Let f(x)=∣∣ ∣∣cosxsinxcosxcox2xsin2x2cos2xcos3xsin3x

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Let

f (x) = ∣ ∣∣ ∣ \begin{matrix} c o s x & s i n x & c o s x c o x 2 x & s i n 2 x & 2 c o s 2 x c o s 3 x & s i n 3 x & 3 c o s 3 x \end{matrix} ∣ ∣∣ ∣

, then

f^{'} (\frac{π}{2})

is equal to

Options:

A . 8

B . 6

C . 4

D . 2

Answer: Option C
:
C

f^{'} (x) = ∣ ∣∣ ∣ \begin{matrix} - s i n x & s i n x & c o s x - 2 s i n 2 x & s i n 2 x & 2 c o s 2 x - 3 s i n 3 x & s i n 3 x & 3 c o s 3 x \end{matrix} ∣ ∣∣ ∣ + ∣ ∣∣ ∣ \begin{matrix} c o s x & c o s x & c o s x c o s 2 x & 2 c o s 2 x & 2 c o s 2 x c o s 3 x & 3 c o s 3 x & 3 c o s 3 x \end{matrix} ∣ ∣∣ ∣ + ∣ ∣∣ ∣ \begin{matrix} c o s x & s i n x & - s i n x c o s 2 x & s i n 2 x & - 4 s i n 2 x c o s 3 x & s i n 3 x & - 9 s i n 3 x \end{matrix} ∣ ∣∣ ∣

f^{'} (\frac{π}{2}) = ∣ ∣∣ ∣ \begin{matrix} - 1 & 1 & 0 0 & 0 & - 2 3 & - 1 & 0 \end{matrix} ∣ ∣∣ ∣ + 0 + ∣ ∣∣ ∣ \begin{matrix} 0 & 1 & - 1 - 1 & 0 & 0 0 & - 1 & 9 \end{matrix} ∣ ∣∣ ∣ = 2 (1 - 3) + 0 + 1 (9 - 1) = - 4 + 8 = 4

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