Processing math: 0%

menu123

Friday, February 15, 2019

Pembahasan Soal UN SMA Materi Limit Trigonometri

Soal 1, UN SMA Tapel 2016/2017 Program Studi IPA
Nilai lim adalah ...
A. -16
B. -4
C. 4
D. 16
E. 32
Pembahasan
\lim_{x\rightarrow 4}\frac{x^{2}-16}{1-\sqrt{x-3}}
=\lim_{x\rightarrow 4}\frac{x^{2}-16}{1-\sqrt{x-3}}\times \frac{1+\sqrt{x-3}}{1+\sqrt{x-3}} [Kalikan dengan sekawannya]
=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{\left ( 1-\sqrt{x-3} \right )\left ( 1+\sqrt{x-3} \right )}
Catatan: (a-b)(a+b)=a^{2}-b^{2}
=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1^{2}-\left ( \sqrt{x-3} \right )^{2}}
=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1-\left ( {x-3} \right )}
=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1-x+3}
=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{4-x}
=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{4-x}
=\lim_{x\rightarrow 4}\frac{(x-4)(x+4)\left ( 1+\sqrt{x-3} \right )}{-(x-4)}
=\lim_{x\rightarrow 4}\frac{(x+4)\left ( 1+\sqrt{x-3} \right )}{-1}
=\frac{(4+4)\left ( 1+\sqrt{4-3} \right )}{-1}
=\frac{8\times2}{-1}=-16
Jawabannya E

Soal 2, UN SMA Tapel 2015/2016 Program Studi IPA
Nilai \lim_{x\rightarrow 0}\frac{1-cos\left ( 4x \right )}{2x.sin \left ( 4x \right )}= ...
A. 1
B. \frac{1}{2}
C. 0
D. -\frac{1}{2}
E. -1
Pembahasan
\lim_{x\rightarrow 0}\frac{1-cos\left ( 4x \right )}{2x.sin \left ( 4x \right )}
Catatan: cos\left ( 4x \right )=1-2sin^{2}\left ( 2x \right )
=\lim_{x\rightarrow 0}\frac{1-\left ( 1-2sin^{2}\left ( 2x \right ) \right )}{2x.sin \left ( 4x \right )}
=\lim_{x\rightarrow 0}\frac{1-1+2sin^{2}\left ( 2x \right )}{2x.sin \left ( 4x \right )}
=\lim_{x\rightarrow 0}\frac{2sin^{2}\left ( 2x \right )}{2x.sin \left ( 4x \right )}
=\frac{2}{2}\times \lim_{x\rightarrow 0}\frac{sin\left ( 2x \right )}{x}\times \lim_{x\rightarrow 0}\frac{sin\left ( 2x \right )}{sin \left ( 4x \right )}
=\frac{2}{2}\times\frac{2}{1} \times\frac{2}{4}=1
Jawabannya A

Soal 3, UN SMA Tapel 2014/2015 Program Studi IPA
Nilai \lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{1-cos^{2}\left ( 2x \right )} adalah ...
A. -1
B. -\frac{1}{2}
C. 0
D. \frac{1}{2}
E. 1
Pembahasan
Catatan:sin^{2}\left ( 2x \right )+cos^{2}\left ( 2x \right )=1\Leftrightarrow 1-cos^{2}\left ( 2x \right )=sin^{2}\left ( 2x \right )
\lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{1-cos^{2}\left ( 2x \right )}
=\lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{sin^{2}\left ( 2x \right )}
=2\lim_{x\rightarrow 0}\frac{x}{sin\left ( 2x \right )}\lim_{x\rightarrow 0}\frac{tan\left ( x \right )}{sin\left ( 2x \right )}
=\frac{2.1}{2.2}=\frac{1}{2}
Jawabannya D

No comments:

Post a Comment