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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.
| 401. |
Evaluate `lim_(xto1)(sqrt(x)+sqrt(sqrt(x))+sqrt(sqrt(sqrtx))+sqrt(sqrt(sqrt(sqrt(x))))-4)/(x-1).` |
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Answer» `underset(xto1)lim(sqrt(x)+sqrt(sqrt(x))+sqrt(sqrt(sqrtx))+sqrt(sqrt(sqrt(sqrt(x))))-4)/(x-1)` `=underset(xto1)lim(x^(1//2)+x^(1//4)+x^(1//8)+x^(1//16)-4)/(x-1)` `=underset(xto1)lim((x^(1//2)-1)/(x-1)+(x^(1//4)-1)/(x-1)+(x^(1//8)-1)/(x-1)+(x^(1//16)-1)/(x-1))` `=1/2+1/4+1/8+1/16=(8+4+2+1)/(16)=15/16` |
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| 402. |
If `f(x)={{:((x)/(sinx)",",x gt0),(2-x",",xle0):}andg(x)={{:(x+3",",xlt1),(x^(2)-2x-2",",1lexlt2),(x-5",",xge2):}` Then the value of `lim_(xrarr0) g(f(x))`A. is `-2`B. is `-3`C. is 1D. does not exist |
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Answer» Correct Answer - B `underset(xrarr0+)(lim)g(f(x))=g(f(0^(+)))=g(1^(+))" "(becauseunderset(xrarr0)(lim)(x)/(sinx)=1^(+))` `" "=1-2(1)-2=3` `underset(xrarr0^(-))(lim)g(f(x))=g(f(0^(-)))=g(2^(+))" "(becauseunderset(xrarr0^(-))(lim)(2-x)=2^(+))` `" "=2-5=-3` `rArr" "underset(xrarr0)(lim)g(f(x))=-3` |
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| 403. |
`lim_(x -> oo) x^n / e^x = 0`, (n is an integer) forA. no vlaue of nB. all values of nC. only negative values of nD. only positive values of n |
| Answer» Correct Answer - B | |
| 404. |
`lim_(xtooo) (2sqrt(x)+3root(3)(x)+4root(4)(x)+...+nroot(n)(x))/(sqrt((2x-3))+root(3)((2x-3))+...+root(n)((2x-3)))` is equal to |
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Answer» Correct Answer - C Since the highest degree of x is `1//2,` divide numerator and denominator by `sqrt(x).` Then we have limit `(2)/(sqrt(2))` or `sqrt(2)`. |
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| 405. |
`lim_(xtooo) {20sqrt(x+sqrt(x+sqrt(x+sqrt(x+sqrt(x+...)}` is equal to |
| Answer» Correct Answer - B | |
| 406. |
If `L=lim_(xto0)(sinx+ae^(x)+be^(-x)+clog_(e)(1+x))/(x^(3))` exists finitely, then Equation `ax^(2)+bx+c=0` has |
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Answer» Correct Answer - D `L=underset(xto0)lim(sinx+ae^(x)+be^(-x)+clog_(e)(1+x))/(x^(3))` `=underset(xto0)lim((x-(x^(3))/(3!))+a(1+(x)/(1!)+(x^(2))/(2!)+(x^(3))/(3!))+b(1-(x)/(1!)+(x^(2))/(2!)-(x^(3))/(3!))+c(x-(x^(2))/(2)+(x^(3))/(3)))/(x^(3))` `=underset(xto0)lim((a+b)+(1+a-b+c)x+((a)/(2)+(b)/(2)-(c)/(2))x^(2)+(-1(1)/(31)+(a)/(3!)-(b)/(3!)+(c)/(3))x^(3))/(x^(3))` `implies a+b=0,1+a-b+c=0,(a)/(2)+(b)/(2)-(c)/(2)=0` and `L=-(1)/(3!)+(a)/(3!)-(b)/(3!)+(c)/(3)` Solving first three equations we get `c=0, a=-1//2,b=1//2.` `:." "L=-1//3` Equation `ax^(2)+bx+c=0` reduces to `x^(2)-x=0impliesx=0,1` `||x+c|-2a|lt4b` reduces to `||x|+1|lt2` `implies-2lt|x|+1lt2` `implies0le|x|lt1` `impliesx in[-1,1]` |
