Question

If $f(x) = \begin{vmatrix} \cos{(x + x^2)} & \sin{(x + x^2)} & -\cos{(x + x^2)} \\ \sin{(x - x^2)} & \cos{(x - x^2)} & \sin{(x - x^2)} \\ \sin{2x} & 0 & \sin{2x^2} \end{vmatrix}$, then

• A. f(-2) = 0
• B. f'(-1/2) = 0
• C. f' (-1) = -2
• D. f ''(0) = 4

Answer 1

Answer: (B), (C), (D)

f'(-1/2) = 0, f' (-1) = -2, f ''(0) = 4

Answer 2

We start with

$$f(x)=\begin{vmatrix} \cos(x+x^2) & \sin(x+x^2) & -\cos(x+x^2) \\ \sin(x-x^2) & \cos(x-x^2) & \sin(x-x^2) \\ \sin2x & 0 & \sin2x^2 \end{vmatrix}.$$

Let

$$u=x+x^2,\quad v=x-x^2,\quad\text{and write }A=\cos u,\;B=\sin u,\;C=\sin v,\;D=\cos v,\;E=\sin2x,\;F=\sin2x^2.$$

Then the matrix becomes

$$\begin{pmatrix} A & B & -A\\[1mm] C & D & C\\[1mm] E & 0 & F \end{pmatrix}.$$

Perform the column operation: replace the 3rd column by (Column 1 + Column 3). Then

$$C_3\to C_1+C_3:\quad \begin{pmatrix} A & B & A-A=0\\[1mm] C & D & C+C=2C\\[1mm] E & 0 & E+F \end{pmatrix}.$$

Now expanding the determinant along the first row (since the 1st row, 3rd entry is 0):

$$f(x)=A\det\begin{pmatrix}D & 2C\\[0.5mm]0 & E+F\end{pmatrix}-B\det\begin{pmatrix} C & 2C\\[0.5mm] E & E+F\end{pmatrix}.$$

This gives

$$f(x)= A\bigl[D(E+F)\bigr]-B\Bigl[C(E+F)-2C\,E\Bigr] = AD(E+F)-BC(F-E).$$

Now using the trigonometric identities

$$\cos u\,\cos v-\sin u\,\sin v=\cos(u+v),\quad \cos u\,\cos v+\sin u\,\sin v=\cos(u-v),$$

we write

$$f(x)= E\bigl[\cos u\,\cos v+\sin u\,\sin v\bigr]+F\bigl[\cos u\,\cos v-\sin u\,\sin v\bigr] = E\cos(u-v)+ F\cos(u+v).$$

But note that

$$u-v=(x+x^2)-(x-x^2)=2x^2,\quad u+v=(x+x^2)+(x-x^2)=2x.$$

So we have

$$f(x)=\sin2x\cos2x^2+\sin2x^2\cos2x.$$

Using the sine addition formula

$$\sin\alpha\,\cos\beta+\cos\alpha\,\sin\beta=\sin(\alpha+\beta),$$

with $\alpha=2x$ and $\beta=2x^2$, we obtain

$$f(x)=\sin(2x+2x^2).$$

Now we check the options:

1. $f(-2)$:
   $f(-2)=\sin(2(-2)+2(-2)^2)=\sin(-4+8)=\sin(4)$ which is generally nonzero.

2. $f'(-\tfrac12)$:
   Differentiating,

   $$f'(x)=\cos(2x+2x^2)\,(2+4x).$$

   Thus,

   $$f'\Bigl(-\tfrac12\Bigr)=\cos\Bigl(2(-\tfrac12)+2(\tfrac14)\Bigr)\,(2+4(-\tfrac12)) =\cos(-1+0.5)\,(2-2) =\cos(-0.5)\cdot 0=0.$$

3. $f'(-1)$:

   $$f'(-1)=\cos(2(-1)+2(1))\,(2+4(-1)) =\cos(-2+2)\,(2-4) =\cos(0)\cdot(-2) = 1\cdot(-2)=-2.$$

4. $f''(0)$:
   First, $f(0)=\sin(0)=0.$
   We have

   $$f'(x)=\cos(2x+2x^2)(2+4x).$$

   Then,

   $$f''(x)=-\sin(2x+2x^2)(2+4x)^2+4\cos(2x+2x^2).$$

   At $x=0$ we get

   $$f''(0)=-\sin(0)\cdot(2)^2+4\cos(0)=0+4\cdot1=4.$$

Thus, the correct statements are options 2, 3, and 4.