Download Question Paper (Link Below)

Question Paper Solution

1. (i) Examine the Convergence of the series:

$\frac{x}{1+x} + \frac{x^2}{1+x^2} + \frac{x^3}{1+x^3} + \cdots$

Here $u_n = \frac{x^n}{1 + x^n}$ and $u_{n+1} = \frac{x^{n+1}}{1 + x^{n+1}}$

$\text{Lt}_{n \to \infty} \frac{u_n}{u_{n+1}} = \text{Lt}_{n \to \infty} \left( \frac{x^n}{x^{n+1}} \cdot \frac{1 + x^{n+1}}{1 + x^n} \right) = \text{Lt}_{n \to \infty} \left( \frac{1 + x^{n+1}}{x + x^n} \right) = \frac{1}{x}, \text{ if } x < 1.$

$$\left[ \text{Since } x^{n+1} \to 0 \text{ as } n \to \infty \right]$$

Also,

$\text{Lt}_{n \to \infty} \frac{u_n}{u_{n+1}} = \text{Lt}_{n \to \infty} \left( \frac{1 + 1/x^{n+1}}{1 + x/x^{n+1}} \right) = 1 \text{ if } x > 1.$

$\text{by Ratio test,} \sum u_n \text{ converges for } x < 1 \text{ and fails for } x \geq 1.$

When $x = 1$, $\sum u_n = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots + \infty$, which is divergent.

Hence the given series converges for $x < 1$ and diverges for $x \geq 1$.

(ii) Discuss the convergence of the series:

$1 - 2x + 3x^2 - 4x^3 + \cdots \quad \left( x < \frac{1}{2} \right)$

To discuss the convergence of the series:

$$1 - 2x + 3x^2 - 4x^3 + \cdots \quad \left( x < \frac{1}{2} \right)$$

We can start by considering the general term and applying appropriate convergence tests.

Analysis of the General Term

The general term of the series can be written as:

$$a_n = (-1)^n (n+1) x^n$$

Ratio Test

The ratio test can be used to determine the convergence of the series. For the ratio test, we examine the limit of the absolute value of the ratio of successive terms:

$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$$

Let's compute this limit:

$\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1} (n+2) x^{n+1}}{(-1)^n (n+1) x^n} \right| = \left| \frac{(n+2) x}{n+1} \right| = \left| x \cdot \frac{n+2}{n+1} \right|$

As $n \to \infty$:

$$\lim_{n \to \infty} \left| x \cdot \frac{n+2}{n+1} \right| = \left| x \cdot 1 \right| = |x|$$

For the series to converge, the ratio must be less than 1:

$$|x| < 1$$

Given the condition $x < \frac{1}{2}$, it is clear that $|x| < \frac{1}{2}$, which is indeed less than 1. Hence, the ratio test confirms the convergence of the series for $x < \frac{1}{2}$.

2. (i) Test the following series for absolute convergence:

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

Given series is $\sum u_n = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ........\infty$

This is an alternating series of which terms go on decreasing and

$$\text{Lt}_{n \to \infty} u_n = \text{Lt}_{n \to \infty} \frac{1}{2n - 1} = 0$$ $$\text{by Leibnitz's rule, } \sum u_n \text{ converges.}$$

The series of absolute terms is $1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots \infty$

Here $u_n = \frac{1}{2n - 1}$. Taking $v_n = \frac{1}{n}$, we have

$\text{Lt}_{n \to \infty} \frac{u_n}{v_n} = \text{Lt}_{n \to \infty} \left( \frac{\frac{1}{2n - 1}}{\frac{1}{n}} \right) = \text{Lt}_{n \to \infty} \left( \frac{n}{2n - 1} \right) = \text{Lt}_{n \to \infty} \left( \frac{1}{2 - \frac{1}{n}} \right) = \frac{1}{2} \ne 0 \text{ and finite.}$

$\text{by Comparison test, }$

$\sum u_n \text{ diverges because,} \sum v_n \text{ diverges.}$

Hence the given series converges and the series of absolute terms diverges, therefore the given series converges conditionally.

(ii) Examine for term by term integration the series which $f_n(x) = nxe^{-n^2 x^2}$ in the intervals $(i) [0, 1]$ and $(ii) [c, 1]$ where $0 < c < 1$.

