1. f(x)=cosx \mathrm{f}(\mathrm{x})=\cos \mathrm{x} f(x)=cosx is a trigonometric function.
If tan−1x+tan−1y=π2 \tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{2} tan−1x+tan−1y=2π, then show that, x=1y x=\frac{1}{y} x=y1, [where \( x>0, y>0,0 ]
If f−1(2x)+f−1(2y)=3π2 f^{-1}(2 x)+f^{-1}(2 y)=\frac{3 \pi}{2} f−1(2x)+f−1(2y)=23π, then show that, x2+y2= x^{2}+y^{2}= x2+y2= 14 \frac{1}{4} 41
If f(x)+3f(π2−x)=2 f(x)+\sqrt{3} f\left(\frac{\pi}{2}-x\right)=\sqrt{2} f(x)+3f(2π−x)=2, then solve the equation.
2. The above figure indicates a conic whose focus is S \mathrm{S} S, vertex is A \mathrm{A} A and MZM′ \mathrm{MZM}^{\prime} MZM′ ' is directriex.
Find the eccentricity of the conic x24−y29+1=0.2 \frac{x^{2}}{4}-\frac{y^{2}}{9}+1=0.2 4x2−9y2+1=0.2
If the equation of the conic given in stem is y2= y^{2}= y2= 6x 6 x 6x and SP=6 S P=6 SP=6, then find the coordinate of point P.
Find the equation of the stem mentioned conic whose focus is (−1,1) (-1,1) (−1,1) and vertex (2,−3) (2,-3) (2,−3).
3.
If P=Q,R=33N P=Q, R=3 \sqrt{3} N P=Q,R=33N and α=60∘ \alpha=60^{\circ} α=60∘, then find two forces.
In scenario-1, if α=3θ \alpha=3 \theta α=3θ, then prove that, R= R= R=P2−Q2Q;(P>Q). \frac{P^{2}-Q^{2}}{Q} ;(P>Q) \text {. } QP2−Q2;(P>Q).
In scenario-2, if the forces are interchanged, then find the distance through which resultant is shifted along AB \mathrm{AB} AB.
4. Stem-1: 4x2+6y2−4x−36y+43=0 4 x^{2}+6 y^{2}-4 x-36 y+43=0 4x2+6y2−4x−36y+43=0.
Stem-2: Two focus points of a conic are (10,5) (10,5) (10,5) and (8,3) (8,3) (8,3). Also eccentricity is 2 \sqrt{2} 2.
Find the length of the latus rectum of the conic 5x2+4y2=1 5 x^{2}+4 y^{2}=1 5x2+4y2=1.
Express the equation described in stemstandard form then find the equation of of the conic.
Determine the equation of the conic described in stem-2.
5. 8x2+2x−(b+4)=0 8 x^{2}+2 x-(b+4)=0 8x2+2x−(b+4)=0 and y2+y+1=0 y^{2}+y+1=0 y2+y+1=0 are two quadratic equations.
Find an equation having roots 2−−3 2-\sqrt{-3} 2−−3.
If one root of first equation is equal to 2 another, then find the value of b b b.
If α \alpha α and β \beta β ind the value of show the roots of second equation, then show that, α2=β \alpha^{2}=\beta α2=β and β2=α \beta^{2}=\alpha β2=α.