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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
`e^{(x-1)-1/2(x-1)^2+((x-1)^3)/3-(x-1)^(4)/4+......}` is eqaul toA. `log(x-1)`B. `logx`C. xD. none of these |
| Answer» Correct Answer - c | |
| 2. |
`1+(log_e n)^2 /(2!) + (log_e n )^4 / (4!)+...=`A. nB. `(1)/(n)`C. `(n+n^(-1))/(2)`D. `(e^(n)+e^(-n))/(2)` |
| Answer» Correct Answer - c | |
| 3. |
If n is not a multiple of 3, then the coefficient of `x^n` in the expansion of `log_e (1+x+x^2)`is : (A) `1/n` (B) `2/n` (C) `-1/n` (D) `-2/n`A. `-(2)/(n)`B. `(1)/(n)`C. `(2)/(n)`D. none of these |
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Answer» Answer: We have Coefficient of `x^(n) in log (1+x+x^(2))` `=(1)/(n)(w^(n)+w^(2n)` `=(1)/(n) [before w^(n)+w^(2n)=-1` when n is not a mutiple of 3] |
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| 4. |
The sum of the series `1/2x^2+2/3x^3+3/4x^4+4/5x^5+`... is :A. `(x)/(1+x)+log(1+x)`B. `(x)/(1-x)+log(1-x)`C. `-(x)/(1+x)+log(1+x)`D. none of these |
| Answer» Correct Answer - b | |
| 5. |
`(1)/(n!)+(1)/(2!(n-2)!)+(1)/(4!(n-4)!)`+….to n terms is equal toA. `(2n-1)/(2!)`B. `(2^(n))/(n+1)!`C. `(2^(n))/(n!)`D. `(2^(n)-2)/(n-1)!` |
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Answer» Answer: We have `(1)n!+(1)/(2!(n-2)!)+(1)/(4!(n-4)!)`+… `=(1)/(n!){(1+.^(n)C_(2)+.^(n)C_(4)+…}` `=(1)/(n!){.^(n)C_(0)+.^(n)C_(2)+.^(n)C_(4)+…}=(1)/(n!)(2^(n-1))=(2^(n-1))/(n!)` |
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| 6. |
The coefficient of `x^(6)` in the expansion of `log{(1+x)^(1+x)(1-x)^(1-x)`}` isA. `(1)/(15)`B. `(1)/(30)`C. `(1)/(10)`D. `(1)/(45)` |
| Answer» Correct Answer - a | |
| 7. |
The value of `9+(16)/(2!)+(27)/(3!)+(42)/(4!)+……infty` isA. `9e-6`B. `11e-6`C. `13e-6`D. `12e-6` |
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Answer» Answer: Consider the series 9+16+27+42+…. We observe that the successive differences of the terms of this series are 7,11,15… Clearly these are in A.P so let its `n^(th)` term be `t_(n)=an^(2)+bn+c` `rarr t_(1)=a+b+c rarr a+b+c=9` `t_(2)=4 a+2b+c rarr 4a+2b+c=16` `t_(3)=9a+3b+c rarr 9a+3b+c=27` Solving these equation we get a=2,b=1 and c =6 `therefore t_(n)=2n^(2)+n+6` Thus we have `9+(16)/(2!)+(27)/(3!)+(42)/(4!)+...infty` `=underset(n=1)overset(infty)Sigma(2n^(2)+n+6)/(n!)` `=2underset(n=1)overset(infty)Sigma(n^(2))/(n!)+underset(n=1)overset(infty)Sigma(n)/(n!)+6 underset(n=1)overset(infty)Sigma(1)/(n!)` =11e-6 |
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| 8. |
Sum of n terms of the series `1/(1.2.3.4.)+1/(2.3.4.5) +1/(3.4.5.6)+....`A. `1/6log_(e)2-(1)/(24)`B. `5/2-log_(e)2`C. `3/2-log_(e)2`D. none of these |
| Answer» Correct Answer - a | |
| 9. |
