Find the value(s) of x for which y = [x(x - 2)]² is an increasing function.

OR

Find the equations of the tangent and normal to the curve at the point

y = [x(x – 2)]²

⇒ y = x²(x – 2)²

⇒ y = x²(x² – 4x + 4)

⇒ y = x⁴ – 4x³ + 4x²

Now for a function y = f(x) to be increasing should be positive that is

Let us first find

y = x⁴ – 4x³ + 4x²

Differentiating with respect to x

As

⇒ 4x³ – 12x² + 8x ≥ 0

⇒ 4x(x² – 3x + 2) ≥ 0

⇒ x(x² – 2x – x + 2) ≥ 0

⇒ x(x(x – 2) – 1(x – 2)) ≥ 0

⇒ x(x – 1)(x – 2) ≥ 0 …(i)

Now this equation (i) is 0 at x = 0, x = 1 and x = 2

So we have to check the intervals x ≤ 0, x is between 0 and 1, x is between 1 and 2 and x ≥ 2

Now observe that for x ≥ 2 equation (i) holds true

Hence one possible solution set for x is x > 2

When x is in between 1 and 2 x is positive (x – 1) is positive but (x – 2) is negative and hence the product x(x – 1)(x – 2) becomes negative and hence equation (i) does not hold true

When x is between 0 and 1 x is positive (x – 1) is negative and (x – 2) is also negative, and hence the product x(x – 1)(x – 2) becomes positive, and equation (i) holds true hence x also can be between 0 and 1

Now when x < 0 that is x negative implies (x – 1) and (x – 2) also negative hence the product x(x – 1)(x – 2) is negative, and the equation (i) does not hold true

Hence for y to be increasing x ≥ 2 or x is between 0 and 1 both included that is x ∈ [0, 1] U [2, ∞).

OR

The tangent and normal are at the point (√2 a,b)

For the equation of tangent, we first need the slope of the tangent which is given by

We have to find the slope at (√2 a,b)

Differentiate with respect to x

As we have to find at put and y = b

Hence slope of tangent through the point is

Using slope point form to write the equation of the line which says that if m is the slope and (x₁, y₁) is a point on line, then the equation of the line is y – y₁ = m(x – x₁)

Hence the equation of tangent will be

Now let us find the equation of normal at the same point

Tangent and normal are perpendicular to each other at the same point

And the product of slopes of two perpendicular lines is -1

Let mbe the slope of normal and slope of the tangent at is given by

Again using slope point form

Hence the equation of the tangent is √2 bx – ay = ab and that of normal is .