If y(θ)=(2 cos θ+cos 2 θ)/(cos 3 θ+4 cos 2 θ+5 cos θ+2), then at θ=π/2, y^(′ ′)+y′+y is equal to :

Solution

Maths sir · 5 steps

Write it all in cos θ and factor: y collapses to 1/4 sec² θ/2.

1. Everything in cos. Put c = cos θ and use cos 2θ = 2c² - 1 and cos 3θ = 4c³ - 3c. The numerator becomes 2c² + 2c - 1. The denominator becomes 4c³ + 8c² + 2c - 2 = 2(2c³ + 4c² + c - 1).
2. Factor and cancel. The cubic factors as 2c³ + 4c² + c - 1 = (c + 1)(2c² + 2c - 1), so the numerator cancels. At θ = π/2 we have c = 0, so the cancelled factor is -1, not zero, and the cancelling is allowed there.

   y = 1/2(1+cos θ) = 1/(4 cos² θ/2) = 1/4 sec² θ/2

3. Differentiate. Let u = θ/2, so each derivative brings a factor 1/2. Then y' = 1/4 sec² u tan u and y'' = 1/8(2 sec² u tan² u + sec⁴ u).
4. Put in numbers. At θ = π/2, u = π/4, so sec² u = 2 and tan u = 1. This gives y = 1/2, y' = 1/2 and y'' = 1/8(4 + 4) = 1.
5. So. y'' + y' + y = 1 + 1/2 + 1/2 = 2, which is option (d).

The idea to keep. Before you differentiate a messy trigonometric fraction twice, simplify it. Convert multiple angles to powers of cos θ, factor, and cancel. The derivatives then take two lines.

Answer from the NTA final answer key. Solution by PrepWiser.