CUET PG 2023 Mathematics PYQ — For , the solution(s) of is/are (A) (B) (C) (D) Choose the correc… | Mathem Solvex | Mathem Solvex
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CUET PG 2023 — Mathematics PYQ
CUET PG | Mathematics | 2023
For 0<\theta<\frac{\pi}{2}, the solution(s) of ∑m=16cosec(θ+(m−1)4π)cosec(θ+4mπ)=42 is/are (A) 4π (B) 6π (C) 12π (D) 125π Choose the correct answer from the option given below:
Choose the correct answer:
A.
A and B only
B.
C and D only
(Correct Answer)
C.
A and C only
D.
B and D only
Correct Answer:
C and D only
Explanation
1. Simplify the General Term
The general term of the summation is:
Tm=csc(θ+(m−1)4π)csc(θ+4mπ)
Let A=θ+(m−1)4π and B=θ+4mπ.
The difference between the two angles is:
B−A=(θ+4mπ)−(θ+4(m−1)π)=4π
We can rewrite Tm as:
Tm=sinAsinB1
Multiply and divide by sin(B−A)=sin4π=21:
Tm=sin4π1⋅sinAsinBsin(B−A)
Tm=2⋅[sinAsinBsinBcosA−cosBsinA]
Tm=2(cotA−cotB)
2. Summation (Telescoping Series)
Substituting A and B back into the sum:
m=1∑6Tm=2m=1∑6[cot(θ+(m−1)4π)−cot(θ+4mπ)]
This is a telescoping sum where most terms cancel out:
Sum=2[cotθ−cot(θ+46π)]
Sum=2[cotθ−cot(θ+23π)]
Since cot(23π+θ)=−tanθ, the equation becomes:
2(cotθ+tanθ)=42
cotθ+tanθ=4
3. Solve for θ
Convert to sine and cosine:
sinθcosθ+cosθsinθ=4
sinθcosθcos2θ+sin2θ=4
sinθcosθ1=4
1=2(2sinθcosθ)
1=2sin2θ⟹sin2θ=21
Given 0 < \theta < \frac{\pi}{2}, we have 0 < 2\theta < \pi.
The values for 2θ are:
2θ=6π⟹θ=12π
2θ=π−6π=65π⟹θ=125π
Conclusion
The solutions are θ=12π and θ=125π.
Matching with the options:
(C) 12π
(D) 125π
Correct Options: (C) and (D)
Explanation
1. Simplify the General Term
The general term of the summation is:
Tm=csc(θ+(m−1)4π)csc(θ+4mπ)
Let A=θ+(m−1)4π and B=θ+4mπ.
The difference between the two angles is:
B−A=(θ+4mπ)−(θ+4(m−1)π)=4π
We can rewrite Tm as:
Tm=sinAsinB1
Multiply and divide by sin(B−A)=sin4π=21:
Tm=sin4π1⋅sinAsinBsin(B−A)
Tm=2⋅[sinAsinBsinBcosA−cosBsinA]
Tm=2(cotA−cotB)
2. Summation (Telescoping Series)
Substituting A and B back into the sum:
m=1∑6Tm=2m=1∑6[cot(θ+(m−1)4π)−cot(θ+4mπ)]
This is a telescoping sum where most terms cancel out:
Sum=2[cotθ−cot(θ+46π)]
Sum=2[cotθ−cot(θ+23π)]
Since cot(23π+θ)=−tanθ, the equation becomes:
2(cotθ+tanθ)=42
cotθ+tanθ=4
3. Solve for θ
Convert to sine and cosine:
sinθcosθ+cosθsinθ=4
sinθcosθcos2θ+sin2θ=4
sinθcosθ1=4
1=2(2sinθcosθ)
1=2sin2θ⟹sin2θ=21
Given 0 < \theta < \frac{\pi}{2}, we have 0 < 2\theta < \pi.
