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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2020 · Mathematics · Application of Derivatives

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

Consider the function f(x)=∣x−1∣/x2,f(x)=|x-1| /x^{2}, then f(x)f(x) is

Options

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

Correct answer

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

Step-by-step solution

f(x)=\frac{|x-1|}{x^{2}}=\left\{1−xx2,x<1,x≠0x−1x2,x>1\begin{array}{l}\frac{1-x}{x^{2}}, x< 1, x \neq 0 \\ \frac{x-1}{x^{2}}, x > 1\end{array}\right. Clearly, f(x)f(x) is continuous for all x∈Rx \in \mathbb{R} except at x=0f^{\prime}(x)=\left\{x−2x3,x<1,x±02−xx3,x>1\begin{array}{l} \frac{x-2}{x^{3}}, x< 1, x \pm 0 \\ \frac{2-x}{x^{3}}, x > 1 \end{array}\right.f^{\prime}(x) > 0 \Rightarrow x< 0 or 1<x<2f′(x)<0⇒0<x<11< x< 2f^{\prime}(x)< 0 \Rightarrow 0< x< 1 or x>2x > 2 Hence, f(x)f(x) is increasing in (−∞,0)∪(1,2)(-\infty, 0) \cup(1,2) and decreasing in (0,1)∪(2,∞)\cup(2, \infty).
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