Find the angle of intersection of the curves x² + y² = 4 and (x – 2)² + y² = 4, at the point in the first quadrant.

OR

Find the intervals in which the function f(x) = –2x³–9x²–12x + 1 is

(i) Strictly increasing (ii) Strictly decreasing.

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,∞)