If f(x) satisfies the relation 2f(x)+f(1-x)=x2 for all real x, then f(x) is

Question

If f(x) satisfies the relation 2f(x)+f(1-x)=x2 for all real x, then f(x) is (A)  (x2+2x-1/6)  (B)  (x2+2x-1/3)  (C)  (x2+4x-1/3)  (D)  (x2-3x+1/6)

Tardigrade Admin · Accepted Answer

Given, 2f(x)+f(1-x)=x2 ...(i) Replacing x by (1-x), we get 2f(1-x)+f(x)=(1-x)2 ⇒ 2f(1-x) +f(x)=1+x2-2x ...(ii) On multiplying Eq. (i) by 2 and subtracting from Eq. (ii), we get 3f(x)=x2+2x-1 ⇒ f(x)=(x2+2x-1/3)

Q. If $f (x)$ satisfies the relation $2 f (x) + f (1 - x) = x^{2}$ for all real $x,$ then $f (x)$ is

$\frac{x ^{2} + 2 x - 1}{6}$

$\frac{x ^{2} + 2 x - 1}{3}$

$\frac{x ^{2} + 4 x - 1}{3}$

$\frac{x ^{2} - 3 x + 1}{6}$

$\frac{x ^{2} + 3 x - 1}{3}$

Q. If f(x) satisfies the relation 2f(x)+f(1−x)=x2 for all real x, then f(x) is

6x2+2x−1​

3x2+2x−1​

3x2+4x−1​

6x2−3x+1​

3x2+3x−1​