Question

If the function f given by $f(x)={{x}^{3}}-3(a-2){{x}^{2}}+3ax+7$ , for some $a\in R$ is increasing in $\text{(0,1]}$ and decreasing in $\text{[1,5)}$, then a root of the equation $\dfrac{f(x)-14}{{{(x-1)}^{2}}}=0,(x\ne 1)$ is

(a) 6

(b) 5

(c) 7

(d) -7

Answer 1

We should read the question backwards and proceed. Simplify the second equation first and then substitute the value of f(x) in the second equation. Find the value of a from the data given in the question and then substitute this value of a in the equation derived by substituting f(x) to obtain the required root.

Complete step-by-step answer:

First we need to write the equation in a simplified way to proceed and try to find its roots.

$\dfrac{f(x)-14}{{{(x-1)}^{2}}}=0,(x\ne 1)$

After substituting f(x) we can write,

$\dfrac{{{x}^{3}}-3(a-2){{x}^{2}}+3ax+7-14}{{{(x-1)}^{2}}}=0$

$\Rightarrow \dfrac{{{x}^{3}}-3(a-2){{x}^{2}}+3ax-7}{{{(x-1)}^{2}}}=0$ ……………….(i)

We need to find the roots of the above equation. For that we will need to find the value of ‘a’ from the data given in the question.

The question says that the function is increasing in (0,1] and decreasing in [1,5).

Now we need to pay close attention to this statement and point x=1. It says that the function f(x) is changing its nature at point x=1. And we know that it happens only at the point of local maxima or minima. And since its nature is first increasing then decreasing from left to right, we can say that x=1 is a point of local maxima as it attains its peak value when changing its nature.

We know that at a point of local maxima or minima $f'(x)=0$ .

Therefore we can write, $f'(1)=0$

Now we need $f'(x)$

$f(x)={{x}^{3}}-3(a-2){{x}^{2}}+3ax+7$

Differentiating both sides we have,

$f'(x)=3{{x}^{2}}-3(a-2).2x+3a.1+0$

On simplifying we get,

$f'(x)=3{{x}^{2}}-6(a-2)x+3a$

Substituting x=1 we have,

$f'(1)={{3.1}^{2}}-6(a-2).1+3a$

$\Rightarrow f'(1)=3-6(a-2)+3a$

We had, $f'(1)=0$ which implies

$3-6(a-2)+3a=0$

On simplification we get,

$3-6a+12+3a=0$

$\Rightarrow -3a+15=0$

Adding 3a both sides we have,

$3a=15$

Dividing both sides by 3 we have,

$a=\dfrac{15}{3}=5$

Hence, we have a=5

$\dfrac{{{x}^{3}}-3(a-2){{x}^{2}}+3ax-7}{{{(x-1)}^{2}}}=0$

Now we substitute the value of ‘a’ in equation (i)

$\dfrac{{{x}^{3}}-3(5-2){{x}^{2}}+3.5.x-7}{{{(x-1)}^{2}}}=0$

On simplifying we get,

$\dfrac{{{x}^{3}}-9{{x}^{2}}+15x-7}{{{(x-1)}^{2}}}=0$

To find the roots of this equation the numerator must be equal to 0 and the denominator must not be equal to 0. So, we can write,

${{x}^{3}}-9{{x}^{2}}+15x-7=0$ ………(ii)

Therefore we need to find a solution to this equation.

By substituting x=1 in LHS we have,

${{1}^{3}}-9.1+15.1-7=0$

Therefore x=1 is a root of this cubic polynomial.

Now if we divide equation (ii) with (x-1) we will have a quadratic polynomial whose roots we can easily find.

Dividing equation (ii) with (x-1) we have,

$\dfrac{{{x}^{3}}-9{{x}^{2}}+15x-7}{x-1}={{x}^{2}}-8x+7$

Therefore, now we have

$(x-1)({{x}^{2}}-8x+7)=0$

Since $x\ne 1$ equation (i) becomes undefined.

${{x}^{2}}-8x+7=0$

Now we split the middle term to do factorisation,

${{x}^{2}}-7x-x+7=0$

Taking x is common from the first two terms and -1 from last two terms we have,

$x(x-7)-1(x-7)=0$

Taking (x-7) we have,

$(x-1)(x-7)=0$

Again $x\ne 1$. Therefore we have,

$x=7$ is the only root of the given equation.

Hence, option (c) is correct.

Note: The major difficulty in the question is finding the value of ‘a’ for that we must know how the nature of the graph is changing and when we are having local minima and maxima. We can also try to graph the function and try to understand from there.