JEE Main 2023 — Mathematics Calculus

Given:

$$f(x)={x}^{2}+{g}^{'}(1)x+{g}^{"}(2)$$f(x)=x2+g′(1)x+g"(2)

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

$$\Rightarrow {f}^{"}(x)=2$$⇒f"(x)=2

Now,

$$g(x)=f(1){x}^{2}+x{f}^{'}(x)+{f}^{"}(x)$$g(x)=f(1)x2+xf′(x)+f"(x)

$$\Rightarrow g(x)=f(1){x}^{2}+x[2x+{g}^{'}(1)]+2$$⇒g(x)=f(1)x2+x[2x+g′(1)]+2

$$\Rightarrow g(x)=[f(1)+2]{x}^{2}+{g}^{'}(1)x+2$$⇒g(x)=[f(1)+2]x2+g′(1)x+2

$$\Rightarrow {g}^{'}(x)=2f(1)x+4x+{g}^{'}(1)....(1)$$⇒g′(x)=2f(1)x+4x+g′(1)....(1)

$$\Rightarrow {g}^{"}(x)=2f(1)+4...(2)$$⇒g"(x)=2f(1)+4...(2)

Put $x=1$x=1 in $(1)$(1), then we get

$$2f(1)+4=0$$2f(1)+4=0

$$\Rightarrow f(1)=-2$$⇒f(1)=−2

$$\therefore {g}^{"}(2)=2f(1)+4=0$$∴g"(2)=2f(1)+4=0

Now,

$$f(x)={x}^{2}+{g}^{'}(1)x+{g}^{"}(2)$$f(x)=x2+g′(1)x+g"(2)

Put $x=1$x=1, then we get

$$f(1)=1+{g}^{'}(1)+{g}^{"}(2)$$f(1)=1+g′(1)+g"(2)

$$\Rightarrow {g}^{'}(1)=-3$$⇒g′(1)=−3

So,

$${f}^{'}(x)=2x-3$$f′(x)=2x−3

$$f(x)={x}^{2}-3x+c$$f(x)=x2−3x+c

Since, $f(1)=-2\Rightarrow c=0$f(1)=−2⇒c=0

So,

$$f(x)={x}^{2}-3x$$f(x)=x2−3x

And,

$$g(x)=f(1){x}^{2}+x{f}^{'}(x)+{f}^{"}(x)$$g(x)=f(1)x2+xf′(x)+f"(x)

$$\Rightarrow g(x)=-2{x}^{2}+x(2x-3)+2$$⇒g(x)=−2x2+x(2x−3)+2

$$\Rightarrow g(x)=-3x+2$$⇒g(x)=−3x+2

Hence, $f(4)-g(4)=4+10=14$f(4)−g(4)=4+10=14