Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

AnswEr:-

$\sf \: = > f( \theta) = ( \sin \theta \: + \cos\theta ) {}^{2} + ( \sin \theta \: - \cos \theta \: ) {}^{4} \\ \\ = > \sin {}^{2} (2 \theta) - \sin(2 \theta) + 2$

$\sf \implies f'(\theta )=2(sin2\theta ).(2cos2\theta )-2cos2 \theta \\\\\\</p><p>= 2cos2\theta (2sin2\theta -1)</p><p>[tex]$

From the graph given in attachment,

Minimum is at \theta = π/12 ,5π/12.

Hence,

$\sf \implies \lambda _{1} + \lambda _{2} = \frac{1}{12}+\frac{5}{12}\\\\\\=6/12 = 1/2$

Hence,

• Answer : - 1/2

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Answer 2

Let the function f : (0, π) → R be defined by

f(θ) = (sinθ + cosθ)² + (sinθ - cosθ)⁴

suppose the function f has a local minimum at θ precisely when θ ∈ {λ₁π , .... λ_r π} , where 0< λ₁ , ....λr < 1 Then the value of λ₁ + .... + λr is ....

solution : given f(θ) = (sinθ + cosθ)² + (sinθ - cosθ)⁴

= sin²θ + cos²θ + 2sinθ cosθ + (sin²θ + cos²θ - 2sinθcosθ)²

= 1 + sin2θ + (1 - sin2θ)²

= 1 + sin2θ + 1 + sin²2θ - 2sin2θ

= 1 + sin²2θ - sin2θ

now differentiating w.r.t θ we get,

f'(θ) = 4sin2θ cos2θ - 2cos2θ

= 2cos2θ(2sin2θ - 1)

at f'(θ) = 0,

cos2θ = 0 ⇒ θ = (2m + 1)π/4

and , sin2θ = 1/2 ⇒θ = 1/12 (2nπ + π) and 1/12(2nπ + 5π)

so, θ = π/12, π/4, 5π/12 ,

again differentiating w.r.t θ we get,

f"(θ) = -4sin2θ(2sin2θ -1) + 2cos2θ × 4cos2θ

at θ = π/12, 5π/12 , f"(θ) > 0

so, f(θ) is minimum at θ = π/12 , 5π/12

Therefore λ₁π = π/12 ⇒λ₁ = 1/12

and λ₂π = 5π/12 ⇒λ₂ = 5/12

now the value of (λ₁ + λ₂) = 1/12 + 5/12 = 1/2