Q. 165.0( 1 Vote )

Find the equation

Answer :

y = (x3 - 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


(x3 – 1) (x – 2) = 0


x3 – 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 = (x3 - 1) (x - 2)


y = x4 – 2x3 – 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 – y1) = m (x – x1) where m is slope and (x1, y1) 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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