For 0 1/2 . If f is differentiable, find ∣ ( a + b ) / c ∣ .

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Mathematics - Continuity - Differentiability Question with Solution | TestHub

MathematicsContinuity - DifferentiabilityDifferentiabilityHard2 minQB

MathematicsHardinteger

For $0 < x < 1/ 2$ , define $f (x) = tan^{- 1} ((a x + b) / c)$ when $x < 1/2, f (1/2) = 0$ , and $f (x) = ln (2 + b x^{2})$ when $x > 1/2$ . If $f$ is differentiable, find $∣ (a + b) / c ∣$ .

Answer:

Solution:

Continuity at $1/2$ gives $a /2 + b = 0$ and $2 + b /4 = 1$ , so $b = - 4, a = 8$ . Equality of one-sided derivatives gives $8/ c = - 4$ , hence $c = - 2$ . Therefore $∣ (a + b) / c ∣ = ∣4/ (- 2) ∣ = 2$ .

Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Differentiability

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