HOTS Zone : Persamaan Garis Singgung (Equation of a Tangent Line)

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Tipe:

Standar SBMPTN HOTS

No.

Jika y=\ln(x) dan y=kx^2+1, k\gt0, berpotongan di satu titik, maka nilai k adalah....

1. \dfrac{e^3}2
2. \dfrac2{e^{\frac13}}
3. \dfrac1{2e^3}

4. 2e^{\frac13}
5. \dfrac1{2e}

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Misal gradien garis singgung di titik potong kedua kurva adalah m.
$$\begin{array}{rl}y& =\ln (x)\\ y^{\prime }& ={\displaystyle \frac{1}{x}}\\ m& ={\displaystyle \frac{1}{x}}\end{array}$$

$$\begin{array}{rl}y& =k{x}^2+1\\ y^{\prime }& =2kx\\ m& =2kx\end{array}$$

$$\begin{array}{rl}{\displaystyle \frac{1}{x}}& =2kx\\ 1& =2k{x}^2\\ x^2& ={\displaystyle \frac{1}{2k}}\end{array}$$

$$\begin{array}{rl}y& =k{x}^2+1\\ & =k\left({\displaystyle \frac{1}{2k}}\right)+1\\ & ={\displaystyle \frac{1}{2}}+1\\ & ={\displaystyle \frac{3}{2}}\end{array}$$

$$\begin{array}{rl}y& =\ln (x)\\ 2y& =2\ln (x)\\ 2y& =\ln \left(x^2\right)\\ 2\left({\displaystyle \frac{3}{2}}\right)& =\ln \left({\displaystyle \frac{1}{2k}}\right)\\ 3& =\ln \left({\displaystyle \frac{1}{2k}}\right)\\ e^3& ={\displaystyle \frac{1}{2k}}\\ k& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{2e^3}}}}}}\end{array}$$

Jadi, k=\dfrac1{2e^3}. JAWAB: C