JEE Main 2024 — Mathematics Calculus

Given: $f(x)={x}^{3}+{x}^{2}{f}^{'}(1)+x{f}^{"}(2)+{f}^{'''}(3)$f(x)=x3+x2f′(1)+xf"(2)+f′′′(3)

$$\Rightarrow {f}^{'}(x)=3{x}^{2}+2x{f}^{'}(1)+{f}^{"}(2)...(i)$$⇒f′(x)=3x2+2xf′(1)+f"(2)...(i)

$$\Rightarrow {f}^{"}(x)=6x+2{f}^{'}(1)...(ii)$$⇒f"(x)=6x+2f′(1)...(ii)

$$\Rightarrow {f}^{'''}(x)=6$$⇒f′′′(x)=6

$$\Rightarrow {f}^{'''}(3)=6...(iii)$$⇒f′′′(3)=6...(iii)

Using $(ii)$(ii),

$$\Rightarrow {f}^{"}(2)=12+2{f}^{'}(1)$$⇒f"(2)=12+2f′(1)

Putting this value in equation $(i)$(i),

$$\Rightarrow {f}^{'}(1)=3+2{f}^{'}(1)+12+2{f}^{'}(1)$$⇒f′(1)=3+2f′(1)+12+2f′(1)

$$\Rightarrow 0=15+3{f}^{'}(1)$$⇒0=15+3f′(1)

$$\Rightarrow {f}^{'}(1)=-5...(iv)$$⇒f′(1)=−5...(iv)

$$\Rightarrow {f}^{"}(2)=12+2(-5)$$⇒f"(2)=12+2(−5)

$$\Rightarrow {f}^{"}(2)=2...(v)$$⇒f"(2)=2...(v)

$$\Rightarrow f(x)={x}^{3}+{x}^{2}\cdot (-5)+x\cdot (2)+6$$⇒f(x)=x3+x2⋅(−5)+x⋅(2)+6

$$\Rightarrow {f}^{'}(x)=3{x}^{2}-10x+2$$⇒f′(x)=3x2−10x+2

$$\Rightarrow {f}^{'}(10)=300-100+2$$⇒f′(10)=300−100+2

$$\Rightarrow {f}^{'}(10)=202$$⇒f′(10)=202