01_7exponential Series and Logarithmic Series

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7. EXPONENTIAL SERIES AND LOGARITHMIC SERIES PREVIOUS EAMCET BITS

1.

1 x 5x ( e + e ) = a 0 + a1x + a 2 x 2 + ..... ⇒ 2a1 + 23 a 3 + 25 a 5 + ....... = e3x 3) 1 1) e 2) e −1 Ans:

[EAMCET 2009] 4) 0

Sol: e −2x + e 2x ⎡ ( 2x )2 ( 2x )4 ⎤ = 2 ⎢1 + + + .......⎥ 2! 4! ⎢⎣ ⎥⎦ ⇒ a1 = 0, a 3 = 0, a 5 = 0..... and so on.

2.

1 1 1 + + + ...... = 1.3 2.5 3.7 1) 2 log e2 − 2 2) 2 − 2 log e2

[EAMCET 2008]

3) 2 log e4

4) log e4

Ans: 2 ∞ 1 1 1 1 + + + ...... = ∑ Sol: 1.3 2.5 3.7 n =1 n ( 2n + 1) ∞

∞ 1 1 ⎤ ⎡1 = 2∑ ⎢ − 2n + 1 ⎥⎦ n =1 2n ( 2n + 1) n =1 ⎣ 2n

= 2∑

⎡1 1 1 1 ⎤ = 2 ⎢ − + − + .....⎥ 2 3 4 5 ⎣ ⎦ ⎡ 1 1 1 ⎤ = 2 − 2 ⎢1 − + − + .....⎥ = 2 − 2 log e2 ⎣ 2 3 4 ⎦ 3.

The coefficient of xk in the expansion of 1− k − k2 k! Ans : 1

1)

Sol: Now

2)

k2 +1 k!

1 − 2x − x 2 is e− x 3)

1− k k!

[EAMCET 2007]

4)

1 − 2x − x 2 = (1 − 2x − x 2 ) e x e− x

⎡ ⎤ x 2 x3 xk = ⎣⎡1 − 2x − x 2 ⎦⎤ ⎢1 + x + + + .... + + ....∞ ⎥ 2! 3! k! ⎣ ⎦

⎡ ⎤ ⎛ ⎞ x3 xk x k +1 x2 xk + + .....∞ ⎥ = ⎜1 + x + + .... + + ....∞ ⎟ −2 ⎢ x + + ..... + 2! 2! k! ( k − 1)! k! ⎝ ⎠ ⎣ ⎦ ⎡ ⎤ x4 xk x k +1 − ⎢ x 2 + x 3 + + .... + + + .....∞ ⎥ 4! ( k − 2 )! ( k − 1)! ⎣ ⎦ 1

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1 k!

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Exponential series and logarithmic series ∴ coefficient of xk is

4.

1 2 1 − − k! ( k − 1) ! ( k − 2 ) !

1 2k k ( k − 1) 1 − k − k 2 − − = k! k! k! k! 1 1 1 1 − + 3− + ... = 2 2 2.2 3.2 4.24 1 ⎛3⎞ 1) 2) log 3 ⎜ ⎟ 4 ⎝4⎠

[EAMCET 2007] ⎛3⎞ 3) log e ⎜ ⎟ ⎝2⎠

⎛2⎞ 4) log e ⎜ ⎟ ⎝3⎠

Ans: 2 Sol: x −

x 2 x3 x 4 + − + ...... = log (1 + x ) 2 3 4 2

3

4

⎛1⎞ ⎛1⎞ ⎛1⎞ 1 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎛ 1⎞ ⎛3⎞ − + − + ..... = log ⎜1 + ⎟ = log 3 ⎜ ⎟ 2 2 3 4 ⎝ 2⎠ ⎝2⎠ 5.

1 − 2x is ex n (1 + 2n ) 2) ( −1) n!

The coefficient of xn in

1 + 2n n! Ans : 1 − 2x Sol: = (1 − 2x ) e− x ex

1)

⎡

(1 − 2x ) ⎢1 − ⎣

n!

6.

n

−

3) ( −1)

n

(1 − 2n ) n!

4) ( −1)

n

(1 + 4n ) n!

n ⎤ x x 2 x3 n x + + + ........ + ( −1) + .....⎥ 1! 2! 3! n! ⎦

Coefficient of xn in

( −1) =

[EAMCET 2006]

2 ( −1)

n

( n − 1)!

