JEE Main 2022 — Mathematics Algebra

Given $f(x)=a{x}^{2}+bx+c$f(x)=ax2+bx+c be such that $f(1)=3,f(-2)=\lambda$f(1)=3,f(−2)=λ and $f(3)=4$f(3)=4. If $f(0)+f(1)+f(-2)+f(3)=14$f(0)+f(1)+f(−2)+f(3)=14, then $\lambda$λ is equal to

Now given $f(0)+f(1)+f(-2)+f(3)=14$f(0)+f(1)+f(−2)+f(3)=14

$$\Rightarrow f(0)+3+\lambda +4=14$$⇒f(0)+3+λ+4=14

$$\Rightarrow f(0)=7-\lambda =c$$⇒f(0)=7−λ=c

Now $f(1)=a+b+c=3...(i)$f(1)=a+b+c=3...(i)

$$f(3)=9a+3b+c=4...(\mathrm{ii})$$f(3)=9a+3b+c=4...(ii)

$$f(-2)=4a-2b+c=\lambda ...(\mathrm{iii})$$f(−2)=4a−2b+c=λ...(iii)

Now subtracting equation $(\mathrm{ii})-(\mathrm{iii})$(ii)−(iii)

We get, $a+b=\frac{4-\lambda }{5}$a+b=54−λ​, now putting in equation $(i)$(i)

We get, $\frac{4-\lambda }{5}+7-\lambda =3$54−λ​+7−λ=3

$$\Rightarrow 6\lambda =24$$⇒6λ=24

$$\Rightarrow \lambda =4$$⇒λ=4