Given equation \(x^7 – 7x – 2 = 0\)
Let \(f(x) = x^7 – 7x – 2\)
\(f′(x) = 7x^6 – 7 = 7(x^6 – 1)\)
and \(f′(x) = 0 ⇒ x = +1\)
and \(f(–1) = –1 + 7 – 2 = 5 > 0\)
\(f(1) = 1 – 7 – 2 = –8 < 0\)
So, number of real roots of \(f(x) = 0\) and \(3\).
Hence, the correct option is (D): \(3\)
∫ √(2x2 - 5x + 2) dx = ∫ (41/60) dx,
and
-1/2 > α > 0, then α = ?
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.