1708\hspace{-20}$Given $\bf{f(x+1)-f(x)=8x+3.......................(1)}$\\\\\\ and $\bf{f(x)=bx^2+cx+d}$\\\\\\ Replace $\bf{x\rightarrow (x+1)\;,}$ we get $\bf{f(x+1)=b(x+1)^2+c(x+1)+d}$\\\\\\ Put into Equation....$\bf(1)$\\\\\\ $\bf{\Rightarrow \left\{b(x+1)^2+c(x+1)+d\right\}-\left\{bx^2+cx+d\right\}=8x+3}$\\\\\\ $\bf{\Rightarrow b(2x+1)+c = 8x+3}$\\\\\\ Now Camparing Coefficients, we get.......\\\\\\ $\bf{\Rightarrow b=4}$ and $\bf{c=-1}$
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shivendu gupta please solve 83,84!
146for the q.83 simply apply distance formula to get the side of the triangle then apply the formula of area of triangle which would give answer as g(x)=±√1-x^2
146my mistake it should be the area of the equilateral triangle
146simply apply the condition for modulus in sign-scheme.i think the answer is possibly D
1708\hspace{-20}(81)\;\; $Given $\bf{2f(x)+f(1-x) = x^2-2x+3............(1)}$\\\\\\ Here $\bf{R.H.S}$ is a Polynomial, So $\bf{L.H.S}$ must be polynomial...\\\\\\ So Here $\bf{f(x)}$ must be in the form of $\bf{(2)\;} Degree.$\\\\\\So Let $\bf{f(x)=ax^2+bx+c}$ and $\bf{f(1-x)=a(1-x)^2+b(1-x)+c}$\\\\\\ Put into $\bf{(1)\;,}$ we get\\\\\\ $\bf{\Rightarrow 2ax^2+2bx+2c+a(1-x)^2+b(1-x)+c = x^2-2x+3}$\\\\\\ Now camparing The Coefficients, we get\\\\\\ $\bf{\Rightarrow a= \frac{1}{3}\;,b=-\frac{4}{3}\;\;,c=\frac{4}{3}}$\\\\\\ So $\bf{f(x)=\frac{1}{3}\left\{x^2-4x+4\right\}=\frac{1}{3}\cdot\left(x-2\right)^2}$
