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

\(\begin{aligned} f'(-1)&=m\\ 2a(-1)+b&=-7\\ -2a+b&=-7 \end{aligned}\)

\(\begin{aligned} f'(1)&=1\\ 2a(1)+b&=1\\ 2a+b&=1 \end{aligned}\)

\(\begin{aligned} -2a+b&=-7\\ 2a+b&=1\qquad+\\\hline 2b&=-6\\ b&=-3 \end{aligned}\)

\(\begin{aligned} 2a+b&=1\\ 2a-3&=1\\ 2a&=4\\ a&=2 \end{aligned}\)

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

\(\begin{aligned} y_1&=-7(-1)+3\\ f(-1)&=10\\ 2(-1)^2-3(-1)+c&=10\\ 2+3+c&=10\\ c&=5 \end{aligned}\)

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

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

\(\begin{aligned} y&=\dfrac{x-1}x\\[8pt] &=1-\dfrac1x\\[8pt] &=1-x^{-1}\\ y'&=x^{-2}\\ &=\dfrac1{x^2}\\[8pt] m&=\dfrac1{a^2} \end{aligned}\)

Persamaan garis singgungnya,
\(\begin{aligned} y-y_1&=m(x-x_1)\\ y-\dfrac{a-1}a&=\dfrac1{a^2}(x-a) \end{aligned}\)
Persamaan garis tersebut melalui (b,-b)
\(\begin{aligned} -b-\dfrac{a-1}a&=\dfrac1{a^2}(b-a)\quad&\color{red}{\times a^2}\\[8pt] -a^2b-a(a-1)&=b-a\\ -a^2b-a^2+a&=b-a\\ -a^2+2a&=a^2b+b\\ 2a-a^2&=(a^2+1)b\\ b&=\boxed{\boxed{\dfrac{2a-a^2}{a^2+1}}} \end{aligned}\)

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

\(\eqalign{ y&=2x+3\\ q&=2p+3 }\)

\(\eqalign{ y'&=2ax-(a+1)\\ &=2ax-a-1\\ 2&=2ap-a-1\\ 3&=2ap-a\\ 3&=a(2p-1)\\ a&=\dfrac3{2p-1} }\)

\(\eqalign{ y &= ax^2 - (a + 1)x + 6\\ q&= ap^2 - (a + 1)p + 6\\ 2p+3&= \left(\dfrac3{2p-1}\right)p^2 - \left(\dfrac3{2p-1}+ 1\right)p + 6\\ 2p+3&=\dfrac{3p^2}{2p-1} - \left(\dfrac{3+2p-1}{2p-1}\right)p + 6\\ 2p+3&=\dfrac{3p^2}{2p-1} - \left(\dfrac{2p+2}{2p-1}\right)p + 6\\ 2p+3&=\dfrac{3p^2}{2p-1} - \dfrac{2p^2+2p}{2p-1} + 6\qquad&{\color{red}\times(2p-1)}\\ (2p+3)(2p-1)&=3p^2-\left(2p^2+2p\right)+6(2p-1)\\ 4p^2+4p-3&=3p^2-2p^2-2p+12p-6\\ 4p^2+4p-3&=p^2+10p-6\\ 3p^2-6p+3&=0\\ p^2-2p+1&=0\\ (p-1)^2&=0\\ p&=1 }\)

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

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

Jadi, p + q =6.
JAWAB: D