For all real values of x, the range of the function f(x) = x2+2x+42x2+4x+9 is
Explanation:
Given, f(x) = x2+2x+42x2+4x+9
⇒ f(x) = 122x2+4x+82x2+4x+9
⇒ f(x) = 122x2+4x+9-12x2+4x+9
⇒ f(x) = 121-12x2+4x+9
f(x) will be minimum when (2x2 + 4x + 9) is minimum.
Now, 2x2 + 4x + 9 will be minimum when x = -(4)/2 × 2 = -1
∴ Minimum value of 2x2 + 4x + 9 = 2(-1)2 + 4(-1) + 9 = 7
∴ Minimum value of f(x) = 121-17 = 37
f(x) will be maximum when (2x2 + 4x + 9) is maximum.
Maximum value of (2x2 + 4x + 9) will be ∞.
∴ Maximum value of f(x) = 121-1∞ = 12
∴ Range of f(x) = 37,12
Upper value of ½ is in open bracket as value of (2x2 + 4x + 9) will never actually be ∞. Hence, option (a).
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