Problems

Solution:

1. Consider \(f(x) = x - \ln (1+ x)\), then \(f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} > 0\) if \(x >0\). Therefore \(f(x)\) is strictly increasing on the positive reals. Since \(f(0) = 0\) we must have \(f(x) > 0\) for all positive \(x\), ie \(x - \ln(1+x)\) is positive for all positive \(x\). \begin{align*} \sum_{k=1}^n \frac1k &\underbrace{>}_{x > \ln(1+x)} \sum_{k=1}^n \ln \left (1 + \frac1k \right ) \\ &= \sum_{k=1}^n \ln \left ( \frac{k+1}{k} \right ) \\ &= \sum_{k=1}^n \left ( \ln (k+1) - \ln (k) \right) \\ &= \ln (n+1) - \ln 1 \\ &= \ln (n+1) \end{align*}
2. Let \(g(x) = x + \ln (1-x)\) ,then \(g'(x) = 1 - \frac{1}{1-x} = \frac{-x}{1-x} < 0\) if \(0 < x < 1\) and \(g(0) = 0\). Therefore \(g(x)\) is decreasing and hence negative on \(0 < x < 1\), in particular \(x < -\ln(1-x) \) \begin{align*} \sum_{k=2}^n \frac1{k^2} &\underbrace{<}_{x < -\ln(1+x)} \sum_{k=2}^n - \ln \left (1-\frac1{k^2} \right) \\ &= -\sum_{k=2}^n \ln \left ( \frac{k^2-1}{k^2}\right) \\ &= \sum_{k=2}^n \l 2 \ln k - \ln(k-1) - \ln(k+1) \r \\ &= \ln n - \ln(n+1) - \ln 0+\ln 2 \\ &= \ln 2 + \ln \frac{n}{n+1} \end{align*} as \(n \to \infty\) we must have \(\displaystyle \sum_{k=2}^{\infty} \frac1{k^2} < \ln 2\) ie \[ \sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2\]

Solution:

1. \(C\) has the equation \(x^2 + y^2 = a^2\). One suitable circle would ideally pass through \((0,a)\) and \((0,-a)\) have a centre on the positive \(x\)-axis, so we would need \(a^2+c^2 = r^2\) so \(c = \sqrt{r^2-a^2}\) and the equation would be \((x-\sqrt{r^2-a^2})^2 + y^2 = r^2\). Clearly a circle with radius \(r < a\) cannot pass through two diametrically opposed points of a circle radius \(a\), since the furthest two points can be on a circle is \(2r\), and diametrically opposed points are \(2a\) apart. Similarly if they are exactly the same radii, then if they pass through diametrically opposed points they must be the same circle.
2. Let the centre of \(D\) be at \((x,y)\), then it must be a distance of \(\sqrt{r^2-a_i}\) from each circle centre, ie \begin{align*} && (x-d)^2+y^2 &= r^2-a_2^2 \\ && (x+d)^2 + y^2 &= r^2-a_1^2 \\ \Rightarrow && 4dx &= a_2^2 - a_1^2 \\ \Rightarrow && x &= \frac{a_2^2-a_1^2}{4d} \\ \Rightarrow && y^2 &= r^2-a_1^2 - \left (\frac{a_2^2-a_1^2}{4d}+d \right)^2 \\ &&&= r^2 - a_1^2 - \frac{(a_2^2-a_1^2+4d^2)^2}{16d^2} \\ &&&= \frac{16d^2r^2-16d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2+8a_1^2d^2-8a_2^2d^2}{16d^2} \\ &&&= \frac{16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2}{16d^2} \\ \Rightarrow && y &= \pm \sqrt{ \frac{16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2}{16d^2}} \end{align*} and we need \begin{align*} && 0 &\leq 16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2 \\ \Rightarrow && 16d^2 d^2 &\geq 8d^2a_1^2 + a_2^4+a_1^4+16d^4+2a_1^2a_2^2+8a_2^2d^2 \\ &&&= (4d^2+(a_2-a_1)^2)(4d^2+(a_2+a_1)^2) \end{align*}

Solution:

