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Let $f(x) = {x^2} + 4x + 1$. Then
A. $f(x) > 0$ for all $x$
B. $f(x) > 1$ when $x \geqslant 0$
C. $f(x) \geqslant 1$ when $x \leqslant - 4$
D. $f(x) = f( - x)$ for all $x$

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Last updated date: 22nd Jul 2024
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Answer
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Hint: Check whether each option is true one by one. Try to take small numbers to prove each option wrong and if it is not possible then find the range of x for the given inequality and check whether the option is true or false.

Formula used: If $f(x) = {x^2} + y + 2x + 3$ then $f(c) = {c^2} + y + 2c + 3$

Complete step-by-step solution:
We need to check whether each option is true one by one.
A. $f(x) > 0$ for all $x$
When $x = - 1$, $f( - 1) = 1 - 4 + 1 = - 2 < 0$
Therefore, $f(x) > 0$ for all $x$ is false.
B. $f(x) > 1$ when $x \geqslant 0$
When $x = 0$, $f(0) = 0 + 0 + 1 = 1$
Therefore, $f(x) > 1$ when $x \geqslant 0$ is false.
C. $f(x) \geqslant 1$ when $x \leqslant - 4$
${x^2} + 4x + 1 \geqslant 1$
$x(x + 4) \geqslant 0$
$x \in ( - \infty , - 4] \cup [0,\infty )$
Therefore, $f(x) \geqslant 1$ when $x \leqslant - 4$ is true.
Therefore, the correct answer is option C. $f(x) \geqslant 1$ when $x \leqslant - 4$.

Note: Since there is only one correct option, we need not check option D once we know that option C is correct. However, let us check option D as well to verify our answer. Let $x = 1$, $f(1) = 1 + 4 + 1 = 6$ while $f( - 1) = 1 - 4 + 1 = - 2$. Clearly $f(1) \ne f( - 1)$. Therefore, $f(x) = f( - x)$ for all $x$ is false.