The sum of series ` 2[ 7^(-1)+3^(-1).7^(-3)+5^(-1).7^(-5)+...]` isA. `log_(e)(4/3)`B. `log_(e)(3/4)`C. `2log_(e)(3/4)`D. `2 loge(4/3)` |
| Answer» Correct Answer - a | |
| 10. |
The value of `log_(e) e- log_(9) e + log_(27) e- log_(81) e+…infty` isA. `log_(2)3`B. `log_(3)2`C. `log_(10)e`D. `log_(e)2` |
| Answer» Correct Answer - b | |
| 11. |
The value of `1+(log_(e)x)+(log_(e)x)^(2)/(2!)+(log_(e)x)^(3)/(3!)+…infty`A. 2B. `1/2`C. `log_(e)3`D. none of these |
| Answer» Correct Answer - a | |
| 12. |
The sum of the series `(x-1)/(x+1)+1/2(x^(2)-1)/((x+1)^(2)+1/3(x^(3)-1)/(x+1)^(3)`+…is equal toA. `log_(e)x`B. `2 log_(e)x`C. `-log_(e)(x+1)`D. none of these |
| Answer» Correct Answer - a | |
| 13. |
Let x and a be positive real numbers Statement 1: The sum of theries `1+(log_(e)x)^(2)/(2!)+(log_(e)x)^(3)/(3!)+(log_(e)x)^(2)+…to infty is x` Statement2: The sum of the series `1+(xlog_(e)a)+(x^(2))/(2!)(log_(e)x))^(2)+(x^(3))/(3!)(log_(e)a)^(3)/(3!)+to infty is a^(x)`A. 1B. 2C. 3D. 4 |
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Answer» Answer: We have `1+(x log_(e)a)+(x^(2))/(2!)(log_(e)a)^()+(x^(3))/(3!)(log_(e)a)^(3)+ to infty` `=e^(xlog_(e) a)=e^(log_(e)a^(x)=a^(x)` So statement 2 is true Replacing x by 1 and a by x in statemnt 2 we obtain statement 1 so statement 1 is true and statement 2 is a correct explantion for statement-1 |
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| 14. |
Statement 1: If `log(1-x+x^(2))=a_(1)x+a_(2)x^(32)+a_(3)x^(3)+..` then `a_(3)+a_(6)+a_(9)+..=2/3log_(e)2` Statement 2: `1-1/2+1/3-11/4+1/5-1/6+..=log_(e)2`A. 1B. 2C. 3D. 4 |
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Answer» Answer: We have `log(1-x+x^(2))=a_(1)x+a_(2)x^(2)+a_(3)x^(3)+...` `rarr log(1-x+x^(2))=x(a_(1)+a_(4)x^(3)+a_(7)x^(6)+..)` `+x^(2)(a_(2)+a_(5)x^(3)+..)+(a_(3)x^(3)+a_(6)x^(6)+…)` Replacing x by 1 w and `w^(2)` respectively we get `0=e_(1)+E_(2)+E_(3)` `log(-2w)=wE_(1)+w^(2)E_(2)+E_(3)` `log(-2w^(2))=w^(2)E_(1)+wE_(2)+E_(3)` When `E_(1)=a_(1)+a_(4)+a_(7)++...` `E_(2)=a_(2)+a_(5)+..` `E_(3)=A_(3)+a_(6)+..` Adding i,ii and iii we get `log4=3 E_(3)rarrE_(3)=2/3log_(e)2` Clearly statement 2 is also true |
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| 15. |
The coeffiecent of `n^(-r)` in the expansion of `log_(10)((n)/(n-1))` isA. `(1)/(rlog_(e)10)`B. `-(1)/(rlog_(e)10)`C. `(1)/(r!log_(e)10)`D. `log_(e)1-(1)(n)/(log_(e))` |
| Answer» Correct Answer - a | |
| 16. |
The coefficient of `x^(n)` in the exansion of `log_(e)(1+3x+2x^(2))` isA. `(-1)^(n)((2^(n)+1)/(n))`B. `((-1)^(n+1))/(n)(2^(n)+1)`C. `(2^(n+1))/(n)`D. none of these |
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Answer» Answer: We have `log(1+3x+2x^(2))` `=log(1+x)+log(1+2x)` `=underset(n=1)overset(infty)Sigma(-1)^(n-1)(x^(n))/(n)+underset(n=1)overset(infty)Sigma(-1)^(n-1)(2x)^(n)/(n)` ltbgt `=underset(n=1)overset(infty)Sigma(-1)^(n-1)((1)/(n)+(2^(n))/(n))x^(n)` `therefore "coefficeient of" x^(n)=(-1)^(n-1)((2^(n)+1)/(n))=(-1)^(n+1)(2^(n)+1)/(n))` |