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| 407. |
If `L=lim_(xtooo) (x+1-sqrt(ax^(2)+x+3))` exists infinetely then The value of a is |
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Answer» Correct Answer - C `L=underset(xtooo)lim(x+1-sqrt(ax^(2)+x+3))` `=underset(xtooo)lim(((x+1)^(2)-(ax^(2)+x+3))/(x+1+sqrt(ax^(2)+x+3)))` `=underset(xtooo)lim(((1-a)x^(2)+x-2)/(x+1+sqrt(ax^(2)+x+3)))` L exists finitely if `1-a=0" or "a=1` `:." "L=underset(xtooo)lim((x-2)/(x+1+sqrt(x^(2)+x+3)))` `=underset(xtooo)lim((1-(2)/(x))/(1+(1)/(x)+sqrt(1+(1)/(x)+(3)/(x^(2)))))` |
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| 408. |
If `f(x)=lim_(nrarroo)((x^(2)+ax+1)+x^(2n)(2x^(2)+x+b))/(1+x^(2n)) and lim_(xrarrpm1)f(x)` exists, then The value of b isA. `-1`B. 1C. 0D. 2 |
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Answer» Correct Answer - C `f(x)=underset(nrarroo)(lim)((x^(2)+ax+1)+x^(2n)(2x^(2)+x+b))/(1+x^(2n))` `={{:(x^(2)+ax+1",",|x|lt1),(2x^(2)+x+b",",|x|gt1),((-a+b+3)/(2)",",x=-1),((a+b+5)/(2)",",x=1):}` `underset(xrarr-1)(lim)f(x)" exists if"` `underset(xrarr -1)(lim)f(x)=underset(xrarr-1^(+))(lim)f(x)` `rArr" "underset(xrarr-1^(-))(lim)(2x^(2)+x+b)=underset(xrarr-1^(+))(lim)(x^(2)+ax+1)` `rArr" "2-1+b=1-a+1` `rArr" "a+b=1" (i)"` `underset(xrarr1)(lim)f(x)" exists if"` `underset(xrarr1^(-))(lim)f(x)=underset(xrarr1^(+))(lim)f(x)` `rArr" "underset(xrarr1^(-))(lim)(x^(2)+ax+1)=underset(xrarr1^(+))(lim)(2x^(2)+x+b)` `rArr 1+a+1=2+1+b` `rArr a-b=1" (ii)"` `"Solving Eqs. (i) and (ii), we get a = 1 and b=0."` |
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| 409. |
If `f(x)=lim_(nrarroo) ((x^(2)+ax+1)+x^(2n)(2x^(2)+x+b))/(1+x^(2n)) and lim_(xrarrpm1) f(x)` exists, then The value of a isA. `-1`B. 1C. 0D. 2 |
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Answer» Correct Answer - B `f(x)=underset(nrarroo)(lim)((x^(2)+ax+1)+x^(2n)(2x^(2)+x+b))/(1+x^(2n))` `={{:(x^(2)+ax+1",",|x|lt1),(2x^(2)+x+b",",|x|gt1),((-a+b+3)/(2)",",x=-1),((a+b+5)/(2)",",x=1):}` `underset(xrarr-1)(lim)f(x)" exists if"` `underset(xrarr -1)(lim)f(x)=underset(xrarr-1^(+))(lim)f(x)` `rArr" "underset(xrarr-1^(-))(lim)(2x^(2)+x+b)=underset(xrarr-1^(+))(lim)(x^(2)+ax+1)` `rArr" "2-1+b=1-a+1` `rArr" "a+b=1" (i)"` `underset(xrarr1)(lim)f(x)" exists if"` `underset(xrarr1^(-))(lim)f(x)=underset(xrarr1^(+))(lim)f(x)` `rArr" "underset(xrarr1^(-))(lim)(x^(2)+ax+1)=underset(xrarr1^(+))(lim)(2x^(2)+x+b)` `rArr 1+a+1=2+1+b` `rArr a-b=1" (ii)"` `"Solving Eqs. (i) and (ii), we get a = 1 and b=0."` |
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| 410. |
`lim_(nrarroo) (1-x+x.root n e)^(n)` is equal toA. `e^(x)`B. `e^(-x)`C. `e^(2x)`D. none of these |
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Answer» Correct Answer - A `underset(nrarroo)(lim)(1-x+x.rootn(e))^(n)" "(1^(oo)"form")` `=e^(underset(nrarroo)(lim)(1-x+xe^(1//n)-1))` `=e^(x)e^(underset(nrarroo)(lim)((e^(1//n)-1))/(1//n))` `=e^(x)` |
