Consider the equation xlogx=a then answer the following questions (a) ...
Solution:
The given equation is xlogx=a.
(a) Range of a for which the equation has no real roots:
For the equation xlogx = a to have real roots, a must be greater than or equal to 0. This is because if a is negative, then the left-hand side of the equation is always negative, while the right-hand side is always positive. Therefore, there are no real values of x that can satisfy the equation.
So, the range of a for which the equation has no real roots is a < />
(b) Range of a for which the equation has two real roots:
Let f(x) = xlogx - a. We want to find the range of a for which f(x) has two real roots.
The derivative of f(x) is given by f'(x) = logx + 1. This is always positive for x > 0. Therefore, f(x) is an increasing function for x > 0.
Also, f(1) = 0 and lim x→0⁺ f(x) = 0, lim x→∞ f(x) = ∞.
Therefore, f(x) has two real roots if and only if there exists a value x > 0 such that f(x) < />
Now, f(x) < 0="" ⇒="" xlogx="" />< a.="" since="" x="" /> 0, we can divide both sides by x to get logx < a/x.="" if="" a="" /> 0, then we can choose x = e^a/2 to get logx = a/2 > a/x. Therefore, f(x) < 0="" for="" all="" x="" /> e^a/2.
So, the range of a for which the equation has two real roots is a > 0.
(c) Range of a for which the equation has only one real root:
If a = 0, then the equation xlogx = 0 has only one real root at x = 1. If a > 0, then the equation xlogx = a has two real roots, and if a < 0,="" then="" the="" equation="" has="" no="" real="" roots.="" therefore,="" the="" equation="" has="" only="" one="" real="" root="" if="" a="" />
So, the range of a for which the equation has only one real root is a = 0.
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