1. If \(m = 1000\), then \(n \geq m \Rightarrow n^2 \geq 1000n \Rightarrow (1000n) \leq (n^2)\)
2. This is false. Let \(s_i = 1,2,1,2,\cdots\) and \(t_i = 2,1,2,1,\cdots\).
3. Suppose that for \(n \geq m_1, s_n \le t_n\) and for \(n \geq m_2, s_t \le u_n\), then for \(n \geq m = \max(m_1, m_2), s_n \leq t_n \leq u_n \Rightarrow s_n \leq u_n \Rightarrow (s_n) \leq (u_n)\)
4. Let \(m = 6\), then if \(n \geq m, 2^n \geq 1 + n + \frac{n(n-1)}{2} + \frac{n(n-1)}{2} + n + 1 = n^2 + n + 2 \geq n^2\), so \((2^n) \geq (n^2)\)

Solution:

1. Notice that \(u^2 = v+2wz\), so \(w,z\) are roots of the quadratic \(t^2 -ut+\frac{u^2-v}{2}\). Therefore they are both real if \(u^2 \geq 2(u^2-v) \Rightarrow 2v \geq u^2\).
2. \begin{align*} && w+z &= u \\ && w^2+z^2 &= u^2 - \tfrac23 \\ && w^3+z^3 &= \lambda(u-1) \\ \\ && wz &= \frac{u^2 - (u^2-\tfrac23)}{2} = \tfrac13\\ \\ && (w+z)(w^2+z^2) &= w^3+z^3+wz(w+z) \\ &&u(u^2-\tfrac23)&= \lambda(u-1)+\frac13u \\ \Rightarrow && u^3-u&= \lambda (u-1) \\ \Rightarrow && 0 &= (u-1)(u(u+1) - \lambda) \\ \Rightarrow && 0 &= (u-1)(u^2+u - \lambda) \end{align*} Therefore there will be at most 3 values for \(u\), unless \(1\) is a root of \(u^2+u-\lambda\), ie \(\lambda = 2\). Suppose \(u = 1\), then we have: \(w+z = 1, wz = 1/3 \Rightarrow w,z = \frac{-1 \pm \sqrt{-1/3}}{2}\) which are clearly complex.

Solution:

1. \(\,\) \begin{align*} && \int_0^b x^2 \d x &= \left ( \int_0^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} &= \left ( \frac{b^2}{2} \right)^2 \\ \Rightarrow && b &= \frac{4}{3} \end{align*}
2. \(\,\) \begin{align*} && \int_1^b x^2 \d x &= \left ( \int_1^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{1}{3} &= \left ( \frac{b^2}{2} - \frac{1}{2} \right)^2 \\ \Rightarrow && 4(b^3 - 1) &= 3(b^2-1)^2 \\ \Rightarrow && 4(b^3-1) &= 3(b^4-2b^2+1) \\ \Rightarrow && 0 &= 3b^4-4b^3-6b^2+7 \\ &&&= (b-1)(3b^3-b^2-7b-7) \\ \Rightarrow && 0 &= 3b^3-b^2-7b-7 \end{align*}
   

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   Let \(f(x) = 3x^3-x^2-7x-7\) then \(f(2) = 3 \cdot 8 - 4 - 14 - 7 = -1 < 0\), \(f(3) = 3 \cdot 27 - 9 - 21 - 7 = 44 > 0\) therefore the root must lie between \(2\) and \(3\).
3. \(,\) \begin{align*} && \int_a^b x^2 \d x &= \left ( \int_a^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{a^3}{3} &= \left ( \frac{b^2}{2} - \frac{a^2}{2} \right)^2 \\ \Rightarrow && 4(b^3 - a^3) &= 3(b^2-a^2)^2 \\ \Rightarrow && 4(b^2+ab+a^2) &= 3(b-a)(b+a)^2 \\ \Rightarrow && 4 \left ( \left ( \frac{p+q}{2}\right)^2+\left ( \frac{p+q}{2}\right)\left ( \frac{p-q}{2}\right)+\left ( \frac{p-q}{2}\right)^2\right) &= 3qp^2 \\ \Rightarrow && 3p^2 + q^2 &= 3qp^2 \\ \Rightarrow && 3p^2(q-1) &= q^2 \\ \Rightarrow && p^2 &= \frac{q^2}{3(q-1)} \\ \Rightarrow && 1 &\leq \frac{1}{3(q-1)} \\ \Rightarrow && 3(q-1) &\leq 1 \\ \Rightarrow && q & \leq \frac{4}{3} \\ \end{align*}

