Applications Practice 01

Square both sides: $\left(x + \dfrac{1}{x}\right)^2 = x^2 + 2 \cdot x \cdot \dfrac{1}{x} + \dfrac{1}{x^2} = x^2 + \dfrac{1}{x^2} + 2$. So $x^2 + \dfrac{1}{x^2} = 3^2 - 2 = 7$.

Pair consecutive terms and use the difference of two squares: $(2k-1)^2 - (2k)^2 = -(4k-1)$. Summing over $k = 1, 2, \ldots, 10$ gives $-\sum_{k=1}^{10}(4k-1) = -(4 \cdot 55 - 10) = -210$.

Substituting gives $(3-1)^2 + (k-2)^2 = 20$, so $(k-2)^2 = 16$ and $k = 2 \pm 4$, i.e. $k = 6$ or $k = -2$. Their sum is $4$. (Equivalently, by symmetry in $y = 2$, the two values sum to $2 \cdot 2 = 4$.)

Square both sides: $(\sin\theta - \cos\theta)^2 = 1 - 2\sin\theta\cos\theta$. So $1 - 2\sin\theta\cos\theta = \dfrac{1}{4}$, giving $\sin\theta\cos\theta = \dfrac{3}{8}$.

$f(g(x)) = f(x^2) = x^2 + 3$ and $g(f(x)) = g(x+3) = (x+3)^2 = x^2 + 6x + 9$. Equating gives $x^2 + 3 = x^2 + 6x + 9$, so $6x = -6$ and hence $x = -1$.

From $\log_a b = 2$, $b = a^2$, so $a = b^{1/2}$. From $\log_b c = 3$, $c = b^3$. Hence $ab^2 = b^{1/2} \cdot b^2 = b^{5/2}$, and since $\log_c b = \dfrac{1}{3}$, $\log_c(ab^2) = \dfrac{5}{2} \cdot \dfrac{1}{3} = \dfrac{5}{6}$.

Let the AP have first term $a$ and common difference $d$. The three terms are $a+d$, $a+3d$ and $a+7d$. For a GP, $(a+3d)^2 = (a+d)(a+7d)$. Expanding: $a^2+6ad+9d^2 = a^2+8ad+7d^2$, so $2d^2 = 2ad$, i.e. $d(d-a)=0$. The GP is non-constant, so $d \neq 0$, giving $d=a$. The three terms become $2a$, $4a$, $8a$, so the common ratio is $2$.

$y' = 3x^2 + 2ax + b$. The gradient at $x=0$ equals $b$, so $b = -9$. The stationary point at $x=1$ gives $3 + 2a + b = 0$, hence $a = 3$. Then $y' = 3x^2 + 6x - 9 = 3(x+3)(x-1)$, so the other stationary point is at $x = -3$. Substituting: $y(-3) = -27 + 27 + 27 = 27$.

The translation by $(-3, 0)$ replaces $x$ by $x+3$, shifting the vertical asymptote from $x = 2$ to $x = -1$; the horizontal asymptote stays at $y = -1$. Reflection in the $x$-axis negates $y$, so $y = -1$ becomes $y = 1$ (vertical asymptote unchanged). The vertical stretch by factor $2$ scales $y$, sending $y = 1$ to $y = 2$. Final asymptotes: $x = -1$ and $y = 2$.

By the factor theorem, $p(2) = 0$: $8 + 4a + 2b + 12 = 0$, so $2a + b = -10$. By the remainder theorem, $p(-1) = -18$: $-1 + a - b + 12 = -18$, so $a - b = -29$. Adding: $3a = -39$, so $a = -13$ and $b = -10 - 2(-13) = 16$. Therefore $a + b = 3$.

Factorise: $y = x(x-2)(x+2)$, so the curve meets the $x$-axis at $x = -2, 0, 2$. It is positive on $(-2,0)$ and negative on $(0,2)$, so the enclosed region splits into two lobes whose areas must be summed in absolute value. With $F(x) = \frac{x^4}{4} - 2x^2$: $\int_{-2}^{0}(x^3-4x)\,dx = F(0) - F(-2) = 0 - (4-8) = 4$ and $\int_{0}^{2}(x^3-4x)\,dx = F(2) - F(0) = -4$. The integrand is odd, so the two lobes have equal area. Total area $= 4 + |-4| = 8$.

Writing $f(\theta) = R\sin(\theta + \phi)$ with $R = 5$ and $\tan\phi = \tfrac{3}{4}$, the maximum occurs when $\theta + \phi = \tfrac{\pi}{2}$, i.e. $\tan\alpha = \cot\phi = \tfrac{4}{3}$. Then $\tan(2\alpha) = \dfrac{2\tan\alpha}{1 - \tan^2\alpha} = \dfrac{8/3}{1 - 16/9} = \dfrac{8/3}{-7/9} = -\dfrac{24}{7}$.

Statement II rearranges to $(a-b)^2 \geq 0$, true for all real $a,b$. Statement III: $\tfrac{1}{a} + \tfrac{1}{b} = \tfrac{a+b}{ab}$, so the hypothesis becomes $(a+b)^2 \leq 4ab$, i.e. $(a-b)^2 \leq 0$, forcing $a = b$. Statement I is the trap: $a+b \geq 2\sqrt{ab}$ is the AM-GM inequality, true for ALL positive $a,b$, not only when $a=b$. So the hypothesis always holds but the conclusion need not (e.g. $a=1, b=4$ gives $5 \geq 4$ with $a \neq b$). Hence II and III only.