To examine the series for term-by-term integration where $f_n(x) = nxe^{-n^2 x^2}$ in the intervals $(i) [0, 1]$ and $(ii) [c, 1]$ where $0 < c < 1$, we need to check whether the series $\sum_{n=1}^{\infty} f_n(x)$ can be integrated term by term within these intervals.

Series Definition

The series is:

$$\sum_{n=1}^{\infty} nxe^{-n^2 x^2}$$

Term-by-Term Integration

For term-by-term integration, we need to ensure uniform convergence of the series within the given intervals.

Interval (i): $[0, 1]$

Let's analyze the behavior of $f_n(x)$ on the interval $[0, 1]$.

1. Pointwise Convergence:

   For each fixed $x \in [0, 1]$:

   $$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} nxe^{-n^2 x^2}$$

   Since $e^{-n^2 x^2}$ decays very rapidly as n increases, $f_n(x) \to 0$ for each $x \in [0, 1]$.

2. Uniform Convergence:

   To check uniform convergence on $[0, 1]$, consider the maximum value of $f_n(x)$:

   $$f_n(x) = nxe^{-n^2 x^2}$$

   Let $g_n(x) = ne^{-n^2 x^2}$. The maximum of $g_n(x)$ occurs at $x = \frac{1}{n}$:

   $$g_n\left(\frac{1}{n}\right) = ne^{-n^2 \left(\frac{1}{n}\right)^2} = ne^{-1}$$

   Thus, for $x \in [0, 1]$:

   $$|f_n(x)| \leq nxe^{-n^2 x^2} \leq ne^{-1}$$

   Since $ne^{-1} \to 0$ as $n \to \infty$, the series converges uniformly to 0 on $[0, 1]$.

Interval (ii): $[c, 1]$ where $0 < c < 1$

1. Pointwise Convergence:

   For each fixed $x \in [c, 1]$:

   $$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} nxe^{-n^2 x^2} = 0$$

   As before, $f_n(x) \to 0$ for each $x \in [c, 1]$.

2. Uniform Convergence:

   On $[c, 1]$, the behavior of $f_n(x)$ is more straightforward since x is bounded away from 0:

   $$f_n(x) = nxe^{-n^2 x^2}$$

   For $x \geq c$:

   $$f_n(x) \leq ne^{-n^2 c^2}$$

   As $n \to \infty$:

   $$ne^{-n^2 c^2} \to 0$$

   Therefore, the series converges uniformly to 0 on $[c, 1]$.

3. (i) Verify Rolle's theorem for the function $f(x) = x(x + 3)e^{\frac{-x}{2}}$ in $[-3, 0]$.

To verify Rolle's Theorem for the function $f(x) = x(x+3)e^{\frac{-x}{2}}$ in the interval $[-3, 0]$, we need to check the hypotheses of Rolle's Theorem and then find the value of $c$ in the interval $(-3, 0)$ where $f'(c) = 0$.

Rolle's Theorem states that if a function is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then:

1. Continuity and Differentiability:
   The given function $f(x) = x(x+3)e^{\frac{-x}{2}}$ is a product of a polynomial and an exponential function, which are both continuous and differentiable everywhere. Hence, $f$ is continuous on $[-3, 0]$ and differentiable on $(-3, 0)$.

2. Checking $f(a) = f(b)$:
   We need to verify that $f(-3) = f(0)$.
   $f(-3) = (-3)(-3+3)e^{\frac{3}{2}} = (-3)(0)e^{\frac{3}{2}} = 0$
   $f(0) = 0(0+3)e^{\frac{-3}{2}} = 0$
   Since $f(-3) = f(0) = 0$, the function satisfies the condition $f(a) = f(b)$.