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| 17. |
In the expansion of `log_(10)(1-x),|x|lt1` the coefficient of `x^(n)` isA. `-(1)/(n)`B. `-(1)/(n)log10^(e )`C. `(1)/(n)`D. `(1)/(n)log10^(e )` |
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Answer» Answer: `log_(10)(1-x)=log_(10)exx(log_(e))(1-x)=-(log_(10)e)underset(r=1)overset(infty)Sigma (x^(r ))/(r )` coefficient of `x^(n) in log_(10)(1-x)=(-log_(10)e)/(n)` |
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| 18. |
2log x-log(x+1)-log(x-1) is equals toA. `x^(2)+1/2x^(4)+1/3x^(6)`+..B. `(1)/(x^(2))+(1)/(2x^(4))+(1)/(3x^(6))`+..C. `-{(1)/(x^(2))+(1)/(2x^(4))+(1)/(3x^(6))`+..D. `-(1)/(n)(w)^(n)+w^(2n)` |
| Answer» Correct Answer - b | |
| 19. |
The sum of the series `1+(2^(4))/(2!)+(3^(4))/(3!)+(4^(4))/(4!)+(5^(4))/(5!)`+…..isA. 12eB. 5eC. 14eD. 15e |
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Answer» Answer: We have `1+(2^(4))/(2!)+(3^(4))/(3!)+(4^(4))/(4!)+(5^(4))/(5!)+…underset(n=1)overset(infty)Sigma (n^(4))/(n!)` Let `n^(4)=a_(0)+a_(1)n+a_(2)n(n-1)` `+a_(3)n(n-1)(n-2)+a_(4)n(n-1)(n-32)(n-3)` On equating the coefficients of like poweres of n on both sides we get `a_(0)=0,a_(1)-a_(2)+2a_(3)-6a_(4)=0,a_(2)-3a_(3)+11a_(4)=0a_(3)-6a_(4) and a_(4)=1` `rarr a_(0=0a_(1)=1 , a_(2)=7 , a_(3)=6,a_(4)=1` substituting the value of `a_(0),a_(1),a_(2),a_(3),a_(4)` (i) we get `n^(4)=n+7n(n-1)+6n(n-1)(n-2)+n(n-21)+n(n-1)(n-2)(n-3)` `therefore 1+(2^(4))/(2!)+(3^(4))/(3!)+(4^(4))/(4!)`+... `=underset(n=1)overset(infty)Sigma(1)/(n-1)!+7 underset(n=2)overset(infty)Sigma(1)/(n-2)!+6 underset(n=3)overset(infty)Sigma(1)/(n-3)!+underset(n=4)overset(infty)Sigma(1)/(n-3)!+underset(n=4)overset(infty)Sigma(1)/(n-4)!` `=e+7e+6e+e=15e` |
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| 20. |
The sum of the series `x+(2^(3))/(2!)x^(2)+(3^(3))/(3!)x^(3)+(4^(3))/(4!)x^(4)`+……..to `infty` isA. `(x+x^(2)+x^(3))e^(x)`B. `(x^(2)+x^(3))e^(x)`C. `(x+3x^(2)+x^(3))e^(x)`D. `x^(3)e^(x)` |
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Answer» Answer: We have `(x+2^(3))/(2!)x^(2)(3^(3))/(3!)x^(3)+(4^(3))/(4!)x^(4)+….` `=underset(n=1)overset(infty)Sigma (n^(3))/(n!)x^(n)` `=underset(n=1)overset(infty)Sigma(n+3n(n-1)+n(n-1)(n-2))/(n!)x^(n)` `underset(n=1)overset(infty)Sigma{(n)/(n!)+(3n(n-1))/(n!)+(n(n-1)(n-2))/(n!)}x^(n)` `=underset(n=1)overset(infty)Sigma(x^(n))/(n-1)!+3underset(n=1)overset(infty)Sigma(x^(n))/(n-2)! underset(n=1)overset(infty)Sigma(x^(n))/(n-3)!` `=xe^(x)+3x^(2)e^(x)+x^(3)e^(x)=(x+3x^(2)+x^(3))e^(x)` |
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| 21. |
The sum of the series `1+(2^(3))/(2!)+(3^(3))/(3!)+(4^(3))/(4!)`+… to `infty` isA. 2eB. 3eC. 4eD. 5e |
