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Question
The set of all real values of ′a′ so that the range of function y=x2+ax+1, x∈R−{−1} is R, is
The set of all real values of ′a′ so that the range of function y=x2+ax+1, x∈R−{−1} is R, is
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Solution
The correct option is B (−∞,−1)
y=x2+ax+1⇒xy+y=x2+a⇒x2−yx+(a−y)=0
As x∈R−{−1},
D≥0⇒y2−4(a−y)≥0 ∀y∈R⇒y2+4y−4a≥0 ∀y∈RD≤0
16+16a≤0⇒a≤−1∴a∈(−∞,−1]
Now, checking on the boundary values, i.e. at a=−1
y=x2−1x+1=x−1
When x=−1, y=−2
So, the range of x2−1x+1 in this case is R−{−2}
So, a≠−1
Hence, a∈(−∞,−1)
y=x2+ax+1⇒xy+y=x2+a⇒x2−yx+(a−y)=0
As x∈R−{−1},
D≥0⇒y2−4(a−y)≥0 ∀y∈R⇒y2+4y−4a≥0 ∀y∈RD≤0
16+16a≤0⇒a≤−1∴a∈(−∞,−1]
Now, checking on the boundary values, i.e. at a=−1
y=x2−1x+1=x−1
When x=−1, y=−2
So, the range of x2−1x+1 in this case is R−{−2}
So, a≠−1
Hence, a∈(−∞,−1)
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