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| 411. |
The value of `lim_(nrarroo) ((sqrtn^(2)+n-1)/(n))^(2sqrt(n^(2)+n-1))` isA. eB. `1//e`C. `e^(2)`D. `e^(-2)` |
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Answer» Correct Answer - B `L=underset(nrarroo)(lim)((sqrt(n^(2)+n)-1)/(n))^(2sqrt(n^(2)+n)-1)` `=e^(underset(nrarroo)(lim)[(sqrt(n^(2)+n)-1-n)/(n)](2sqrt(n^(2)+n)-1))` `=e^(underset(nrarroo)(lim)([(n^(2)+n)-(1+n)^(2)](2sqrt(n^(2)+n)-1))/(n{sqrt(n^(2)+n)+(1+n)}))` `=e^(underset(nrarroo)(lim)((-n-1)(2sqrt(n^(2)+n)-1))/(n{sqrt(n^(2)+n)+(1+n)}))` `=e^(underset(nrarroo)(lim)((-1-(1)/(n))(2sqrt(1+(1)/(n))-(1)/(n))n^(2))/(n^(2){sqrt(1+(1)/(n))+(1)/(n)+1})=e^(-1))` |
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| 412. |
If `f(x)=lim_(nrarroo) (cos(x)/(sqrtn))^(n)`, then the value of `lim_(xrarr0) (f(x)-1)/(x)` is |
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Answer» Correct Answer - A `f(x)=underset(nrarroo)(lim)(cos.(x)/(sqrtn))^(n)` `" "=e^(underset(nrarroo)(lim)(cos.(x)/(sqrtn)-1).n)` `" "e^(-(2underset(nrarroo)(lim)(x^(2))/(4)(sin^(2).(x)/(2sqrtn))/((x^(2))/(4sqrtn))))` `" "=e^(-(1)/(2)x^(2))` `rArr" "underset(xrarr0)(lim)(f(x)-1)/(x)=underset(xrarr0)(lim)(e^(-(1)/(2)x^(2))-1)/(x)=0` |
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| 413. |
If `k in I` such that `lim_(nrarroo) (cos.(kpi)/(4))^(2n)-(cos.(kpi)/(6))^(2n)=0,` thenA. k must bot be divisible by 24B. k is divisible by 24 or k is divisible neither by 4 nor by 6C. k must be divisible by 12 but not necessarity by 24D. none of these |
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Answer» Correct Answer - B `underset(nrarroo)(lim)(cos.(kpi)/(4))^(2n)-(cos.(kpi)/(6))^(2n)=0` holds good if `"Case I": cos .(kpi)/(4)=cos.(kpi)/(6)=1` i.e., `(kpi)/(4)=2mpi and (kpi)/(6)=2p pi,m,p in Z` i.e., k = 8m and k = 12 p i.e., k is divisible by both 8 and 12 i.e., k is divisible by 24. Case II: `-1 lt cos.(kpi)/(4),cos.(kpi)/(6) lt 1` i.e., k is not divisible by 4 and k is not divisible by 6. Case III: `cos.(kpi)/(4)=(2m+1)pi and (kpi)/(6)=(2p+1)pi` `k=4 (2m+1) and k=6(2p+1)` This is not possible. |
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| 414. |
The value of `(lim_(x rarr 0) (tanx^((1)/(5)))/((tan^(-1)sqrtx)^(2))(log(1+5x))/(e^(3root5x)-1)` isA. `(3)/(5)`B. `(5)/(3)`C. 1D. none of these |
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Answer» Correct Answer - B `L=underset(xrarr0^(+))(lim)(tanx^((1)/(5)))/(tan^(-1)sqrtx)(log(1+5x))/((e^(3root5x)-1))` `=underset(xrarr0^(+))(lim)(tan(x^((1)/(5))))/(x^((1)/(5))).x^((1)/(5))((sqrtx)/(tan^(-1)sqrtx))^(2).(1)/(x).5x.((log(1+5x))/(5x))/(((e^(3.x^((1)/(5)))-1)/(3x^((1)/(5)))).3x^((1)/(5)))` `=1.1.(loge)/(1).(5)/(3)=(5)/(3)` |
