Question

If there are two linear functions $f$ and $g$ which map $[1,2]$ on $[4,6]$ and in a $△ABC,c=f\left(1\right)+g\left(1\right)$ and $a$ is the maximum value of $r^2,$ where $r$ is the distance of a variable point on the curve $x^2+y^2-xy=10$ from the origin, then $\sin A:\sin C$ is

• A. $1:2$
• B. $2:1$
• C. $1:1$
• D. none of these

Answer 1

Answer: (C)

$1:1$

Let linear function is $F\left(x\right)=Ax+B$
$\because [1,2]\to [4,6]$
$\Rightarrow F\left(1\right)=4\Rightarrow A+B=4$
$\Rightarrow F\left(2\right)=6\Rightarrow 2A+B=6$
On solving, $2A+B-A-B=6-4\Rightarrow A=2$
Substituting $A=2$ in $A+B=4$ we get
$B=4-A=4-2=2$
$\therefore A=2,B=2$
Then, one function is $F\left(x\right)=2x+2=f\left(x\right)$(say)
$F\left(1\right)=6\Rightarrow A+B=6$
$F\left(2\right)=4\Rightarrow 2A+B=4$
On solving, we get
$2A+B-A-B=4-6\Rightarrow A=-2$
Substituting $A=-2$ in $A+B=6$ we get
$B=6-A=6-(-2)=6+2=8$
$\therefore A=-2,B=8$
then other function is $F\left(x\right)=-2x+8=g\left(x\right)$(say)
$\therefore c=f\left(1\right)+g\left(1\right)=4+6=10$
Now, $x^2+y^2-xy=10$
$\Rightarrow \frac{{(x-\frac{y}{2})}^2}{{\left(\sqrt{10}\right)}^2}+\frac{y^2}{{\left(\frac{\sqrt{10}}{\sqrt{3}}\right)}^2}=1$
is an ellipse whose centre is $(0,0)$
Maximum distance from origin on any point on ellipse$=$Semi-major axis$=\sqrt{10}$
$\therefore r=\sqrt{10}$
Then, $a=r^2=10$
$\because a=c=10$
$\therefore \sin A:\sin C=1:1$ (using sine rule)