1.
Explain, what is the resolved part of force.
In scenario-1, if F1∝cosP,F2∝cosQ \mathrm{F}_{1} \propto \cos \mathrm{P}, \mathrm{F}_{2} \propto \cos \mathrm{Q} F1∝cosP,F2∝cosQ and resultant of F1 \mathrm{F}_{1} F1 and F2 F_{2} F2 be F F F, then show that R−φ=12(R+Q−P) R-\varphi=\frac{1}{2}(R+Q-P) R−φ=21(R+Q−P)
In scenario-2, if the forces Q,R,S Q, R, S Q,R,S are in equilibrium, then show that S2=R(R−Q) S^{2}=R(R-Q) S2=R(R−Q).
2.
Show that cos(2tan1yx)=x2−y2x2+y2 \cos \left(2 \tan 1 \frac{y}{x}\right)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}} cos(2tan1xy)=x2+y2x2−y2.
In the stem, if A+P=φ A+P=\varphi A+P=φ prove that x2−2xycosφ+y2=r2 x^{2}-2 x y \cos \varphi+y^{2}=r^{2} x2−2xycosφ+y2=r2 sin2φ \sin ^{2} \varphi sin2φ.
If f(θ)=rx f(\theta)=\frac{r}{x} f(θ)=xr solve the equation f(2θ)−f(θ)=2 f(2 \theta)-f(\theta)=2 f(2θ)−f(θ)=2 in the limit π≤x≤π \pi \leq x \leq \pi π≤x≤π.
3. Z=−2−23i \mathrm{Z}=-2-2 \sqrt{3} \mathrm{i} Z=−2−23i is a complex term.
If x+iy=p+iqr+is x+i y=\sqrt{\frac{p+i q}{r+i s}} x+iy=r+isp+iq, show that (x2+y2)2=p2+q2r2+s2 \left(x^{2}+y^{2}\right)^{2}=\frac{p^{2}+q^{2}}{r^{2}+s^{2}} (x2+y2)2=r2+s2p2+q2.
Determine, Arg(z) \operatorname{Arg}(\sqrt{\mathrm{z}}) Arg(z).
If a root of a cubic equation is z z z and the product of the roots be 80 , form the equation.
4.
Find the equation of directrix of the conic 9x2−4y2=36. 9 x^{2}-4 y^{2}=36 \text {. } 9x2−4y2=36.
Determine the equation of parabola, taking A A A as vertex and S \mathrm{S} S as focus.
In the stem, if OB′=4 \mathrm{OB}^{\prime}=4 OB′=4 and AS=A′S \mathrm{AS}=\mathrm{A}^{\prime} \mathrm{S} AS=A′S, determine the equation of the lotus rectum of the ellipse, taking BB′ \mathbf{B B}^{\prime} BB′ as major axis and AA′ \mathrm{AA}^{\prime} AA′ as minor axis.
5. (1+2y)m (1+2 y)^{m} (1+2y)m is an algebraic expression.
Find the co-efficient of xn x^{n} xn in the expansion of (42x2−13x+1)−12 \left(42 x^{2}-13 x+1\right)^{-1} 2 (42x2−13x+1)−12
If m=2n,n∈Z m=2 n, n \in Z m=2n,n∈Z show that the co-efficient of the middle term in-the expansion of the expression in the stem is 1.3.5…(2n−1)n!⋅22n⋅yn \frac{1.3 .5 \ldots(2 n-1)}{n !} \cdot 2^{2 n} \cdot y^{n} n!1.3.5…(2n−1)⋅22n⋅yn.
If m=20 m=20 m=20, the ratio of the co-efficient of two consecutive terms in the expansion of the expression in the stem be 11:20, find the two terms.
