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Algebra Level 3

If a , b ϵ ( 0 , 2 ) a,b\epsilon(0,2) and the equation x 2 + 5 2 = x 2 c o s ( a x + b ) \frac{x^{2}+5}{2}=x-2cos(ax+b) has atleast one solution then a + b = a+b=

π \pi π 2 \frac{\pi}{2} e e 2

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1 solution

Tanishq Varshney
Feb 12, 2015

Just put x = 1 x=1

wrong option selected.

We can make conclusion from this.

x 2 2 x + 5 2 = 2 cos ( a x + b ) \dfrac{x^2 - 2x + 5}{2} = -2\cos (ax + b)

( x 1 ) 2 + 4 = 4 cos ( a x + b ) (x - 1)^2 + 4 = -4\cos (ax + b)

( x 1 ) 2 = 4 ( 1 + cos ( a x + b ) (x - 1)^2 = -4( 1 + \cos (ax + b)

( x 1 ) 2 = 8 cos 2 a x + b 2 (x - 1)^2 = -8 \cos^2 \dfrac{ax + b}{2}

Thus for x R x \in \mathbb R , the only posibble solution is x = 1 , R H S = 0 a + b 2 = π 2 x =1 , \implies RHS = 0 \implies \dfrac{a + b}{2} = \dfrac{\pi}{2}

U Z - 6 years, 4 months ago

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a + b 2 = π 2 \frac{a+b}{2}=\frac{\pi}{2}

so a + b = π a+b=\pi

Tanishq Varshney - 6 years, 4 months ago

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That mistake only I did while solving , and too writing it , thanks edited

U Z - 6 years, 4 months ago

Actually the answer can be any odd multiple of pi.

Raghav Vaidyanathan - 6 years, 3 months ago

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