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Answer» Answer: We have `1+(2^(3))/(2!)+(3^(3))/(3!)+(4^(3))/(4!)+…underset(n=1)overset(infty)Sigma(n^(3))/(n!)` Let `n^(3)=a_(0)+a_(1)n+a_(2)n(n-1)+a+_(3)n(n-1)(n-2)` By comparing the coefficients of like powers of n on both sides we get Substituting the value in (i) we get `n^(3)=n+3n(n-1)+n((n-1)(n-2)` `therefore 1+(2^(3))/(2!)+(3^(3))/(3!)+(4^(3))/(4!)+...` `-underset(n=1)overset(infty)Sigma(n^(3))/(n!)` `=underset(n=1)overset(infty)Sigma(n+3n(n-1)+n(n-1)(n-2))/(n!)` =e+3e+e=5e |
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| 22. |
The sum of the series `x+(2^(2))/(2!)x^(2)+(3^(2))/(3!)x^(3)`+…. to `infty` isA. `x^(2)e^(x)`B. `(x+x^(2))e^(x)`C. `(x+1)e^(x)`D. `(2x+x^(2))e^(x)` |
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Answer» Answer: We have `x+(2^(2))/(2!)x^(2)+(3)^(2)/(3!)x^(3)+….infty` `=underset(n=1)overset(infty)Sigma (n+n(n-1))/(n!).x^(n)` `=xunderset(n=1)overset(infty)Sigma(x^(n-1))/(n-1)!+x^(2)underset(n=2)overset(infty)Sigma(x^(n-2))/(n-2)!` `=xe^(x)+x^(2)e^(x)=(x+x^(2))e^(x)` |
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| 23. |
The sum of the series `(1^(2))/(1!)+(2^(2))/(2!)+(3^(2))/(3!)`+……to `infty` isA. eB. 2eC. 3eD. none of these |
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Answer» Answer: We have `(1)^(2)/(1!)+(2)^(2)/(2!)+(3)^(2)/(3!)+…….underset(n=1)overset(infty)Sigma(n)^(2)/(n!)` Let `n^(2)=a_(0)+a_(1)=a_(2)=1` By comparing the coefficeints of like poweres of n both sides we get `a_(0)=0,a_(1)-a_(2)=0 and a_(2)=1` `rarr a_(0)=0, a_(1)=a_(2)=1` Putting the value of `a_(0,a_(1),a_(2)` (i) we get `n^(2)=n+n(n-1)` `therefore underset(n=1)overset(infty)Sigma(n^(2))/(n=1)(n+n(n-1))/(n!)` `rarr underset(n=1)overset(infty)Sigma(n^(2))/(n=1){(n)/(ni+n(n-1))/(ni)}` |
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| 24. |
`(2)/(1!)+(4)/(3!)+(6)/(5!)+….infty` is equal toA. e+1B. e-1C. `e^(-1)`D. e |
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Answer» Answer: We have `(2)/(1!)+(4)/(3!)+(6)/(5!)+…infty` `=underset(n=1)overset(infty)Sigma(2n)/(2n-1)!=underset(n=1)overset(infty)Sigma((2n-1)+(1))/(2n-1)!` `=underset(n=1)overset(infty)Sigma{(1)/(2n-2)!+(1)/(2n-1)!}=(1)/(0!)+(1)/(1!)+(1)/(2!)+(1)/(3!)`+...=e |
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| 25. |
`1+(2x)/(1!)+(3x^(2))/(2!)+(4x^(3))/(3!)+..infty` is equal toA. `xe^(x)`B. `(x+1)e^(x)`C. `xe^(-x)`D. `(x+1)e^(-x)` |
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Answer» Answer: We have `1+(2x)/(1!)+(3x^(2))/(2!)+(4x^(3))/(3!)+..infty` `=underset(n=0)overset(infty)Sigma (n+1)x^(n)/(n!)=underset(n=0)overset(infty)Sigma(n)/(n!)x^(n)+underset(n=0)overset(infty)Sigma(x^(n))/(n!)` `=xunderset(n=1)overset(infty)Sigma (x^(n-1))/(n-1)!+ underset(n=0)overset(infty)Sigma (x^(n))/(n!)=xe^(x)+e^(x)=(x+1)e^(x)` |
