JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 1)
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 :
a$$\alpha $$2 + b$$\alpha $$ - a = -2$$\alpha $$2
a$$\alpha $$2 + b$$\alpha $$ + a = 0
a$$\alpha $$2 - b$$\alpha $$ - a = 0
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
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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