Solution:

1. Note that \(f(x) = (x+2a)^3 - 27a^2x\) so \(f'(x) = 3(x+2a)^2-27a^2 = 3x^2+12ax-15a^2 = (x-a)(3x+15a)\) so the turning points are at \(x = a, x = -5a\). But \(f(a) = 0\), so the curve just touches the \(x\)-axis. Note also that \(f(0) = 8a^3 \geq 0\) so:
   

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2. \(\,\) \begin{align*} && 0 &\leq (x+2y)^3 - 27y^2 x \\ &&& \leq 3^3 - 27 y^2 x \\ &&&= 27(1-y^2x) \\ \Rightarrow && xy^2 &\leq1 \end{align*} with equality when \(x = y = 1\)
3. Notice that \begin{align*} && 0 &\leq (p+q+r)^3 - 27\left (\frac{q+r}{2}\right)^2 p \\ \Rightarrow && (p+q+r)^3 &\geq 27\left (\frac{q+r}{2}\right)^2 p \\ &&&\underbrace{\geq}_{AM-GM} 27\left (\sqrt{qr}\right)^2 p \\ &&&= 27pqr \end{align*} Equality can only hold if \(p = \frac{q+r}{2}\), but by symmetry we must also have \(q = \frac{r+p}{2}, r = \frac{p+q}{2}\) ie \(p = q = r\). And indeed equality does hold in this case.

Solution: \begin{align*} && v_{\uparrow} &= U\sin \theta - g t \\ \Rightarrow && T_H &= \frac{U \sin \theta}{g} \\ \\ && s_{\uparrow} &= U \sin \theta t - \frac12 g t^2 \\ \Rightarrow && 0 &= U\sin \theta T_L - \frac12 g T_L^2 \\ && T_L &= \frac{2 U \sin \theta}{g} \end{align*} \(T = U\cos \theta / (kg)\) is the point when the particle's horizontal motion is reversed.

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Solution:

1. \(\,\)
   

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   \begin{align*} \text{N2}(\uparrow, m): && T - mg &= ma_1 \\ \text{N2}(\uparrow, M): && T-Mg &= -Ma_1 \\ \Rightarrow && (M-m)g &= a_1(m+M) \\ \Rightarrow && a_1 &= \frac{M-m}{M+m}g \\ && T &= mg + ma_1 \\ &&&= \frac{2mM}{M+m}g \end{align*}
2. System II is the same as system I, but with \(m\) replaced with \(2\frac{T}{g} = \frac{4mM}{M+m}\). In particular, this means that: \begin{align*} && a_2 &= \frac{M - \frac{4m_1m_2}{m_1+m_2}}{M + \frac{4m_1m_2}{m_1+m_2}} g \\ &&&= \frac{M-4\mu}{M+4\mu}g \end{align*} If \(m = m_1 + m_2\) then \begin{align*} && a_1 &= a_2 \\ \Leftrightarrow && \frac{M-m_1-m_2}{M+m_1+m_2} &= \frac{M - \frac{4m_1m_2}{m_1+m_2}}{M + \frac{4m_1m_2}{m_1+m_2}} \\ \Leftrightarrow && \frac{M-m_1-m_2}{M+m_1+m_2} &= \frac{M(m_1+m_2) -4m_1m_2}{M(m_1+m_2) + 4m_1m_2} \\ \Leftrightarrow && M^2(m_1+m_2)+4m_1m_2M &- M(m_1+m_2)^2 - 4m_1m_2(m_1+m_2) \\ &&\quad \quad = M^2(m_1+m_2) - 4m_1m_2M &+M(m_1+m_2)^2-4m_1m_2(m_1+m_2) \\ \Leftrightarrow && 8m_1m_2M&= 2M(m_1+m_2)^2 \\ \Leftrightarrow && 0 &= (m_1-m_2)^2 \\ \Leftrightarrow && m_1 &= m_2 \end{align*}

