Question

If the function $f(x)={{x}^{3}}-6{{x}^{2}}+ax+b$ defined on [1,3], satisfies the Rolle’s Theorem for $c=\dfrac{2\sqrt{3}+1}{\sqrt{3}}$ , then
(a) $a=11,\ b=6$
(b) $a=-11,\ b=6$
(c) \[~a=11,\text{ }b~\in R\]
(d) None of these

Answer 1

Hint: We may need to check conditions required for a function to satisfy Rolle’s Theorem and form equations to find values of a and b.

Complete step-by-step answer:

First we need to recall Rolle’s Theorem,
Rolle’s Theorem says that if-
1.f(x) is continuous in [a,b]
2.f(x) is differentiable in (a,b)
3.f(a)=f(b)
Then there is at least one ‘c’ for which f’(c)=0
Our question says that the given function satisfies Rolle’s Theorem means the above conditions are already satisfied in the given interval [1,3]. Now we need to make use of these conditions to form equations.
As the given function is a polynomial therefore, the function is differentiable and continuous everywhere. The first two conditions are satisfied for any interval of x for the polynomial.
Since, it is given that the function $f(x)={{x}^{3}}-6{{x}^{2}}+ax+b$ satisfies Rolle’s Theorem in the interval [1,3] then it must satisfy the third condition also. Therefore, we have
Therefore, f(1)=f(3) $\Rightarrow {{1}^{3}}-6({{1}^{2}})+a\cdot 1+b={{3}^{3}}-6({{3}^{2}})+a\cdot 3+b$
On further simplification we have,
$\Rightarrow 1-6+a+b=27-54+3a+b$
$\Rightarrow -5+a+b=-27+3a+b$
Cancelling b both sides and simplifying to get ‘a’ we have,
$\Rightarrow -2a=-22$
$\Rightarrow a=11$
Now we have the value of ‘a’. Now we the last condition to make use of i.e. $f'(c)=0$
For this we need evaluate $f'(x)$ .
$f(x)={{x}^{3}}-6{{x}^{2}}+ax+b$
Differentiating both sides we have,
$f'(x)=\dfrac{d}{dx}({{x}^{3}})-6\dfrac{d}{dx}({{x}^{2}})+a\dfrac{d}{dx}(x)+\dfrac{d}{dx}(b)$
           $=3{{x}^{2}}-6.2x+a.1+0$
           $=3{{x}^{2}}-12x+a$
Therefore we have, $f'(x)=3{{x}^{2}}-12x+a$ ……….(i)
From the question we have $c=\dfrac{2\sqrt{3}+1}{\sqrt{3}}$ .
Now we have,
$f'(c)=0$
But now we will not proceed with this equation because we can see that if we substitute c in equation (i) we will have everything as constant and no variable since ‘b’ cancels out after differentiation and is already known. Therefore, there is no need to proceed. This also implies that for any value of ‘b’ the equation will satisfy i.e. $b\in R$ .
Hence, our answer is option (c).

Note: We must always proceed with the question sequentially otherwise it will only get chaotic.
We could have skipped the third condition used $f'(c)=0$ to find ‘a’. In that case the calculations would have been long. And at last we could also have proceeded without noticing that we are not getting the value of ‘b’ and again solved to get ‘a’ wasting our time.
But we can also check by solving $f'(c)=0$ to see if the value of ‘a’ is consistent or not.