JEE Main 2023 — Mathematics Calculus

Given functions are:

$${f}^{"}(x)={g}^{"}(x)+6x\ldots (1)$$f"(x)=g"(x)+6x…(1)

$${f}^{'}(1)=4{g}^{'}(1)-3=9\ldots (2)$$f′(1)=4g′(1)−3=9…(2)

$$f(2)=3g(2)=12\ldots (3)$$f(2)=3g(2)=12…(3)

By integrating $(1)$(1), we get

$${f}^{'}(x)={g}^{'}(x)+3{x}^{2}+C$$f′(x)=g′(x)+3x2+C

At $x=1$x=1

$${f}^{'}(1)={g}^{'}(1)+3+C$$f′(1)=g′(1)+3+C

$$\Rightarrow 9=3+3+C\Rightarrow C=3$$⇒9=3+3+C⇒C=3

$$\therefore {f}^{'}(x)={g}^{'}(x)+3{x}^{2}+3$$∴f′(x)=g′(x)+3x2+3

Again by integrating, we get

$$f(x)=g(x)+{x}^{3}+3x+D$$f(x)=g(x)+x3+3x+D

At $x=2$x=2, we get

$$f(2)=g(2)+8+3(2)+D$$f(2)=g(2)+8+3(2)+D

$$\Rightarrow 12=4+8+6+D\Rightarrow D=-6$$⇒12=4+8+6+D⇒D=−6

So,

$$f(x)=g(x)+{x}^{3}+3x-6$$f(x)=g(x)+x3+3x−6

Option $(A)$(A):

At $x=-2$x=−2,

$$\Rightarrow g(-2)-f(-2)=20$$⇒g(−2)−f(−2)=20

So, this option is true.

Option $(B)$(B)

If $-1<x<2$−1<x<2

Let $h(x)=f(x)-g(x)={x}^{3}+3x-6$h(x)=f(x)−g(x)=x3+3x−6

$$\Rightarrow {h}^{'}(x)=3{x}^{2}+3$$⇒h′(x)=3x2+3

$\Rightarrow h'(x)>0$⇒h′(x)>0 for all values of $x$x.

So, $h(-1)<h(x)<h(2)$h(−1)<h(x)<h(2)

$$\Rightarrow -10<h(x)<8$$⇒−10<h(x)<8

$$\Rightarrow |h(x)|<10$$⇒∣h(x)∣<10

So, this option is NOT true.

Option $(C)$(C)

$${h}^{'}(x)={f}^{'}(x)-{g}^{'}(x)=3{x}^{2}+3$$h′(x)=f′(x)−g′(x)=3x2+3

If $|{h}^{'}(x)|<6\Rightarrow |3{x}^{2}+3|<6$∣h′(x)∣<6⇒∣3x2+3∣<6

$\Rightarrow 3{x}^{2}+3<6$⇒3x2+3<6 and $-6<3{x}^{2}+3$−6<3x2+3

$\Rightarrow {x}^{2}<1$⇒x2<1 and ${x}^{2}>-3(\mathrm{always}\mathrm{true})$x2>−3(alwaystrue)

$$\Rightarrow -1<x<1$$⇒−1<x<1

So, If $x\in (-1,1)$x∈(−1,1) then $|{f}^{'}(x)-{g}^{'}(x)|<6$∣f′(x)−g′(x)∣<6

So, this option is true.

Option $(D)$(D)

$$f(x)-g(x)=0$$f(x)−g(x)=0

$$\Rightarrow {x}^{3}+3x-6=0$$⇒x3+3x−6=0

$$h(x)={x}^{3}+3x-6$$h(x)=x3+3x−6

Here, $h(1)=-\mathrm{ve}$h(1)=−ve and $h(\frac{3}{2})=+\mathrm{ve}$h(23​)=+ve

So, there exists ${x}_{0}\in (1,\frac{3}{2})$x0​∈(1,23​) such that $f({x}_{0})=g({x}_{0})$f(x0​)=g(x0​)

So, this option is true.