A value of `alpha` such that `int_(alpha)^(alpha+1) (dx)/((x+a)(x+alpha+1))="loge"((9)/(8))` isA. `-2`B. `(1)/(2)`C. `-(1)/(2)`D. 2

A value of `alpha` such that `int_(alpha)^(alpha+1) (dx)/((x+a)(x+alph

1.	A value of `alpha` such that `int_(alpha)^(alpha+1) (dx)/((x+a)(x+alpha+1))="loge"((9)/(8))` isA. `-2`B. `(1)/(2)`C. `-(1)/(2)`D. 2
Answer» Correct Answer - A Let I `int_(alpha)^(alpha+1)(dx)/((x+a)(x+alpha1))` `int_(alpha)^(alpha+1)((x+alpha+1)-(x+alpha))/((x+alpha)(x+alpha+1))`dx `=int_(alpha)^(alpha+1)((1)/(x+alpha)-(1)/(x+alpha+1)) dx` `=[log_(e)(x+alpha)-log_(e)(x+alpha+1)]_(alpha)^(alpha+1)` `=[log_(e)((x+alpha)/(x+alpha+1))]_(alpha)^(alpha+1)` `="log"_(e)(2alpha+1)/(2alpha+2)-"log"_(e)(2alpha)/(2alpha+1)` `=log_(e)((2alpha+1)/(2alpha+2)xx(2alpha+1)/(2alpha))="log"_(e)((9)/(8))` (given) `rArr((2alpha+1)^(2))/(4alpha(alpha+1))=(9)/(8)rArr8 [4alpha+1] = 36 (alpha^(2)+alpha)` `rArr8alpha^(2)+8alpha+2=9alpha^(2)+9alpha` `rArralpha^(2)+alpha-2=0` ` rArr(alpha+2)(alpha-1)=0` `rArralpha=1,-2` From the options we get `alpha -=2`

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Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.
`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`
`lim_(nrarroo) (((n+1)(n+2)"........"3n)/(n^(2n)))^(1//n)` is equal to :A. `(27)/(e^(2))`B. `(9)/(e^(2))`C. `3 log3-2`D. `(18)/(e^(4))`