Which of the following are correct? (A) The function f(x) = 3x+12 is increasing on R. (B) The function f(x) = e^2x is decreasing on R. (C) The…

The correct option is:
$$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0$$(1−x2)dx2d2y​−xdxdy​−a2y=0

Step-by-Step Solution:

1. Rearrange the given expression:
   $$x = \cos\left(\frac{1}{a} \log y\right)$$x=cos(a1​logy)
   $$\cos^{-1}x = \frac{1}{a} \log y$$cos−1x=a1​logy
   $$a \cos^{-1}x = \log y$$acos−1x=logy
   $$y = e^{a \cos^{-1}x}$$y=eacos−1x

2. Differentiate with respect to $x$x:
   $$\frac{dy}{dx} = e^{a \cos^{-1}x} \cdot \frac{-a}{\sqrt{1-x^2}}$$dxdy​=eacos−1x⋅1−x2​−a​
   $$\frac{dy}{dx} = \frac{-ay}{\sqrt{1-x^2}}$$dxdy​=1−x2​−ay​

3. Square both sides to clear the square root:
   $$\left(\frac{dy}{dx}\right)^2 = \frac{a^2y^2}{1-x^2}$$(dxdy​)2=1−x2a2y2​
   $$(1-x^2)\left(\frac{dy}{dx}\right)^2 = a^2y^2$$(1−x2)(dxdy​)2=a2y2

4. Differentiate again with respect to $x$x using the product rule:
   $$(1-x^2) \cdot 2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \cdot (-2x) = a^2 \cdot 2y\frac{dy}{dx}$$(1−x2)⋅2(dxdy​)dx2d2y​+(dxdy​)2⋅(−2x)=a2⋅2ydxdy​

5. Divide the entire equation by $2\frac{dy}{dx}$2dxdy​:
   $$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = a^2y$$(1−x2)dx2d2y​−xdxdy​=a2y
   $$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0$$(1−x2)dx2d2y​−xdxdy​−a2y=0