JEE Main 2022 — Sequences And Series Question with Solution

Given,

$$f(x)=a_0{x}^2+a_{1}x+a_2$$

$$f^{\prime }(0)=1$$

$$f^{\prime }(1)=0$$

a₀, a₁, a₂ are in A. G. P

Common difference of $$AP=1$$

Common ratio of $$GP=2$$

A.P terms = a, a + 1, a + 2

G.P terms = y, ry, r²y

$$\therefore$$ AGP terms = ay, (a+1)ry, (a+2)r²y

$$\therefore$$ $$a_0=ay$$

$$a_1=(a+1)ry=(a+1)2y$$

$$a_2=(a+2)r^{2}y=(a+2)4y$$

Now, $$f^{\prime }(x)=2x{a}_0+a_1$$

$$\therefore$$ $$f^{\prime }(0)=a_1=1$$

and $$f^{\prime }(1)=2a_0+a_1=0$$

$$\Rightarrow 2a_0+1=0$$

$$\Rightarrow a_0=-\frac{1}{2}$$

$$\therefore$$ $$ay=-\frac{1}{2}$$

and $$(a+1)2y=1$$

$$\Rightarrow 2ay+2y=1$$

$$\Rightarrow 2\times \left(-\frac{1}{2}\right)+2y=1$$

$$\Rightarrow 2y=+2$$

$$\Rightarrow y=+1$$

$$\therefore$$ $$a=-\frac{1}{2}$$

$$\therefore$$ $$a_2=(a+2)4y$$

$$=\left(-\frac{1}{2}+2\right)\times 4.1$$

$$=6$$

$$\therefore$$ $$f(x)=-\frac{1}{2}{x}^2+x+6$$

$$\therefore$$ $$f(4)=-\frac{1}{2}(4{)}^2+4+6$$

$$=-8+10$$

$$=2$$