JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(f\left( x \right) = \frac{x}{{\sqrt {{a^2} + {x^2}} }} - \frac{{d - x}}{{\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},x \in R\,\), where \(a, b\) and \(d\) are non -zero real constant. Then
- A \(f\) is an increasing function of \(x\)
- B \(f\) is a decreasing function of \(x\)
- C \(f\) is not a continuous function of \(x\)
- D \(f\) is neither increasing nor decreasing function of \(x\)
Answer & Solution
Correct Answer
(A) \(f\) is an increasing function of \(x\)
Step-by-step Solution
Detailed explanation
\(f(x)=x\left(x^{3}+a^{2}\right)^{\frac{-1}{2}}-(d-x)\left(b^{2}+(d-x)^{2}\right)^{-\frac{1}{2}}\) \(f^{\prime}(x)=\frac{a^{2}}{\left(x^{2}+a^{2}\right) \sqrt{x^{2}+a^{2}}}+\frac{b^{2}}{\left(b^{2}+(d-x)^{2}\right) \sqrt{b^{2}+(d-x)^{2}}}\) \(=+v e\)
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