First slide

Trigonometric equations

Question

$T h e n u m b e r o f p o i n t s i n t e r s e c t i o n o f t h e t w o c u r v e s y = 2 \sin x a n d y = 5 x^{2} + 2 x + 3 i s$

Easy

A

0
B

1
C

2
D

$\infty$

Solution

$W e h a v e y = 5 x^{2} + 2 x + 3 = 5 [x^{2} + \frac{2}{5} x + \frac{3}{5}]$
$= 5 [{(x + \frac{1}{5})}^{2} + \frac{3}{5} - \frac{1}{25}] = 5 {(x + \frac{1}{5})}^{2} + \frac{14}{5} > 2$
$S i n c e y = 2 \sin x \leq 2, s o t h e r e c a n n o t b e a n y p o i n t o f i n t e r s e c t i o n .$

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