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Let f(x) = tan-1 {φ(x)}, where φ(x)is monotonically increasing for 0 < x < π/2. Then f(x) is
  • a)
    increasing in (0, π/2)
  • b)
    decreasing in (0, π/2)
  • c)
    increasing in (0, π/4) and decreasing in (π/4, π/2)
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let f(x) = tan-1 {φ(x)}, where φ(x)is monotonically increasing for 0 &...
f x = φ x 1 + { φ x } 2 > 0 for 0 < x < π 2 because φ x > 0, φ x being monotonically increasing
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Most Upvoted Answer
Let f(x) = tan-1 {φ(x)}, where φ(x)is monotonically increasing for 0 &...
To find the derivative of f(x), we can use the chain rule:

f'(x) = d/dx [tan-1 {φ(x)}] = [1/(1 + φ(x)^2)] * d/dx [φ(x)]

Since φ(x) is monotonically increasing for 0 < x="" />< π/2,="" we="" know="" that="" φ'(x)="" /> 0. Therefore, f'(x) > 0 for all x in the given interval.

This means that f(x) is monotonically increasing for 0 < x="" />< π/2,="" since="" its="" derivative="" is="" always="" positive.="" π/2,="" since="" its="" derivative="" is="" always="" />
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Let f(x) = tan-1 {φ(x)}, where φ(x)is monotonically increasing for 0 < x < π/2. Then f(x) isa) increasing in (0, π/2) b) decreasing in (0, π/2) c) increasing in (0, π/4) and decreasing in (π/4, π/2) d) none of these Correct answer is option 'A'. Can you explain this answer?
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