Question
Answers

The equation $sin x(sin x+cos x)=k$ has real solutions then :
a) $0 \leq k \leq \dfrac{1+ \sqrt{2}}{2}$
b) $2- \sqrt{3} \leq k \leq 2+ \sqrt{3}$
c) $0 \leq k \leq 2- \sqrt{3}$
d) $ (\dfrac{1- \sqrt{2}}{2}) \leq k \leq (\dfrac{1+ \sqrt{2}}{2})$

Answer Verified Verified
Hint: Solve the equation by using trigonometric formulae like $sin 2x= 2 sin x cos x$ and $cos 2x= 1- 2 sin ^2x$ and find the range of k .

Complete step-by-step answer:
Given , the equation $sin x(sin x+cos x)=k$ has real solutions .
By solving the equation we get ,
$sin x(sin x+cos x)=k$
$\Rightarrow sin^2 x + sinx cos x=k……(i)$
We know that ,
$sin 2x= 2 sin x cos x$
$\Rightarrow sin x cos x = \dfrac{sin 2x}{2}…..(ii)$
Again ,
$cos 2x=1-2sin^2x$
$\Rightarrow 2sin^2x =1-cos 2x$
$\Rightarrow sin^2x = \dfrac{1}{2}-\dfrac{cos 2x}{2}…..(iii)$
Substituting the values of (ii) and (iii) in the equation (i) we get ,
$\dfrac{1}{2}-\dfrac{cos 2x}{2}+ \dfrac{sin 2x}{2}=k$
$\Rightarrow 1-cos 2x+sin2x=2k $
$\Rightarrow sin2x -cos 2x =2k-1……(iv)$
Now , let $f(x)=sin x- cos x$
$=\sqrt{2} \times \dfrac{1}{ \sqrt{2}}(sinx-cosx)$
$=\sqrt{2} (\dfrac{1}{ \sqrt{2}}sinx- \dfrac{1}{ \sqrt{2}}cosx$
$=\sqrt{2} (cos 45 ^{\circ} sin x - sin 45 ^{\circ} cos x )$
$=\sqrt{2}sin(x-\dfrac{\pi}{4})$
Range of the sin function is [ -1 , 1 ]
The range of f(x) will be $[- \sqrt{2}, \sqrt{2}]$----(v)
Putting the range in equation (iv) we get ,
$ sin2x -cos 2x =2k-1$
$\Rightarrow - \sqrt{2}\leq 2k-1 \leq \sqrt{2} $
$\Rightarrow (1- \sqrt{2})\leq 2k \leq (1+\sqrt{2}) $
Dividing both sides by 2 we get ,
$\Rightarrow (\dfrac{1- \sqrt{2}}{2}) \leq k \leq (\dfrac{1+ \sqrt{2}}{2})$
The range of k is $ (\dfrac{1- \sqrt{2}}{2}) \leq k \leq (\dfrac{1+\sqrt{2}}{2})$

Thus , the correct option is d) $ (\dfrac{1- \sqrt{2}}{2}) \leq k \leq (\dfrac{1+ \sqrt{2}}{2})$.

Note: Before starting the problem , the student needs to be properly acquainted with the trigonometric formulae . Now , after using the formulae , the equation can be easily solved . The student needs to find the range from the trigonometric functions in the equation (iv) . After that , they can put the values of the range and find the range of k . Students go wrong in understanding the problem and choosing the required formulae . Once they understand the concept properly , the sum can be solved easily .
Bookmark added to your notes.
View Notes
×