Solution:

1. If \(f(x) = \sin nx\) then \(f'(x) = n \cos n x\) and so \begin{align*} && LHS &= \int_0^\pi \sin^2 n x \d x \\ &&&= \left [ \frac{x+\frac1{2n}\sin 2n x}{2} \right ]_0^{\pi} \\ &&&= \frac{\pi}{2} \\ \\ && RHS &= \int_0^{\pi} n^2 \cos^2 n x \d x \\ &&&= n^2 \left [ \frac{\frac{1}{2n}\sin 2n x + x}{2} \right]_0^{\pi} \\ &&&= n^2\frac{\pi}{2} \geq LHS \end{align*} [\(f(0) = 0, f(\pi) \neq 0\)] Suppose \(f(x) = x\) then \(f'(x) = 1\) and \(LHS = \frac{\pi^3}{3} > \pi = RHS\). [\(f(0) \neq 0, f(\pi) = 0\)] Suppose \(f(x) = \pi - x\) then \(f'(x) = -1\) and \(LHS = \frac{\pi^3}{3} > \pi = RHS\)
2. Suppose \(f(x) = x(\pi - x)\) then \(f'(x) = \pi - 2x\) and so \begin{align*} && \int_0^\pi x^2(\pi-x)^2 \d x &\leq \int_0^\pi (\pi-2x)^2 \d x \\ \Leftrightarrow && \left [\pi^2 \frac{x^3}{3} - 2\pi \frac{x^4}{4} + \frac{x^5}{5} \right]_0^{\pi} &\leq \left [ \pi^2x - 4\pi \frac{x^2}{2} + \frac{4x^3}{3} \right]_0^{\pi} \\ \Leftrightarrow && \pi^5 \left (\frac13 - \frac12+\frac15 \right) &\leq \pi^3 \left ( 1 - 2+\frac43 \right) \\ \Leftrightarrow && \pi^2 \frac{1}{30} &\leq \frac13 \\ \Leftrightarrow && \pi^2 &\leq 10 \end{align*} Suppose \(f(x) = p\sin \tfrac12 x + q \cos \tfrac12 x + r\), so \(f(0) = q + r\) and \(f(\pi) = p + r\), so say \(p = q = 1, r = -1\) \begin{align*} && LHS &= \int_0^{\pi} \left ( \sin \tfrac12 x + \cos \tfrac12 x-1\right)^2 \d x \\ &&&=\int_0^\pi \left ( \sin^2 \tfrac12 x + \cos^2 \tfrac12 x+1-2\sin \tfrac12 x - 2\cos \tfrac12 x+ \sin x \right)\\ &&&= \left [2x + 4\cos \tfrac12 x - 4\sin \tfrac12 x - \cos x \right]_0^{\pi} \\ &&&= \left ( 2\pi -4+1 \right) - \left ( 4-1 \right) \\ &&&= 2\pi -6\\ \\ && RHS&= \int_0^{\pi} \left ( \tfrac12 \cos \tfrac12 x -\tfrac12 \sin \tfrac12 x\right)^2 \d x \\ &&&= \int_0^{\pi} \left ( \tfrac14 \cos^2 \tfrac12 x +\tfrac14 \sin^2 \tfrac12 x-\tfrac14 \sin x\right) \d x \\ &&&= \frac{\pi}{4} - \frac12 \\ \Rightarrow && 2\pi -6 &\leq \frac{\pi}{4} - \frac12 \\ \Rightarrow && \frac{7\pi}{4} &\leq \frac{11}{2} \\ \Rightarrow && \pi &\leq \frac{22}{7} \end{align*} \(22^2/7^2 = 484/49 < 10\) therefore \(\pi \leq \frac{22}{7}\) is the better estimate.

