Question

Let a integral is given as ${a_n} = \int_{ - \pi }^\pi {\left| {x - 1} \right|\cos nxdx}$ for all natural numbers n. Then the sequence ${\left( {{a_n}} \right)_{n \geqslant 0}}$ satisfies.
$\left( a \right)\mathop {\lim }\limits_{n \to \infty } {a_n} = \infty$
$\left( b \right)\mathop {\lim }\limits_{n \to \infty } {a_n} = - \infty$
$\left( c \right)\mathop {\lim }\limits_{n \to \infty } {a_n}$ exists and is positive.
$\left( d \right)\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$

Answer 1

In this particular question use the concept of breaking the integration limit by using the property that $\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( x \right)dx} + \int_b^c {f\left( x \right)dx}$ where, $a < c < b$, we break this because of the modulus function, so we have to break the integration limit into which modulus function is negative and into which modulus function is positive, so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given integral:
${a_n} = \int_{ - \pi }^\pi {\left| {x - 1} \right|\cos nxdx}$
Now substitute x – 1 = 0
Therefore, x = 1.
So from $\left( { - \pi {\text{ to 1}}} \right)$ |x – 1| is negative and from $\left( {{\text{1 to }}\pi } \right)$ |x – 1| is positive.
So break the integration limits accordingly we have,
$\Rightarrow {a_n} = \int_{ - \pi }^1 { - \left( {x - 1} \right)\cos nxdx} + \int_1^\pi {\left( {x - 1} \right)\cos nxdx}$
$\Rightarrow {a_n} = \int_{ - \pi }^1 {\left( {1 - x} \right)\cos nxdx} + \int_1^\pi {\left( {x - 1} \right)\cos nxdx}$
Now as we know the formula of integration by parts which is given as, $\int {\left( {{I_1}} \right)\left( {{I_2}} \right)dx} = {I_1}\int {{I_2}dx} - \int {\left( {\dfrac{d}{{dx}}{I_1}\int {{I_2}dx} } \right)dx} + C$, where C is some arbitrary integration constant, so use this property in the above equation we have,
Where, in first integral ${I_1} = \left( {1 - x} \right),{I_2} = \cos nx$ and in second integral ${I_1} = \left( {x - 1} \right),{I_2} = \cos nx$ so we have,
\Rightarrow {a_n} = \left[ {\left\\{ {\left( {1 - x} \right)\int {\cos nxdx} } \right\\}_{ - \pi }^1 - \int\limits_{ - \pi }^1 {\left( {\dfrac{d}{{dx}}\left( {1 - x} \right)\int {\cos nxdx} } \right)} dx} \right] + \left[ {\left\\{ {\left( {x - 1} \right)\int {\cos nxdx} } \right\\}_1^\pi - \int\limits_1^\pi {\left( {\dfrac{d}{{dx}}\left( {x - 1} \right)\int {\cos nxdx} } \right)} dx} \right]
Now as we know that $\int {\cos nxdx} = \dfrac{{\sin nx}}{n} + C,\dfrac{d}{{dx}}1 = 0,\dfrac{d}{{dx}}x = 1$ so use these properties in the above equation we have,
\Rightarrow {a_n} = \left[ {\left\\{ {\left( {1 - x} \right)\left( {\dfrac{{\sin nx}}{n}} \right)} \right\\}_{ - \pi }^1 - \int\limits_{ - \pi }^1 {\left( {\left( {0 - 1} \right)\dfrac{{\sin nx}}{n}} \right)} dx} \right] + \left[ {\left\\{ {\left( {x - 1} \right)\left( {\dfrac{{\sin nx}}{n}} \right)} \right\\}_1^\pi - \int\limits_1^\pi {\left( {\left( {1 - 0} \right)\dfrac{{\sin nx}}{n}} \right)} dx} \right]
\Rightarrow {a_n} = \left[ {\left\\{ {\left( {1 - x} \right)\left( {\dfrac{{\sin nx}}{n}} \right)} \right\\}_{ - \pi }^1 + \int\limits_{ - \pi }^1 {\dfrac{{\sin nx}}{n}} dx} \right] + \left[ {\left\\{ {\left( {x - 1} \right)\left( {\dfrac{{\sin nx}}{n}} \right)} \right\\}_1^\pi - \int\limits_1^\pi {\dfrac{{\sin nx}}{n}} dx} \right]
Now as we know that $\int {\sin nxdx} = \dfrac{{ - \cos nx}}{n} + C$ so we have,
\Rightarrow {a_n} = \left[ {\left\\{ {\left( {1 - x} \right)\left( {\dfrac{{\sin nx}}{n}} \right)} \right\\}_{ - \pi }^1 + \left[ {\dfrac{{ - \cos nx}}{{{n^2}}}} \right]_{ - \pi }^1} \right] + \left[ {\left\\{ {\left( {x - 1} \right)\left( {\dfrac{{\sin nx}}{n}} \right)} \right\\}_1^\pi - \left[ {\dfrac{{ - \cos nx}}{{{n^2}}}} \right]_1^\pi } \right]
\Rightarrow {a_n} = \left[ {\left\\{ {\left( {1 - x} \right)\left( {\dfrac{{\sin nx}}{n}} \right)} \right\\}_{ - \pi }^1 - \left[ {\dfrac{{\cos nx}}{{{n^2}}}} \right]_{ - \pi }^1} \right] + \left[ {\left\\{ {\left( {x - 1} \right)\left( {\dfrac{{\sin nx}}{n}} \right)} \right\\}_1^\pi + \left[ {\dfrac{{\cos nx}}{{{n^2}}}} \right]_1^\pi } \right]
Now apply integration limits we have,
$\Rightarrow {a_n} = \left[ { \left( {1 + \pi } \right)\left[ {\dfrac{{\sin \left( { - n\pi } \right)}}{n}} \right] - \left[ {\dfrac{{\cos n}}{{{n^2}}} - \dfrac{{\cos \left( { - n\pi } \right)}}{{{n^2}}}} \right]} \right] + \left[ {\left( {\pi - 1} \right)\left[ {\dfrac{{\sin n\pi }}{n}} \right] + \left[ {\dfrac{{\cos n\pi }}{{{n^2}}} - \dfrac{{\cos n}}{{{n^2}}}} \right]} \right]$
