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Grade 11IIT JEE Entrance Exam

The number of all possible triplets (a1,a2,a3)such that
a1 + a2(Cos2x) +a3Sin²(x) = 0 for all values of x

Abhinav Acharya

12 Years agoGrade 11

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello Student,
Please find answer to your question below,
$a_{1} + a_{2}cos2x + a_{3}sin^{2}x = 0$
$a_{1} + a_{2}(1-2sin^{2}x) + a_{3}sin^{2}x = 0$
$a_{1} + a_{2}=sin^{2}x(2a_{2}-a_{3})$
$sin^{2}x = \frac{a_{1}+a_{2}}{2a_{2}-a_{3}}$
Since x belong to R, we have
$0\leq sin^{2}x \leq 1$
$0\leq \frac{a_{1}+a_{2}}{2a_{2}-a_{3}}\leq 1$
$0\leq a_{1}+a_{2}\leq 2a_{2} - a_{3}$
$a_{2}\geq a_{1} + a_{3}$
a₁, a₂& a₃belong to R with above cond. satisfied.

Let S={x:x belongs to R(√3+√2)^(x^2-4) - (√3-√2)^(x^2-4)=10}. Then n(S) is equal to-

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