Question

If y = a sin x + b cos x, then \[{{y}^{2}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}\] is a
(a) Function of x
(b) Function of y
(c) Function of x and y
(d) Constant

Answer 1

Hint: First of all, find the value of \[\dfrac{dy}{dx}\] by using \[\dfrac{d}{d\theta }\left( \sin \theta \right)=\cos \theta \text{ and }\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta \]. Now, substitute the value of y and \[\dfrac{dy}{dx}\] in \[E={{y}^{2}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}\] and simplify it by using \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta \]. Finally, observe the final value and mark the option accordingly.
Complete step-by-step answer:
We are given that y = a sin x + b cos x. We have to find the value of \[{{y}^{2}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}\] and accordingly mark the answer. Let us consider the expression of y given in the question.
\[y=a\sin x+b\cos x....\left( i \right)\]
We know that, \[\dfrac{d}{d\theta }\left( \sin \theta \right)=\cos \theta \text{ and }\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta \]. By using this and differentiating both sides of the above equation, with respect to x, we get,
\[\dfrac{dy}{dx}=a\dfrac{d}{dx}\left( \sin x \right)+b\dfrac{d}{dx}\left( \cos x \right)\]
\[\dfrac{dy}{dx}=a\cos x+b\left( -\sin x \right)\]
\[\dfrac{dy}{dx}=a\cos x-b\sin x....\left( ii \right)\]
Let us now find the value of
\[E={{y}^{2}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}\]
By substituting the values of y and \[\dfrac{dy}{dx}\] from equation (i) and (ii) respectively, we get,
\[E={{\left( a\sin x+b\cos x \right)}^{2}}+{{\left( a\cos x-b\sin x \right)}^{2}}\]
We know that
\[{{\left( x\pm y \right)}^{2}}={{x}^{2}}+{{y}^{2}}\pm 2xy\]
By using this in the above equation, we get,
\[E={{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x+2\left( a\sin x \right)\left( b\cos x \right)+{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x-2\left( a\cos x \right)\left( b\sin x \right)\]
By rearranging the terms of the above equation, we get,
\[E=\left( {{a}^{2}}{{\sin }^{2}}x+{{a}^{2}}{{\cos }^{2}}x \right)+\left( {{b}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x \right)+2ab\sin x\cos x-2ab\sin x\cos x\]
\[E={{a}^{2}}\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)+{{b}^{2}}\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\]
We know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. By using this, we get,
\[E={{a}^{2}}\left( 1 \right)+{{b}^{2}}\left( 1 \right)\]
\[E={{a}^{2}}+{{b}^{2}}\]
So, we get the value of \[{{y}^{2}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}\text{ as }{{a}^{2}}+{{b}^{2}}\] which is constant as a and b are constants.
Therefore, option (d) is the correct answer.

Note: In this question, some students write the differentiation of cos x as sin x while actually, it is – sin x. So, this must be taken care of as it will lead to the wrong answer. Also, in the above question, many concepts of Mathematics like Calculus, Algebra, trigonometry, etc. are getting used simultaneously. So, students are required to practice multiple concepts time by time to be able to do these types of questions easily. Also, note that in the above question, we must assume that a and b are constants while x and y are variable as \[\dfrac{dy}{dx}\] is used.