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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Mathematical Reasoning

BITSAT 2009 · Mathematics · Mathematical Reasoning

If (f(t)=(1-t)/(1+t)), then (f^ (1 / t)) is equal to

If (f(t)=\frac{1-t}{1+t}), then (f^{\prime}(1 / t)) is equal to

Options

  1. A(\frac{1}{(1+t)^2})
  2. B(\frac{1}{(t-1)^2})
  3. C(\frac{-2 t^2}{(t+1)^2})
  4. D(\frac{2}{(t-1)^2})

Correct answer

C. (\frac{-2 t^2}{(t+1)^2})

Step-by-step solution

(f′(t)=ddt[1−t1+t]=(1+t)(−1)−(1−t)×(1)(1+t)2=−1−t−1+t(1+t)2=−2(1+t)2\begin{aligned} & f^{\prime}(t)=\frac{d}{d t}\left[\frac{1-t}{1+t}\right]=\frac{(1+t)(-1)-(1-t) \times(1)}{(1+t)^2} \\ & =\frac{-1-t-1+t}{(1+t)^2}=\frac{-2}{(1+t)^2} \end{aligned}) (f^{\prime}[1 / t]=\frac{-2}{\left(1+\frac{1}{t}\right)^2}=\frac{-2 t^2}{(t+1)^2})
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