1. Scenario-1: f(y)=ay2+by+c f(\mathrm{y})=a y^{2}+\mathrm{by}+\mathrm{c} f(y)=ay2+by+c.
Scenario-2: A \mathrm{A} A be the vertex and S \mathrm{S} S be the focus
Find the equation of directrix of the conic y2=8x+5 y^{2}=8 x+5 y2=8x+5
From scenario-1, the vertex of the conic x=f(y) \mathrm{x}=f(\mathrm{y}) x=f(y) is at the point (3,−2) (3,-2) (3,−2) and it passes through the point (5,0) (5,0) (5,0), find the value of a,b,c a, b, c a,b,c.
From scenario-2, find the equation of the parabola.
2. f(x)=sin−1x,g(x)=cosx f(x)=\sin ^{-1} x, g(x)=\cos x f(x)=sin−1x,g(x)=cosx.
Solve: tan2θtanθ=1 \tan 2 \theta \tan \theta=1 tan2θtanθ=1.
Show that, f(2g(π2−θ))+f(g(2θ))=π2 f\left(\sqrt{2} g\left(\frac{\pi}{2}-\theta\right)\right)+f(\sqrt{g(2 \theta)})=\frac{\pi}{2} f(2g(2π−θ))+f(g(2θ))=2π.
Solve: g(x)+3 g′(x)=2 \mathrm{g}(\mathrm{x})+\sqrt{3} \mathrm{~g}^{\prime}(\mathrm{x})=\sqrt{2} g(x)+3 g′(x)=2, when −π<x<π -\pi<\mathrm{x}<\pi −π<x<π.
3. z1=−1−i3,z2=3−i \mathrm{z}_{1}=-1-\mathrm{i} \sqrt{3}, \mathrm{z}_{2}=\sqrt{3}-\mathrm{i} z1=−1−i3,z2=3−i.
Find the square root of z1 z_{1} z1.
Show that, Arg(z1z2)=Argz1−Argz2 \operatorname{Arg}\left(\frac{z_{1}}{z_{2}}\right)=\operatorname{Arg} z_{1}-\operatorname{Arg} z_{2} Arg(z2z1)=Argz1−Argz2.
Prove that, (12zˉ1)n+(12z1)n=2 \left(\frac{1}{2} \bar{z}_{1}\right)^{n}+\left(\frac{1}{2} z_{1}\right)^{n}=2 (21zˉ1)n+(21z1)n=2, when n n n is divisible by 3 or, -1 , when n n n is any other integer.
4. f(x)=a+bx+cx2,g(x)=px2+qx+r f(x)=a+b x+c x^{2}, g(x)=p x^{2}+q x+r f(x)=a+bx+cx2,g(x)=px2+qx+r.
Find the value of −814 \sqrt[4]{-81} 4−81.
If f(1)=0 f(1)=0 f(1)=0, then prove that, {f(ω)}3+{f(ω2)}3=27abc \{f(\omega)\}^{3}+\left\{f\left(\omega^{2}\right)\right\}^{3}=27 \mathrm{abc} {f(ω)}3+{f(ω2)}3=27abc, when ω \omega ω is one of the complex cube roots of unity.
If the roots of the equation g(x)=0 \mathrm{g}(\mathrm{x})=0 g(x)=0 are γ \gamma γ and δ \delta δ, then express the roots of rp(x2+1)−(q2−2rp)x=0 \mathrm{rp}\left(\mathrm{x}^{2}+1\right)-\left(\mathrm{q}^{2}-2 \mathrm{rp}\right) \mathrm{x}=0 rp(x2+1)−(q2−2rp)x=0 in terms of γ,δ \gamma, \delta γ,δ
5.
If the resultant of the forces a \mathrm{a} a and b(a>b) \mathrm{b}(\mathrm{a}>\mathrm{b}) b(a>b) trisects the angle between them, find the angle between two forces.
If the forces P,Q,R \mathrm{P}, \mathrm{Q}, \mathrm{R} P,Q,R act at the incentre I \mathrm{I} I are in equilibrium show that, P:Q:R=sin(π2−A2):sin(π2−B2):sin \mathrm{P}: \mathrm{Q}: \mathrm{R}=\sin \left(\frac{\pi}{2}-\frac{\mathrm{A}}{2}\right): \sin \left(\frac{\pi}{2}-\frac{\mathrm{B}}{2}\right): \sin P:Q:R=sin(2π−2A):sin(2π−2B):sin (π2−C2) \left(\frac{\pi}{2}-\frac{C}{2}\right) (2π−2C).
If the forces P,Q,R \mathrm{P}, \mathrm{Q}, \mathrm{R} P,Q,R of the stem are not acted, only three like parallel forces U,V,W U, V, W U,V,W act at the vertrices A,B,C A, B, C A,B,C and their resultant force passes through the circumcentre O O O, prove that, U:V:W=acosA:bcosB:cosC \mathrm{U}: \mathrm{V}: \mathrm{W}=\operatorname{acos} \mathrm{A}: \mathrm{b} \cos \mathrm{B}: \cos \mathrm{C} U:V:W=acosA:bcosB:cosC.