For x2 -(a+3)x+ 4= 0 to have real solutions, the range of a is :

|x|+4|x|=a+3⇒a+3≥2|x|⋅4|x|=4 [A.M. ≥ G.M. ⇒ a≥1∴ Range of a is [1,∞).

For x2 -(a+3)x+ 4= 0 to have real solutions, the range of a is :

|x|+4|x|=a+3⇒a+3≥2|x|⋅4|x|=4 [A.M. ≥ G.M. ⇒ a≥1∴ Range of a is [1,∞).

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For x² -(a+3) $|x|$ + 4= 0 to have real solutions, the range of a is :

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$(- \infty, - 7] \cup [1, \infty)$

$(- \infty, - 7]$

$[1, \infty)$

$[- 7, 1]$

answer is C.

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Detailed Solution

$\begin{array}{l} | x | + \frac{4}{| x |} = a + 3 \\ \Rightarrow a + 3 \geq 2 \sqrt{| x | \cdot \frac{4}{| x |}} = 4 [A.M. \geq G.M.] \\ \Rightarrow a \geq 1 \end{array}$

$∴ Range of a is [1, \infty) .$

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For x² -(a+3) $|x|$ + 4= 0 to have real solutions, the range of a is :