If f(x) satisfies f(7 - x) = f(7 + x) for all | vedclass

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

Question

(C) The condition $f(7 - x) = f(7 + x)$ implies that the function $f(x)$ is symmetric about the line $x = 7$.Let the $5$ distinct real roots of $f(x)

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The condition $f(7 - x) = f(7 + x)$ implies that the function $f(x)$ is symmetric about the line $x = 7$.Let the $5$ distinct real roots of $f(x) = 0$ be $x_1, x_2, x_3, x_4, x_5$.Since the function is symmetric about $x = 7$,if $x_i$ is a root,then $14 - x_i$ must also be a root.For $5$ distinct roots,one root must be the axis of symmetry itself,so $x_3 = 7$.The remaining roots must form pairs symmetric about $7$,such that $\frac{x_1 + x_5}{2} = 7$ and $\frac{x_2 + x_4}{2} = 7$.This gives $x_1 + x_5 = 14$ and $x_2 + x_4 = 14$.The sum of the roots $S = x_1 + x_2 + x_3 + x_4 + x_5 = (x_1 + x_5) + (x_2 + x_4) + x_3 = 14 + 14 + 7 = 35$.Therefore,$S/7 = 35/7 = 5$.

If $f(x)$ satisfies $f(7 - x) = f(7 + x)$ for all $x \in R$ such that $f(x)$ has exactly $5$ real roots which are all distinct,and the sum of the real roots is $S$,then $S/7$ is equal to:

Similar Questions

Let $f$ be a function such that $3f(x) + 2f\left(\frac{m}{19x}\right) = 5x$,$x \neq 0$,where $m = \sum_{i=1}^9 (i)^2$. Then $f(5) - f(2)$ is equal to

If $f(10-x)=3x^2+4x-5$ and $f(x)=px^2+qx+r$,then find the value of $p+q+r$.

Let $\sum\limits_{k = 1}^{10} {f(a + k)} = 16(2^{10} - 1),$ where the function $f$ satisfies $f(x + y) = f(x)f(y)$ for all natural numbers $x, y$ and $f(1) = 2.$ Then the natural number $a$ is

Let $f(x + y) = f(x) + f(y)$ for all $x, y \in R.$ Then:

If $f(x) + 2f(1/x) = 3x$ for $x \neq 0$ and $S = \{x \in R : f(x) = f(-x)\}$,then $S$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module