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BITSAT 2023Paper 1

Exam: BITSAT
Year: 2023
Paper: Paper 1
Subject: Mathematics
Chapter: Quadratic Equation

BITSAT 2023 · Paper 1 · Mathematics · Quadratic Equation

The range of the function f(x)= 3 x^2-4 x+5 is

The range of the function

f(x)=\sqrt{3 x^2-4 x+5}

is

Options

A $\left(-\infty, \sqrt{\frac{11}{3}}\right]$
B $\left(-\infty, \sqrt{\frac{11}{5}}\right)$
C $\left[\sqrt{\frac{11}{3}}, \infty\right)$
D $\left(\sqrt{\frac{11}{5}}, \infty\right)$ .

Correct answer

C. $\left[\sqrt{\frac{11}{3}}, \infty\right)$

Step-by-step solution

f(x)

is defined if

3 x^2-4 x+5 \geq 0

\Rightarrow 3\left[x^2-\frac{4}{3} x+\frac{5}{3}\right] \geq 0 \Rightarrow 3\left[\left(x-\frac{2}{3}\right)^2+\frac{11}{9}\right] \geq 0

Which is true for all real

x \therefore

Domain of

f(x)=(-\infty, \infty)

Let

y=\sqrt{3 x^2-4 x+5}

\Rightarrow y^2=3 x^2-4 x+5

i.e.

3 x^2-4 x+\left(5-y^2\right)=0

For

x

to be real ,

16-12\left(5-y^2\right) \geq 0 \Rightarrow y \geq \sqrt{\frac{11}{3}}

\therefore \quad

Range of

y= \left[\sqrt{\frac{11}{3}}, \infty\right)

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