Functions & Graphs: CAT previous year questions

Solution

$g$ is undefined at $x = 1$; $f$ is undefined at $x = \dfrac{1}{2}$.

$f(g(x))$ also fails where $g(x) = \dfrac{1}{2}$: $\dfrac{x}{x - 1} = \dfrac{1}{2} \Rightarrow x = -1$.

$g(f(x))$ also fails where $f(x) = 1$: $\dfrac{x}{2x - 1} = 1 \Rightarrow x = 1$.

Excluded values: $-1, \dfrac{1}{2}$ and $1$.

Solution

Let $y = f(x)$: $2yx^2 + (4y - 2)x + (3 - 6y) = 0$ must have a real solution.

Discriminant $\geq 0$: $(4y - 2)^2 - 8y(3 - 6y) \geq 0 \Rightarrow 64y^2 - 40y + 4 \geq 0 \Rightarrow (8y - 1)(2y - 1) \geq 0$.

$y \leq \dfrac{1}{8}$ or $y \geq \dfrac{1}{2}$.

Range: $\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty\right)$.

Solution

Let $g(x) = f(x) + 1$. Then $g(xy) = f(xy) + 1 = (f(x) + 1)(f(y) + 1) = g(x)g(y)$, and $g(p) = 2$ for every prime.

$160000 = 2^8 \times 5^4$, which has 12 prime factors (with repetition), so $g(160000) = 2^{12} = 4096$.

$f(160000) = 4096 - 1 = 4095$.

Solution

Replace $x$ by $\dfrac{1}{x}$: $f\left(\dfrac{1}{x}\right) + 2f(x) = \dfrac{3}{x}$.

Solve the two equations: $3f(x) = \dfrac{6}{x} - 3x \Rightarrow f(x) = \dfrac{2}{x} - x$.

$f(x) = 3 \Rightarrow 2 - x^2 = 3x \Rightarrow x^2 + 3x - 2 = 0$.

Sum of the values $= -3$.

Solution

Let $u = 3x + 2y$ and $v = 2x - 5y$. Then $5u + 2v = 19x$, so $f(u, v) = 5u + 2v$.

$f(x, 2x) = 5x + 4x = 9x = 27 \Rightarrow x = 3$.

Solution

$f(x) + f(x - 1) = 1$ for all integers means $f$ alternates: $f(x + 2) = f(x)$.

$x^2 - x$ is always even, so $f(\text{even}) = 5$ and $f(\text{odd}) = 1 - 5 = -4$.

$f(g(5)) = f(25) = -4$. $g(f(5)) = g(-4) = 16$.

Sum $= -4 + 16 = 12$.

Solution

For $x < r$: $f(x) = r$ and $f(r) = 2r - r = r$, so $f(f(x)) = f(x)$ ✓.

For $x \geq r$: $f(x) = 2x - r \geq r$, so $f(f(x)) = 2(2x - r) - r = 4x - 3r$. This equals $2x - r$ only when $x = r$.

So the equation holds exactly for $x \leq r$.

Solution

Let $t = x^2 + 2x + 4 = (x + 1)^2 + 3 \geq 3$. The denominator is $2t + 1$.

$f = \dfrac{t}{2t + 1} = \dfrac{1}{2} - \dfrac{1}{2(2t + 1)}$, which increases with $t$.

At $t = 3$: $\dfrac{3}{7}$. As $t \to \infty$: approaches $\dfrac{1}{2}$ but never reaches it.

Range $= \left[\dfrac{3}{7}, \dfrac{1}{2}\right)$.

Solution

$f(g(x)) = (x + 3)^2 - 7(x + 3) = x^2 - x - 12$.

$f(g(x)) - 3x = x^2 - 4x - 12 = (x - 2)^2 - 16$.

Minimum $= -16$.

Solution

$f(5 + x) = f(5 - x)$ means the graph is symmetric about $x = 5$.

Roots come in pairs $5 + a$ and $5 - a$, each pair adding to 10.

Four roots: sum $= 2 \times 10 = 20$.

Solution

$g(x) = f(x + 1) - f(x - 1) = [(x + 1)^2 - (x - 1)^2] + 2a = 4x + 2a$.

$g(20) = 80 + 2a = 72 \Rightarrow a = -4$.

$f(x) \geq 0$ for all $x$ needs $a^2 - 4b \leq 0 \Rightarrow b \geq 4$.

Smallest $b = 4$.

Solution

$f(0) = f(0)^2 \Rightarrow f(0) = 1$ (since $f(5) \neq 0$).

$f(10) = f(5)^2 = 16$, and $f(10)f(-10) = f(0) = 1 \Rightarrow f(-10) = \dfrac{1}{16}$.

$f(10) - f(-10) = 16 - 0.0625 = 15.9375$.