1 − 2x ex

⇒ (1)

If |x| < 1 and y = x −

n

(1 − 2n ) n!

x 2 x3 x 4 + − + ...... then x is equal to 2 3 4

1) y +

y 2 y3 + + ...... 2 3

2) y −

y 2 y3 y 4 + − + ..... 2 3 4

3) y +

y 2 y3 + + ..... 2! 3!

4) y −

y 2 y3 y 4 + − + ..... 2! 3! 4!

Ans: x 2 x3 x 4 Sol: y = x − + − + ..... 2 3 4 y = log e (1 + x ) ⇒ 1 + x = e y

2

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[EAMCET 2006]

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Exponential series and logarithmic series y2 + ..... 2!

⇒ 1+ x = 1+ y + ∴x = y +

y 2 y3 + + ..... 2! 3!

2n 2 + n + 1 = ∑ n! n =1 ∝

7.

[EAMCET 2005]

1) 2e − 1 Ans: 3

2) 2e + 1

3) 6e − 1

4) 6e + 1

2n 2 + n + 1 ∝ ⎡ 2n 2 n 1 ⎤ = ∑⎢ + + ⎥ n! n! n!⎦ n =1 n =1 ⎣ n! ∝

Sol: Let s = ∑

∝ ⎡ 2 3 1⎤ = ∑⎢ + + ⎥ ( n − 1)! n!⎦ n =1 ⎣ ( n − 2 ) !

⎡ 1 1 1 ⎤ ⎡ 1 1 ⎤ ⎡1 1 1 ⎤ = 2 ⎢1 + + + + .... ∝ ⎥ + 3 ⎢1 + + + .... ∝ ⎥ + ⎢ + + + ... ∝ ⎥ ⎣ 1! 2! 3! ⎦ ⎣ 1! 2! ⎦ ⎣1! 2! 3! ⎦ = 2e + 3e + e − 1 = 6e – 1 ∝

8.

ak If a < 1, b = ∑ , then a is equal to n =1 k ∝

1)

∑

( −1)

k

∝

bk

2)

k

n =1

∑ k =1

( −1)

k −1

[EAMCET 2005]

( −1) b k ∑ k =1 ( k − 1) ! ∝

bk

3)

k!

k

( −1) b k ∑ k =1 ( k + 1) ! ∝

4)

k −1

Ans: 2 ∝

ak a a 2 a3 = + + + .... ∝ 1 2 3 k =1 k!

Sol: b = ∑

b = − log e (1 − a ) ⎡ b b 2 b3 ⎤ e − b = 1 − a ⇒ a = 1 − e − b = 1 − ⎢1 − + − + .... ∝ ⎥ ⎣ 1! 2! 3! ⎦ ∝

∴a = ∑ k =1

( −1)

k −1

bk

k! ∝

9.

The value of the series ∴ a = ∑ 1) cosh ( x log ae )

k =1

( −1)

k −1

bk

[EAMCET 2004]

k!

2) coth ( x log ae )

3) sin h ( x log ae )

Ans: 3.

e x loge − e − x loge Sol: 2 a

a

⎡ e x − e− x ⎤ x3 x5 x = + + + ....⎥ ∵ ⎢ 2 3! 5! ⎣ ⎦

= sinh ⎡⎣ x log ae ⎤⎦

3

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4) tan h ( x log ae )

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Exponential series and logarithmic series 10.

Coefficient of x10 in the expansion of ( 2 + 3x ) e − x is 1) −

26 (10 )!

2) −

28 (10 )!

3) −

30 (10 )!

[EAMCET 2004]

4) −

32 (10 )!

Ans: 2 Sol:

⎡

( 2 + 3x ) e− x = ( 2 + 3x ) ⎢1 −

x x 2 x3 x 9 x10 ⎤ ..... + − ...... + 1! 2! 3! 9! 10! ⎥⎦

⎣ ∴ Coefficient of x in the above series 2 3 1 28 = − = ( 2 − 30 ) ⇒ − 10! 9! 10! 10! 1 1+ 2 1+ 2 + 3 11. + + + .... is equal to 2! 3! 4! e e 1) 2) 2 3 Ans: 1 ∝ 1 + 2 + 3 + ..... + n Sol: = ∑ ( n + 1)! n =1 10

[EAMCET 2003]

3)

e 4

4)

e 5

n ( n + 1) 1 ∝ 1 = ∑ 2 n =1 ( n − 1) ! n =1 2 ( n + 1) n ( n − 1) ! ∝

=∑ = 12.

