CBSE Class 12 Sample Paper for 2018 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

### Evaluate Definite Integral ∫ -2 to 2 (3x 2  - 2x + 4) as the limit of a sum.

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/

### 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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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.