Skip to main content
Back to Differentiation
JEE Main 2023
Differentiation
Differentiation
Easy

Question

If f(x)=2cos4x2sin4x3+sin22x3+2cos4x2sin4xsin22x2cos4x3+2sin4xsin22x,f(x)=\left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x \end{array}\right|, then 15f(0)=\frac{1}{5} f^{\prime}(0)= is equal to :

Options

Solution

2cos4x2sin4x3+sin22x3+2cos4x2sin4xsin22x2cos4x3+2sin24xsin22xR2R2R1,R3R3R12cos4x2sin4x3+sin22x303033f(x)=45f(x)=0\begin{aligned} & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x \end{array}\right| \\ & \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\ & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{array}\right| \\ & \mathrm{f}(\mathrm{x})=45 \\ & \mathrm{f}^{\prime}(\mathrm{x})=0 \\ & \end{aligned}

Practice More Differentiation Questions

View All Questions