Question

If \[a{x^2}\; + {\text{ }}bx{\text{ }} + {\text{ }}4\]attains its minimum value \[ - {\text{ }}1\]at \[x{\text{ }} = {\text{ }}1\], then the values of a and b are respectively
(A) \[5,{\text{ }} - 10\]
(B) \[5,{\text{ }} - 5\]
(C) \[10,{\text{ }} - 5\]
(D) \[10,{\text{ }}10\]

Answer 1

Hint: In this question, we are given the minimum value of the given quadratic equation is \[ - 1\] at \[x = 1\]. Hence the first derivative is zero at \[x = 1\]. By substituting \[x = 1\] in the first derivative we obtain an equation. By substituting \[x = 1\] in the given quadratic equation and equating it to \[ - 1\] we get another equation. On solving both the equations obtained we can find the value of a and b.

Complete step by step solution: 
Let \[f\left( x \right) = a{x^2}\; + {\text{ }}bx{\text{ }} + {\text{ }}4\]
Then \[f'\left( x \right) = 2ax{\text{ }} + {\text{ }}b\]
Let us put \[{\text{ }}f'\left( x \right) = 0\]
Then we have, \[f'\left( x \right) = 2ax{\text{ }} + {\text{ }}b = 0\]
\[ \Rightarrow \,\,2ax = - b\]
\[x = \frac{{ - b}}{{2a}}\]
We are given that when \[x{\text{ }} = {\text{ }}1\], the minimum value is \[ - 1\]
Therefore, substituting the value of x as 1 in the above equation we get, \[1 = \frac{{ - b}}{{2a}}\]
\[2a = - b\]
\[ \Rightarrow \,\,2a + b = 0\]…\[(1)\]
Substituting the value of x as 1 in\[f(x)\] we get, \[f\left( 1 \right) = a\; + b + 4\]
Since when \[x{\text{ }} = {\text{ }}1\], the minimum value is \[ - 1\]
\[f\left( 1 \right) = a\; + b + 4 = - 1\]
\[ \Rightarrow \,\,a\; + b + 4 + 1 = 0\]
\[ \Rightarrow \,\,a\; + b = - 5\]…\[(2)\]
On solving equation \[(1)\]and \[(2)\]we get,
\[a{\text{ }} = {\text{ }}5,{\text{ }}b{\text{ }} = - 10\]
So the correct answer is option(A).

Note: It is important to note that the first derivative is zero at the maximum and minimum point. This is because at the maximum and minimum point, the line becomes parallel to the x axis and hence its slope becomes 0.