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Question: If \(f\) is defined in \([1,3]\) by \(f(x)={{x}^{3}}+b{{x}^{2}}+ax\) , such that \(f(1)-f(3)=0\) and...

If ff is defined in [1,3][1,3] by f(x)=x3+bx2+axf(x)={{x}^{3}}+b{{x}^{2}}+ax , such that f(1)−f(3)=0f(1)-f(3)=0 and f′(c)=0f'(c)=0 where c=2+13c=2+\dfrac{1}{\sqrt{3}} , then (a,b)=(a,b)=
A. (−6,11)(-6,11)
B. (2−13,2+13)\left( 2-\dfrac{1}{\sqrt{3}},2+\dfrac{1}{\sqrt{3}} \right)
C. (11,−6)(11,-6)
D. (6,11)(6,11)

Explanation

Solution

Hint: We can see that there are two unknowns in the question. To find these unknowns we need two equations which are given in the question. Upon solving these two equations, we get the values of a and b. Therefore we can easily find a and b in this question.

“Complete step-by-step answer:”
First of all we need to make use of the equation f(1)−f(3)=0f(1)-f(3)=0 . We substitute 1 place of x to find f(1) and the same for f(3).
We have, f(x)=x3+bx2+axf(x)={{x}^{3}}+b{{x}^{2}}+ax substituting 1 we get, f(1)=13+b(12)+a⋅1f(1)={{1}^{3}}+b({{1}^{2}})+a\cdot 1 similarly we have, f(3)=33+b(32)+a⋅3f(3)={{3}^{3}}+b({{3}^{2}})+a\cdot 3
Now we compute f(1)−f(3)=0⇒f(1)=f(3)f(1)-f(3)=0\Rightarrow f(1)=f(3)
f(1)=f(3) ⇒13+b(12)+a⋅1=33+b(32)+a⋅3 \begin{aligned} & f(1)=f(3) \\\ & \Rightarrow {{1}^{3}}+b({{1}^{2}})+a\cdot 1={{3}^{3}}+b({{3}^{2}})+a\cdot 3 \\\ \end{aligned}
On simplifying we get,
1+b+a=27+9b+3a1+b+a=27+9b+3a
On further simplification we have
8b+2a=−26 ⇒4b+a=−13 \begin{aligned} & 8b+2a=-26 \\\ & \Rightarrow 4b+a=-13 \\\ \end{aligned}
Therefore we can write,
a=−13−4ba=-13-4b …(i)
Now we will find f′(x)f'(x) .
f′(x)=ddx(x3+bx2+ax) =3⋅x2+b⋅2⋅x+a⋅1 \begin{aligned} & f'(x)=\dfrac{d}{dx}({{x}^{3}}+b{{x}^{2}}+ax) \\\ & =3\cdot {{x}^{2}}+b\cdot 2\cdot x+a\cdot 1 \\\ \end{aligned}
On simplifying we have,
f′(x)=3x2+2bx+af'(x)=3{{x}^{2}}+2bx+a …(ii)
We are given that, f′(c)=0f'(c)=0 where c=2+13c=2+\dfrac{1}{\sqrt{3}} .
f′(c)=0⇒f′(2+13)=0f'(c)=0\Rightarrow f'\left( 2+\dfrac{1}{\sqrt{3}} \right)=0
Substituting ‘c’ in equation (ii) we have,
3(2+13)2+2b(2+13)+a=03{{\left( 2+\dfrac{1}{\sqrt{3}} \right)}^{2}}+2b\left( 2+\dfrac{1}{\sqrt{3}} \right)+a=0
On squaring we have,
3(4+13+43)+2b(2+13)+a=0 ⇒3(133+43)+2b(2+13)+a=0 \begin{aligned} & 3\left( 4+\dfrac{1}{3}+\dfrac{4}{\sqrt{3}} \right)+2b\left( 2+\dfrac{1}{\sqrt{3}} \right)+a=0 \\\ & \Rightarrow 3\left( \dfrac{13}{3}+\dfrac{4}{\sqrt{3}} \right)+2b\left( 2+\dfrac{1}{\sqrt{3}} \right)+a=0 \\\ \end{aligned}
Substituting ‘a’ from equation (i) we have,
3(133+43)+2b(2+13)+(−13−4b)=0 ⇒2b(2+13)−4b=13−3(133+43) ⇒2b3=−123 ⇒2b=−12 ⇒b=−6 \begin{aligned} & 3\left( \dfrac{13}{3}+\dfrac{4}{\sqrt{3}} \right)+2b\left( 2+\dfrac{1}{\sqrt{3}} \right)+(-13-4b)=0 \\\ & \Rightarrow 2b\left( 2+\dfrac{1}{\sqrt{3}} \right)-4b=13-3\left( \dfrac{13}{3}+\dfrac{4}{\sqrt{3}} \right) \\\ & \Rightarrow \dfrac{2b}{\sqrt{3}}=-\dfrac{12}{\sqrt{3}} \\\ & \Rightarrow 2b=-12 \\\ & \Rightarrow b=-6 \\\ \end{aligned}
Now we can calculate ‘a’ from equation (i)
a=−13−4ba=-13-4b
Substituting the value of ‘b’ we have,
a=−13−4×−6 ⇒a=−13+24 ⇒a=11 \begin{aligned} & a=-13-4\times -6 \\\ & \Rightarrow a=-13+24 \\\ & \Rightarrow a=11 \\\ \end{aligned}
Therefore (a,b)=(11,−6)(a,b)=(11,-6)
Hence, option C is the correct option.

Note: The information given in the question hints us to consider 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.
The question could have been asked in some other way also. If the value of any one of ‘a’ and ‘b’ had been given then the question could have asked to find the value of ‘c’ and the remaining one of ‘a’ and ‘b’.