Question

Let \([t]\) denote the greatest integer function. If \(\int_0^1 [1 + x^2 + x^4]\,\mathrm{d}x = a\), then \(36a - 25a^2 + 8a^3 - a^4\) is equal to

• A. 19
• B. \(-19\)
• C. 18
• D. \(-21\)

Hint:

The integrand is 1 until \(x^4 + x^2 = 1\) and 2 afterwards on [0, 1), so \(a = 2 - x_0\) where \(x_0\) is the root of \(x^4 + x^2 - 1 = 0\).

Answer 1

The Correct Option is A

For \(0 \le x < 1\), \(0 \le x^2 + x^4 < 2\), so \(1 + x^2 + x^4\) lies in [1, 3). Its integer part is 1 when \(x^4 + x^2 < 1\) and 2 when \(x^4 + x^2 \ge 1\).
Let \(x_0 \in (0,1)\) satisfy \(x_0^4 + x_0^2 = 1\). Then \(a = \int_0^{x_0} 1\,dx + \int_{x_0}^{1} 2\,dx = x_0 + 2(1 - x_0) = 2 - x_0\), so \(x_0 = 2 - a\).
Put \(x = x_0\); then \(x^4 = 1 - x^2\) and \(a = 2 - x\).
\(a^2 = 4 - 4x + x^2\), \(a^3 = 8 - 12x + 6x^2 - x^3\), \(a^4 = 16 - 32x + 24x^2 - 8x^3 + x^4 = 17 - 32x + 23x^2 - 8x^3\).
\(36a = 72 - 36x\); \(-25a^2 = -100 + 100x - 25x^2\); \(8a^3 = 64 - 96x + 48x^2 - 8x^3\); \(-a^4 = -17 + 32x - 23x^2 + 8x^3\).
Adding: constants 72 - 100 + 64 - 17 = 19; the coefficients of \(x\), \(x^2\) and \(x^3\) all cancel to 0.
So the expression equals 19.
Answer: (1) 19