1. Scenario-1: The roots of the equation 5x3−4x2+1=0 5 x^{3}-4 x^{2}+1=0 5x3−4x2+1=0 are α,β \alpha, \beta α,β and γ \gamma γ.
Scenario-2: Complex roots of the equation x3−1=0 x^{3}-1=0 x3−1=0 are a a a and b b b.
Show that, sec2(cot−112)+cosec2(tan−113)=15 \sec ^{2}\left(\cot ^{-1} \frac{1}{2}\right)+\operatorname{cosec}^{2}\left(\tan ^{-1} \frac{1}{3}\right)=15 sec2(cot−121)+cosec2(tan−131)=15.
From scenario-1, find the value of ∑α2β \sum \alpha^{2} \beta ∑α2β.
From scenario-2, prove that, an+bn=2 a^{n}+b^{n}=2 an+bn=2 or -1 , where the value of n n n is respectively divisible by 3 or any other integer.
2.
If the square of resultant of two equal forces is equal to the thrice of the product of that two forces, then find the angle between the forces.
If three forces P,Q \mathrm{P}, \mathrm{Q} P,Q and R \mathrm{R} R of scenario-1 are in equilibrium, then prove that, Pa2( b2+c2−a2)=Qb2(a2+c2−b2)=Rc2(a2+b2−c2) \frac{\mathrm{P}}{\mathrm{a}^{2}\left(\mathrm{~b}^{2}+\mathrm{c}^{2}-\mathrm{a}^{2}\right)}=\frac{\mathrm{Q}}{\mathrm{b}^{2}\left(\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}\right)}=\frac{\mathrm{R}}{\mathrm{c}^{2}\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}\right)} a2( b2+c2−a2)P=b2(a2+c2−b2)Q=c2(a2+b2−c2)R.
In scenario-2 if the forces P \mathrm{P} P and Q \mathrm{Q} Q are both increased by S \mathrm{S} S and the resultant is displaced from the point C \mathrm{C} C to D \mathrm{D} D, then show that, CD=SP−QAB C D=\frac{S}{P-Q} A B CD=P−QSAB.
3.
If a particle projected with the velocity u u u at an angle α \alpha α to the horizon, then prove that, greatest height, H=u2sin2α2g. \mathrm{H}=\frac{u^{2} \sin ^{2} \alpha}{2 g} . H=2gu2sin2α.
In scenario-1, if a swimmer requires t t t seconds to cross AB A B AB distance and t′ t^{\prime} t′ seconds to cross AC distance, then show that, t:t′=u+v:u−v t: t^{\prime}=\sqrt{u+v}: \sqrt{u-v} t:t′=u+v:u−v.
From scenario-2, find the horizontal range of projectile.
4. A=sin−123, B=cos−134,C=tan−115 \mathrm{A}=\sin ^{-1} \frac{2}{3}, \mathrm{~B}=\cos ^{-1} \frac{3}{4}, \mathrm{C}=\tan ^{-1} \frac{1}{\sqrt{5}} A=sin−132, B=cos−143,C=tan−151 and f(x)=sinx f(\mathrm{x})=\sin \mathrm{x} f(x)=sinx.
Prove that, cos−1x=2cos−11+x2 \cos ^{-1} x=2 \cos ^{-1} \sqrt{\frac{1+x}{2}} cos−1x=2cos−121+x.
Prove that, A−12B+C=tan−1(35−17+5) A-\frac{1}{2} B+C=\tan ^{-1}\left(\frac{\sqrt{35}-1}{\sqrt{7}+\sqrt{5}}\right) A−21B+C=tan−1(7+535−1).
Find the solution of the equation f(x)+f(2x)+f(3x)=0 f(\mathrm{x})+f(2 \mathrm{x})+f(3 \mathrm{x})=0 f(x)+f(2x)+f(3x)=0 from the stem where 0≤x≤π 0 \leq \mathrm{x} \leq \pi 0≤x≤π.
5. Scenario-1: Two roots of the equation 3x2−4x+1=0 3 x^{2}-4 x+1=0 3x2−4x+1=0 are a \mathrm{a} a and b b b.
Scenario-2: Two roots of the equation x2−qx+r=0 \mathrm{x}^{2}-\mathrm{qx}+\mathrm{r}=0 x2−qx+r=0 are α \alpha α and β \beta β.
If 9x2−(k+2)x+4 9 x^{2}-(k+2) x+4 9x2−(k+2)x+4 is a perfect square, then find the value of k k k.
From scenario-1, find the quadratic equation whose roots are a+1 b \mathrm{a}+\frac{1}{\mathrm{~b}} a+ b1 and b+1a \mathrm{b}+\frac{1}{\mathrm{a}} b+a1.
By using α \alpha α and β \beta β of scenario-2, express the two roots of the equation r(x2+1)−(q2−2r)x=0 \mathrm{r}\left(\mathrm{x}^{2}+1\right)-\left(\mathrm{q}^{2}-2 \mathrm{r}\right) \mathrm{x}=0 r(x2+1)−(q2−2r)x=0 in terms of α \alpha α and β \beta β.