# Ex 5 3 FSC Part1 Ver3 1

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www.mathcity.org FSC-I / Ex 5.3  - 1 MathCity.org Merging man and maths   Exercise 5.3 (Solutions) Textbook of Algebra and Trigonometry for Class XI Available online @ http://www.mathcity.org, Version: 3.1   Question # 1   2 97(1)(3)  x  x x  −+ +   Solution 2 97(1)(3)  x  x x  −+ +  Resolving it into partial fraction. 22 97(1)(3)(1)(3)  x Ax B C  x x x x  − += ++ + + +  Multiplying both sides by 2 (1)(3)  x x  + + . 2 97()(3)(1)...  x Ax B x C x  − = + + + + (i) Put 303  x x  + =  ⇒  = −  in equation (i). ( )( )  ( ) 2 9(3)7(3)0(3)1  A B C  − − = − + + − +   ( ) 277091 C  ⇒  − − = + +  3410 C  ⇒  − =  3410 C  ⇒  =−  175 C  ⇒  =−  Now equation ( i  ) can be written as 22 97(3)(3)(1)  x A x x B x C x  − = + + + + +  Comparing the coefficients of 2 ,  x x   and 0  x  . 0  A C  = +  … (ii) 93  B = +  … (iii) 73 B C  − = + +  … (iv) Putting value of C in equation (ii) 1705  A = −  175  A ⇒  =  Now putting value of  A  in equation (iii) 17935 B   = +     5195 B ⇒  = +  5195 B ⇒  − =  65 B ⇒  = −  Hence 22 17617555 97(1)(3)1(3)  x  x  x x x x  − −−= ++ + + +   2 1761755 1(3)  x   x x  − = −+ +   2 176175(1)5(3)  x  x x  −= −+ +    Answer    www.mathcity.org FSC-I / Ex 5.3  - 2  Question # 2 2 1(1)(1)  x x  + +   Solution   2 1(1)(1)  x x  + +  Now Consider 22 1(1)(1)11  Ax B C  x x x x  += ++ + + +  Multiplying both sides by 2 (1)(1)  x x  + + . 2 1()(1)(1).........  Ax B x C x  = + + + + (i) Put 101  x x  + =  ⇒  = −  in equation (i) ( ) 2 10(1)1 C  = + − +  12 C  ⇒  =  12 C  ⇒  =  Now eq. ( i  ) can be written as 22 1()(1)(1)  A x x B x C x  = + + + + +  Comparing the coefficients of 2 ,  x x   and 0  x  . 0  A C  = +  … (ii) 0  A B = +  … (iii) 1  A C  = +  … (iv) Putting value of C   in equation (ii) 102  A = +  12  A ⇒  = −  Putting value of  A  in equation (iii) 102 B = − +  12 B ⇒  =  Hence 22 111222 1(1)(1)11  x  x x x x  − + = ++ + + + 2 1122 11  x   x x  − + = ++ +   2 112(1)2(1)  x  x x  − += ++ + 2 112(1)2(1)  x  x x  −= ++ +    Answer Question # 3   2 37(4)(3)  x  x x  ++ +   Solution   2 37(4)(3)  x  x x  ++ +  Resolving it into partial fraction. 22 37(4)(3)43  x Ax B C  x x x x  + += ++ + + +      www.mathcity.org FSC-I / Ex 5.3  - 3 3322131313 ,,  Now do yourself you will get  A B and C   −    = = =      Question # 4 22 15(25)(1)  x  x x x  ++ + −   Solution   22 15(25)(1)  x  x x x  ++ + −  Resolving it into partial fraction. 222 15(25)(1)251  x Ax B C  x x x x x x  + += ++ + − + + −   22 15()(1)(25)...  x Ax B x C x x  ⇒  + = + − + + + (i) Put 101  x x  − =  ⇒  =  in equation (i) ( )  ( ) 22 (1)15(1)(0)(1)2(1)5  A B C  + = + + + +   ( ) 1150125 C  ⇒  + = + + +  168 C  ⇒  =  168 C  ⇒  =   2 C  ⇒  =   Now equation ( i  ) can be written as 222 15()(1)(25)  x A x x B x C x x  + = − + − + + +  Comparing the coefficients of 2 ,  x x   and 0  x  . 1  A C  = +  … (ii) 02  A B C  = − + +  … (iii) 155 B C  = − +  … (iv) Putting value of C   in equation (ii). 12 = +  12  A ⇒  − =  1  A ⇒  = −  Putting value of  A  and C   in equation (iii) 0(1)2(2) B = − − + +  014 B ⇒  = + +  05 B ⇒  = +  5 B ⇒  =−  Hence 222 15(1)52(25)(1)251  x x  x x x x x x  + − −= ++ + − + + −   2 52251  x  x x x  − −= ++ + −    Answer [ Question # 5 22 (4)(2)  x  x x  + +   Solution   22 (4)(2)  x  x x  + +  Resolving it into partial fraction.    www.mathcity.org FSC-I / Ex 5.3  - 4 222 (4)(2)42  x Ax B C  x x x x  += ++ + + +   1122 ,,1  Now do yourself you will get  A B and C   −    = = − =      Question # 6 23 11  x  x  ++   Solution   23 11  x  x  ++   22 1(1)(1)  x  x x x  +=+ − +   32 1(1)(1)  x x x x  + = + − + ∵  Now consider 222 1(1)(1)11  x A Bx C  x x x x x x  + += ++ − + + − +  … (A) ( )  ( )( ) 22 111  x A x x Bx C x  + = − + + + +   ( ) ( ) 222 11  x A x x Bx Bx Cx C  + = − + + + + +   ( ) ( ) ( ) 22 1  x A B x B C A x A C  + = + + + − + +  By comparing the coefficients of 2 ,  x x   and 0  x   1  A B + =  … (1) 0 B C A + − =  … (2) 1  A C  + =  … (3) From (3), We have 1 C A = −  Put the value of C   in (2). 21  A B − + =−  … (4) Subtract (1) from (4), We have 2323  A A =  ⇒  =  Put the value of  A  in (1) and (3), we have 11,33 B C  = =  Put the values of ,  A B andC   in (A), We have ( )( ) ( ) 222 112(1)(1)3131  x  x  x x x x  x x  ++= ++ − + + − +   Question # 7   22 22(3)(1)(1)  x x  x x x  + ++ + −  Solution 22 22(3)(1)(1)  x x  x x x  + ++ + −

Sep 22, 2019

#### Procedure Lab 3

Sep 22, 2019
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