First slide

Trigonometric equations

Question

$\begin{array}{l} I f 0 \leq a \leq 3, 0 \leq b \leq 3 a n d t h e e q u a t i o n x^{2} + 4 + 3 \cos (a x + b) = 2 x \\ h a s a t l e a s t s o l u t i o n . T h e n t h e v a l u e s o f a + b i s \end{array}$

Moderate

A

0
B

$\frac{π}{2}$
C

$π$
D

None of these

Solution

$W e h a v e$
$\begin{array}{l} x^{2} - 2 x + 4 = - 3 \cos (a x + b) \\ \Rightarrow {(x - 1)}^{2} + 3 = - 3 \cos (a x + b) \\ A s - 1 \leq \cos (a x + b) \leq 1 a n d {(x - 1)}^{2} \geq 0 \\ T h e a b o v e i s p o s s i b l e o n l y i f \cos (a x + b) = - 1 w h e r e x = 1 \\ \Rightarrow \cos (a + b) = - 1 \\ \Rightarrow a + b = π, 3 π, 5 π e t c; \\ \Rightarrow a + b = π (∵ a + b \leq 6) \end{array}$

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