3. Finding $c$ where $f'(c) = 0$:
   According to Rolle's Theorem, there exists a value of $c$ such that $f'(c) = 0$.

   The function is given as:
   $f(x) = x(x+3)e^{\frac{-x}{2}}$
   Expanding it:
   $f(x) = (x^2 + 3x)e^{\frac{-x}{2}}$

   Using the product rule for differentiation:
   $f'(x) = (2x + 3)e^{\frac{-x}{2}} + (x^2 + 3x)\left(-\frac{1}{2}\right)e^{\frac{-x}{2}}$
   Simplifying:
   $f'(x) = e^{\frac{-x}{2}} \left[(2x + 3) - \frac{1}{2}(x^2 + 3x)\right]$
   $f'(x) = e^{\frac{-x}{2}} \left[\frac{4x + 6 - x^2 - 3x}{2}\right]$
   $f'(x) = e^{\frac{-x}{2}} \left[-x^2 + x + 6\right]$

   Solving $f'(x) = 0$:
   $e^{\frac{-x}{2}}[x^2 - x - 6] = 0$
   Since $e^{\frac{-x}{2}} \neq 0$, we solve:
   $x^2 - x - 6 = 0$
   Factoring:
   $x^2 - 3x + 2x - 6 = 0$
   $x(x - 3) + 2(x - 3) = 0$
   $(x + 2)(x - 3) = 0$
   $x = -2 \quad \text{or} \quad x = 3$

   Since $x = -2$ lies in the interval $[-3, 0]$, we take $c = -2$.

Hence, Rolle's Theorem is verified.

(ii) Use Cauchy's Mean Value theorem to evaluate:

$\lim_{x \to 1} \frac{\cos\left(\frac{\pi x}{2}\right)}{\log\left(\frac{1}{x}\right)}$

Cauchy's Mean Value Theorem states that if functions f and g are continuous on $[a, b]$ and differentiable on $(a, b)$, and $g'(x) \neq 0$ for all $x \in (a, b)$, then there exists a point $c \in (a, b)$ such that:

$$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$

We need to evaluate the following limit using CMVT:

$$\lim_{x \to 1} \frac{\cos\left(\frac{\pi x}{2}\right)}{\log\left(\frac{1}{x}\right)}$$

Steps to Solve:

1. Rewrite the limit in a suitable form:

   We rewrite the limit as:

   $$\lim_{x \to 1} \frac{\cos\left(\frac{\pi x}{2}\right)}{\log\left(\frac{1}{x}\right)}$$

   Notice that $\log\left(\frac{1}{x}\right) = -\log(x)$, so the limit becomes:

   $$\lim_{x \to 1} \frac{\cos\left(\frac{\pi x}{2}\right)}{-\log(x)}$$

   This is in the form $\frac{f(x)}{g(x)}$, where $f(x) = \cos\left(\frac{\pi x}{2}\right)$ and $g(x) = -\log(x)$.

2. Apply CMVT:

   By CMVT, there exists a point $c \in (a, b)$ such that:

   $$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$

   Let $a = x$ and $b = 1$.

   So,

   $$\frac{f'(c)}{g'(c)} = \frac{\cos\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi x}{2}\right)}{-\log(1) - (-\log(x))}$$

   Simplifying this,

   $\frac{f'(c)}{g'(c)} = \frac{0 - \cos\left(\frac{\pi x}{2}\right)}{0 - (-\log(x))} = \frac{-\cos\left(\frac{\pi x}{2}\right)}{\log(x)}$

   So, the point $c$ satisfies:

   $$\lim_{x \to 1} \frac{-\cos\left(\frac{\pi c}{2}\right)}{\log(c)}$$

3. Evaluate the derivatives:

   To use CMVT properly, let's calculate the derivatives $f'(x)$ and $g'(x)$:

   $f(x) = \cos\left(\frac{\pi x}{2}\right) \implies f'(x) = -\frac{\pi}{2} \sin\left(\frac{\pi x}{2}\right)$

   $$g(x) = -\log(x) \implies g'(x) = -\frac{1}{x}$$

4. Use the derivatives in the limit:

   Now, applying these derivatives:

   $\lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{-\frac{\pi}{2} \sin\left(\frac{\pi x}{2}\right)}{-\frac{1}{x}} = \lim_{x \to 1} \frac{\frac{\pi}{2} \sin\left(\frac{\pi x}{2}\right)}{\frac{1}{x}} = \lim_{x \to 1} \frac{\pi x}{2} \sin\left(\frac{\pi x}{2}\right)$

5. Simplify and evaluate the final limit:

   As $x \to 1$,

   $\frac{\pi x}{2} \to \frac{\pi}{2} \quad \text{and} \quad \sin\left(\frac{\pi x}{2}\right) \to \sin\left(\frac{\pi}{2}\right) = 1$

   Therefore,

   $$\lim_{x \to 1} \frac{\pi x}{2} \sin\left(\frac{\pi x}{2}\right) = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2}$$

4. (i) Find the length of the arc of the parabola $y^2 - 4y + 2x = 0$ which lies in the first quadrant.

