Application of DerivativeshardFree

Application of Derivatives: Function

JEE Maths question with a full step-by-step solution.

Question
The function
f(x)=x2−21+x2f(x) = \frac{x^{2}-2}{\sqrt{1+x^{2}}}
Ais always increasing
Bis always decreasing
Chas exactly one point of minimacorrect
Dhas exactly one point of maxima
Solution
Step 1: Write ff in a form convenient for differentiation.
f(x)=(x2−2)(1+x2)−1/2.f(x) = \left(x^{2}-2\right)\left(1+x^{2}\right)^{-1/2} .
 Step 2: Differentiate by the product rule.
f′(x)=2x(1+x2)−1/2+(x2−2)(−12)(1+x2)−3/2(2x).f'(x) = 2x\left(1+x^{2}\right)^{-1/2} + \left(x^{2}-2\right)\left(-\tfrac12\right)\left(1+x^{2}\right)^{-3/2}(2x) .
 Step 3: Tidy the second term.
f′(x)=2x(1+x2)−1/2−x(x2−2)(1+x2)−3/2.f'(x) = 2x\left(1+x^{2}\right)^{-1/2} - x\left(x^{2}-2\right)\left(1+x^{2}\right)^{-3/2} .
 Step 4: Take out the common factor x(1+x2)−3/2x\left(1+x^{2}\right)^{-3/2}.
f′(x)=x(1+x2)3/2[2(1+x2)−(x2−2)].f'(x) = \frac{x}{\left(1+x^{2}\right)^{3/2}}\left[2\left(1+x^{2}\right) - \left(x^{2}-2\right)\right].
 Step 5: Simplify the bracket.
2+2x2−x2+2=x2+4.2 + 2x^{2} - x^{2} + 2 = x^{2}+4 .
f′(x)=x(x2+4)(1+x2)3/2.f'(x) = \frac{x\left(x^{2}+4\right)}{\left(1+x^{2}\right)^{3/2}} .
 Step 6: Determine the sign. The factor x2+4x^{2}+4 is always positive and so is (1+x2)3/2\left(1+x^{2}\right)^{3/2}, so
sign⁡(f′(x))=sign⁡(x).\operatorname{sign}\left(f'(x)\right) = \operatorname{sign}(x) .
 Step 7: Read off the behaviour. f′<0f' < 0 for x<0x < 0 and f′>0f' > 0 for x>0x > 0, so ff decreases then increases, with a single turning point at x=0x = 0 - a minimum. Step 8: So (1) and (2) fail (the function does both), (4) fails (there is no maximum), and (3) holds, with the minimum value f(0)=−21=−2f(0) = \dfrac{-2}{1} = -2. Answer: (3).
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