Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If f(g(x)) = 8x2 – 2x and g(f(x)) = 4x2 + 6x + 1, then the value of f(2) + g(2) is ____________ .
The correct answer is 18
f(g(x)) = 8x2 – 2x
g(f(x)) = 4x2 + 6x + 1
let f(x) = cx2 + dx + e
g(x) = ax + b
f(g(x)) = c(ax + b)2 + d(ax + b) + e ≡ 8x2 – 2x
g(f(x)) = a(cx2 + dx + e) + b ≡ 4x2 + 6x + 1
Therefore ac = 4 ad = 6 ae + b = 1
a2c = 8 2abc + ad = –2 cb2 + bd + e = 0
By solving, we get
a = 2 , b = –1, c = 2 , d = 3, e = 1
∴ f(x) = 2x2 + 3x +1
g(x) = 2x – 1
f(2) + g(2) = 2(2)2 + 3(2) + 1 + 2(2) – 1
= 18
Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then
\((\frac{1}{p} + \frac{1}{q} )^{-2}\)
is equal to _______.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A polynomial that has two roots or is of 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.
Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.
The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)
Read More: Nature of Roots of Quadratic Equation