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

Berikut ini adalah kumpulan soal mengenai Persamaan Garis Singgung tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

No. 1

Diketahui f adalah fungsi kuadrat yang mempunyai garis singgung {y=-7x+3} di titik {x=-1}, jika {f'(1)=1}, maka f(3)= ....

1. 1
2. 2
   
3. 3
   

4. 4
5. 5
   

Misal f(x)=ax^2+bx+c
f'(x)=2ax+b

x_1=-1

$\begin{array}{rl}{f}^{\prime }(-1)& =m\\ 2a(-1)+b& =-7\\ -2a+b& =-7\end{array}$

$\begin{array}{rl}{f}^{\prime }(1)& =1\\ 2a(1)+b& =1\\ 2a+b& =1\end{array}$

$\begin{array}{rl}-2a+b& =-7\\ 2a+b& =1+\\ 2b& =-6\\ b& =-3\end{array}$

$\begin{array}{rl}2a+b& =1\\ 2a-3& =1\\ 2a& =4\\ a& =2\end{array}$

f(x)=2x^2-3x+c

$\begin{array}{rl}{y}_1& =-7(-1)+3\\ f(-1)& =10\\ 2(-1{)}^2-3(-1)+c& =10\\ 2+3+c& =10\\ c& =5\end{array}$

$\begin{array}{rl}f(3)& =2(3{)}^2-3(3)+5\\ & =18-9+5\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 14}}}}\end{array}$

Jadi, f(3)=14.
JAWAB: C

No. 2

Jika garis singgung dari kurva {y=\dfrac{x-1}x} pada {x=a} memotong garis {y=-x} di titik (b,-b) maka {b=}

1. \dfrac{a^2-a}{a^2+1}
2. \dfrac{2a-a^2}{a^2+1}
3. \dfrac{a^2-a}{a^2+1}

4. \dfrac{a^2+a}{a^2-1}
5. \dfrac{2a^2-a}{a^2+1}

x_1=a
y_1=\dfrac{a-1}a

$\begin{array}{rl}y& ={\displaystyle \frac{x-1}{x}}\\ & =1-{\displaystyle \frac{1}{x}}\\ & =1-x^{-1}\\ y^{\prime }& =x^{-2}\\ & ={\displaystyle \frac{1}{x^2}}\\ m& ={\displaystyle \frac{1}{a^2}}\end{array}$

Persamaan garis singgungnya,
$\begin{array}{rl}y-y_1& =m(x-x_1)\\ y-{\displaystyle \frac{a-1}{a}}& ={\displaystyle \frac{1}{a^2}}(x-a)\end{array}$
Persamaan garis tersebut melalui (b,-b)
$\begin{array}{rlr}-b-{\displaystyle \frac{a-1}{a}}& ={\displaystyle \frac{1}{a^2}}(b-a)& \times a^2\\ -a^{2}b-a(a-1)& =b-a\\ -a^{2}b-a^2+a& =b-a\\ -a^2+2a& =a^{2}b+b\\ 2a-a^2& =(a^2+1)b\\ b& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2a-a^2}{a^2+1}}}}}}\end{array}$

Jadi, b=\dfrac{2a-a^2}{a^2+1}.
JAWAB: B

No. 3

Jika garis singgung kurva {y = ax^2 - (a + 1)x + 6}, a\neq 0 di titik (p, q) adalah {y = 2x + 3}, maka nilai p + q =

1. 1
2. 2
3. 4

4. 6
5. 8

SKMP Juli 2020

$\begin{array}{rl}y& =2x+3\\ q& =2p+3\end{array}$

$\begin{array}{rl}{y}^{\prime }& =2ax-(a+1)\\ & =2ax-a-1\\ 2& =2ap-a-1\\ 3& =2ap-a\\ 3& =a(2p-1)\\ a& ={\displaystyle \frac{3}{2p-1}}\end{array}$

$\begin{array}{rl}y& =a{x}^2-(a+1)x+6\\ q& =a{p}^2-(a+1)p+6\\ 2p+3& =\left({\displaystyle \frac{3}{2p-1}}\right)p^2-({\displaystyle \frac{3}{2p-1}}+1)p+6\\ 2p+3& ={\displaystyle \frac{3p^2}{2p-1}}-\left({\displaystyle \frac{3+2p-1}{2p-1}}\right)p+6\\ 2p+3& ={\displaystyle \frac{3p^2}{2p-1}}-\left({\displaystyle \frac{2p+2}{2p-1}}\right)p+6\\ 2p+3& ={\displaystyle \frac{3p^2}{2p-1}}-{\displaystyle \frac{2p^2+2p}{2p-1}}+6& \times (2p-1)\\ (2p+3)(2p-1)& =3p^2-(2p^2+2p)+6(2p-1)\\ 4p^2+4p-3& =3p^2-2p^2-2p+12p-6\\ 4p^2+4p-3& =p^2+10p-6\\ 3p^2-6p+3& =0\\ p^2-2p+1& =0\\ (p-1{)}^2& =0\\ p& =1\end{array}$

{q=2(1)+3=5}

{p+q=1+5=\boxed{\boxed{6}}}

Jadi, p + q =6.
JAWAB: D