Question

If $a,b\in \\{1,2,3,4,5,6\\}$, find the number of ways a, b, c can be selected if $f(x)={{x}^{3}}+a{{x}^{2}}+bx+c$ is an increasing function.

Answer 1

Hint: We will first differentiate $f(x)={{x}^{3}}+a{{x}^{2}}+bx+c$ and this will be greater than 0 and equal to 0 that is ${{f}^{'}}(x)\ge 0$. Also we will use the determinant formula $D={{b}^{2}}-4ac$ to get a relation between a and b as $D\le 0$ for an increasing function. Finally we will apply the product rule to get the answer.

Complete step-by-step answer:
We know that for $f(x)$ to be an increasing function ${{f}^{'}}(x)\ge 0$.
So $f(x)={{x}^{3}}+a{{x}^{2}}+bx+c.....(1)$
Now differentiating equation (1) we get,
${{f}^{'}}(x)=3{{x}^{2}}+2ax+b.....(2)$
Now from equation (2) and definition of increasing function we get,
$\Rightarrow 3{{x}^{2}}+2ax+b\ge 0.....(3)$
So to satisfy equation (3) the discriminant(D) of the quadratic equation should be equal to zero or less than zero that is $D\le 0$.
Now we know the formula of the discriminant is $D={{b}^{2}}-4ac$ and here from equation (3) a is 3, b is 2a and c is b. So substituting these in the discriminant formula we get,
$D=4{{a}^{2}}-12b\le 0.......(4)$
Now rearranging the terms in equation (4) we get,

& \Rightarrow 4{{a}^{2}}\le 12b \\\ & \Rightarrow {{a}^{2}}\le 3b.......(5) \\\ \end{aligned}$$ Now we will draw the table to see which pairs of (a, b) satisfies equation (5), a is| Condition: $${{a}^{2}}\le 3b$$| then b can be| No of pairs (a, b) ---|---|---|--- 1| $$1\le 3b$$| 1, 2, 3, 4, 5, 6| 6 2| $$4\le 3b$$| 2, 3, 4, 5, 6| 5 3| $$9\le 3b$$| 3, 4, 5, 6| 4 4| $$16\le 3b$$| 6| 1 5| $$25\le 3b$$| No value of b holds| 0 6| $$36\le 3b$$| No value of b holds| 0 Hence here we get the total (a, b) pairs as $$6+5+4+1=16$$. And there is no restriction on c. So c can be selected in 6 different ways. So from the product rule the number of ways a, b, c can be selected is $$16\times 6=96$$. Note: Remembering the concept of differentiation and increasing function is the key here. Also knowing the relation of discriminant for an increasing function is important. Here we may get confused about how to get the different ways of selecting c but we need to understand that there is no restriction on c and hence it can be any number between 1 to 6 and hence 6 ways.