Question

The most general values of x for which $\sin x + \cos x = {\min _{a \in R}}\\{ 1,{a^2} - 4a + 6\\}$are given by
$(a){\text{ 2n}}\pi \\\ (b){\text{ 2n}}\pi + \dfrac{\pi }{2} \\\ (c){\text{ n}}\pi + {( - 1)^n}\dfrac{\pi }{4} - \dfrac{\pi }{4} \\\ (d){\text{ none of these}} \\\$

Answer 1

Hint – In this question first of all convert the quadratic ${a^2} - 4a + 6$ into a perfect square form so as to determine that which amongst 1 or ${a^2} - 4a + 6$ is minimum. Then equate it to the L.H.S part that is $\sin x + \cos x$, then try and convert this into the standard trigonometric form of $\sin (A + B) = \sin A\cos B + \cos A\sin B$, this will get the value of x, consider the general solution to get the right option.

Complete step-by-step answer:

${a^2} - 4a + 6$
Now make this complete square by add and subtract by half of square of coefficient of (a) so we have,
$\Rightarrow {a^2} - 4a + 6 + {\left( {\dfrac{4}{2}} \right)^2} - {\left( {\dfrac{4}{2}} \right)^2}$
Now simplify the above equation we have,
$\Rightarrow {a^2} - 4a + 6 + 4 - 4$
$\Rightarrow {a^2} - 4a + 4 + 6 - 4$
$\Rightarrow {\left( {a - 2} \right)^2} + 2$
Now as we know square term is always positive or zero it cannot be negative.
Therefore, $\left[ {{{\left( {a - 2} \right)}^2} + 2} \right] \geqslant 2$
So, \min \left\\{ {1,{a^2} - 4a + 6} \right\\} = 1
Therefore, the given equation becomes
$\Rightarrow \sin x + \cos x = 1$
Now multiply and divide by $\sqrt 2$ in L.H.S we have,
$\Rightarrow \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} \times \sin x + \dfrac{1}{{\sqrt 2 }} \times \cos x} \right) = 1$
Now as we know that $\sin {45^0} = \cos {45^0} = \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow \sqrt 2 \left( {\cos {{45}^0} \times \sin x + \sin {{45}^0} \times \cos x} \right) = 1$
$\Rightarrow \left( {\cos {{45}^0} \times \sin x + \sin {{45}^0} \times \cos x} \right) = \dfrac{1}{{\sqrt 2 }} = \sin {45^0} = \sin \dfrac{\pi }{4}$, $\left[ {{{45}^0} = \dfrac{\pi }{4}} \right]$
Now as we know that $\sin (A + B) = \sin A\cos B + \cos A\sin B$ so use this property in above equation we have,
$\Rightarrow \sin \left( {x + {{45}^0}} \right) = \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4}$.............. (1)
Now as we know sin is positive in first and second quadrant as
$\sin \left( {\pi - \theta } \right) = \sin \theta$ and $\sin \left( {2\pi + \theta } \right) = \sin \theta$
So in general we can say that
$\Rightarrow \sin \left( {n\pi + {{\left( { - 1} \right)}^n}\theta } \right) = \sin \theta$, where $n \in N$
So use this property in equation (1) we have,
$\Rightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \left( {n\pi + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right)$
Now cancel out sin from both sides we have,
$\Rightarrow \left( {x + \dfrac{\pi }{4}} \right) = \left( {n\pi + {{\left( { - 1} \right)}^n}\dfrac{\pi }{4}} \right)$
$\Rightarrow x = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{4} - \dfrac{\pi }{4}$
So this is the required most general solution of the given equation.
Hence option (C) is correct.

Note – If a quadratic is not getting converted into a perfect square form initially then the trick is to add and subtract the square of the half of the coefficient of term x in a quadratic equation of the form $a{x^2} + bx + c = 0$. It is advised to remember basic trigonometric identities as it helps saving a lot of time while solving problems of this kind.