theta+i\sin n\theta
=(\cos k\theta \cos \theta-\sin k\theta \sin\theta)+i(\cos k\theta \sin\theta+\sin k\thewta \cos \theta)
x^3\cos3y+2x^2 \cos2y+2x\cosy=\mathrm{\ Re}((xz)^3+2(xz)^2+2xz) \mathrm{\ and\ }x^3\sin3y+2x^2\sin2y+2x\siny= \mathrm{\ Im}((xz)^3+2(xz)^2+2xz)
a=\frac{V^2}{2g}+\frac{V^2\sqrt}{2g}=\frac{V^2}{2g}(1+\sqrt3)
\phi=\tan^{-1}\Left( \frac{V}{\sqrt2} \times \frac{2\sqrt2}{V(\sqrt3-L1)}\Right)=\tan^{-1}\frac{2}{\sqrt3-1}=\tan^{-1}(\sqrt3+1)
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