JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the tangent to the curve, \(y =f( x )= x \log _{ e } x\) \((x>0)\) at a point \((c, f(c))\) is parallel to the line segement joining the points \((1,0)\) and \(( e , e ),\) then \(c\) is equal to
- A \(\frac{1}{ e -1}\)
- B \(e^{\left(\frac{1}{1-e}\right)}\)
- C \(e^{\left(\frac{1}{e-1}\right)}\)
- D \(\frac{ e -1}{ e }\)
Answer & Solution
Correct Answer
(C) \(e^{\left(\frac{1}{e-1}\right)}\)
Step-by-step Solution
Detailed explanation
\(f(x)=x \log _{e} x\) \(\left.f^{\prime}(x)\right|_{(c, f(c))}=\frac{e-0}{e-1}\) \(f^{\prime}(x)=1+\log _{e} x\) \(\left.f^{\prime}(x)\right|_{(c, f(c))}=1+\log _{c} c=\frac{e}{e-1}\) \(\log _{e} c=\frac{e-(e-1)}{e-1}=\frac{1}{e-1} \Rightarrow c=e^{\frac{1}{e-1}}\)
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