GTU 3110014 Mathematics - 1 (Maths 1) Summer 2026 Solved Paper & Solutions

At $x = 0$ both the top and the bottom are $0$, so the form is $\dfrac{0}{0}$. Use L'Hopital's rule repeatedly: differentiate the top and the bottom until the form is no longer $\dfrac{0}{0}$.

The series $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$ is a geometric series. Its first term is $a = 1$ and its common ratio is $r = \dfrac{1}{2}$.

A geometric series converges when $|r| < 1$, and then its sum is $\dfrac{a}{1 - r}$. Here $|r| = \dfrac{1}{2} < 1$, so the series converges:

(i) The integrand $\dfrac{1}{\sqrt{4 - x^2}}$ becomes infinite at the upper limit $x = 2$, so this is an improper integral of type II. Define it as a limit and use $\displaystyle\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a}$:

The limit is finite, so the integral converges to $\dfrac{\pi}{2}$.

(ii) The solid is made by revolving the area under $y = xe^x$ from $x = 0$ to $x = 1$ about the $x$-axis. Use the disc method: $V = \pi\displaystyle\int_0^1 y^2\,dx$.

Integrate by parts twice. The result is $\displaystyle\int x^2e^{2x}\,dx = e^{2x}\left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right)$:

Sandwich theorem (also called the squeeze theorem): if $b_n \le a_n \le c_n$ for all $n$, and both $b_n$ and $c_n$ tend to the same limit $L$, then $a_n$ also tends to $L$.

For $a_n = \dfrac{\sin^2 n}{n}$, we know $0 \le \sin^2 n \le 1$. Dividing by $n > 0$ keeps the inequality:

As $n \to \infty$, both $0$ and $\dfrac{1}{n}$ tend to $0$. So the middle term is squeezed between two sequences that go to the same limit, and

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

Both ratios, $\dfrac{5}{7}$ and $\dfrac{6}{7}$, are less than $1$. So both series converge, and so does their sum. Each sum is $\dfrac{r}{1 - r}$ for a series starting at $n = 1$: $\dfrac{5/7}{2/7} = \dfrac{5}{2}$ and $\dfrac{6/7}{1/7} = 6$.

The function $f(x) = \dfrac{\pi - x}{2}$ is given on $(0, 2\pi)$, so we use the Fourier formulas over this interval. The series is $f(x) = \dfrac{a_0}{2} + \sum\left(a_n\cos nx + b_n\sin nx\right)$. The deduction in the last step needs the value of the series at one chosen point.

So the Fourier series is

Differentiate $u = \log(x^2 + y^2)$ first with respect to $x$ and then with respect to $y$. Then do it in the opposite order, treating the other variable as a constant each time. We must show that the two mixed partial derivatives are equal.

Both are equal to $-\dfrac{4xy}{(x^2 + y^2)^2}$. This agrees with Clairaut's theorem: the mixed derivatives are equal when they are continuous. Hence $\dfrac{\partial^2 u}{\partial x\,\partial y} = \dfrac{\partial^2 u}{\partial y\,\partial x}$.

The rate of change of a function in a given direction is the directional derivative, $\nabla f\cdot\hat{n}$. Here $\hat{n}$ is the unit vector normal to the surface $F = x^2y + xy^2 + yz^2 = 3$. The point $P(1, 1, 1)$ lies on it, as $1 + 1 + 1 = 3$.

We maximise $f = x + y + z$ with the condition $\dfrac{4}{x} + \dfrac{9}{y} + \dfrac{16}{z} = 25$. Use Lagrange's method with

Note on "maximum". For positive $x, y, z$ this stationary value is in fact the least value of $x + y + z$. By the Cauchy-Schwarz inequality, $(x + y + z)\left(\dfrac{4}{x} + \dfrac{9}{y} + \dfrac{16}{z}\right) \ge (2 + 3 + 4)^2 = 81$, so $x + y + z \ge \dfrac{81}{25}$. The sum has no upper limit for positive values. Lagrange's method gives the same values of $x, y, z$ for the question as asked.

Integrate with respect to $y$ first. For a fixed $x$, write $a^2 = 1 + x^2$, so the upper limit of $y$ is $a$ and the integrand is $\dfrac{1}{a^2 + y^2}$.

Now integrate with respect to $x$. The standard integral is $\displaystyle\int\frac{dx}{\sqrt{1 + x^2}} = \sinh^{-1}x = \ln\left(x + \sqrt{1 + x^2}\right)$:

The question asks us to look at the order of integration. In the given order, $y$ goes from $0$ to $2x$ and then $x$ from $0$ to $1$. The region is the triangle with vertices $(0, 0)$, $(1, 0)$ and $(1, 2)$, bounded by the line $y = 2x$, the line $x = 1$ and the $x$-axis. If the order is changed, $y$ goes from $0$ to $2$, and for each $y$, $x$ goes from $\dfrac{y}{2}$ to $1$:

But $\displaystyle\int e^{x^2}\,dx$ has no closed form, so the changed order cannot be evaluated. The order as given, with $y$ first, is the one that works. The inner integral has $e^{x^2}$ as a constant:

Now put $t = x^2$, so $dt = 2x\,dx$ and $t$ goes from $0$ to $1$:

Region. The region $R$ is bounded by the $x$-axis ($y = 0$), the line $x = 2a$ and the parabola $x^2 = 4ay$, that is $y = \dfrac{x^2}{4a}$. The parabola passes through the origin and meets the line $x = 2a$ at $y = a$.

Limits. $x$ goes from $0$ to $2a$, and for each $x$, $y$ goes from $0$ up to the parabola, $y = \dfrac{x^2}{4a}$. We integrate $y$ first, because the upper limit of $y$ depends on $x$ and the limits of $x$ are constants.

Inner integral.

Outer integral.