JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(P\) and \(Q\) be any points on the curves \(( x-1)^{2}+(y+1)^{2}=1\) and \(y=x^{2}\), respectively. The distance between \(P\) and \(Q\) is minimum for some value of the abscissa of \(P\) in the interval
- A \(\left(0, \frac{1}{4}\right)\)
- B \(\left(\frac{1}{2}, \frac{3}{4}\right)\)
- C \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
- D \(\left(\frac{3}{4}, 1\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(Q =\left( t , t ^{2}\right)\) \(m_{C Q}=m_{\text {normal }}\) \(\frac{ t ^{2}+1}{ t -1}=-\frac{1}{2 t }\) Let \(f(t)=2 t^{3}+3 t-1\) \(f \left(\frac{1}{4}\right) f \left(\frac{1}{3}\right)<0 \Rightarrow t \in\left(\frac{1}{4}, \frac{1}{3}\right)\)…
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