1. Rewrite the Equation:

   Given equation:

   $$y^2 - 4y + 2x = 0$$

   Completing the square for the y-terms:

   $$y^2 - 4y = -2x$$ $$(y^2 - 4y + 4) - 4 = -2x$$ $$(y - 2)^2 - 4 = -2x$$ $$(y - 2)^2 = 2(x + 2)$$

   Standard form:

   $$(y - 2)^2 = 2(x + 2)$$

2. Range of y Values:

   In the first quadrant, both x and y are positive. For the arc in the first quadrant, we consider $y \geq 2$.

3. Set Up the Arc Length Integral:

   The formula for the arc length L of a curve $y = f(x)$ from x = a to x = b is:

   $$L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$$

   Solve for x in terms of y:

   $$(y - 2)^2 = 2(x + 2)$$ $$x + 2 = \frac{(y - 2)^2}{2}$$ $$x = \frac{(y - 2)^2}{2} - 2$$ $$\frac{dx}{dy} = \frac{d}{dy} \left( \frac{(y - 2)^2}{2} - 2 \right)$$ $$\frac{dx}{dy} = \frac{1}{2} \cdot 2(y - 2)$$ $$\frac{dx}{dy} = y - 2$$

   Arc length integral in terms of y:

   $$L = \int_{y_1}^{y_2} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy$$

   Assume $y_1 = 2$ and $y_2 = 4$:

   $$L = \int_{2}^{4} \sqrt{1 + (y - 2)^2} \, dy$$

4. Evaluate the Integral:

   Let $u = y - 2$. Then $du = dy$, and the bounds change from $y = 2$ to $y = 4$ to $u = 0$ to $u = 2$:

   $$L = \int_{0}^{2} \sqrt{1 + u^2} \, du$$

   Using trigonometric substitution, let $u = \sinh(t)$, then $du = \cosh(t) \, dt$:

   $L = \int_{0}^{\sinh^{-1}(2)} \sqrt{1 + \sinh^2(t)} \cosh(t) \, dt$

   Since $\sqrt{1 + \sinh^2(t)} = \cosh(t)$:

   $$L = \int_{0}^{\sinh^{-1}(2)} \cosh^2(t) \, dt$$

   Using the identity $\cosh^2(t) = \frac{1 + \cosh(2t)}{2}$:

   $$L = \int_{0}^{\sinh^{-1}(2)} \frac{1 + \cosh(2t)}{2} \, dt$$

   $L = \frac{1}{2} \int_{0}^{\sinh^{-1}(2)} 1 \, dt + \frac{1}{2} \int_{0}^{\sinh^{-1}(2)} \cosh(2t) \, dt$

   Evaluate the integrals:

   $L = \frac{1}{2} \left[ t \right]_{0}^{\sinh^{-1}(2)} + \frac{1}{4} \left[ \sinh(2t) \right]_{0}^{\sinh^{-1}(2)}$

   $L = \frac{1}{2} \left[ \sinh^{-1}(2) \right] + \frac{1}{4} \left[ \sinh(2 \sinh^{-1}(2)) - \sinh(0) \right]$

   $$L = \frac{1}{2} \sinh^{-1}(2) + \frac{1}{4} \left[ 2 \cdot 2 \sqrt{1 + 4} \right]$$ $$L = \frac{1}{2} \sinh^{-1}(2) + \frac{1}{4} \cdot 4\sqrt{5}$$ $$L = \frac{1}{2} \sinh^{-1}(2) + \sqrt{5}$$

   Thus, the arc length is:

   $$\frac{1}{2} \sinh^{-1}(2) + \sqrt{5}$$