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| 26. |
If `x-x^2/2+x^3/3+x^4/4+.....` to `oo=y` , then `y+y^2/(2!)+y^3/(3!)+....` + to `oo` is equal toA. `e^(y)-1`B. `log_(e)(1+y)`C. `x^(3)=e^(y)`D. `x=1+e^(y)` |
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Answer» Answer: We have `y=x-(x^(2))/(2)+(x^(3))/(3)-(x^(4))/(4)`+.. `rarr y= log_(e)(1+x)rarre^(y)=1+xrarrx=e^(y)-1` |
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| 27. |
`(2)/(2!)+(2+4)/(3!)+(2+4+6)/(4!)+….infty` is equal toA. `e-2`B. `e-1`C. eD. `e^(-1)` |
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Answer» Answer: Let `T_(n)` be the `n^(th)` term of the given series then `T_(n)=(2+4+6+..+2n)/(n+1)!,n=1,2,3………` `rarr T_(n)=(n(n+1))/(n+1)!=(1)/(n-1)!,n=1,2…` `therefore "Sum of the series" =underset(n=1)overset(infty)Sigma T_(n)=underset(n=1)overset(infty)Sigma (1)/(n-1)!=e` |
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| 28. |
The sum of the series `(1)/(1!)x+(2)/(2!)x^(2)+(3)/(3!)x^(3)`….. To `infty` isA. `e^(x)`B. `x e^(x)`C. `x e^(x)-1`D. none of these |
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Answer» Answer: We have `1+(1)/(1!)x+(2)/(2!)x^(2)+(3)/(3!)x^(3)…infty` `=underset(n=1)overset(infty)Sigma (n)/(n!)x^(n)` `=underset(n=1)overset(infty)Sigma(1)/(n-1)!x^(n)=xunderset(n=1)overset(infty)Sigma(x^(n)-1)/(n-1)!=xe^(x)` |
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| 29. |
The sum of the series `(1)/(2!)-(1)/(3!)+(1)/(4!)-(1)/(5!)`+….up to `infty` isA. `e^(-1//2)`B. `e^(1//2)`C. `e^(-2)`D. `e^(-1)` |
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Answer» Answer: We have `(1)/(2!)-(1)/(3!)+(1)/(4!)-(1)/(5!)`+… `=1-(1)/(1!)+(1)/(2!)-(1)/(3!)+(1)/(4!)-(1)/(5!)+….e^(-1)` |
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| 30. |
The sum of the series `((a-b)/(a))+1/2((a-b)/(a))^(2)+1/3((a-b)/(a))^(3)+….infty` isA. `log_(a)(b)`B. `log_(a-b)(a)`C. `log_(b)(a)`D. `log_(a-b)(b)` |
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Answer» Answer: We have `(a-b)/(a)+1/2(a-b)/(a)^(2)+1/3((a-b)/(a))^(3)+…infty` `={-((a-b)/(a)-1/2(a-b)/(a))^(2)-1/3((a-b)/(a))^(3)-…infty}` `=-log_(e )(b/a)=log_(e )(a/b)` |
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| 31. |
The sum of the series `1+(1+a)/(2!)+(1+a+a^(2))/(3!)+(1+a+a^(2)+a^(3))/(4!)`+…..isA. `(e^(a)-e)/(a-1)`B. `(e^(a)-e)/(a+1)`C. `(e^(2a)+1)/(a-1)`D. `(e^(a)+e)/(a-1)` |
| Answer» Correct Answer - a | |
| 32. |
The sum of the series `(1)/(1!)+(2)/(2!)+(3)/(3!)+(4)/(4!)`+…..to `infty` isA. eB. 2eC. `(1)/(2)e`D. none of these |
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Answer» Answer: We have `1+(1)/(1!)+(2)/(2!)+(3)/(3!)+…= underset(n=1)overset(infty)Sigma(nb)/(n!)=underset(n=1)overset(infty)Sigma(1)/(n-1)!=e` |