Solution: \begin{align*} && \sin(r + \tfrac12)x - \sin(r - \tfrac12) x &= \sin rx \cos \tfrac12x + \cos r x\sin\tfrac12x - \sin r x \cos \tfrac12 x + \cos rx \sin \tfrac12 x \\ &&&= 2\cos r x \sin\tfrac12 x \\ \\ && S &= \cos x + \cos 2x + \cdots + \cos n x \\ && 2\sin \tfrac12 x S &= \sin(1 + \tfrac12)x - \sin \tfrac12 x + \\ &&&\quad+ \sin(2+\tfrac12)x - \sin(2- \tfrac12)x + \\ &&&\quad+ \sin(3+\tfrac12)x - \sin(3 - \tfrac12)x + \\ &&& \quad + \cdots + \\ &&&\quad + \sin(n+\tfrac12)x - \sin(n-\tfrac12)x \\ &&&=\sin(n+\tfrac12)x - \sin\tfrac12 x \\ \Rightarrow && S &= \frac{\sin(n+\tfrac12)x - \sin\tfrac12 x}{2 \sin \tfrac12 x} \end{align*}

1. \(\,\) \begin{align*} && S_2(x) &= \sin x + \tfrac12 \sin 2 x \\ && S'_2(x) &= \cos x + \cos 2x \\ &&&= \cos x + 2\cos^2 x - 1 \\ &&&= (2\cos x -1)(\cos x + 1) \\ \end{align*} Therefore the turning points are \(\cos x= \frac12 \Rightarrow x = \frac{\pi}{3}\) and \(\cos x = -1 \Rightarrow x = \pi\)
   

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2. Suppose \(S_n(x)\) has a stationary point at \(x_0\), then $$ therefore \begin{align*} &&0 &= S_n'(x_0) \\ &&&= \cos x_0 + \cos 2x_0 + \cdots + \cos n x_0 \\ &&&= \frac{\sin(n+\tfrac12)x_0 - \sin \tfrac12x_0}{2 \sin \tfrac12 x_0} \\ \Rightarrow &&\sin\tfrac12 x_0&= \sin nx_0 \cos \tfrac12 x_0 + \cos nx_0 \sin \tfrac12x_0 \\ \Rightarrow && \sin nx_0 &= (1-\cos nx_0)\tan \tfrac12 x_0 \end{align*} Therefore \(S_n(x_0) -S_{n-1}(x_0) = \tfrac1n \sin n x_0 = \tfrac1n \underbrace{(1-\cos nx_0)}_{\geq 0}\underbrace{\tan\tfrac12 x_0}_{\geq 0} \geq 0\). Therefore if \(S_{n-1}(x) > 0\) for all \(x\) on \(0 < x < \pi\) then since \(S_n(x) > S_{n-1}(x)\) at the turning points and since they agree at the end points, it must be larger at all points inbetween.
3. Notice that \(S_1(x) = \sin x \geq 0\) for all \(x \in [0,1]\) and by our previous argument we can show \(S_n > S_{n-1}\) inside the interval and equal on the boundary we must have \(S_n(x) \geq 0\) for \(x \in [0, \pi]\)

Solution:

1. \(\,\)
   

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   \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\) forms a rectangle.
2. There are no solutions if \(a < c < b\):
   

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   There is one solution if \(a=c\) or \(a = b\)
   

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   And there are two solution if \(c \not \in [a,b]\)
   

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   There is exactly one solution unless....
   

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   ... there are infinitely many solutions when the gradients line up perfectly, ie when \(a+b=c+d\)
   

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Solution: \(c_r = \binom{n}{r}a^{n-r}b^r\) \begin{align*} && c_m &\geq c_{m+1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m+1} a^{n-m-1}b^{m+1} \\ \Rightarrow && \frac{1}{(n-m)}a &\geq \frac{1}{m+1}b \\ \Rightarrow && (m+1)a &\geq (n-m)b \\ \Rightarrow && m(a+b) &\geq nb -a \\ \Rightarrow && m &\geq \frac{n(b+1)-a-b}{a+b}=\frac{n(b+1)}{a+b} - 1 \\ \\ && c_m &\geq c_{m-1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m-1} a^{n-m+1}b^{m-1} \\ \Rightarrow && \frac{1}m b &\geq \frac{a}{(n-m+1)} \\ \Rightarrow && (n-m+1)b &\geq ma \\ \Rightarrow && (n+1)b &\geq m(a+b) \\ \Rightarrow && m &\leq \frac{(n+1)b}{a+b} \end{align*} Since \(m\) lies between two values \(1\) apart is is either equal to one of those two values or is the unique integer between them. Let \(\displaystyle G(n,a,b) = \left \lfloor \frac{b(n+1)}{a+b} \right \rfloor\), so

