JEE 2023 — Mathematics PYQ

1. Equation ko Simplify karna

Di gayi equation hai:

Yahan ${\sin }^6\theta$ ko ${\cos }^2\theta$ ke terms mein badalte hain:

${\sin }^6\theta =({\sin }^2\theta {)}^3=(1-{\cos }^2\theta {)}^3$

Maan lijiye ${\cos }^2\theta =t$. Yahan $t$ ki value $0\le t\le 1$ ke beech honi chahiye.

Ab equation ban jayegi:

2. Equation ko Solve karna

$(1-t{)}^3$ ko expand karte hain: $(1-t{)}^3=1-3t+3t^2-t^3$

Ab terms ko combine karein:

Yahan se $t$ common nikalne par:

Toh hamare paas $t$ ki teen values aayi hain:

1. $t=0$

2. $t=1/2$

3. $t=1$

3. $\theta$ ki values nikalna $[0,2\pi ]$ mein

• Case 1: $t=0⟹{\cos }^2\theta =0⟹\cos \theta =0$

  $\theta =\frac{\pi }{2},\frac{3\pi }{2}$ (2 values)

• Case 2: $t=1⟹{\cos }^2\theta =1⟹\cos \theta =\pm 1$

  $\theta =0,\pi ,2\pi$ (3 values)

• Case 3: $t=1/2⟹{\cos }^2\theta =1/2⟹\cos \theta =\pm \frac{1}{\sqrt{2}}$

  $\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ (4 values)

4. Kul (Total) Elements

Total values $=2+3+4=9$