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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Functions

BITSAT 2024 · Paper 1 · Mathematics · Functions

The domain of the real valued function f(x)= (2 x^2-7 x+5)/(3 x^2-5...

The domain of the real valued function f(x)=2x2−7x+53x2−5x−2f(x)=\sqrt{\frac{2 x^2-7 x+5}{3 x^2-5 x-2}} isThe domain of the real valued function f(x)= (2 x^2-7 x+5)/(3 x^2-5 x-2) is

Options

  1. A(−∞,−13)∪[1,2)∪[52,∞)\left(-\infty,-\frac{1}{3}\right) \cup[1,2) \cup\left[\frac{5}{2}, \infty\right)
  2. B(−∞,1)∪(2,∞)(-\infty, 1) \cup(2, \infty)
  3. C(−13,52]\left(-\frac{1}{3}, \frac{5}{2}\right]
  4. D(−∞,−13]∪[52,∞)\left(-\infty, \frac{-1}{3}\right] \cup \left[\frac{5}{2}, \infty\right)

Correct answer

A. (−∞,−13)∪[1,2)∪[52,∞)\left(-\infty,-\frac{1}{3}\right) \cup[1,2) \cup\left[\frac{5}{2}, \infty\right)

Step-by-step solution

Given function f(x)=2x2−7x+53x2−5x−2f(x)=\sqrt{\frac{2 x^2-7 x+5}{3 x^2-5 x-2}} Here, f(x)\mathrm{f}(\mathrm{x}) should be greater than or equal to 0 . So, 2x2−7x+5=0⇒2x2−5x−2x+5=0⇒(x−1)(2x−5)=0⇒x=1,523x2−5x−2=0⇒3x2−6x+x−2=03x(x−2)+1(x−2)=0⇒(x−2)(3x+1)=0⇒x=2,−132 x^2-7 x+5=0\Rightarrow \quad 2 x^2-5 x-2 x+5=0\Rightarrow \quad(x-1)(2 x-5)=0\Rightarrow \quad x=1, \frac{5}{2}3 x^2-5 x-2=0 \Rightarrow 3 x^2-6 x+x-2=03 x(x-2)+1(x-2)=0 \Rightarrow(x-2)(3 x+1)=0\Rightarrow \quad x=2, \frac{-1}{3} The domain of the real valued function f(x)= (2 x^2-7 x+5)/(3 x^2-5 x-2) is When we include x=−13,2thenx=\frac{-1}{3}, 2 thenf(x)wouldgivenotdefinevalueso,wewillexcludethesevaluesfromthedomain.Whenwetakevaluesbetweenwould give not define value so, we will exclude these values from the domain. When we take values between\left(-\frac{1}{3}, 1\right)andSo,domainisand So, domain is\left(-\infty, \frac{-1}{3}\right) \cup[1,2) \cup\left[\frac{5}{2}, \infty\right)
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