Question

In a $△ABC$, $a,b,c$ are the sides opposite to the angles $A,B,C$ respectively. The corresponding values of sides are $a=x^2,b=4x^2+x-1$ and $c=\frac{x}{3}$, where $x\in R$. If $\sin A+\sin C=2\sin B$, then $x$ can be expressed as $\frac{\alpha }{\beta }(0<\alpha ,\beta <10).$
A function is defined as $f\left(t\right)=\{\begin{array}{c}{k}_1\alpha +tt>0\\ k_2\beta +t+2t\le 0\end{array},$
where $k_1,k_2$ are integers.
If the function is continous for all $x\in R$, then $k_1+k_2$ can be equal to

• A. $33$
• B. $34$
• C. $35$
• D. $36$

Answer 1

The correct option is B $34$

We know that, if $\sin A+\sin C=2\sin B$, then $a+c=2b$

$\Rightarrow x^2+\frac{x}{3}=2(4x^2+x-1)\Rightarrow 21x^2+5x-6=0\Rightarrow (7x-3)(3x+2)=0$
$\Rightarrow x=\frac{3}{7}$ (since, sides can't be negative)

So, $x=\frac{3}{7}=\frac{\alpha }{\beta }$
$\Rightarrow \alpha =3,\beta =7(\because 0<\alpha ,\beta <10)$
Now, the function can be written as,
$f\left(t\right)=\{\begin{array}{c}3k_1+tt>0\\ 7k_2+t+2t\le 0\end{array}$
If the function is continous then,
$3k_1=7k_2+2\Rightarrow k_1=\frac{7k_2+2}{3}$
So, $k_2=3n+1$, where $n$ is an integer.
$\Rightarrow k_1=7n+3$
$\therefore k_1+k_2=10n+4$

Hence, the only possible solution from the options is $34$.