Question

$\frac{d^{20}(2\cos x\cos 3x)}{d{x}^{20}}=$

• A. $2^{20}(\cos 2x–2^{20}\cos 4x)$
• B. $2^{20}(\cos 2x+2^{20}\cos 4x)$
• C. $2^{20}(\sin 2x+2^{20}\sin 4x)$
• D. $2^{20}(\sin 2x–2^{20}\sin 4x)$

Answer 1

The correct option is B

$2^{20}(\cos 2x+2^{20}\cos 4x)$

Explanation for the correct option:

Step 1.simplify the given function :

Let, $y=2\cos \left(x\right)\cos \left(3x\right)$

$\begin{array}{rcl}& =& 2\cos \left(x\right)\cos \left(3x\right)\\ & =& \cos \left(3x-x\right)+\cos \left(3x+x\right)\\ & =& \cos \left(2x\right)+\cos \left(4x\right)\end{array}$ $\left[\mathbf{\because }\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\left(A\right)\mathit{c}\mathit{o}\mathit{s}\left(B\right)\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\left(A+B\right)\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\left(A-B\right)\right]$

Step 2. Differentiate with respect to ‘$\mathit{x}$’

$\begin{array}{rcl}\frac{dy}{dx}& =& -2\sin \left(2x\right)-4\sin \left(4x\right)\\ & =& -2\sin \left(2x\right)-4\sin \left(4x\right)\end{array}$

Step 3. Again differentiating with respect to ‘$\mathit{x}$’

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& -4\cos \left(2x\right)-16\cos \left(4x\right)\\ & =& -(2^2\cos \left(2x\right)+4^2\cos \left(4x\right))\\ & =& -4\left(\cos \left(2x\right)+2^2\cos \left(4x\right)\right)\end{array}$

$\begin{array}{rcl}\frac{d^{3}y}{d{x}^3}& =& -4\left(-2\sin \left(2x\right)-16\sin \left(4x\right)\right)\\ & =& 8\sin \left(2x\right)+64\sin \left(4x\right)\\ & =& 2^3\sin \left(2x\right)+4^3\sin \left(4x\right)\\ & =& 4^{3}sin\left(4\mathrm{x}\right)+2^{3}sin\left(2\mathrm{x}\right)\\ \frac{{\mathrm{d}}^4\mathrm{y}}{{\mathrm{dx}}^4}& =& 4^4\cos \left(4\mathrm{x}\right)+2^4\cos \left(2\mathrm{x}\right)\end{array}$

Similarly

$\frac{d^{4n}y}{d{x}^{4n}}=4^{4n}\cos \left(4x\right)+2^{4n}\cos \left(2x\right)$

Put $n=5$

$\begin{array}{rcl}\therefore \frac{d^{20}y}{d{x}^{20}}& =& 4^{20}\cos \left(4x\right)+2^{20}\cos \left(2x\right)\\ & =& 2^{20}\left(2^{20}\cos \left(4x\right)+\cos \left(2x\right)\right)\\ & \mathbf{=}& 2^{20}(\cos \left(2\mathrm{x}\right)+2^{20}\cos \left(4\mathrm{x}\right))\end{array}$

$\therefore \frac{{\mathbf{d}}^{\mathbf{20}}\mathbf{(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{4}\mathbf{x}\mathbf{)}}{\mathbf{d}{\mathbf{x}}^{\mathbf{20}}}\mathbf{=}{\mathbf{2}}^{\mathbf{20}}\mathbf{(}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathbf{+}{\mathbf{2}}^{\mathbf{20}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{4}\mathit{x}\mathbf{)}$

Hence, option (B) is correct.