Q52 / 160 · TS EAMCET 2019 (06 May Shift 1) · Mathematics
Consider the curves given by the following quadratic function.
$\begin{array}{ll}f_1(x)=5 x^2+2 x+1, & f_2(x)=5 x^2+6 x+1 \\ f_3(x)=x^2-7 x+6, & f_4(x)=64 x^2+48 x+9\end{array}$
If $A_1, A_2, A_3$ and $A_4$ denote the lengths of the intercepts on the $X$-axis made by the above curve respectively, then which of the following is true?
- A$A_1>A_2>A_3>A_4>0$
- Correct answerB$A_4 < A_2 < A_3$
- C$A_3 < A_2 < A_4$
- D$A_2 < A_4 < A_3$
Solution
Since, length of the intercept on the $X$-axis
made by the curve $f(x)=a x^2+b x+c$ is
$A=\left|x_1-x_2\right|=\frac{\sqrt{D}}{|a|}$, where $D=b^2-4 a c$
So, for curve $f_1(x)=5 x^2+2 x+1$,
$A_1=\frac{\sqrt{4-20}}{5}=\frac{\sqrt{-16}}{5}$, not exist
For curve $f_2(x)=5 x^2+6 x+1$,
$A_2=\frac{\sqrt{36-20}}{5}=\frac{4}{5}$
For curve $f_3(x)=x^2-7 x+6$,
$A_3=\sqrt{49-24}=5$
For curve $f_4(x)=64 x^2+48 x+9$
$A_4=\frac{\sqrt{2304-2304}}{64}=0$
So, $A_3>A_2>A_4$
Hence, option (b) is correct.
Hence, the correct answer is option: 'B'.