JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Find the value of:
\max_{0 \le x < \pi} \left\{ x - 2\sin x \cos x + \frac{1}{3}\sin 3x \right\}
Choose the correct answer:
- A.
0
- B.
2
- C.
65π+2+33
(Correct Answer)
65π+2+33
Explanation
Solution:
Let f(x)=x−sin2x+31sin3x. We find the maximum using derivatives.
1. Find the derivative f′(x):
f′(x)=1−2cos2x+cos3x
2. Set f′(x)=0 to find critical points:
Using identities cos2x=2cos2x−1 and cos3x=4cos3x−3cosx:
1−2(2cos2x−1)+(4cos3x−3cosx)=0
4cos3x−4cos2x−3cosx+3=0
(4cos2x−3)(cosx−1)=0
3. Identify x in the range [0,π):
-
cosx=1⟹x=0
-
cos2x=43⟹cosx=±23⟹x=6π,65π
4. Check x=65π for maximum:
f(65π)=65π−sin(35π)+31sin(25π)
f(65π)=65π−(−23)+31(1)
f(65π)=65π+23+31=65π+33+2
Correct Answer: Option (3)
Explanation
Solution:
Let f(x)=x−sin2x+31sin3x. We find the maximum using derivatives.
1. Find the derivative f′(x):
f′(x)=1−2cos2x+cos3x
2. Set f′(x)=0 to find critical points:
Using identities cos2x=2cos2x−1 and cos3x=4cos3x−3cosx:
1−2(2cos2x−1)+(4cos3x−3cosx)=0
4cos3x−4cos2x−3cosx+3=0
(4cos2x−3)(cosx−1)=0
3. Identify x in the range [0,π):
-
cosx=1⟹x=0
-
cos2x=43⟹cosx=±23⟹x=6π,65π
4. Check x=65π for maximum:
f(65π)=65π−sin(35π)+31sin(25π)
f(65π)=65π−(−23)+31(1)
f(65π)=65π+23+31=65π+33+2
Correct Answer: Option (3)