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Let S=x-(x^(3))/(3!)+(x^(5))/(5!)… and C...

Let `S=x-(x^(3))/(3!)+(x^(5))/(5!)`… and `C=1-(x^(2))/(2!)+(x^(4))/(4!)` Then,

A

`C^(2)+S^(2)` is not independent of x

B

2CS = sin 2x

C

`C^(2)-S^(2)` is independent of x

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
a,b,c
We have
`C+is =(1-(x^(2))/(2!)+(x^(4))/(4!)+…)+i(x-(x^(3))/(3!)+(x^(5))/(5!)..)`
`rarr C+is =e^(ix)=cos x +I sin x`
`therefore C= cos x and s= sin x`
`rarr 2CS = sin 2x , C^(2)+S^(2)=1 and C^(2)-S^(2)=cos 2x`
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OBJECTIVE RD SHARMA ENGLISH-EXPONENTIAL AND LOGARITHMIC SERIES-Section I - Solved Mcqs
  1. In the expansion of log(10)(1-x),|x|lt1 the coefficient of x^(n) is

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  2. Sum of series 9/(1!)+19/(2!)+35/(3!)+57/(4!)+... (A) 7e-3 (B) 12e-5...

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  3. The constant term in the expansion of (3^(x)-2^(x))/(x^(2)) is

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  4. Sigma(n=1)^(oo) (x^(2n))/(2n-1) is equal to

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  5. Then sum of the series 1+(1+3)/(2!)x+(1+3+5)/(3!)x^(2)+(1+3+5+7)/(4...

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  6. 1/(e^(3x))(e^x+e^(5x))=a0+a1x+a2x^2+........=>2a1+2^3a3+2^5a5+......=

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  7. Let S=x-(x^(3))/(3!)+(x^(5))/(5!)… and C=1-(x^(2))/(2!)+(x^(4))/(4!) T...

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  8. The sum of series 2/(3!)+4/(5!)+6/(7!)+...........oo is :

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  9. The sum of the series S=Sigma(n=1)^(infty)(1)/(n-1)! is

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  10. The sum of the series loge(3)+(loge(3))^3/(3!)+(loge(3))^5/(5!)+....+ ...

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  11. The value of 1+(log(e)x)+(log(e)x)^(2)/(2!)+(log(e)x)^(3)/(3!)+…inft...

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  12. (1+3)loge3+(1+3^2)/(2!)(loge3)^2+(1+3^3)/(3!)(loge 3)^3+....oo= (a)28...

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  13. If agt0 and x in R, then 1+(xlog(e)a)+(x^(2))/(2!)(log(e)a)^(2)+(x^(...

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  14. (2)/(2!)+(2+4)/(3!)+(2+4+6)/(4!)+….infty is equal to

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  15. 1+(2x)/(1!)+(3x^(2))/(2!)+(4x^(3))/(3!)+..infty is equal to

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  16. 1/2(1/2+1/3)-1/4((1)/(2^(2))+(1)/(3^(2)))+1/6((1)/(2^(3))+(1)/(3^(3)))...

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  17. The sum of series (1)^(2)/(1.2!)+(1^(2)+2^(2))/(2.3!)+(1^(2)+2^(2)+3...

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  18. The sum of the series (1)/(2!)+(1)/(4!)+(1)/(6!)+..to infty is

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  19. If 0ltylt2^(1//3) and x(y^(3)-1)=1 then (2)/(x)+(2)/(3x^(3))+(2)/(5x...

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  20. The sum of the series (1)/(2!)+(1+2)/(3!)+(1+2+3)/(4!)+..to infty is...

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