Set, Relations and FunctionsMTP Dec 22 - Series IQuestion 1994 of 217
All Questions

The domain of the function f(x)=x2+3x+5x2−5x+4\displaystyle f(x)=\frac{x^2+3x+5}{x^2-5x+4} is:

Options

AR
BR−{1,4}\displaystyle R - \{1, 4\}
CR−{1}\displaystyle R - \{1\}
D(1,4)\displaystyle (1, 4)
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Correct Answer

✅ Option b — R−{1,4}\displaystyle R - \{1, 4\}

All Options:

  • AR
  • BR−{1,4}\displaystyle R - \{1, 4\}
  • CR−{1}\displaystyle R - \{1\}
  • D(1,4)\displaystyle (1, 4)

Detailed Solution & Explanation

A rational function f(x)=P(x)Q(x)\displaystyle f(x) = \frac{P(x)}{Q(x)} is defined for all real numbers except where the denominator Q(x)=0\displaystyle Q(x) = 0.

Given the function:
f(x)=x2+3x+5x2−5x+4f(x) = \frac{x^2+3x+5}{x^2-5x+4}
We set the denominator equal to zero to find the excluded values of x\displaystyle x:
x2−5x+4=0x^2 - 5x + 4 = 0
We factor the quadratic equation:
x2−4x−x+4=0x^2 - 4x - x + 4 = 0
x(x−4)−1(x−4)=0x(x - 4) - 1(x - 4) = 0
(x−1)(x−4)=0(x - 1)(x - 4) = 0
This gives the roots:
x=1orx=4x = 1 \quad \text{or} \quad x = 4
Therefore, the function is undefined at x=1\displaystyle x = 1 and x=4\displaystyle x = 4. The domain of the function is all real numbers R\displaystyle \mathbb{R} except 1\displaystyle 1 and 4\displaystyle 4:
Domain=R−{1,4}\text{Domain} = \mathbb{R} - \{1, 4\}

Hence, **Option B** is the correct answer.

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