JEE Main 2023 — Mathematics Calculus

Since, f(x)={\begin{matrix}3{x}^{2}+k\sqrt{x+1};0<x<1 \\ m{x}^{2}+{k}^{2};x\geq 1\end{matrix} is differentiable at $x=1$x=1, so function must be continuous at $x=1$x=1, hence

$$\mathrm{LHL}=\mathrm{RHL}$$LHL=RHL

$$3+k\sqrt{2}=m+{k}^{2}$$3+k2​=m+k2

$$\Rightarrow {k}^{2}-k\sqrt{2}+m-3=0....(1)$$⇒k2−k2​+m−3=0....(1)

And,

{f}^{'}(x)={\begin{matrix}6x+\frac{k}{2\sqrt{x+1}};0<x<1 \\ 2mx;x>1\end{matrix}

So,

$${f}^{'}({1}^{-})={f}^{'}({1}^{+})$$f′(1−)=f′(1+)

$$\Rightarrow 6+\frac{k}{2\sqrt{2}}=2m$$⇒6+22​k​=2m

$$\Rightarrow m=3+\frac{k}{4\sqrt{2}}...(2)$$⇒m=3+42​k​...(2)

Putting in $(1)$(1), we get

$${k}^{2}-k\sqrt{2}+3+\frac{k}{4\sqrt{2}}-3=0$$k2−k2​+3+42​k​−3=0

$$\Rightarrow {k}^{2}-(\sqrt{2}-\frac{1}{4\sqrt{2}})k=0$$⇒k2−(2​−42​1​)k=0

$$\Rightarrow {k}^{2}-(\frac{7\sqrt{2}}{8})k=0$$⇒k2−(872​​)k=0

$$\Rightarrow k(k-\frac{7\sqrt{2}}{8})=0$$⇒k(k−872​​)=0

$$\Rightarrow k=\frac{7\sqrt{2}}{8}$$⇒k=872​​

So,

$$m=1+\frac{k}{12\sqrt{2}}=1+\frac{7\sqrt{2}}{96\sqrt{2}}=\frac{103}{96}$$m=1+122​k​=1+962​72​​=96103​

Hence,

{f}^{'}(x)={\begin{matrix}6x+\frac{7\sqrt{2}}{16\sqrt{x+1}};0<x<1 \\ \frac{103x}{16};x>1\end{matrix}

So,

$$\frac{{f}^{'}(8)}{{f}^{'}(\frac{1}{8})}=\frac{103\times 8}{16}\times \frac{1}{(\frac{6}{8}+\frac{7\sqrt{2}}{16\sqrt{\frac{9}{8}}})}$$f′(81​)f′(8)​=16103×8​×(86​+1689​​72​​)1​

$$\Rightarrow \frac{{f}^{'}(8)}{{f}^{'}(\frac{1}{8})}=\frac{103}{2}\times \frac{3}{4}$$⇒f′(81​)f′(8)​=2103​×43​

$$\Rightarrow \frac{8{f}^{'}(8)}{{f}^{'}(\frac{1}{8})}=309$$⇒f′(81​)8f′(8)​=309