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| 415. |
If `L=lim_(xto0)(sinx+ae^(x)+be^(-x)+clog_(e)(1+x))/(x^(3))` exists finitely, then The solutions set of `||x+c|-2a|lt4b` isA. `{:(a,b,c,d),(s,r,q,p):}`B. `{:(a,b,c,d),(q,s,r,p):}`C. `{:(a,b,c,d),(s,r,p,q):}`D. `{:(a,b,c,d),(s,p,q,r):}` |
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Answer» Correct Answer - C `L=underset(xto0)lim(sinx+ae^(x)+be^(-x)+clog_(e)(1+x))/(x^(3))` `=underset(xto0)lim((x-(x^(3))/(3!))+a(1+(x)/(1!)+(x^(2))/(2!)+(x^(3))/(3!))+b(1-(x)/(1!)+(x^(2))/(2!)-(x^(3))/(3!))+c(x-(x^(2))/(2)+(x^(3))/(3)))/(x^(3))` `=underset(xto0)lim((a+b)+(1+a-b+c)x+((a)/(2)+(b)/(2)-(c)/(2))x^(2)+(-1(1)/(31)+(a)/(3!)-(b)/(3!)+(c)/(3))x^(3))/(x^(3))` `implies a+b=0,1+a-b+c=0,(a)/(2)+(b)/(2)-(c)/(2)=0` and `L=-(1)/(3!)+(a)/(3!)-(b)/(3!)+(c)/(3)` Solving first three equations we get `c=0, a=-1//2,b=1//2.` `:." "L=-1//3` Equation `ax^(2)+bx+c=0` reduces to `x^(2)-x=0impliesx=0,1` `||x+c|-2a|lt4b` reduces to `||x|+1|lt2` `implies-2lt|x|+1lt2` `implies0le|x|lt1` `impliesx in[-1,1]` |
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| 416. |
Let `< a_n >` be a sequence such that `lim_(x->oo)a_n=0.` Then `lim_(n->oo)(a_1+a_2++a_n)/(sqrt(sum_(k=1)^n k)),` is |
| Answer» Correct Answer - C | |
| 417. |
The value of `lim_(xrarr oo)(2x^(1//2)+3x^(1//3)+4x^(1//4)+....nx^(1//n))/((2x-3)^(1//2)+(2x-3)^(1//3)+....+(2x-3)^(1//n))`is |
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Answer» Correct Answer - C Let `l=lim_(xto oo)(2x^(1//2)+3x^(1//3)+4x^(1//4)+....+nx^(1//n))/((2x-3)^(1//2)+(2x-3)^(1//3)+....+(2x-3)^(1//n))` Then, , `l=lim_(xto oo)(2h^(-1//2)+3h^(-1//3)+4h^(-1//4)+....+nx^(-1//n))/((2-3h)^(1//2)+(h)^(-1//2)+(2-3h)^(1//3)+h^(-1//3)....+(2-3h)^(1//n)h^(-1//n)) "where" h=(1)/(x)` `rArrl=lim_(hto0)(2+3h^(((1)/(2)-(1)/(3)))+4h^(((1)/(2)-(1)/(4)))+.....+nh^(((1)/(2)-(1)/(n))))/((2-3h)^(1/2)+(2-3h)^(1/3)+h^(((1)/(2)-(1)/(4)))+....+(2-3h)^(1/n)h^(((1)/(2)-1/n)))` `rArr l=(2)/(sqrt(2))=sqrt(2)` |
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| 418. |
If `A_l=(x-a_i)/(|x-a_i|)30, i=1,2,..., n` and if `a_1 lt a_2 lt a_3lt ..... lta_n`. Then, `lim_(xtoa_m) (A_1A_2.....A_n), 1 le mle n`A. is equal ot `(-1)^m`B. is equal to `(-1)^m+1`C. is equal to `(-1)^m-1`D. does not exist. |
| Answer» Correct Answer - D | |
| 419. |
Let `lt a_n gt` be a sequence such that `a_1=1 and a_n+1 =cos a_n, n gt 1 `. If `a=lim_(xtooo) a_n`, then a belongs to the intervalA. `(0,pi//6)`B. `(pi//6,pi//3)`C. `(pi//3,pi//2)`D. `(pi//2,4pi//3)` |
| Answer» Correct Answer - B | |
| 420. |
IF `f(x)={{:(x,"if",x,"is rational"),(1-x,"if",x, "is irrational"):},"then find" lim_(xto1//2) f(x)` if exists. |
