Let $ f $ and $ g $ be two functions defined by $ f(x) = |x + |x|| $ and $ g(x) = \frac{1 {x} $ for $ x \neq 0 $. If $ f(a) + g(f(a)) = 13/6 $ for some real $ a $, then the maximum possible value of $ f(g(a)) $ is:}

Question

cdquestions Admin · Accepted Answer

Step 1: Evaluating $ f(x) $. The function $ f(x) $ is defined as: $$ f(x) = |x + |x|| $$ - If $ x \geq 0 $, then $ |x| = x $, so $ f(x) = |x + x| = |2x| = 2x $. - If $ x<0 $, then $ |x| = -x $, so $ f(x) = |x + (-x)| = |0| = 0 $. Step 2: Evaluating $ g(f(a)) $. Since $ f(a) + g(f(a)) = \frac{13}{6} $, we set $ f(a) = y $, which gives: $$ y + \frac{1}{y} = \frac{13}{6} $$ Multiplying both sides by $ y $: $$ y^2 - \frac{13}{6} y + 1 = 0 $$ Solving for $ y $, we get: $$ y = 2 \quad \text{or} \quad y = \frac{1}{3} $$ Step 3: Finding Maximum $ f(g(a)) $. Since $ f(g(a)) = f\left(\frac{1}{a}\right) $, we substitute $ a = 3 $ (corresponding to $ y = \frac{1}{3} $): $$ f(g(3)) = f\left(\frac{1}{3}\right) = 2 \times \frac{1}{3} = \frac{2}{3} $$ For $ a = \frac{1}{2} $ (corresponding to $ y = 2 $): $$ f(g(1/2)) = f(2) = 2 \times 2 = 4 $$ Thus, the maximum possible value is 6.

Let \( f \) and \( g \) be two functions defined by \( f(x) = |x + |x|| \) and \( g(x) = \frac{1{x} \) for \( x \neq 0 \). If \( f(a) + g(f(a)) = 13/6 \) for some real \( a \), then the maximum possible value of \( f(g(a)) \) is:}

Show Hint

Solution and Explanation

Top Questions on Algebra

Questions Asked in IPMAT exam