KVPY 2019 SB SX Exam Previous Papers

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Question : 90 of 120

Marks: +1, -0

Let f : [0, 1] → [0, 1] be a continuous function such that

x^2+(\f(x))^2≤1

for all x ∈ [0, 1] and

∫↖{1}↙{0}\f(x)dx=π/4

Then

∫↖{1/√2}↙{1/2}{\f(x)}/{1-x^2}dx

equals

$π/12$
$π/15$
${√2-1}/2π$
$π/10$

Solution:

x^2 + y^2 ≤ 1

and x, y ∈ [0, 1]

∫↖{1}↙{0}\f(x)dx=π/4

⇒ area =

π/4

⇒

y=√{1-x^2}

⇒

∫↖{1\/√2}↙{1\/2}{\f(x)}/{1-x^2}dx= ∫↖{1\/√2}↙{1\/2}{dx}/√{1-x^2}

=|\sin^{-1}x| ^{1\/√2}_{1\/2}=π/4-π/6=π/12

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