Given,
fλ(x)=4λx3−36λx2+36x+48
Differentiating w.r.t x both side we get,
fλ′(x)=12λx2−72λx+36
fλ(x)′=12(λx2−6λx+3)≥0
∴λ>0 and D≤0
36λ2−4×λ×3≤0
9λ2−3λ≤0
3λ(3λ−1)≤0
λ∈[0,31]
So, λlargest=31
f(x)=34x3−12x2+36x+48
So,f(1)+f(−1)=72
JEE Main 2022 — Mathematics Calculus
Let λ∗ be the largest value of λ for which the function fλ(x)=4λx3−36λx2+36x+48 is increasing for all x∈R. Then fλ∗(1)+fλ,∗(−1) is equal to:
Held on 24 Jun 2022 · Verified 6 Jul 2026.
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