Find the intervals in which the function f x = sin 3 x , x ∈ 0 , π 2 is decreasing.

For the given function f x = sin 3 x On differentiating both sides w.r.t. x , we get f ' x = d d x sin 3 x f ' x = cos 3 x d d x 3 x f ' x = 3 cos 3 x On considering f ' x = 0 ,we get 3 cos 3 x = 0 cos 3 x = 0 x = π 6 , π 2 Which divides the limit 0 , π 2 into the limits 0 , π 6 and π 6 , π 2 Where, f ' x > 0 for x ∈ 0 , π 6 f ' x < 0 x ∈ π 6 , π 2 Hence, the function is decreasing for x ∈ π 6 , π 2 .

Find the intervals in which the function f x = sin 3 x , x ∈ 0 , π 2 is decreasing.

For the given function f x = sin 3 x On differentiating both sides w.r.t. x , we get f ' x = d d x sin 3 x f ' x = cos 3 x d d x 3 x f ' x = 3 cos 3 x On considering f ' x = 0 ,we get 3 cos 3 x = 0 cos 3 x = 0 x = π 6 , π 2 Which divides the limit 0 , π 2 into the limits 0 , π 6 and π 6 , π 2 Where, f ' x > 0 for x ∈ 0 , π 6 f ' x < 0 x ∈ π 6 , π 2 Hence, the function is decreasing for x ∈ π 6 , π 2 .

Find the equation of the tangent to the curve y = f ( x ) at x 0 , y 0 , if d y d x does not exist at this point.

Given, y = f ( x ) d y d x does not exist implies line is parallel to y - axis. ∴ Equation of tangent is line parallel to y - axis passing through x 0 , y 0 . ∴ Equation of tangent to the given curve is y = y 0 .

Find the absolute maximum and minimum values of the function f given by f x = c o s 2 x + s i n x , x ∈ 0 , π .

Given that: f x = c o s 2 x + s i n x Differentiating both sides with respect to x we get, f ' x = d d x cos 2 x + sin x = 2 cos x - sin x + cos x = - 2 s i n x c o s x + c o s x Now to know the absolute maxima or minima, we need to equate f ' ( x ) to 0 . Therefore, f ' x = 0 ⇒ 2 s i n x c o s x - c o s x = 0 ⇒ c o s x 2 s i n x - 1 = 0 ⇒ s i n x = 1 2 or c o s x = 0 ⇒ x = π 6 , or π 2 a s x ∈ 0 , π f π 6 = c o s 2 π 6 + s i n π 6 = 3 2 2 + 1 2 = 5 4 f 0 = c o s 2 0 + s i n 0 = 1 + 0 = 1 f π = c o s 2 π + s i n π = - 1 2 + 0 = 1 f π 2 = c o s 2 π 2 + s i n π 2 = 0 + 1 = 1 Therefore, the absolute maximum value of f is 5 4 occurring at x = π 6 and the absolute minimum value of f is 1 occurring at x = 0 , π 2 and π .

Prove that the curves x = y 2 and x y = k cut at right angles if 8 k 2 = 1 .

The equations of the given curves are given as x = y 2 and x y = k . Putting x = y 2 in x y = k , we get: y 3 = k ⇒ y = k 1 3 Now, substituting y i n x = y 2 , we get: x = k 2 3 Thus, the point of intersection of the given curves is k 2 3 , k 1 3 . Differentiating x = y 2 with respect to x , we have: 1 = 2 y d y d x ⇒ d y d x = 1 2 y Hence, the slope of the tangent to the curve x = y 2 at k 2 3 , k 1 3 is d y d x k 2 3 , k 1 3 = 1 2 k 1 3 On differentiating x y = k with respect to x , we have: x d y d x + y = 0 ⇒ d y d x = - y x ∴ Slope of the tangent to the curve x y = k at k 2 3 , k 1 3 is given by, d y d x k 2 3 , k 1 3 = - y x k 2 3 , k 1 3 = - k 1 3 k 2 3 = - 1 k 1 3 We know that two curves intersect at right angles, if the tangents to the curves at the point of intersection i.e., at k 2 3 , k 1 3 are perpendicular to each other. This implies that we should have the product of the slopes of tangents as - 1 . Hence, the given two curves cut at right angles if the product of the slopes of their respective tangents at k 2 3 , k 1 3 is - 1 . ⇒ (Slope of tangent to the curve x = y 2 ) × ( Slope of tangent to the curve x y = k ) = - 1 i.e., 1 2 k 1 3 - 1 k 1 3 = - 1 - 1 2 k 1 3 × k 1 3 = - 1 1 2 k 1 3 + 1 3 = 1 1 2 k 2 3 = 1 ⇒ 2 k 2 3 = 1 ⇒ 2 k 2 3 3 = 1 3 ⇒ 8 k 2 = 1 Hence proved.

Find the equation of the tangent to the curve y = f ( x ) at x 0 , y 0 , if d y d x does not exist at this point.

Given, y = f ( x ) d y d x does not exist implies line is parallel to y - axis. ∴ Equation of tangent is line parallel to y - axis passing through x 0 , y 0 . ∴ Equation of tangent to the given curve is y = y 0 .