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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2024 · Mathematics · Application of Derivatives

Consider the function f(x)=(|x-1|)/(x^2) , then f(x) is

Consider the function f(x)=∣x−1∣x2f(x)=\frac{|x-1|}{x^2}, then f(x)f(x) isConsider the function f(x)=(|x-1|)/(x^2) , then f(x) is

Options

  1. Aincreasing in (0,1)∪(2,∞)(0,1) \cup(2, \infty)
  2. Bincreasing in (−∞,0)∪(0,1)(-\infty, 0) \cup(0,1)
  3. Cdecreasing in (−∞,0)∪(2,∞)(-\infty, 0) \cup(2, \infty)
  4. Ddecreasing in (0,1)∪(2,∞)(0,1) \cup(2, \infty)

Correct answer

D. decreasing in (0,1)∪(2,∞)(0,1) \cup(2, \infty)

Step-by-step solution

Given, f(x)=\left\{x−1x2;x≥1−x+1x2;x<1\begin{aligned} \frac{x-1}{x^2} ; & x \geq 1 \\ \frac{-x+1}{x^2} ; & x < 1\end{aligned}\right.f(x)={1x−1x2;x≥1−1x+1x2;x<1f′(x)={−1x2+2x3;x≥11x2−2x3;x<1f′(x)={2−xx3;x≥1x−2x3;x<1\begin{aligned} & f(x)= \begin{cases}\frac{1}{x}-\frac{1}{x^2} ; & x \geq 1 \\ -\frac{1}{x}+\frac{1}{x^2} ; & x < 1\end{cases} \\ & f^{\prime}(x)= \begin{cases}\frac{-1}{x^2}+\frac{2}{x^3} ; & x \geq 1 \\ \frac{1}{x^2}-\frac{2}{x^3} ; & x < 1\end{cases} \\ & f^{\prime}(x)= \begin{cases}\frac{2-x}{x^3} ; & x \geq 1 \\ \frac{x-2}{x^3} ; & x < 1\end{cases} \end{aligned}Byobservation,By observation,f^{\prime}(x)willbe<imgsrc="/api/proxy−image?url=httpsDecreasinginwill be<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fbitsat%252FmZpWNRtakll5g4R_qPQjO8-Xlgu9YWyJyuF1K2KsqOc.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FmZpWNRtakll5g4R_qPQjO8-Xlgu9YWyJyuF1K2KsqOc.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FmZpWNRtakll5g4R_qPQjO8-Xlgu9YWyJyuF1K2KsqOc.original.fullsize.png?alt=media"/> Decreasing in(0,1) \cup(2, \infty)andincreasinginand increasing in(-\infty, 0) \cup(1,2)$
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