WASSCE Further Maths: Functions, Polynomials & Equations Practice Questions & Answers

For the function to be defined, the expression inside the square root must be strictly positive (since it's in the denominator and must avoid square roots of negative numbers). Thus, $x^2 - 4 > 0$x2−4>0. Factoring gives $(x-2)(x+2) > 0$(x−2)(x+2)>0. Testing intervals reveals the inequality holds for $x < -2$x<−2 or $x > 2$x>2.

Let $y = 3x - 5$y=3x−5. To find the inverse, swap $x$x and $y$y yielding $x = 3y - 5$x=3y−5. Solve for $y$y: $x + 5 = 3y$x+5=3y, so $y = \frac{x + 5}{3}$y=3x+5​. Thus, $f^{-1}(x) = \frac{x + 5}{3}$f−1(x)=3x+5​.

To find $f(g(x))$f(g(x)), substitute $g(x)$g(x) into $f(x)$f(x). $f(g(x)) = (2x - 3)^2 + 1$f(g(x))=(2x−3)2+1. Expanding this gives $(4x^2 - 12x + 9) + 1 = 4x^2 - 12x + 10$(4x2−12x+9)+1=4x2−12x+10.

First find $(f \circ g)(x) = f(g(x)) = \frac{\frac{1}{x}+1}{\frac{1}{x}-1} = \frac{1+x}{1-x}$(f∘g)(x)=f(g(x))=x1​−1x1​+1​=1−x1+x​. To find the inverse, set $y = \frac{1+x}{1-x}$y=1−x1+x​, rewrite as $x = \frac{1+y}{1-y}$x=1−y1+y​. Then $x(1-y) = 1+y \implies x - xy = 1 + y \implies x - 1 = y(x+1) \implies y = \frac{x-1}{x+1}$x(1−y)=1+y⟹x−xy=1+y⟹x−1=y(x+1)⟹y=x+1x−1​.

For a square root to be real, the radicand must be non-negative. Therefore, $5 - x \ge 0 \implies 5 \ge x$5−x≥0⟹5≥x, which means $x \le 5$x≤5.

$f(g(x)) = f(x^2) = 2x^2$f(g(x))=f(x2)=2x2. Additionally, $g(f(x)) = g(2x) = (2x)^2 = 4x^2$g(f(x))=g(2x)=(2x)2=4x2. Setting them equal: $2x^2 = 4x^2 \implies 2x^2 = 0 \implies x^2 = 0 \implies x = 0$2x2=4x2⟹2x2=0⟹x2=0⟹x=0.

Set $y = \frac{2x}{x+3}$y=x+32x​. Swap to find inverse: $x = \frac{2y}{y+3}$x=y+32y​. Cross-multiply: $x(y+3) = 2y \implies xy + 3x = 2y \implies 3x = y(2-x) \implies y = \frac{3x}{2-x}$x(y+3)=2y⟹xy+3x=2y⟹3x=y(2−x)⟹y=2−x3x​. Thus, $f^{-1}(x) = \frac{3x}{2-x}$f−1(x)=2−x3x​.

According to the Remainder Theorem, the remainder when $P(x)$P(x) is divided by $(x - c)$(x−c) is $P(c)$P(c). Substituting $x = 1$x=1: $P(1) = 2(1)^3 - (1)^2 + 3(1) - 4 = 2 - 1 + 3 - 4 = 0$P(1)=2(1)3−(1)2+3(1)−4=2−1+3−4=0.

By the Factor Theorem, if $x+2$x+2 is a factor, then $P(-2) = 0$P(−2)=0. Substituting $x = -2$x=−2: $(-2)^3 + k(-2)^2 - 3(-2) + 6 = -8 + 4k + 6 + 6 = 4k + 4$(−2)3+k(−2)2−3(−2)+6=−8+4k+6+6=4k+4. Setting this to zero: $4k + 4 = 0 \implies 4k = -4 \implies k = -1$4k+4=0⟹4k=−4⟹k=−1.

Since 1 is a root, $(x-1)$(x−1) is a factor. Performing polynomial division of $x^3 - 6x^2 + 11x - 6$x3−6x2+11x−6 by $(x-1)$(x−1) yields a quotient of $x^2 - 5x + 6$x2−5x+6. Factoring the quadratic gives $(x-2)(x-3)$(x−2)(x−3). Setting this to zero gives the remaining roots: $x = 2$x=2 and $x = 3$x=3.