KVPY 2013 - 11

Algebra Level 3

Let f ( x ) = 2 x x 2 f(x)=\sqrt{2-x-x^2} and g ( x ) = cos x g(x)=\cos x .

Which of the following statements are true?

  • ( I ) (I) Domain of f ( ( g ( x ) ) 2 ) f((g(x))^2) = Domain of f ( g ( x ) ) f(g(x))
  • ( I I ) (II) Domain of f ( g ( x ) ) + g ( f ( x ) ) f(g(x))+g(f(x)) = Domain of g ( f ( x ) ) g(f(x))
  • ( I I I ) (III) Domain of f ( g ( x ) ) f(g(x)) = Domain of g ( f ( x ) ) g(f(x))
  • ( I V ) (IV) Domain of g ( ( f ( x ) ) 3 ) g((f(x))^3) = Domain of f ( g ( x ) ) f(g(x))

Try the whole set KVPY 2013 SB/SX here.

Only I I I III and I V IV Only I I and I V IV Only I I Only I I and I I II

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1 solution

f ( g ( x ) ) = 2 cos x cos 2 x f(g(x)) = \sqrt{2-\cos x-\cos^2 x}

Now, since cos x 1 cos x + cos 2 x 2 2 cos x cos 2 x 0 \cos x \le 1 \implies \cos x + \cos^2 x \le 2 \implies 2- \cos x - \cos^2 x \ge 0

Hence , the domain of f ( g ( x ) ) f(g(x)) is R \mathbb{R}

Similarly , the domain of f ( g ( x ) ) 2 f(g(x))^2 is also R \mathbb{R}

Clearly , domain of g ( f ( x ) ) g(f(x)) = domain of f ( x ) f(x)

So, for domain of f ( x ) f(x) :

2 x x 2 0 2 < x < 1 2-x-x^2 \ge 0 \implies -2<x<1

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