JEE Main 2021 — Mathematics Calculus

$$f(x)={x}^{3}-3{x}^{2}-\frac{3{f}^{"}(2)}{2}x+{f}^{"}(1)\ldots (i)$$f(x)=x3−3x2−23f"(2)​x+f"(1)…(i)

$$\Rightarrow {f}^{'}(x)=3{x}^{2}-6x-\frac{3}{2}{f}^{"}(2)\ldots (\mathrm{ii})$$⇒f′(x)=3x2−6x−23​f"(2)…(ii)

$$\Rightarrow {f}^{"}(x)=6x-6\ldots (\mathrm{iii})$$⇒f"(x)=6x−6…(iii)

Then,

$${f}^{"}(2)=12-6=6$$f"(2)=12−6=6

$${f}^{"}(1)=0$$f"(1)=0

Now,

$${f}^{'}(x)=3{x}^{2}-6x-\frac{3}{2}{f}^{"}(2)$$f′(x)=3x2−6x−23​f"(2)

$$\Rightarrow {f}^{'}(x)=3{x}^{2}-6x-\frac{3}{2}\times 6$$⇒f′(x)=3x2−6x−23​×6

$$\Rightarrow {f}^{'}(x)=3{x}^{2}-6x-9$$⇒f′(x)=3x2−6x−9

For maxima/minima, we have

$${f}^{'}(x)=0$$f′(x)=0

$$\Rightarrow 3{x}^{2}-6x-9=0$$⇒3x2−6x−9=0

$$\Rightarrow {x}^{2}-2x-3=0$$⇒x2−2x−3=0

$$\Rightarrow {x}^{2}-3x+x-3=0$$⇒x2−3x+x−3=0

$\Rightarrow x=-1$⇒x=−1 and $3$3

Now,

$${f}^{"}(x)=6x-6$$f"(x)=6x−6

${f}^{"}(-1)=-12<0$f"(−1)=−12<0 (maxima)

${f}^{"}(3)=12>0$f"(3)=12>0 (minima)

Again

$$f(x)={x}^{3}-3{x}^{2}-\frac{3{f}^{"}(2)}{2}x+{f}^{"}(1)$$f(x)=x3−3x2−23f"(2)​x+f"(1)

$$\Rightarrow f(x)={x}^{3}-3{x}^{2}-\frac{3}{2}\times 6x+0$$⇒f(x)=x3−3x2−23​×6x+0

$$\Rightarrow f(x)={x}^{3}-3{x}^{2}-9x$$⇒f(x)=x3−3x2−9x

$$\Rightarrow f(3)=27-27-9\times 3=-27$$⇒f(3)=27−27−9×3=−27