Question

If a_n =$\int_{0}^{\pi/2}{\frac{1 - \cos 2nx}{1 - \cos 2x}dx}$ then the value of determinant

$\left| \begin{matrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{matrix} \right|$is –

• A. 1
• B. –1
• C. 2
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

Here a_n + a_n+2

= $\int _ { 0 } ^ { \pi / 2 } \frac { 1 - \cos 2 n x } { 1 - \cos 2 x } d x$ +

$\int _ { 0 } ^ { \pi / 2 } \frac { 1 - \cos 2 ( \mathrm { n } + 2 ) \mathrm { x } } { 1 - \cos 2 \mathrm { x } } \mathrm { dx }$

=

= $\int_{0}^{\pi/2}{\frac{2 - 2\cos 2(n + 1)x.\cos 2x}{1 - \cos 2x}dx}$

Also 2.a_n+1 = 2$\int_{0}^{\pi/2}{\frac{1 - \cos 2(n + 1)x}{1 - \cos 2x}dx}$

\ a_n + a_n+2 – 2a_n+1 =

$2\int_{0}^{\pi/2}{\frac{\left\{ 1 - \cos 2(n + 1)x.\cos 2x - 1 + \cos 2(n + 1)x \right\}}{1 - \cos 2x}dx}$=$2\int_{0}^{\pi/2}{\frac{\cos 2(n + 1)x.\{ 1 - \cos 2x\}}{1 - \cos 2x}dx}$=$2\int_{0}^{\pi/2}{\cos 2(n + 1)xdx}$

\ a_n + a_n+2 – 2a_n+1 = 2 $\left\lbrack \frac{\sin 2(n + 1)x}{2(n + 1)} \right\rbrack_{0}^{\pi/2}$

= $\frac{1}{n + 1}$ [0 – 0] = 0, …(1)

for all n.

\ a_n+1 = $\frac{a_{n} + a_{n + 2}}{2}$

\ a_{n + 1} is the AM between a_n, a_n+2.

Now, D = $\left| \begin{matrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{matrix} \right|$ = $\frac { 1 } { 2 }$ $\left| \begin{matrix} a_{1} & 2a_{2} & a_{3} \\ a_{4} & 2a_{5} & a_{6} \\ a_{7} & 2a_{8} & a_{9} \end{matrix} \right|$

= $\frac { 1 } { 2 }$ $\left| \begin{matrix} a_{1} & 2a_{2} - (a_{1} + a_{3}) & a_{3} \\ a_{4} & 2a_{5} - (a_{4} + a_{6}) & a_{6} \\ a_{7} & 2a_{8} - (a_{7} + a_{9}) & a_{9} \end{matrix} \right|$, C₂ ® C₂, – C₁ – C₃

= $\frac { 1 } { 2 }$ $\left| \begin{matrix} a_{1} & 0 & a_{3} \\ a_{4} & 0 & a_{6} \\ a_{7} & 0 & a_{9} \end{matrix} \right|$, using (1)

\ D = 0.