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Question: If \[f(x) = 3{x^4} + 4{x^3} - 12{x^2} + 12\] then \[f(x)\] is A.Increasing in \[\left( { - \infty...

If f(x)=3x4+4x312x2+12f(x) = 3{x^4} + 4{x^3} - 12{x^2} + 12 then f(x)f(x) is
A.Increasing in (,2)\left( { - \infty , - 2} \right) and in (1,)\left( {1,\infty } \right)
B.Increasing in (2,0)\left( { - 2,0} \right) and in (1,)\left( {1,\infty } \right)
C.Decreasing in (2,0)\left( { - 2,0} \right) and in (0,1)\left( {0,1} \right)
D.Decreasing in (,2)\left( { - \infty , - 2} \right) and in (1,)\left( {1,\infty } \right)

Explanation

Solution

Hint : Firstly find the derivative of the function as it can be at times used to determine whether a function is increasing or decreasing on any interval in its domain. If f(x)>0f'(x) > 0 in an interval I, then the function is said to be increasing on I and if f(x)<0f'(x) < 0 in an interval I, then the function is said to be decreasing on I.

Complete step-by-step answer :
If f(x)>0f'(x) > 0 then ff is increasing on the interval, and if f(x)<0f'(x) < 0 then ff is decreasing on the interval.
Following steps are involved in the process of finding the intervals of increasing and decreasing function:
Firstly, differentiate the given function with respect to the constant variable.
Then solve f(x)=0f'(x) = 0 .
We are given f(x)=3x4+4x312x2+12f(x) = 3{x^4} + 4{x^3} - 12{x^2} + 12
Differentiating both the sides with respect to xx we get ,
f(x)=12x3+12x224xf'(x) = 12{x^3} + 12{x^2} - 24x
Putting f(x)=0f'(x) = 0 i.e. 12x3+12x224x=012{x^3} + 12{x^2} - 24x = 0
On simplification we get 12x(x2+x2)=012x({x^2} + x - 2) = 0
Now splitting the middle term in the quadratic equation in bracket above we get
12x(x2+2xx2)=012x({x^2} + 2x - x - 2) = 0
12x(x(x+2)1(x+2))=012x\left( {x\left( {x + 2} \right) - 1\left( {x + 2} \right)} \right) = 0
Hence we get 12x(x2)(x+1)=012x(x - 2)(x + 1) = 0
Hence x=0,1,2x = 0, - 1,2
Hence we get the intervals (,1),(1,0),(0,2),(2,)\left( { - \infty , - 1} \right),\left( { - 1,0} \right),\left( {0,2} \right),\left( {2,\infty } \right)
In the interval (,1)\left( { - \infty , - 1} \right) f(x)<0f'(x) < 0 hence decreasing.
In the interval (1,0)\left( { - 1,0} \right) f(x)>0f'(x) > 0 hence increasing.
In the interval (0,2)\left( {0,2} \right) f(x)<0f'(x) < 0 hence decreasing.
In the interval (2,)\left( {2,\infty } \right) f(x)>0f'(x) > 0 hence increasing.
Therefore f(x)f(x) is increasing in (1,0)\left( { - 1,0} \right) and (2,)\left( {2,\infty } \right) and f(x)f(x) is decreasing in (,1)\left( { - \infty , - 1} \right) and (0,2)\left( {0,2} \right) .
Therefore f(x)f(x) is Increasing in (2,0)\left( { - 2,0} \right) and in (1,)\left( {1,\infty } \right) .
So, the correct answer is “Option B”.

Note : After solving the equation of the first derivative and finding the points of discontinuity we get the open intervals with the value of xx , through which the sign of the intervals can be taken into consideration.
If the sign of the interval in their first derivative form gives more than 00 then the function is said to be increasing in nature, while if the sign of the intervals in their first derivative form gives less than 00 then the function is said to be decreasing in nature.