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| 33. |
The value of `sqrt(c )` rounded off of three decimal places isA. 1.648B. 1.65C. 1.652D. none of these |
| Answer» Correct Answer - a | |
| 34. |
If `e^(x)=y+sqrt(1+y^(2)` then the value of y isA. `e^(x)-e^(-x)`B. `1/2(e^(x)-e^(-x))`C. `e^(x)+e^(-x)`D. `1/2(e^(x)+e^(-x))` |
| Answer» Correct Answer - d | |
| 35. |
The coefficient of `x^(n)` in the expansion of `(a+bx+cx^(2))/(e^(x))` isA. `(-1)^(n)/(n!){cn^(2)-(b+c)(n+a}`B. `(-1)^(n)/(n!){cn^(2)+(b+c)(n+a}`C. `(-1)^(n)/(n!){cn^(2)+(b+c)(n-a}`D. none of these |
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Answer» Answer: We have `(a+bx+cx^(2))/(e^(x))=ae^(-x)+bxe^(-x)+cx^(2)e^(-x)` =coefficient of `x^(n) "in" (a+bx+cx^(2)+ex)` =coefficient of `x^(n) in (ae^(-x)+bxe^(-x)+cx^(2)e^(-x)` `=a("coeff of" x^(n)) in e^(-x))+b("coeff of" x^(n-1)) in e^(-x)+("coeff of" x^(n)-2) in e^(-x))` `=a(-1)^(n)/(n!)-bn(-1)^(n)/(n!)+c(-1)^(n)n(n-1)/(n!)` `=(-1)^(n)/(n!){a-bn+cn(n-1)}` |
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| 36. |
The coefficent of `x^(n)` in the expansion of `(1+(x^(2))/(2!)+(x^(4))/(4!)+…)^(2)` When n is odd isA. `(2^(n))/(n!)`B. `(2^(2n))/(2n)!`C. 0D. `(2^(2x))/(n!)` |
| Answer» Correct Answer - b | |
| 37. |
The coefficient of `x^(n)` in the expansion of `e^(x)` isA. `underset(r=n)overset(infty)Sigma (r^(n))/(r!)`B. `(1)/(n!)underset(r=1)overset(infty)Sigma (r^(n))/(r!)`C. `(1)/(n)underset(r=1)overset(infty)Sigma (r^(n))/(r!)`D. none of these |
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Answer» Answer: Let `e^(x)=z` Then `e^(e^(x))=e^(Z)=underset(k=0)overset(infty)Sigma (z^(k))/(k!)=underset(k=0)overset(infty)Sigma(e^(x^(k)))/(k!)=underset(k=0)overset(infty)Sigma (e^(kx))/(k!)` `rarr e^(e^(x))=(1+(e^(x)))/(1!)+(e^(2x))/(2!)+(e^(3x))/(3!)+…infty` `rarr e^(e^(x))=1+(1)/(1!)+underset(n=0)overset(infty)Sigma(x^(n))/(n)+(1)/(2!) underset(n=0)overset(1)Sigma underset(n=0)overset(infty)Sigma (2x)^(n)/(n!)` `+(1)/(3!)underset(n=0)overset(infty)Sigma(3x)^(n)/(n!)+...infty` `therefore` coefficient of `x^(n) in e^(e^(x))` `=(1)/(1!)(1)/(n!)+(1)/(2!)(2^(n))/(n!)+(1)/(3!)+...infty` `(=(1)/(ni)(1)/(1!)+(2^(n))/(2!)+(3^(n))/(3!)+........infty)` `=(1)/(ni)underset(r=1)overset(infty)Sigma(r^(n))/(r!)` |
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| 38. |
Prove that :`(log)_e((x+1)/x)=2[1/((2x+1))+1/(3(2x+1)^3)+1/(5(2x+1)^5)+]`A. `log(x)/(x+1)`B. `log(x+1)/(x)`C. `log(2x+1)`D. `log(1)/(2x+1)` |
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Answer» Answer: We have `2{(1)/(2x+1)+(1)/3(2x+1)^(3)+(1)/5(2x+1)^(5)`+….} `log(1+(1)/(2x+1)/(1-(1))/(2x+1)=log(2x+2)/(2)=log(x+1)/(x)` |
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| 39. |
