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The most general values of x for which $\sin x + \cos x = \mathop {\min }\limits_{a \in {\Bbb R}} \left\{ {1,{a^2} - 4a + 6} \right\}$ are given by:
        $\eqalign{
  & A.\,\,2n\pi ;n \in {\Bbb Z} \cr
  & B.\,\,2n\pi + \dfrac{\pi }{2};n \in {\Bbb Z} \cr
  & C.\,\,n\pi + {( - 1)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{4};n \in {\Bbb Z} \cr
  & D.\,\,2n\pi - \dfrac{\pi }{2};n \in {\Bbb Z} \cr} $

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Last updated date: 22nd Jul 2024
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Answer
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Hint: First we have to find the minimum value, that is we have to simplify the right hand side.Then simplify the equation to find the range of x there so that the equation holds.

Complete step by step solution:
Step1: We know that the value of a perfect square is always non-negative. So the minimum value may be zero.
$\eqalign{
  & {a^2} - 4a + 6 \cr
  & = {(a - 2)^2} + 2 \cr} $
  Then taking the value of the perfect square to zero.
$\eqalign{
  & {a^2} - 4a + 6\,\,is\,\,\operatorname{minimum} \,\,for\,\,a = 2 \cr
  & so,\mathop {\min }\limits_{a \in { R}} \left\{ {{a^2} - 4a + 6} \right\} = {(2 - 2)^2} + 2 = 2 \cr} $.
Step2: Simplifying the right hand side, we get
$Now,\mathop {\,\,\min }\limits_{a \in {l R}} \left\{ {1,{a^2} - 4a + 6} \right\} = 1$

Step3: Now from the equation, we have

 $$\eqalign{
  & \sin x + \cos x = 1 \cr
  & or,\dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{1}{{\sqrt 2 }},\,\,both\,\,side\,\,dividing\,\,by\,\,\sqrt 2 \cr
  & or,\sin x\cos \dfrac{\pi }{4} + \cos x\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} \cr
  & or,\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} \cr
  & or,\sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4} \cr
  & or,x + \dfrac{\pi }{4} = n\pi + {( - 1)^n}.\dfrac{\pi }{4};n \in {\Bbb Z} \cr
  & or,x = n\pi + {( - 1)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{4};n \in {\Bbb Z} \cr} $$
Hence,The most general values of x are here, $n\pi + {( - 1)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{4};n \in {\Bbb Z}$

Therefore,option C) is correct.

Note:
Here we use the formula $\sin x = \sin \alpha \Rightarrow x = n\pi + {( - 1)^n}\alpha ;n \in {\Bbb Z},\,the\,\,set\,\,of\,\,all\,\,\operatorname{integers} $. These are general values of x for which $\sin x = \sin \alpha $. If we want to solve such types of equations, we have to find out general values or all values satisfying that trigonometric equation.