For p, q in R,consider the real-valued functi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The roots of $f(x) = (x - p)^2 - q = 0$ are $x = p \pm \sqrt{q}$.The absolute difference between the roots is $|(p + \sqrt{q}) - (p - \sqrt{q})| =

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) The roots of $f(x) = (x - p)^2 - q = 0$ are $x = p \pm \sqrt{q}$.The absolute difference between the roots is $|(p + \sqrt{q}) - (p - \sqrt{q})| = 2\sqrt{q}$.Given $a_1, a_2, a_3, a_4$ are in arithmetic progression with mean $p$ and common difference $d > 0$,we have $a_1 = p - \frac{3d}{2}, a_2 = p - \frac{d}{2}, a_3 = p + \frac{d}{2}, a_4 = p + \frac{3d}{2}$.Since $|f(a_i)| = 500$,we have $|(a_i - p)^2 - q| = 500$.For $i=4$,$|(\frac{3d}{2})^2 - q| = 500 \Rightarrow |\frac{9d^2}{4} - q| = 500$.For $i=1$,$|(-\frac{3d}{2})^2 - q| = 500 \Rightarrow |\frac{9d^2}{4} - q| = 500$.For $i=2$,$|(-\frac{d}{2})^2 - q| = 500 \Rightarrow |\frac{d^2}{4} - q| = 500$.Since $q > 0$ and $d > 0$,$\frac{9d^2}{4} - q = 500$ and $q - \frac{d^2}{4} = 500$ (as $\frac{d^2}{4} Adding these equations: $(\frac{9d^2}{4} - q) + (q - \frac{d^2}{4}) = 500 + 500$ $\Rightarrow 2d^2 = 1000$ $\Rightarrow d^2 = 500$.Substituting $d^2 = 500$ into $q - \frac{d^2}{4} = 500$,we get $q - 125 = 500 \Rightarrow q = 625$.The absolute difference between the roots is $2\sqrt{q} = 2\sqrt{625} = 2 	imes 25 = 50$.

Similar Questions

If $a_1, a_2, a_3, ......., a_n$ are in $A.P.$,where $a_i > 0$ for all $i$,then the value of $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = $

The sum of the common terms of the following three arithmetic progressions:
$3, 7, 11, 15, \ldots, 399$
$2, 5, 8, 11, \ldots, 359$ and
$2, 7, 12, 17, \ldots, 197$,is equal to $................$.

If the roots of the equation $x^3 - 9x^2 + \alpha x - 15 = 0$ are in $A.P.$,then $\alpha$ is:

In an $A.P.$,the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that the $20^{th}$ term is $-112$.

Let $x_n, y_n, z_n, w_n$ denote the $n^{th}$ terms of four different arithmetic progressions with positive terms. If $x_4 + y_4 + z_4 + w_4 = 8$ and $x_{10} + y_{10} + z_{10} + w_{10} = 20$,then the maximum value of $x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module