Question

Can anyone answer this ?

Answer 1

$\displaystyle\large\boxed{\sf{(D)\quad\!}\bf{6}}$

Given,

$\displaystyle\longrightarrow\sf{(a+b)^2=c^2+ab}$

$\displaystyle\longrightarrow\sf{a^2+2ab+b^2=c^2+ab}$

$\displaystyle\longrightarrow\sf{a^2+ab+b^2=c^2}$

But by cosine rule,

$\displaystyle\longrightarrow\sf{a^2+ab+b^2=a^2+b^2-2ab\cos C}$

$\displaystyle\longrightarrow\sf{\cos C=-\dfrac{1}{2}}$

This implies C is an obtuse angle, and therefore,

$\displaystyle\Longrightarrow\sf{C=120^{\circ}}$

Now,

$\displaystyle\longrightarrow\sf{8\cos A\cos B=8\cdot\dfrac{1}{2}[\cos(A+B)+\cos(A-B)]}$

$\displaystyle\longrightarrow\sf{8\cos A\cos B=4[\cos(A+B)+\cos(A-B)]\quad\quad\dots(1)}$

But since A, B and C are angles in a triangle,

$\displaystyle\longrightarrow\sf{A+B+C=180^{\circ}}$

$\displaystyle\longrightarrow\sf{A+B=180^{\circ}-C}$

$\displaystyle\longrightarrow\sf{\cos(A+B)=\cos(180^{\circ}-C)}$

$\displaystyle\longrightarrow\sf{\cos(A+B)=\dfrac{1}{2}\quad\quad\left[\because\ C=120^{\circ}\right]}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{8\cos A\cos B=4\left[\dfrac{1}{2}+\cos(A-B)\right]}$

$\displaystyle\longrightarrow\sf{8\cos A\cos B=2+4\cos(A-B)\quad\quad\dots(2)}$

To get maximum value of $\displaystyle\sf{8\cos A\cos B,\quad\cos(A-B)}$ should be maximum and hence $\displaystyle\sf{|A-B|}$ should be minimum.

[This is because value of $\displaystyle\sf{\cos\theta}$ decreases on increasing the magnitude of $\displaystyle\sf{\theta,}$ but for $\displaystyle\sf{\theta\in\left[2n\pi,\ (2n+1)\pi\right],\ n\in\mathbb{Z}.}$ ]

Thus the minimum possible value for $\displaystyle\sf{|A-B|}$ is 0 and hence $\displaystyle\sf{\max(\cos(A-B))=1.}$

[Remember, $\displaystyle\sf{|x|\geq0.}$ ]

It also implies that $\displaystyle\sf{A=B=30^{\circ}.}$

Therefore (2) will be,

$\displaystyle\longrightarrow\sf{\max(8\cos A\cos B)=2+4\times1}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\max(8\cos A\cos B)=\bf{6}}}}$

Hence the answer is $\displaystyle\boxed{\sf{(D)\quad\!}\bf{6}}$