Sep 2019 CSAT mock test for Sciences, #30

$(2x+1)f '(x^2+x+1)=f(1)\times (2x+1)\pi\sin\pi x +f(3)x(2x+1)+5x^2(2x+1)$

Integrate both sides,

$f(x^2+x+1)=\left\{-(2x+1)\cos\pi x+\frac{2}{\pi}\sin\pi x\right\}f(1)+\left(\dfrac{2}{3}x^3+\dfrac{1}{2}x^2\right)f(3)+\dfrac{5}{2}x^4+\dfrac{5}{3}x^3+C.$

Substitute $x=0$ and $x=-1,$ (since then $x^2+x+1=1$ )

$\begin{aligned}x=0:&~~f(1)=-f(1)+C,~\boxed{2f(1)=C}. \\ \text{} \\ x=-1:&~~f(1)=-f(1)-\dfrac{1}{6}f(3)+\dfrac{5}{6}+C,~2f(1)=-\dfrac{1}{6}f(3)+\dfrac{5}{6}+C.\end{aligned}$

It is immediate that $f(3)=5.$ Substitute $x=1,$

$x=1:~f(3)=3f(1)-\dfrac{10}{3}f(3)+\dfrac{80}{3}+C,~\boxed{3f(1)=-C-5}.$

Solve the two boxed equations, and you get $f(1)=-1,~C=-2.$ Hence,

$f(x^2+x+1)=(2x+1)\cos\pi x-\frac{2}{\pi}\sin\pi x+\dfrac{5}{2}x^4+5x^3+\dfrac{5}{2}x^2-2.$

Substitute $x=2,$

$x=2:~f(7)=5+40+40+10-2=\boxed{93}.$

Given that:

$\begin{aligned} f'(x^2+x+1) & = \pi f(1) \sin \pi x + f(3) x + 5x^2 & \small \color{#3D99F6} \text{Putting }x = 0 \\ f'(1) & = 0 & \small \color{#3D99F6} \text{Putting }x = -1 \\ f'(1) & = 0 - f(3) + 5 \\ \implies -f(3)+5 & = 0 \\ f(3) & = 5 \end{aligned}$

$\begin{aligned} f'(x^2+x+1) & = \pi f(1) \sin \pi x + 5 x + 5x^2 & \small \color{#3D99F6} \text{Let }u = x^2 + x + 1 \implies du = (2x+1)\ dx \\ \int f'(u) \ du & = \int \left(\pi f(1) \sin \pi x + 5 x + 5x^2\right)(2x+1)\ dx & \small \color{#3D99F6} \text{Integrate both sides w,r,t, }u \\ & = \int \pi f(1) (2x+1) \sin \pi x + 10x^3 + 15x^2 + 5x \ dx \end{aligned}$

Then the equation becomes:

$\begin{aligned} \int_1^3 f'(u) \ du & = \int_0^1 \pi f(1) (2x+1) \sin \pi x + 10x^3 + 15x^2 + 5x \ dx \\ f(3) - f(1) & = {\color{#3D99F6} - f(1)(2x+1)\cos \pi x + \int 2f(1) \cos \pi x \ dx} + \frac 52 x^4 + 5x^3 + \frac 52 x^2 \ \bigg|_0^1 & \small \color{#3D99F6} \text{Integration by parts} \\ & = 4f(1) + \frac {2f(1)\sin \pi x}\pi \ \bigg|_0^1 + \frac 52 + 5 + \frac 52 = 4f(1) + 10 \\ \implies 5f(1) & = f(3) - 10 = - 5 \\ f(1) & = -1 \end{aligned}$

Similarly,

$\begin{aligned} f(7) - f(1) & = (2x+1)\cos \pi x - \frac {2\sin \pi x}\pi + \frac 52 x^4 + 5x^3 + \frac 52 x^2 \ \bigg|_0^2 \\ f(7) +1 & = 4 - 0 + 40 + 40 + 10 \\ \implies f(7) & = \boxed{93} \end{aligned}$