no. of solns. of 2cos x =|sin x|
in[-2Ï€,5Ï€]
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UP 0 DOWN 0 0 1
1 Answers
1133It will be 14.
Draw the two curves and find the intersection points.
|sinx| should be easy to draw.
For 2cosx you need to do some analysis first.
let f(x) = 2cosx
→ 1/2 ≤ f(x) ≤ 2
Also note that f(x) is periodic with period 2Ï€
f'(x) > 0 when π < x < 2π
f'(x) < 0 when 0 < x < π
Also f(x) attains maxima at 0, 2π and minima at π. You can use this much info only to predict a rough sketch of f(x).
Note that I haven't calculated f''(x). You should do that to find the concavity and convexity to predict a more accurate sketch.
After this you will find 4 intersection points of f(x) and |sinx| b/w 0 and 2Ï€.
Anik Chatterjee at x=Ï€/2,LHS=RHS=1...so there are only 2 solutions in 0 & 2Ï€.so the answer must be 7 solutions?Upvote·0· Reply ·2014-05-11 07:51:49- Anik Chatterjee even at x=3Ï€/2,both the curves touch
Himanshu Giria yes i too got only 2 sol ns ... II/2 and 3 II /2
...- Sourish Ghosh You are missing two solutions between pi/2 and 3pi/2.
- Himanshu Giria wich 2 sol ns ...
- Himanshu Giria ???
Soumyadeep Basu Note that f(x) is decreasing at x=(pi/2) and has its minimum value of 1/2 at x=(pi). But mod. sin(pi) has slope zero at x=(pi/2) and is itself zero at x=(pi). So there must be a point of intersection between (pi/2) and (pi).