GTU 3110014 Mathematics - 1 (Maths 1) Winter 2025 Solved Paper & Solutions

As $x \to 0$, $\cos x \to 1$ and $\cot x \to \infty$, so the form is $1^{\infty}$. Take the logarithm to turn the power into a product. Let $L = \lim_{x \to 0}(\cos x)^{\cot x}$.

Apply L'Hopital's rule by differentiating the top and the bottom:

So $\ln L = 0$, and $L = e^0$.

The series is $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$. Split each term by partial fractions:

The partial sum $S_n$ has terms that cancel in pairs (a telescoping series):

As $n \to \infty$, $S_n \to 1$. The partial sums have a finite limit, so the series converges, and the limit is its sum.

The limit is not finite, so the integral diverges.

The expression under the root is a perfect square:

For large $n$, the term $\dfrac{2n + 1}{(n+1)^2}$ behaves like $\dfrac{2n}{n^2} = \dfrac{2}{n}$. So compare with $v_n = \dfrac{1}{n}$ using the limit comparison test:

The limit is finite and not zero, so $\sum u_n$ and $\sum v_n$ behave in the same way. The series $\sum\dfrac{1}{n}$ is the harmonic series, which diverges (a p-series with $p = 1$). Hence the given series is divergent.

(1) Split the term into two geometric series. A geometric series $\sum r^n$ converges when $|r| < 1$.

Both ratios, $\dfrac{2}{3}$ and $\dfrac{5}{6}$, are less than $1$. So both series converge, and so does their sum. The sum is $\dfrac{2/3}{1/3} + \dfrac{5/6}{1/6} = 2 + 5 = 7$.

(2) For $u_n = \dfrac{2^n}{n^3 + 1}$, check the general term. The exponential $2^n$ grows faster than any power of $n$, so $u_n \to \infty$. A series can converge only if its terms tend to $0$ (the n-th term test).

A function on $(-\pi, \pi)$ has the Fourier series

Here $f(x) = 0$ on $(-\pi, 0)$ and $f(x) = x$ on $(0, \pi)$. The part where $f$ is zero adds nothing, so every integral runs only over $(0, \pi)$. The function is neither even nor odd, so both cosine and sine terms are present.

Let $p = x - y$, $q = y - z$ and $r = z - x$, so that $u = f(p, q, r)$. Use the chain rule to find each partial derivative. Each variable $x$, $y$, $z$ appears in two of the three arguments of $f$, so each derivative has two terms. Here $f_p$, $f_q$ and $f_r$ are the partial derivatives of $f$ with its three arguments.

Add the three equations. Each of $f_p$, $f_q$, $f_r$ appears once with $+$ and once with $-$, so they cancel:

Hence proved.

Write the surface as $F(x, y, z) = x^3 + y^3 + z^3 - 3 = 0$. The point $(1, 1, 1)$ lies on it, as $1 + 1 + 1 = 3$. The gradient of $F$ is a vector normal to the surface.

The normal vector is $(3, 3, 3)$, which is parallel to $(1, 1, 1)$.

Tangent plane: $1(x - 1) + 1(y - 1) + 1(z - 1) = 0$, that is $x + y + z = 3$.

Normal line: it passes through the point in the direction of the normal.

Here the extreme values of $f = xy$ are required on the ellipse $\dfrac{x^2}{8} + \dfrac{y^2}{2} = 1$, which is a condition on $x$ and $y$. Use Lagrange's method with

The ellipse is a closed curve, so $xy$ has a largest and a smallest value on it, and these are they.

The inner integral is with respect to $y$. Treat $x$ as a constant. Since $\dfrac{y}{x}$ is the exponent, the antiderivative of $e^{y/x}$ is $x\,e^{y/x}$:

Region. The curve $y = x^2$ meets the line $y = 4$ at $x = \pm 2$. So $x$ goes from $-2$ to $2$, and for each $x$, $y$ goes from the parabola $y = x^2$ up to $y = 4$.

Inner integral:

Outer integral. The result is an even function of $x$, so integrate from $0$ to $2$ and double:

Region of integration. For $y$ from $0$ to $1$, $x$ goes from the left half of the circle $x = -\sqrt{1 - y^2}$ to the line $x = 1 - y$, that is $x + y = 1$.

Change the order. Now $x$ is the outer variable and goes from $-1$ to $1$. The upper boundary is different on the two sides of the $y$-axis:

• for $-1 \le x \le 0$: $y$ goes from $0$ to $\sqrt{1 - x^2}$ (the circle),
• for $0 \le x \le 1$: $y$ goes from $0$ to $1 - x$ (the line).

Evaluate. The inner integral of $y$ gives $\dfrac{y^2}{2}$:

Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation.

For $A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}$ the characteristic equation is