$\Rightarrow {a_n} = \left[ {\left( {1 + \pi } \right)\left[ {\dfrac{{\sin n\pi }}{n}} \right] - \left[ {\dfrac{{\cos n}}{{{n^2}}} - \dfrac{{\cos n\pi }}{{{n^2}}}} \right]} \right] + \left[ {\left( {\pi - 1} \right)\left[ {\dfrac{{\sin n\pi }}{n}} \right] + \left[ {\dfrac{{\cos n\pi }}{{{n^2}}} - \dfrac{{\cos n}}{{{n^2}}}} \right]} \right]$,
$\because \sin (-x) = - \sin x, \cos (-x) = \cos x.$
Now simplify it we have,
$\Rightarrow {a_n} = \dfrac{{\sin n\pi }}{n}\left( {1 + \pi + \pi - 1} \right) + \dfrac{{\cos n}}{{{n^2}}}\left( { - 1 - 1} \right) + \dfrac{{\cos n\pi }}{{{n^2}}}\left( {1 + 1} \right)$
$\Rightarrow {a_n} = 2\pi \dfrac{{\sin n\pi }}{n} - 2\dfrac{{\cos n}}{{{n^2}}} + 2\dfrac{{\cos n\pi }}{{{n^2}}}$
Now it is given that n belongs to natural number, so $\sin n\pi = 0$ (always)
$\Rightarrow {a_n} = 2\pi \dfrac{0}{n} - 2\dfrac{{\cos n}}{{{n^2}}} + 2\dfrac{{\cos n\pi }}{{{n^2}}}$
$\Rightarrow {a_n} = - \dfrac{2}{{{n^2}}}\left( {\cos n - \cos n\pi } \right)$
Now as we know that $\cos C - \cos D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{D - C}}{2}} \right)$ so we have,
$\Rightarrow {a_n} = - \dfrac{2}{{{n^2}}}\left( {2\sin \left( {\dfrac{{n + n\pi }}{2}} \right)\sin \left( {\dfrac{{n\pi - n}}{2}} \right)} \right)$
$\Rightarrow {a_n} = - \dfrac{4}{{{n^2}}}\left( {\sin \left( {\dfrac{{n + n\pi }}{2}} \right)\sin \left( {\dfrac{{n\pi - n}}{2}} \right)} \right)$
$\Rightarrow {a_n} = - 4\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi + 1}}{2}} \right)}}{n}} \right]\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi - 1}}{2}} \right)}}{n}} \right]$
The above equation is also written as,
$\Rightarrow {a_n} = - 4\left[ {\left( {\dfrac{{\pi + 1}}{2}} \right)\dfrac{{\sin n\left( {\dfrac{{\pi + 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi + 1}}{2}} \right)}}} \right]\left[ {\left( {\dfrac{{\pi - 1}}{2}} \right)\dfrac{{\sin n\left( {\dfrac{{\pi - 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi - 1}}{2}} \right)}}} \right]$
$\Rightarrow {a_n} = - 4\left( {\dfrac{{\pi + 1}}{2}} \right)\left( {\dfrac{{\pi - 1}}{2}} \right)\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi + 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi + 1}}{2}} \right)}}} \right]\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi - 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi - 1}}{2}} \right)}}} \right]$
$\Rightarrow {a_n} = - \left( {{\pi ^2} - 1} \right)\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi + 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi + 1}}{2}} \right)}}} \right]\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi - 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi - 1}}{2}} \right)}}} \right]$, $\left[ {\because \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)} \right]$
$\Rightarrow {a_n} = \left( {1 - {\pi ^2}} \right)\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi + 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi + 1}}{2}} \right)}}} \right]\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi - 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi - 1}}{2}} \right)}}} \right]$
Now apply $\mathop {\lim }\limits_{n \to \infty }$ on both sides we have,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } \left( {1 - {\pi ^2}} \right)\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi + 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi + 1}}{2}} \right)}}} \right]\left[ {\dfrac{{\sin n\left( {\dfrac{{\pi - 1}}{2}} \right)}}{{n\left( {\dfrac{{\pi - 1}}{2}} \right)}}} \right]$
Now as we know that $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\sin x}}{x} = 0,\mathop {\lim }\limits_{n \to 0} \dfrac{{\sin x}}{x} = 1$, so use this property we have,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {a_n} = \left( {1 - {\pi ^2}} \right)\left[ 0 \right]\left[ 0 \right]$
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } {a_n} = 0$
So this is the required answer.
Hence option (d) is the correct answer.

Note: Whenever we face such types of problems the key concept we have to remember is that always recall the formula of integration by parts which is stated above and also recall the basic integration and differentiation properties such as $\int {\cos nxdx} = \dfrac{{\sin nx}}{n} + C,\dfrac{d}{{dx}}1 = 0,\dfrac{d}{{dx}}x = 1$ along with standard limit properties such as $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\sin x}}{x} = 0,\mathop {\lim }\limits_{n \to 0} \dfrac{{\sin x}}{x} = 1$.