Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(t)=\frac{t}{2}+\frac{1}{4} \log (2 t-1)$, then $f^{\prime}\left(\frac{t+1}{2 t+1}\right)=$
Options:
Solution:
2367 Upvotes
Verified Answer
The correct answer is:
$1+t$
$f(t)=\frac{t}{2}+\frac{1}{4} \log (2 t-1)$
$\begin{aligned} & f^{\prime}(t)=\frac{1}{2}+\frac{1}{2(2 t-1)} \\ & \Rightarrow f^{\prime}\left(\frac{t+1}{2 t+1}\right)=\frac{1}{2}+\frac{1}{2\left(\frac{2 t+2}{2 t+1}-1\right)}=\frac{1}{2}\left[1+\frac{2 t+1}{1}\right] \\ & =\frac{1}{2}[1+2 t+1]=t+1\end{aligned}$
$\begin{aligned} & f^{\prime}(t)=\frac{1}{2}+\frac{1}{2(2 t-1)} \\ & \Rightarrow f^{\prime}\left(\frac{t+1}{2 t+1}\right)=\frac{1}{2}+\frac{1}{2\left(\frac{2 t+2}{2 t+1}-1\right)}=\frac{1}{2}\left[1+\frac{2 t+1}{1}\right] \\ & =\frac{1}{2}[1+2 t+1]=t+1\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.