1⎡ 1 1 1 e ⎤ 1 1 + + + + .....⎥ = e = ⎢ 2 ⎣ 1! 2! 3! 2 ⎦ 2

If 0 < y < 21/ 3 and x ( y3 − 1) = 1 , then

⎛ y3 ⎞ 1) log ⎜ 3 ⎟ ⎝ 2− y ⎠ Ans: 1 1 Sol: y3 − 1 = x

⎛ y3 ⎞ 2) log ⎜ 3 ⎟ ⎝1− y ⎠

2 2 2 + 3 + 5 + ...... is equal to x 3x 5x ⎛ 2y3 ⎞ 3) log ⎜ 3 ⎟ ⎝ 1− y ⎠

[EAMCET 2003]

⎛ y3 ⎞ 4) log ⎜ 3 ⎟ ⎝ 1 − 2y ⎠

x3 x5 1 ⎛ 1+ x ⎞ x + + + ...... = log ⎜ ⎟ 3 5 2 ⎝ 1− x ⎠ 3 5 ⎡ ⎤ ⎛1⎞ ⎛1⎞ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ 1 1 x x ⎛ 1 + 1/ x ⎞ = 2 ⎢ + ⎝ ⎠ + ⎝ ⎠ + .....⎥ ⇒ 2 × log ⎜ ⎟ ⎢x ⎥ 3 5 2 ⎝ 1 − 1/ x ⎠ ⎢ ⎥ ⎣ ⎦

⎡1 + y 3 − 1 ⎤ ⎡ y3 ⎤ = log ⎢ = log ⎥ ⎢ 2 − y3 ⎥ 3 ⎣1 − y + 1 ⎦ ⎣ ⎦

13.

3 x2 x3 a 2 1 + x log + ( log e ) + ( log ae ) + ...... ( a > 0, x ∈ R ) is equal to 2! 3! a e

4

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[EAMCET 2002]

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Exponential series and logarithmic series

Sol: 1 +

x

2) ax

1) a Ans: 2

3) a loge

4) x

x x2 x3 + + + ........ = e x 1! 2! 3!

a ax x log ae ( x log e ) 1+ + + ........ = e x loge = eloge = a x 1! 2!

a 2

14.

1+

1 + 2 1 + 2 + 22 + + ..... is equal to 2! 3!

1) e 2 + e Ans: 4

Sol:

2

∑

3) e 2 − 1

2) e 2

1 + 2 + 2 + ..... + 2 n! n =1 ∝

n −1

[EAMCET 2002]

∝

=∑ n =1

(2 1

n

4) e 2 − e

− 1)

2 −1 n!

2 −1 2 1 = ∑ −∑ n! n =1 n =1 n! n =1 n! ∝

n

=∑

∝

n

∝

= ( e 2 − 1) − ( e − 1) = e2 − e 15.

2 2+4 2+4+6 + + + ...... 2! 3! 4! 1) e 2) e −1 Ans: 1

[EAMCET 2001] 3) e−2

4) e−3

∝ 2 (1 + 2 + 3 + .....n ) 2 + 4 + 6 + ......... + 2n ⇒∑ ( n + 1)! ( n + 1)! n =1 n =1 ∝

Sol: = ∑

∝ n ( n + 1) 1 ⇒∑ n =1 ( n + 1) n ( n − 1) ! n =1 ( n − 1) ! ∝

=∑ = 1+ 16.

1 1 1 + + + ....... = e 1! 2! 3!

|x| < 1, the coefficient x3 in the expansion of log (1 + x + x 2 ) in ascending powers of x is [EAMCET 2001]

2 3 Ans: 3

1)

2)

4 3

3) −

2 3

4) −

⎡ (1 + x + x 2 ) (1 − x ) ⎤ ⎡1 − x 3 ⎤ ⎥ = log ⎢ Sol: We have log (1 + x + x 2 ) = log ⎢ ⎥ 1− x ⎢⎣ ⎥⎦ ⎣ 1− x ⎦ ⎡ ⎤ ⎡ ⎤ x6 x9 x2 x3 = log ⎡⎣1 − x 3 ⎤⎦ − log [1 − x ] = − ⎢ x 3 + + + ......⎥ + ⎢ x + + + ......⎥ 2 3 2 3 ⎣ ⎦ ⎣ ⎦

5

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4 3

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Exponential series and logarithmic series 1 2 Coefficient of x3 is −1 + = − 3 3

6

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