Question

Let x be a number that exceeds its square by the greatest possible quantity. Then x is equal to
A. \[\dfrac{1}{2}\]
B. \[\dfrac{1}{4}\]
C. \[\dfrac{3}{4}\]
D.None of the above

Answer 1

Hint: First we assume a function for the condition that x exceeds its square by the greatest possible quantity, now we will find at which the function will be maximum by finding the first derivative and double derivative of the function.
Now the value of x at which the function will give its maximum value will be the required value of x.

Complete step-by-step answer:
Given data: x exceeds its square by the greatest possible quantity
We know that square of any number is always greater than the number if the number is greater or equal to 1, but here it is greater than its square so it is clear that the number is smaller than 1
Now it is given that x exceeds its square by the greatest possible quantity i.e. $x - {x^2}$ is maximum
Now let $f(x) = x - {x^2}$,
Now we will find the maximum value of the function and at which the function will give its maximum value will be the value of x
Now differentiating the function $f(x) = x - {x^2}$ , we get,
 $ \Rightarrow f'(x) = 1 - 2x$
Substituting $f'(x) = 0$, we get,
 $ \Rightarrow 0 = 1 - 2x$
On simplification we get,
 $\therefore x = \dfrac{1}{2}$
Now checking the double derivative of the function at $x = \dfrac{1}{2}$ , as we know that if at the values of x at which the first derivative of the function is zero, the double derivative results negative then the function has a maxima at that point.
On differentiating the first derivative $f'(x) = 1 - 2x$, we get,
$ \Rightarrow f'(x) = - 2$, since it is negative for all values of x, the function has a maxima.
Therefore, $x = \dfrac{1}{2}$
Hence, Option (A) is correct.

Note: Most of the students just find the first derivative and guess from that only that at the value of x at which the first derivative is zero, but that is incorrect as we should always check the double derivative of the function to fully assure that the function will be maximum at that point.