Solveeit Logo

Question

Question: A function \(f\left( x \right)={{x}^{2}}+bx+c\) where \(b,c\in R\). If \(f\left( x \right)\) is a fa...

A function f(x)=x2+bx+cf\left( x \right)={{x}^{2}}+bx+c where b,cRb,c\in R. If f(x)f\left( x \right) is a factor of both x4+6x2+25{{x}^{4}}+6{{x}^{2}}+25 and 3x4+4x2+28x+53{{x}^{4}}+4{{x}^{2}}+28x+5 then least value of f(x)f\left( x \right) is
a) 2
b) 3
c) 52\dfrac{5}{2}
d) 4

Explanation

Solution

To obtain the least value of f(x)f\left( x \right)we will find the value of f(x)f\left( x \right) by the two equations given and use a second derivative test. Firstly we will use the two given equations to get the value of f(x)f\left( x \right) in x2+bx+c{{x}^{2}}+bx+c form by using the property. To check which one of the roots is least we will differentiate the function twice and use the second derivative to obtain the desired value.

Complete step by step answer:
The equation given is:
f(x)=x2+bx+cf\left( x \right)={{x}^{2}}+bx+c
Where b,cRb,c\in R
The other two equations given are:
g(x)=x4+6x2+25g\left( x \right)={{x}^{4}}+6{{x}^{2}}+25…..(1)\left( 1 \right)
h(x)=3x4+4x2+28x+5h\left( x \right)=3{{x}^{4}}+4{{x}^{2}}+28x+5……(2)\left( 2 \right)
We will start by finding factor of g(x)g\left( x \right) which is:
x4+6x2+25=(x2+2x+5)(x22x+5){{x}^{4}}+6{{x}^{2}}+25=\left( {{x}^{2}}+2x+5 \right)\left( {{x}^{2}}-2x+5 \right)…..(3)\left( 3 \right)
Next we will find factor of h(x)h\left( x \right) which is:
3x4+4x2+28x+5=(x22x+5)(3x2+6x+1)3{{x}^{4}}+4{{x}^{2}}+28x+5=\left( {{x}^{2}}-2x+5 \right)\left( 3{{x}^{2}}+6x+1 \right)…..(4)\left( 4 \right)
From equation (3) and (4) we get the common factor as below:
f(x)=x22x+5\therefore f\left( x \right)={{x}^{2}}-2x+5………(5)\left( 5 \right)
Next we will differentiate equation (5) with respect to xx and get,
f(x)=2x2{f}'\left( x \right)=2x-2…….(6)\left( 6 \right)
Now, put the above equation equal to 0,
f(x)=2x2 0=2x2 2x=2 x=1 \begin{aligned} & \Rightarrow {f}'\left( x \right)=2x-2 \\\ & \Rightarrow 0=2x-2 \\\ & \Rightarrow 2x=2 \\\ & \therefore x=1 \\\ \end{aligned}
Next again differentiating equation (6) we get,
f(x)=2{f}''\left( x \right)=2
As 2>02>0 the function have minima at x=1x=1
Put x=1x=1 in equation (3) we get,
f(1)=122×1+5 f(1)=12+5 f(1)=4 \begin{aligned} & \Rightarrow f\left( 1 \right)={{1}^{2}}-2\times 1+5 \\\ & \Rightarrow f\left( 1 \right)=1-2+5 \\\ & \therefore f\left( 1 \right)=4 \\\ \end{aligned}
Hence, the correct option is (d).

Note: The second derivative test is used to find out the maxima or minima of a function. When we double differentiate a function we are basically measuring the instantaneous rate of change of the first derivative. Depending on the sign of the second derivative we can find whether the function is increasing or decreasing. The second derivative test doesn’t apply if we get the value of the first derivative equal to 0.