Answer :

Solving x^{2} + y^{2} = 4 and (x – 2)^{2} + y^{2} = 4 we get point of intersection as

Now, on differentiating x^{2} + y^{2} = 4 at , we get

Differentiating (x – 2)^{2} + y^{2} = 4 at , we get

Angle of intersection between two curves is given as

Hence the angle between the curves x^{2} + y^{2} = 4 and

(x–2)^{2} + y^{2} = 4 is equal to .

OR

f(x) = –2x^{3}–9x^{2}–12x + 1

On differentiating with respect to we get,

f’(x) = –6x^{2}–18x–12

On equating to zero we get

f’(x) = –6x^{2}–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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Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

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Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1