GTU BE01000041 Mathematics-I Winter 2025 Solved Paper & Solutions

An improper integral of the second kind is an integral $\int_a^b f(x)\,dx$ in which $f$ is unbounded at one or more points of $[a, b]$. It is defined as a limit. If the limit is finite, the integral converges. Otherwise it diverges.

The integrand $\dfrac{5}{(x-1)^2}$ is unbounded at $x = 1$, which is the upper limit.

As $t \to 1^-$, $t - 1 \to 0^-$, so $\dfrac{-1}{t-1} \to +\infty$.

The gamma function is defined, for $n > 0$, by

Proof: start from the definition and change the variable. Put $t = x^2$. Then $dt = 2x\,dx$. The limits stay $0$ to $\infty$, since $t = 0$ gives $x = 0$ and $t \to \infty$ gives $x \to \infty$.

Maclaurin's series of $f(x)$ is

In general, $f^{(n)}(x) = \dfrac{n!}{(1-x)^{n+1}}$. The pattern is easy to see after the first three derivatives.

This is a geometric series with ratio $x$, which converges for $|x| < 1$.

The beta function is defined, for $m, n > 0$, by

Proof: put $x = 1 - t$. Then $dx = -dt$. When $x = 0$, $t = 1$, and when $x = 1$, $t = 0$.

The last integral is the definition of $\beta(n, m)$.

Put $f'(x) = 0$. Then $x = 1$ or $x = -1$.

• At $x = 1$: $f''(1) = 12 > 0$, so $f$ has a local minimum. $f(1) = 2 - 6 = -4$.
• At $x = -1$: $f''(-1) = -12 < 0$, so $f$ has a local maximum. $f(-1) = -2 + 6 = 4$.

A geometric series is a series in which each term is a fixed multiple $r$ of the previous term.

Convergence: the series converges if $|r| < 1$, and its sum is $\dfrac{a}{1-r}$. It diverges if $|r| \ge 1$ and $a \ne 0$.

(i) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{4n^2 - 1}$

Factorise the denominator and use partial fractions.

The partial sum telescopes, because every term cancels except the first and last.

The series converges, and its sum is $\dfrac{1}{2}$.

(ii) $\displaystyle\sum_{n=0}^{\infty} 2^{n/2}$

Write $2^{n/2} = (\sqrt{2})^n$. This is a geometric series with $a = 1$ and $r = \sqrt{2} \approx 1.414$.

Since $r > 1$, the series diverges.

As $x \to 0$, both $\dfrac{1}{x}$ and $\dfrac{1}{\sin x}$ become infinite, so the form is $\infty - \infty$. Combine the fractions first.

This has the form $\dfrac{0}{0}$. Use L'Hospital's rule.

This is again $\dfrac{0}{0}$. Use the rule once more.

When the curve $y = f(x)$ is revolved about the $x$-axis, the area of the surface generated is

Here $y = x$, so $\dfrac{dy}{dx} = 1$ and $\sqrt{1 + 1} = \sqrt{2}$.

For the power series $\displaystyle\sum_{n=0}^{\infty} a_n x^n$ with $a_n = \dfrac{n}{2^n}$, use the ratio test.

• At $x = 2$: the series is $\sum n$. The terms do not go to zero, so it diverges.
• At $x = -2$: the series is $\sum (-1)^n n$. The terms do not go to zero, so it diverges.

Let $f(x, y) = \dfrac{\sqrt{x} - \sqrt{y}}{\sqrt{x} + \sqrt{y}}$. If the limit exists, it must be the same along every path to $(0, 0)$.

Path 1: along the $x$-axis ($y = 0$, $x > 0$)

Path 2: along the $y$-axis ($x = 0$, $y > 0$)

The two limits are different.

Let $F(x, y, z) = xy^2 + 2yz^2 + 3zx^2 - 6$. The point $(1, 1, 1)$ is on the surface, since $1 + 2 + 3 = 6$.

The gradient $(7, 4, 7)$ is normal to the surface.

We find the points of the sphere $x^2 + y^2 + z^2 = 1$ whose distance from $P(1, 1, 1)$ is the least and the greatest. Use the method of Lagrange multipliers.

In the same way, $y = \dfrac{1}{1+\lambda}$ and $z = \dfrac{1}{1+\lambda}$. So $x = y = z = t$.

• For $t = \dfrac{1}{\sqrt{3}}$: distance $= \sqrt{3}\left(1 - \dfrac{1}{\sqrt{3}}\right) = \sqrt{3} - 1 \approx 0.732$.
• For $t = -\dfrac{1}{\sqrt{3}}$: distance $= \sqrt{3}\left(1 + \dfrac{1}{\sqrt{3}}\right) = \sqrt{3} + 1 \approx 2.732$.

Check: the point $(1,1,1)$ is at distance $\sqrt{3}$ from the centre, and the radius is $1$. So the shortest distance is $\sqrt{3} - 1$ and the longest is $\sqrt{3} + 1$.

Expand $(x+y)^2 = x^2 + 2xy + y^2$. Integrate first with respect to $y$ from $0$ to $2$.

Now integrate with respect to $x$ from $0$ to $1$.

The circle $r = a\cos\theta$ passes through the origin. Its upper half is the part with $y \ge 0$. So $\theta$ goes from $0$ to $\dfrac{\pi}{2}$, and $r$ goes from $0$ to $a\cos\theta$.

The limits give the region: $x$ goes from $0$ to $\infty$, and for each $x$, $y$ goes from $x$ to $\infty$. This is the region $0 \le x \le y < \infty$, which is the part above the line $y = x$ in the first quadrant.

Change the order: for a fixed $y$, $x$ goes from $0$ to $y$, and $y$ goes from $0$ to $\infty$.

The original order cannot be integrated directly, because $e^{-y^2}$ has no elementary antiderivative. After the change, the inner integral is easy.