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Let f(x) = loge(sin x), (0 < x < $$\pi $$) and g(x) = sin–1 (e–x ), (x $$ \ge $$ 0). If $$\alpha $$ is a positive real number such that a = (fog)'($$\alpha $$) and b = (fog)($$\alpha $$), then :

JEE · Math · previous-year question

  1. A.a$$\alpha $$2 + b$$\alpha $$ - a = -2$$\alpha $$2
  2. B.a$$\alpha $$2 + b$$\alpha $$ + a = 0
  3. C.a$$\alpha $$2 - b$$\alpha $$ - a = 0
  4. D.a$$\alpha $$2 - b$$\alpha $$ - a = 1correct

Answer

D. a$$\alpha $$2 - b$$\alpha $$ - a = 1

Explanation

f(x) = ln(sin x), g(x) = sin–1 (e–x) f(g(x)) = ln(sin(sin–1 e–x)) = -x f(g($$\alpha $$)) = – $$\alpha $$ = b As f(g(x)) = – x $$ \therefore $$ (f(g(x)))' = – 1 $$ \Rightarrow $$ (f(g($$\alpha $$)))' = – 1 = a $$ \therefore $$ b = – $$\alpha $$, a = – 1 $$ \therefore $$ a$$\alpha $$2 - b$$\alpha $$ - a = - $$\alpha $$2 + $$\alpha $$2 + 1 = 1

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