JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q20

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MathematicsSingle correct+4 / -1Question 20 of 90

Let $\lambda^*$ be the largest value of $\lambda$ for which the function $f_\lambda(x)=4\lambda x^3-36\lambda x^2+36x+48$ is increasing for all $x\in\mathbb R$ . Then $f_{\lambda^*}(1)+f_{\lambda^*}(-1)$ is equal to :

Answer and worked solution

Official answer: D

Increasing everywhere. $f'(x)=12\lambda x^2-72\lambda x+36\ge0$ for all $x$ needs $\lambda>0$ and $72^2\lambda^2-4(12\lambda)(36)\le0$ , i.e. $\lambda\le\frac13$ . So $\lambda^*=\frac13$ .

Function. $f(x)=\frac43x^3-12x^2+36x+48$ .

Values. $f(1)=\frac43+72$ and $f(-1)=-\frac43+0$ , so the sum is 72.

(D)

Topics:Monotonicity Quadratic expressions

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JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 20