f:R→R fx2+x+3+2fx2−3x+5=6x2−10x+17 ∀x∈R then the value of f(100) is
obviously f is a linear polynomial (Since degree of LHS and RHS is same)
letf(x)=ax+b hence fx2+x+3+2fx2−3x+5≡6x2−10x+17⇒ax2+x+3+b+2ax2−3x+5+b≡6x2−10x+17comparing coeff of x2 and coeff. of x⇒ a+2a=6......(1)⇒ a−6a=−10......(2) a=2comparing constants 3a+b+10a+2b=17⇒ 6+b+20+2b=17⇒ b=−3∴ f(x)=2x−3⇒f(100)=197