First slide

Trigonometric equations

Question

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

Moderate

A

$\frac{π}{2}$
B

$\frac{π}{4}$
C

$\frac{π}{3}$
D

$π$

Solution

$W e h a v e$ $x^{2} - 2 x + 4 = - 3 \cos (a x + b)$
$\Rightarrow {(x - 1)}^{2} + 3 = - 3 \cos (a x + b)$
$L H S \geq 3 a n d R H S \leq 3 \Rightarrow L H S = R H S = 3$
$\Rightarrow x = 1$
Also we have $3 = - 3 \cos (a x + b) = - 3 \cos (a + b) (∵ x = 1)$
$\cos (a + b) = - 1 = cosπ \Rightarrow a + b = π$

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