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QJEEM23Mab817: JEE Main 2023 Mathematics - Class 12 JEEM23Mab817 Solution: Let and be positive real numbers such that the function . Then…

Let k\mathrm{k} and m\mathrm{m} be positive real numbers such that the function f(x)={3x2+kx+1,amp;0lt;xlt;1mx2+k2,amp;x≥1f(x)=\left\{\begin{array}{cc}3 x^{2}+k \sqrt{x+1}, & 0 < x < 1 \\ m x^{2}+k^{2}, & x \geq 1\end{array}\right. is differentiable for all x > 0 . Then 8f′(8)f′(18)\frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)} is equal to ____________.

Step-by-Step Verified Solution

Here, f(x)={3x2+kx+1,amp;0lt;xlt;1mx2+k2,amp;x≥1f(x)=\left\{\begin{array}{cc}3 x^{2}+k \sqrt{x+1}, & 0 < x < 1 \\ m x^{2}+k^{2}, & x \geq 1\end{array}\right.

∵f(x)\because f(x) is differentiable at x>0

So, f(x)f(x) is differentiable at x=1x=1

f(1−)=f(1)=f(1+)3+k2=m+k2......(i)\begin{gathered}f\left(1^{-}\right)=f(1)=f\left(1^{+}\right) \\\\3+k \sqrt{2}=m+k^2 ......(i)\end{gathered}

amp;f′(1−)=f′(1+)amp;6(1)+k21+1=2m(1)amp;⇒6+k22=2m......(ii)\begin{aligned}& f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right) \\\\& 6(1)+\frac{k}{2 \sqrt{1+1}}=2 m(1) \\\\& \Rightarrow 6+\frac{k}{2 \sqrt{2}}=2 m ......(ii)\end{aligned}

Using (i) and (ii),

amp;3+k2=3+k42+k2amp;⇒k2+k[142−2]=0\begin{aligned}& 3+k \sqrt{2}=3+\frac{k}{4 \sqrt{2}}+k^2 \\\\& \Rightarrow k^2+k\left[\frac{1}{4 \sqrt{2}}-\sqrt{2}\right]=0\end{aligned}

⇒k[k+1−842]=0⇒k=0,742\Rightarrow k\left[k+\frac{1-8}{4 \sqrt{2}}\right]=0 \Rightarrow k=0, \frac{7}{4 \sqrt{2}}

As the problem specifies k to be a positive real number, we can rule out k = 0. Hence, k = 742\frac{7}{4 \sqrt{2}}

amp; for k=742,m=3+74242amp;=3+732=96+732=10332\begin{aligned}& \text { for } k=\frac{7}{4 \sqrt{2}}, m=3+\frac{\frac{7}{4 \sqrt{2}}}{4 \sqrt{2}} \\\\& =3+\frac{7}{32}=\frac{96+7}{32}=\frac{103}{32}\end{aligned}

 So, 8f′(8)f′(18)=8×[2×10332×8]6×18+742×2918\text { So, } \frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}=\frac{8 \times\left[2 \times \frac{103}{32} \times 8\right]}{6 \times \frac{1}{8}+\frac{7}{4 \sqrt{2}} \times 2 \sqrt{918}}

=41234+712=4129+712=412×1216=309=\frac{412}{\frac{3}{4}+\frac{7}{12}}=\frac{412}{\frac{9+7}{12}}=\frac{412 \times 12}{16}=309

Concept :

f(x)f(x) is differentiable at x=ax=a , if f′(a−)=f′(a+)f^{\prime}\left(a^{-}\right)=f'\left(a^{+}\right)

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