Answer :

y = (x^{3} - 1) (x - 2) is the equation of curve

First let us find the points where the curve intersects on x axis

On x axis y = 0 hence put y = 0 in the curve equation

⇒ (x^{3} – 1) (x – 2) = 0

⇒ x^{3} – 1 = 0 and x – 2 = 0

⇒ x = 1 and x = 2

(1, 0) and (2, 0) are the points on Y-axis where the curve intersects

To find equation of the tangents find the slope.

Let us now find the slope of tangent at those points that is at x = 1 and x = 2 for

y = (x^{3} - 1) (x - 2)

⇒ y = x^{4} – 2x^{3} – x + 2

Differentiate with respect to x

Put x = 1

Put x = 2

Slope of tangents passing through (1, 0) and (2,0) are -3 and 7 respectively

Using slope point form of equation which is (y – y_{1}) = m (x – x_{1}) where m is slope and (x_{1}, y_{1}) is point

Hence the equations of tangents are:

⇒ (y – 0) = -3(x – 1) and (y – 0) = 7(x – 2)

⇒ y = - 3x + 3 and y = 7x – 14

Hence equations of tangents to the curve are x + y = 3 and 7x – y = 14

**OR**

let us first write the domain of f(x)

domain is the set of permissible values of x

for the term log (1 + x), x should be greater than -1 and for the term log (2 + x), x should be greater than -2 and for the term x can be anything other than -2

Combining the criteria for all three terms we observe that x should be greater than -1

Hence the domain of f(x) is (-1, ∞)

We know that if f’(x) > 0 then f(x) is strictly increasing and if f’(x) < 0 then f(x) is strictly decreasing

Let us find f’(x)

Put f^{’}(x) = 0

⇒ x (4+x) = 0

⇒ x = 0 and (4 + x) = 0

⇒ x = 0 and x = -4

Now f’(x) is 0 at x = 0 and x = -4 but x = -4 is not in the domain

hence, we will only check for (-1, 0) and (0, ∞)

Consider the interval (-1, 0)

x (4 + x) is negative in (-1, 0) as x < 0 and (4 + x) > 0 in the interval ( -1 , 0)

(1 + x) (2 + x)^{2} is positive in (-1, 0) as (1+x) > 0 and (2 + x)^{2}>0 in the interval ( -1, 0).

Hence

Hence f’(x) is negative in (-1, 0) that is f’(x) < 0 and hence f(x) is strictly decreasing in (-1, 0)

Now consider the interval (0, ∞)

x (4 + x) is positive in (0, ∞) as x > 0 and (4+x)>0 in the interval (0, ∞).

(1 + x) (2 + x)^{2} is positive in (0, ∞) as (1 + x)>0 and (2 + x)^{2}>0 in the interval (0, ∞).

Hence

Hence f’(x) is positive in (0, ∞) that is f’(x) > 0 and hence f(x) is strictly increasing in (0, ∞)

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