1. \(\,\) \begin{align*} && G(9,1,3) &= \left \lfloor \frac{3(9+1)}{1+3} \right \rfloor \\ &&&= \left \lfloor \frac{30}{4} \right \rfloor \\ &&&= 7 \\ \\ && G(9,2,3) &= \left \lfloor \frac{3(9+1)}{2+3} \right \rfloor \\ &&&= \left \lfloor \frac{30}{5} \right \rfloor \\ &&&= 6 \end{align*}
2. \(\,\) \begin{align*} && G(2k, a, a) &= \left \lfloor \frac{a(2k+1)}{a+a} \right \rfloor \\ && &= \left \lfloor \frac{2k+1}{2} \right \rfloor \\ &&&= k \\ \\ && G(2k-1, a, a) &= \left \lfloor \frac{a(2k-1+1)}{a+a} \right \rfloor \\ && &= \left \lfloor k\right \rfloor \\ &&&= k \\ \end{align*}
3. \(G(n,a,b)\) is decreasing in \(a\), therefore take \(a = 1\).
4. For fixed \(n\), we are looking at \(\frac{a(n+1)}{a+b}\) and we want this to be as large as possible. By considering \((n+1) - \frac{b(n+1)}{a+b}\) we can see this is increasing in \(b\) and the largest value possible is \(n\). This is achieved when \begin{align*} && \frac{b(n+1)}{a+b} & \geq n \\ \Leftrightarrow && bn + b &\geq na + bn \\ \Leftrightarrow && b& \geq na \end{align*}

Solution:

1. \begin{align*} && I - I_1 &= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) \d x - \int_0^1 \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) - \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left (-y^2-2yy' \tan x - y^2 \tan^2 x \right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2(1+ \tan^2 x )\right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2 \sec^2 x\right) \d x\\ &&&= \int_0^1 -\frac{\d}{\d x} \left (y^2 \tan x \right) \d x\\ &&&= \left [-y^2 \tan x \right]_0^1 \\ &&&= 0 \\ \\ \Rightarrow && I &= I_1 = \int_0^1 \left ( y' + y \tan x \right)^2 d x \geq 0 \end{align*} The only way \(I_0 = 0\) is is \(y' + y \tan x =0\), so \begin{align*} && \frac{\d y}{\d x} &= - y \tan x \\ \Rightarrow && \int \frac{1}{y} &= \int -\tan x \d x \\ \Rightarrow && \ln |y| &= \ln |\cos x| + C \\ \Rightarrow && y &= A \cos x \\ \Rightarrow && A &= 0 \Rightarrow y = 0 \end{align*}
2. Let \(J_1 = \int_0^1 (y'+ay\tan ax)^2 \d x\), then \begin{align*} && J-J_1 &= \int_0^1 \left ( \left ( y' \right)^2 - a^2y^2 \right) - \left ( y' + ya \tan ax \right)^2 \d x\\ &&&= \int_0^1 \left (-a^2y^2-2yy' a \tan a x-y^2a^2 \tan^2 ax \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2(1+\tan^2 ax) \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2\sec^2 ax \right) \d x \\ &&&= \left [ - a y^2 \tan a x \right]_0^1 = 0 \end{align*} This is true if \(a < \frac{\pi}{2}\), since otherwise we might care about the order of the zero for \(y\) at \(x = 1\). Consider \(y = \cos \frac{\pi}{2} x\), then \(y' = -\frac{\pi}{2} \sin^2\frac{\pi}{2} x\) and \begin{align*} && \int_0^1 \frac{\pi^2}{4} \left (\sin^2 \frac{\pi}{2}x - \cos^2 \frac{\pi}{2} x \right) \d x &= -\frac{\pi^2}{4} \int_0^1 \cos(\pi x) \d x \\ &&&= 0 \end{align*}