Question
If x is real, then the maximum and minimum values of expression x2+14x+9x2+2x+3will be
Answer: Option A
:
A
Let y=x2+14x+9x2+2x+3
⇒ y(x2+2x+3)−x2−14x−9=0
⇒ (y−1)x2+(2y−14)x+3y−9=0
For real x, its discriminant≥ 0
i.e. 4(y−7)2−4(y−1)3(y−3)≥0
⇒ y2+y−20≤0or(y−4)(y+5)≤0
Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:
Case 1: y−4≥0ory≥4y+5≤0ory≤−5
This is not possible.
Case 2:y−4≤0ory≤4y+5≥0ory≥−5 Both of these are satisfied if -5≤y≤4
hence maximum value of y is 4 and minimum value is -5
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:
A
Let y=x2+14x+9x2+2x+3
⇒ y(x2+2x+3)−x2−14x−9=0
⇒ (y−1)x2+(2y−14)x+3y−9=0
For real x, its discriminant≥ 0
i.e. 4(y−7)2−4(y−1)3(y−3)≥0
⇒ y2+y−20≤0or(y−4)(y+5)≤0
Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:
Case 1: y−4≥0ory≥4y+5≤0ory≤−5
This is not possible.
Case 2:y−4≤0ory≤4y+5≥0ory≥−5 Both of these are satisfied if -5≤y≤4
hence maximum value of y is 4 and minimum value is -5
Was this answer helpful ?
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