If `x =1 +2+(4)/(2!)+(8)/(3!)+(16)/(4!)+…` then `x^(-1)` is equal toA. `e^(-2)`B. `e^(2)`C. `e^(1//2)`D. none of these |
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Answer» Answer: We have `x=1+(2)/(1!)+(2^(2))/(2!)+(2^(3))/(3!)+(2^(4))/(4!)`+….. `rarr x=e^(2)rarr x^(-1) =e^(-2)` |
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| 40. |
The value of `(x+y)(x-y)+1/(2!)(x+y)(x-y)(x^2+y^2)+1/(3!)(x+y)(x-y)(x^4+y^4+x^2y^2)+`... is :A. `e^(x^(2))-e^(y^(2))`B. `e^(x^(2))+e^(y^(2))`C. `e^(x^(2))-(y^(2))`D. `e^(x^(2))+(y^(2))` |
| Answer» Correct Answer - b | |
| 41. |
The coefficient of `x^(10)` in the series of `e^(2x)` isA. `(2^(9))/(9!)`B. `(2^(10))/(10!)`C. `(1)/(10!)`D. none of these |
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Answer» Answer: We have `e^(x)=1+(x)/(1!)+(x^(2))/(2!)+(x^(3))/(3!)+…= underset(n=0)overset(infty)Sigma(2x)^(n)/(n!)` Replacing x by 2x we get `e^(2x)=1+(2x)/(1!)+(2x)^(2)/(2!)+(2x)^(3)/(3!)+….underset(n=0)overset(infty)Sigma(2x)^(n)/(n!)` Clearly `T_(n+1)=(n+1)^(th) term =(2x)^(n)/(n!)=(2^(n))/(n!)x^(n)` If `(n+1)^(th)` term contains `x^(10)` then we must have n =10 ltbvrgt `therefore` Coefficient of `x^(10)=(2^(10))/(10!)` |
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| 42. |
The sum of the infite series `(1/3)^(2)+1/3(1/3)^(4)+1/5(1/3)^(6)`+……..isA. `1/4log_(e)2`B. `1/2log_(e )2`C. `1/6 log_(e ) 2`D. `1/4 log_(e )3/2` |
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Answer» Answer: We have `(1/2)^(2)+1/3^(4)+1/5^(1)(3)^(6)`+… `=1/3{(1/3+1/3(1/3)^(3)+1/5(1/3)^(5)+…)}` `1/3xx1/2log(1+(1))/(3)(12-(1))/(3) before x+(x^(3))/(3)+(x^(5))/(5)+..=1/2 log(1+x)/(1-x)` `=1/6log2` |
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| 43. |
The sum of the series `(1^(2).2^(2))/(1!)+(2^(2).3^(2))/(2!)+(3^(2).4^(2))/(3!)`+.. IsA. 27eB. 24eC. 28eD. 25e |
| Answer» Correct Answer - a | |
| 44. |
Cofficient of `x^(4)` in the expansion of `(1-3x+x^(2))/(e^(x))` isA. `(5)/(24)`B. `(4)/(25)`C. `(24)/(25)`D. `(25)/(24)` |
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Answer» Answer: We have `(1-3x+x^(2))/(e^(x))` `=(1-3x+x^(2))e^(-x)` `=e^(-x)-3xe^(-x)+x^(2)e^(-x)` `=underset(n=0)overset(infty)Sigma(-1)^(n)(x^(n))/(n!)-3xunderset(n=0)overset(infty)Sigma(-1)^(n)(x^(n))/(n!)+x^(2)underset(n=0)overset(infty)Sigma(-1)^(n)(x^(n))/(n!)` `therefore` coefficient of `x^(4)=(-1)^(4)/(4!)+3(-1)^(4)/(3!)+(-1)^(4)/(2!)` `rarr` coefficient of `x^(4)=1/24+1/2+1/2=25/24` |
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| 45. |
If `alpha` and `beta` are the roots of the equation `x^2-px + q = 0` then the value of `(alpha+beta)x -((alpha^2+beta^2)/2)x^2+((alpha^3+beta^3)/3)x^3+...` isA. `log(1+px+qx^(2))`B. `log(1+qx+px^(2))`C. `log(x^(2)+px+q)`D. none of these |