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Answer» `underset(xto1//2^(+))limf(x)=underset(xto1//2^(+))limx" "`(if `1//2^(+)`rational) `=1//2` `underset(xto1//2^(-))limf(x)=underset(xto1//2^(+))lim(1-x)" "`(if `1//2^(+)`is irrational) `=1-1//2=1//2` So, `underset(xto1//2^(+))limf(x)=1//2` in any case. Similarly, we get `underset(xto1//2^(-))limf(x)=1//2` `:." "underset(xto1//2)limf(x)=1//2` |
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| 421. |
If `S_n=sum_(k=1)^n a_k and lim_(n->oo)a_n=a ,` then `lim_(n->oo)(S_(n+1)-S_n)/sqrt(sum_(k=1)^n k)` is equal to |
| Answer» Correct Answer - A | |
| 422. |
`lim_(xtooo) (x^(2)"tan"(1)/(x))/(sqrt(8x^(2)+7x+1))` is equal toA. 4B. `(1//2`C. 2D. `1//4` |
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Answer» Correct Answer - A `underset(xtooo)lim(x^(2)"tan"(1)/(x))/(sqrt(8x^(2)+7x+1))=underset(xtooo)lim(x^(2)"tan"(1)/(x) )/(-xsqrt(8+(7)/(x)+(1)/(x^(2))))` `=underset(xtooo)lim("tan"(1)/(x) )/((1)/(x)sqrt(8+(7)/(x)+(1)/(x^(2))))=-(1)/(2sqrt(2))` |
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| 423. |
Which of the following is/are correct?A. always 1B. always -1C. `(-1)^(m+1)`D. `(-1)^(n-m)` |
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Answer» Correct Answer - A::B::C `(1)underset(xto0^(+))lim([x+|x|])/(x)=underset(xto0^(+))lim([x+x])/(x)=underset(xto0^(+))lim([2x])/(x)=0` `underset(xto0^(-))lim([x+|x|])/(x)=underset(xto0^(-))lim([x-x])/(x)=underset(xto0^(-))lim(0)/(x)=0` `(2)underset(xto0^(+))lim(xe^((1)/(x)))/(1+e^((1)/(x)))=underset(xto0^(+))lim(x)/(1+e^(-(1)/(x)))=0` `underset(xto0^(-))lim(xe^((1)/(x)))/(1+e^((1)/(x)))=0` `(3)underset(xto3^(+))lim(x-3)^((1)/(5))sgn(x-3)=underset(xto3^(+))lim(x-3)^((1)/(5))xx1=0` `underset(xto3^(-))lim(x-3)^((1)/(5))sgn(x-3)=underset(xto3^(-))lim(x-3)^((1)/(5))xx(-1)=0` `(4)underset(xto0^(+))lim(tan^(-1)|x|)/(x)=underset(xto0^(+))lim(tan^(-1)x)/(x)=1` `underset(xto0^(-))lim(tan^(-1)|x|)/(x)=underset(xto0^(+))limtan^(-1)(-x)/(x)=-1` |
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| 424. |
If `a_(n)` and `b_(n)` are positive integers and `a_(n)+sqrt2b_(n)=(2+sqrt2))^(n)`, then `lim_(nrarroo) ((a_(n))/(b_(n)))`=A. `sqrt2`B. 2C. `e^(sqrt2)`D. `e^(2)` |
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Answer» Correct Answer - A We have `a_(n)+sqrt2b_(n)=(2+sqrt2)^(n)` `rArr" "a_(n)-sqrt2b_(n)=(2-sqrt2)^(n)` Solving we get `a_(n)=(1)/(2)[(2+sqrt2)^(n)+(2-sqrt2)^(n)]` and `b_(n)=([(2+sqrt2)^(n)+(2-sqrt2)^(n)])/([(2+sqrt2)^(n)-(2-sqrt2)^(n)])` `" "=sqrt2([1+((2-sqrt2)/(2+sqrt2))^(n)])/([1-((2-sqrt2)/(2+sqrt2))^(n)])` Hence, `underset(nrarroo)(lim)((a_(n))/(b_(n)))=sqrt2((1+0)/(1-0))=sqrt2" "(because(2-sqrt2)/(2+sqrt2)lt1)` |
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| 425. |
Let `alpha,beta(alpha < beta)` be the roots of the equation `ax^2+bx+c=0.` If `lim_(x->oo)(|ax^2+bx+c|)/(ax^2+bx+c)=1` thenA. `alt 0 and alpha lt m lt beta `B. `agt 0 and m gt 1 `C. a gt0 and m lt 1`D. all the above |
| Answer» Correct Answer - D | |
| 426. |
