Question

The value of $x$ which satisfy $\sin \theta +\sqrt{3}\cos \theta =6x-x^2-11,X\in R$ is

• A. $2$
• B. $3$
• C. $1$
• D. Infinitely many

Answer 1

Answer: (B)

$3$

$\sin \theta +\sqrt{3}\cos \theta =6x-x^2-11$

$-\sqrt{a^2+b^2}⩽a\sin \theta +b\cos \theta ⩽\sqrt{a^2+b^2}$

So,

$-\sqrt{1+3}⩽\sin \theta +\sqrt{3}\cos \theta ⩽\sqrt{1+3}$

$-2⩽\sin \theta +\sqrt{3}\cos \theta ⩽2$ ................ (1)

Now,

$6x-x^2-11=-(x^2-6x)-11$

$=-(x^2-6x+{\left(\frac{6}{2}\right)}^2-{\left(\frac{6}{2}\right)}^2)-11$

$=-(x^2-6x+9)-11+9$

$=-(x-3{)}^2-2$

$⩽-2$

$6x-x^2-11$$⩽-2$ ................ (2)

From (1) and (2),

$6x-x^2-11=-2$

$x^2-6x+9=0$

$(x-3{)}^2=0$

$x=3$