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Answer» Answer: Since `alpha,beta` are the roots of the equation `x^(2)-px+q=0` `therefore alpha+beta=p and alpha beta=q` Now `(alpha+beta)x-(alpha^(2)+beta^(2))/(2)x^(2)+(alpha^(3)+beta^(3))/(3)x^(3)`… `=((alpha x-alpha^(2))x^(2))/(2)+(alpha^(3)x^(3))/(3)+(betax-(beta^(2)x^(2))/(2)+(beta^(3)x^(3))/(3)` `=log(1+alphax)+log(1+betax)` `=log{(1+alphax)(1+betax)}` `log{(1+alpha+beta)}x+alphabetax^(2)}=log(1+px+qx^(2))` |
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| 46. |
The coefficient of `x^(n)` in the expansion of `e^(a+bx)` in power of x isA. `(b^(n))/(n!)`B. `e^(a)(b^(n+1))/(n+1)!`C. `e^(a)(b^(n))/(n!)`D. none of these |
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Answer» Answer: We have `e^(a+bx)=e^(a).e^(bx)=e^(a){underset(n=0)overset(infty)Sigma(bx)^(n)/(n!)}` `rarr e^(a+bx)=e^(a){1+(bx)/(1!)+(bx)^(2)/(2!)+(bx)^(3)/(n!)+….}` `therefore` coefficient of `x^(n) in e^(a+bx)=e^(a)(b^(n))/(n!)` |
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| 47. |
The coefficent of `x^(n)` in the series `1+(a+bx)/(1!)+(a+bx)^(2)/(2!)+(a+bx)^(3)/(3!)`+…isA. `(b^(n))/(n1)`B. `e^(b)(b^(n))/(n!)`C. `e^(a)(b^(n))/(m!)`D. `(e^(b)a^(n))/(n!)` |
| Answer» Correct Answer - a | |
| 48. |
The coefficient of `x^(4)` in the expansion of `(1-ax-x^(2))/(e^(x))` isA. `(-1)r/(r!){-r^(2)+r(a+1)+1}`B. `(-1)r/(r!){-r^(2)-r(a+1)+1}`C. `(-1)r/(r!){-r^(2)-r(a+1)+1}`D. none of these |
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Answer» Answer: We have `(1-ax-x^(2))/(e^(x))=e^(-x)-axe^(x)-x^(2)e^(-x)` `therefore` coefficient of `x^(r ) in (1-ax-x^(2))/(e^(x))` =coefficient of `x^(r ) in e^(-x)-axe^(-x)-x^(2)e^(-x)` =coefficient of `x^(r ) "in" e^(-x)-a xx "coefficient" of x^(r-1) "in" e^(-x)-"coefficient of" x^(r-2) "in" e^(x)` `=(-1)^(r)/(r!)-a(-1)^(r-1)/(r-1)!-(-1)^(r-2)/(r-2)!` `=(-1)^(r)/(r!){1+ar-r(r-1)}` |
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| 49. |
The constant term in the expansion of `(3^(x)-2^(x))/(x^(2))` isA. `log_(e)3`B. `(log_(e)6)xx{(log_(e))(3/2)}`C. `1/2(log_(e)6)xx{(log_(e))(3/2)}`D. none of these |
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Answer» Answer: We have `(3^(X)-2^(y))/(x^(2))=(1)/(x^(2))[{1+x(log_(e)3)+(xlog_(e)3)^(2)/(2!)+….}` `-{1+x(log_(e)2)+(xlog_(e)2)^(2)/(2!)+…}]` `therefore "constant term" =(log_(e)3)^(2)/(2!)-(log_(e)2)6(2)/(2!)` `=1/2(log_(e)3+log_(e)2)(log_(e)3-log_(e)2)=1/2(log_(e)6)log_(e)(3/2)` |
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| 50. |
The expansion of `(1+(x^(2))/(2!)+(x^(4))/(4!)..)^(2)` in ascending powers of x isA. `1+(x^(2))/(2!)+(x^(4))/(4!)+(x^(6))/(6!)`+…B. `1+(2^(2)x^(2))/(2!)+(2^(4)x^(4))/(4!)`C. `1+(2x^(2))/(2!)+(2^(3)x^(4))/(x=4!)+(2^(5)x^(5))/(6!)`D. none of these |
| Answer» Correct Answer - c | |