# Find the an

Solving x2 + y2 = 4 and (x – 2)2 + y2 = 4 we get point of intersection as Now, on differentiating x2 + y2 = 4 at , we get Differentiating (x – 2)2 + y2 = 4 at , we get  Angle of intersection between two curves is given as    Hence the angle between the curves x2 + y2 = 4 and

(x–2)2 + y2 = 4 is equal to .

OR

f(x) = –2x3–9x2–12x + 1

On differentiating with respect to we get,

f’(x) = –6x2–18x–12

On equating to zero we get

f’(x) = –6x2–18x–12

x = –2, –1

We know that if a function is strictly increasing in an interval then its first derivative is greater than or equal to zero in that interval and if it is strictly decreasing then its first derivative is less than or equal to zero in that interval.

Clearly, f' (x)>0 in (–2,–1) and f' (x)<0 in (–∞,–2)U(–1,∞).

By definition,

f(x) is increasing in (–2,–1)and decreasing in (–∞,–2)U(–1,∞)

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