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Question: The equation \(sin x(sin x+cos x)=k\) has real solutions then : a) \(0 \leq k \leq \dfrac{1+ \sqr...

The equation sinx(sinx+cosx)=ksin x(sin x+cos x)=k has real solutions then :
a) 0≤k≤1+220 \leq k \leq \dfrac{1+ \sqrt{2}}{2}
b) 2−3≤k≤2+32- \sqrt{3} \leq k \leq 2+ \sqrt{3}
c) 0≤k≤2−30 \leq k \leq 2- \sqrt{3}
d) (1−22)≤k≤(1+22) (\dfrac{1- \sqrt{2}}{2}) \leq k \leq (\dfrac{1+ \sqrt{2}}{2})

Explanation

Solution

Solve the equation by using trigonometric formulae like sin2x=2sinxcosxsin 2x= 2 sin x cos x and cos2x=1−2sin2xcos 2x= 1- 2 sin ^2x and find the range of k .

Complete step-by-step answer:
Given , the equation sinx(sinx+cosx)=ksin x(sin x+cos x)=k has real solutions .
By solving the equation we get ,
sinx(sinx+cosx)=ksin x(sin x+cos x)=k
⇒sin2x+sinxcosx=k……(i)\Rightarrow sin^2 x + sinx cos x=k……(i)
We know that ,
sin2x=2sinxcosxsin 2x= 2 sin x cos x
⇒sinxcosx=sin2x2…..(ii)\Rightarrow sin x cos x = \dfrac{sin 2x}{2}…..(ii)
Again ,
cos2x=1−2sin2xcos 2x=1-2sin^2x
⇒2sin2x=1−cos2x\Rightarrow 2sin^2x =1-cos 2x
⇒sin2x=12−cos2x2…..(iii)\Rightarrow sin^2x = \dfrac{1}{2}-\dfrac{cos 2x}{2}…..(iii)
Substituting the values of (ii) and (iii) in the equation (i) we get ,
12−cos2x2+sin2x2=k\dfrac{1}{2}-\dfrac{cos 2x}{2}+ \dfrac{sin 2x}{2}=k
⇒1−cos2x+sin2x=2k\Rightarrow 1-cos 2x+sin2x=2k
⇒sin2x−cos2x=2k−1……(iv)\Rightarrow sin2x -cos 2x =2k-1……(iv)
Now , let f(x)=sinx−cosxf(x)=sin x- cos x
=2×12(sinx−cosx)=\sqrt{2} \times \dfrac{1}{ \sqrt{2}}(sinx-cosx)
=2(12sinx−12cosx=\sqrt{2} (\dfrac{1}{ \sqrt{2}}sinx- \dfrac{1}{ \sqrt{2}}cosx
=2(cos45∘sinx−sin45∘cosx)=\sqrt{2} (cos 45 ^{\circ} sin x - sin 45 ^{\circ} cos x )
=2sin(x−π4)=\sqrt{2}sin(x-\dfrac{\pi}{4})
Range of the sin function is [ -1 , 1 ]
The range of f(x) will be [−2,2][- \sqrt{2}, \sqrt{2}]----(v)
Putting the range in equation (iv) we get ,
sin2x−cos2x=2k−1sin2x -cos 2x =2k-1
⇒−2≤2k−1≤2\Rightarrow - \sqrt{2}\leq 2k-1 \leq \sqrt{2}
⇒(1−2)≤2k≤(1+2)\Rightarrow (1- \sqrt{2})\leq 2k \leq (1+\sqrt{2})
Dividing both sides by 2 we get ,
⇒(1−22)≤k≤(1+22)\Rightarrow (\dfrac{1- \sqrt{2}}{2}) \leq k \leq (\dfrac{1+ \sqrt{2}}{2})
The range of k is (1−22)≤k≤(1+22) (\dfrac{1- \sqrt{2}}{2}) \leq k \leq (\dfrac{1+\sqrt{2}}{2})

Thus , the correct option is d) (1−22)≤k≤(1+22) (\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 .