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

f (- 2) = 0

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f^{'} (- \frac{1}{2}) = 0

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f^{'} (- 1) = - 2

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f^{''} (0) = 4

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Solution

The correct option is D $f^{''} (0) = 4$
$f (x) = sin 2 x [sin (x + x^{2}) sin (x - x^{2}) + cos (x - x^{2}) cos (x + x^{2})]$
$+ sin 2 x^{2} [cos (x + x^{2}) cos (x - x^{2}) - sin (x + x^{2}) sin (x - x^{2})]$
$\Rightarrow f (x) = sin 2 x \cdot cos 2 x^{2} + sin 2 x^{2} \cdot cos 2 x$
$\Rightarrow f (x) = sin (2 x + 2 x^{2})$
$\Rightarrow f (- 2) = sin 4 \neq 0$
$\Rightarrow f^{'} (x) = (2 + 4 x) cos (2 x + 2 x^{2})$
$\Rightarrow f^{'} (- \frac{1}{2}) = 0$
$\Rightarrow f^{'} (- 1) = - 2$
$\Rightarrow f^{''} (x) = 4 cos (2 x + 2 x^{2}) - (2 + 4 x)^{2} sin (2 x + 2 x^{2})$
$\Rightarrow f^{''} (0) = 4 - 0 = 4$

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