The coefficient of $x^r$ in $(1+ax)^n$ is $\binom{n}{r} a^r$, so $\binom{n}{2} a^2 = 60$ and $\binom{n}{3} a^3 = 160$. Dividing the second by the first eliminates the awkward power of $a$: $\dfrac{n-2}{3}\, a = \dfrac{8}{3}$, hence $(n-2)a = 8$. Substituting $a = \dfrac{8}{n-2}$ into $\binom{n}{2} a^2 = 60$ gives $\dfrac{n(n-1)}{2} \cdot \dfrac{64}{(n-2)^2} = 60$, which simplifies to $7n^2 - 52n + 60 = 0$, i.e. $(7n - 10)(n - 6) = 0$. Since $n$ is a positive integer, $n = 6$ and $a = 2$. (Check: $\binom{6}{2} \cdot 4 = 60$ and $\binom{6}{3} \cdot 8 = 160$.) So $a + n = 8$.

Let $g(x) = x^2 - 6x + 5 = (x-1)(x-5)$. The parabola has roots at $x = 1, 5$ and a minimum at $x = 3$, where $g(3) = -4$. Reflecting the portion below the $x$-axis gives $f$: it touches zero at $x = 1, 5$, rises to a local maximum of $4$ at $x = 3$, and increases to $+\infty$ as $|x| \to \infty$. The horizontal line $y = k$ meets this graph in: $0$ points if $k < 0$; $2$ points if $k = 0$; $4$ points if $0 < k < 4$; $3$ points if $k = 4$ (the bump's peak counts once); $2$ points if $k > 4$. So $f(x) = k$ has exactly four solutions iff $0 < k < 4$.

If $\ell: y = mx + c$ is tangent to $y = f(x)$ at $x = a$ and $x = b$, then $f(x) - (mx + c)$ has double roots at both points, so $x^4 - 8x^2 + 5x - (mx + c) = (x-a)^2(x-b)^2.$ Expanding the RHS as $x^4 - 2(a+b)x^3 + [(a+b)^2 + 2ab]x^2 - 2ab(a+b)x + a^2b^2$ and matching coefficients: the $x^3$ term forces $a + b = 0$, so $b = -a$; the $x^2$ term gives $-2a^2 = -8$, so $a^2 = 4$; the $x^1$ term gives $5 - m = 0$, so $m = 5$; the constant term gives $-c = a^2 b^2 = 16$, hence $c = -16$. (Check: tangent points $(2, -6)$ and $(-2, -26)$ both have $y' = 5$.)

Rewrite RHS as $\log_2(2(x+1))$, so the equation becomes $x^2 + a = 2x + 2$, i.e.\ $x^2 - 2x + (a-2) = 0$, with domain $x > -1$. (The $x^2 + a > 0$ condition is automatic on solutions, since there $x^2 + a = 2(x+1) > 0$.) Roots: $x = 1 \pm \sqrt{3-a}$, real iff $a \le 3$. Case 1: double root $a = 3$ gives $x = 1$, valid. Case 2: two distinct roots, exactly one in domain, the larger root $1 + \sqrt{3-a} > -1$ always; we need the smaller $1 - \sqrt{3-a} \le -1$, i.e.\ $\sqrt{3-a} \ge 2$, i.e.\ $a \le -1$. (At $a = -1$, the smaller root is $-1$, excluded by the strict $x > -1$.) So exactly one solution iff $a \le -1$ or $a = 3$.

Any $x$ with $f(x) = x$ also satisfies $f(f(x)) = x$, so $f(x) - x$ divides $f(f(x)) - x$. Now $f(x) = x$ gives $x^2 - 2x - 1 = 0$, with roots $x = 1 \pm \sqrt{2}$. Expanding, $f(f(x)) = x^4 - 2x^3 - 2x^2 + 3x + 1$, so $f(f(x)) - x = x^4 - 2x^3 - 2x^2 + 2x + 1$. Polynomial division by $x^2 - 2x - 1$ gives quotient $x^2 - 1$, hence $f(f(x)) - x = (x^2 - 2x - 1)(x^2 - 1).$ The second factor contributes $x = \pm 1$, which form a $2$-cycle of $f$ (since $f(1) = -1$ and $f(-1) = 1$). All four roots are real and distinct.

Let $I = \int_0^2 \frac{x^3}{x^3 + (2-x)^3}\, dx$. Substituting $u = 2 - x$ (so $du = -dx$ and the limits swap) gives $I = \int_0^2 \frac{(2-u)^3}{(2-u)^3 + u^3}\, du.$ Renaming $u$ to $x$ and adding to the original, $2I = \int_0^2 \frac{x^3 + (2-x)^3}{x^3 + (2-x)^3}\, dx = \int_0^2 1 \, dx = 2,$ so $I = 1$. (A direct attack expanding $(2-x)^3$ reduces the integrand to $\frac{x^3}{2(3x^2 - 6x + 4)}$, requiring polynomial division and a logarithmic term, the symmetry pairing collapses it instantly.)

If $\log_a b = p/q$ in lowest terms, then $b^q = a^p$. By unique factorisation, $a$ and $b$ must be integer powers of a common integer $c \ge 2$. Each integer $\ge 2$ has a unique \emph{primitive base} $c$ (not itself a perfect power), so $\log_a b$ is rational iff $a$ and $b$ share the same primitive base. Enumerate by primitive base $c$ with $c^{p_{\max}} \le 100$: valid pairs are $(c^q, c^p)$ with $1 \le q < p \le p_{\max}$, giving $\binom{p_{\max}}{2}$. We get $c = 2$: $\binom{6}{2} = 15$; $c = 3$: $\binom{4}{2} = 6$; $c = 5, 6, 7, 10$: $\binom{2}{2} = 1$ each. (Bases $4, 8, 9$ are perfect powers, already counted; $c \ge 11$ non-power gives $c^2 > 100$.) Total $= 15 + 6 + 4 = 25$.