Question 27 - CBSE Sample Paper Class 12 - 2017-18
Last updated at May 29, 2018 by Teachoo

Transcript

Question 27
Evaluate the following
1_(( )/4)^( /4) ( + /4)/(2 cos 2 )
1_(( )/4)^( /4) ( + /4)/(2 cos 2 )
1_(( )/4)^( /4) /(2 cos 2 )
1_(( )/4)^( /4) ( /4)/(2 cos 2 )
Let f(x) = /(2 cos 2 )
f( x) = ( )/(2 cos ( 2 ) )
As cos ( x) = cos x
f( x) = ( )/(2 cos )
f( x) = f(x)
Let f(x) = ( /4)/(2 cos 2 )
f( x) = ( /4)/(2 cos ( 2 ) )
As cos ( x) = cos x
f( x) = ( /4)/(2 cos )
f( x) = f(x)
Using property
If f(-x) = f(x)
_( )^ ( ) =0
Using property
If f(-x) = f(x)
_( )^ ( ) =2 _0^ ( )
1_(( )/4)^( /4) /(2 cos 2 )
1_(( )/4)^( /4) ( /4)/(2 cos 2 )
Thus,
I = I1 + I2
I
1_0^( /4) ( /4)/(2 cos 2 )
1_0^( /4) 1/(2 cos 2 )
1_0^( /4) 1/(2 (1 2 sin^2 ) )
1_0^( /4) 1/(1+2 sin^2 )
1_0^( /4) (1/cos^2 )/((1+2 sin^2 )/cos^2 )
1_0^( /4) sec^2 /(1+tan^2 +2 tan^2 )
Let tan x = t
sec2 x dx = dt
As x = 0 t = tan 0 = 0
As x = /4, t = tan /4 = 1
1_0^1 /3(1/3+t^2 )
1_0^1 /((1/ 3)^2+t^2 )
Using 1 /( ^2 + ^2 )= 1/ tan^( 1) /
= /6 [ 1/((1/ 3) ) tan^( 1) /((1/ 3) ) ]_0^1
= ( 3 )/6 [tan^( 1) 3(1) tan^( 1) 3(0) ]
= ( 3 )/6 [tan^( 1) 3 tan^( 1) 0 ]
= ( 3 )/6 [ /3 0]
= ( 3 )/6 [ /3]
= ( ^ )/
Question 27
Evaluate as the limit of a sum.
1_( 2)^2 (3 ^2 2 +4)
We know that
1 ( )
=( ) ( ) ( ) 1/ ( ( )+ ( + )+ ( +2 ) + ( +( 1) ))
Putting
= 2
=2
=(2 ( 2))/ =(2 + 2)/ =4/
Now,
( )=3 ^2 2 +4
( 2)=3( 2)^2 2( 2)+4=3(4)+4+4=20
( 2+ )=3 ( 2+ ) ^2 2( 2+ )+4 =3(4+ ^2 4 )+4 2 +4
=3 ^2 14 +20
( 2+2 )=3 ( 2+2 ) ^2 2( 2+2 )+4
=3(4+ 4 ^2 8 )+4 4 +4
" "=12 ^2 28 +20
( 2+3 )=3 ( 2+3 ) ^2 2( 2+3 )+4
=3(4+ 9 ^2 12 )+4 6 +4
" "=27 ^2 42 +20
.
( 2+( 1) )=3 ( 2+( 1) ) ^2 2( 2+( 1) )+4
=3(4+ ( 1)^2 ^2 4( 1) )+4 2( 1) +4
" "= 3( 1) ^2 ^2 14( 1) +20
Hence we can write it as
=(2 ( 2)) ( ) ( ) 1/
20+(3 ^2 14 +20)+(12 ^2 28 +20)+
20+(3 ^2 14 +20)+(12 ^2 28 +20)+
(27 ^2 42 +20) +
. +( 3( 1) ^2 ^2 14( 1) +20)
+ (3 ^2+12 ^2+27 ^2+ (3( 1)^2 ^2 )
(14 +28 +42 + ..( 1) )
+ 3h^2 (1+4+9+ +( 1)^2 )
14h(1+2+3+ ..+( 1))
+ 3h^2 (1^2+2^2+3^3+ +( 1)^2 )
14h(1+2+3+ ..+( 1))
We know that
1^2+2^2+ + ^2= ( ( + 1)(2 + 1))/6
1^2+2^2+ +( 1)^2
= (( 1) ( 1 + 1)(2( 1) + 1))/6
= (( 1) (2 2 + 1) )/6
= ( ( 1) (2 1) )/6
We know that
1+2+3+ + = ( ( + 1))/2
1+2+3+ +( 1)
= (( 1) ( 1 + 1))/2
= ( ( 1) )/2
+3h^2 ( ( 1)(2 1))/6 14 ( ( 1))/2
Putting h = 4/
=4 ( ) ( ) 1/
=4 ( ) ( ) 1/
=4 ( ) ( )
=4 ( ) ( )
=4
=4
=4
=4(20+16 28)
=4 8
=

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