Q43 / 160 · AP EAMCET 2020 (18 Sep Shift 2) · Mathematics
The curves \(y=4 x^2+2 x-8\) and \(y=x^3-x+13\) touch each other at the point
- A\((34,3)\)
- Correct answerB\((3,37)\)
- C\((-3,34)\)
- D\((-34,3)\)
Solution
\(y=4 x^2+2 x-8\)
differentiate w.r.t. ' \(x\) ' on both sides,
\(\begin{aligned}
\frac{d y}{d x} & =4(2 x)+2(1) \\
\frac{d y}{d x} & =8 x+2 \quad \ldots (i) \\
y & =x^3-x+13
\end{aligned}\)
differentiate w.r.t. ' \(x\) ' on both sides,
\(\frac{d y}{d x}=3 x^2-1 \quad \ldots (ii)\)
Since, curves are touch each other
\(\begin{aligned}
\Rightarrow & & 8 x+2 & =3 x^2-1 \\
\Rightarrow & & 3 x^2-8 x-3 & =0 \\
\Rightarrow & & (x-3)(3 x+1) & =0 \\
\Rightarrow & & x & =3 \text { (or) } \frac{-1}{3}
\end{aligned}\)
Put, \(x=3\) in \(y=x^3-x+13\)
\(\begin{aligned}
& y=3^3-3+13 \\
& y=37
\end{aligned}\)
\(\therefore\) Point of contact \(P=(3,37)\)
Hence, the correct answer is option: 'B'.