Let `L=lim_(xtoa) (|2sinx-1|)/(2sinx-1)`. ThenA. `1//2`B. `-1//3`C. `-1//6`D. 3 |
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Answer» Correct Answer - A::B::C `L=underset(xtoa)lim(|2sinx-1|)/(2sinx-1)` For `a=pi//6`. `L.H.L.=underset(xto(pi^(-))/(6))lim(1-2sinx)/(2sinx-1)=-1` `R.H.L.=underset(xto(pi^(+))/(6))lim(2sinx-1)/(2sinx-1)=1` Hence, the limit does not exist. For `a=pi,underset(xto pi)lim(1-2sinx)/(2sinx-1)=-1" "`(as in neighborhood of `pi, sinx` is less than `1/2`). For `a=(pi)/(2),underset(xto pi//2)lim(2sinx-1)/(2sinx-1)=1" "`(as in neighborhood of `pi/2, sin x` approaches 1). |
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| 427. |
If `alpha,beta` are the roots of the equation `ax^2+bx+c=0`, then `lim_(xrarralpha)(ax^2+bx+c+1)^(1//x-alpha) ` is equal toA. `2a(alpha-beta)`B. `-2In |a(alpha-beta)|`C. `e^2a(alpha-beta)`D. `e^(a^(2))|alpha-beta|` |
| Answer» Correct Answer - C | |
| 428. |
Let `alpha,betainR` be such that `lim_(xto0) (x^(2)sin(betax))/(alphax-sinx)=1`. Then `6(alpha+beta)` equals___________. |
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Answer» Correct Answer - `(7)` `underset(xto0)lim(x^(2){betax-((betax)^(3))/(3!)+...})/(alphax-(x-(x^(3))/(3!)+...))=1` `implies" "underset(xto0)lim(x^(3)(beta-(beta^(3)x^(2))/(3!)+...))/((alpha-1)x+(x^(3))/(3!)+(x^(5))/(5!)+...)=1` `implies" "underset(xto0)lim(x^(2)(beta-(beta^(3)x^(2))/(3!)+...))/((alpha-1)+(x^(2))/(3!)+(x^(4))/(5!)+...)=1` `implies" "underset(xto0)lim(beta-(beta^(3))/(3!)x^(2)...)/((1)/(3!)-(x^(2))/(5!)+...)=1` `:." "beta=(1)/(3!)=(1)/(6)` `:." "6(alpha+beta)=6(1+(1)/(6))=7` |
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| 429. |
Evaluate: ` lim_(x->oo) { ((a_1)^(1/x)+(a_2)^(1/x)+.... +(a_n)^(1/x))/n}^(nx)`A. `a_1+a_2+...a_n`B. `e^(a_1+a_2+..+a_n)`C. `(a_1+a_2+...+a_n)/(n)`D. `a_1a_2.......a_n` |
| Answer» Correct Answer - D | |
| 430. |
If `|cos^-1(1)/(n)|lt (pi)/(2)`, then `lim_(n to oo) {(n+1)(2)/(pi)cos^-1.(1)/(n)-n}`A. `(2-pi)/(pi)`B. `(pi-2)/(pi)`C. `1`D. `0` |
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Answer» Correct Answer - B `lim_(n to oo) {(n+1)(2)/(pi)cos^-1.(1)/(n)-n}` `=lim_(n to oo) (2)/(pi){(n+1)cos^-1.(1)/(n)-(pi)/(2)n}` `=lim_(n to oo) (2)/(pi){n(cos^-1.(1)/(n)-(pi)/(2))+cos^-1.(1)/(n)}` `=lim_(n to oo) (2)/(pi){nsin^-1.(1)/(n)-(pi)/(2)+cos^-1.(1)/(n)}` `=lim_(n to oo) (2)/(pi){(sin^-1.(1)/(n))/((1)/(n))+cos^-1.(1)/(n)}=(2)/(pi)(-1+(pi)/(2))=(pi-2)/(pi)` |
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| 431. |
Evaluate `lim_(ntooo) nsin(2pisqrt(1+n^(2))),(n inN).` |
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Answer» L= `underset(ntooo)limnsin(2pisqrt(1+n^(2)))` `underset(ntooo)limnsin(2pisqrt(1+n^(2))-2npi)` `underset(ntooo)limnsin{(2pi(sqrt(1+n^(2))-n))/((sqrt(1+n^(2))+n))(sqrt(1+n^(2))+n)}` `=underset(ntooo)lim{(nsin((2pi)/(sqrt(1+n^(2))+n)))/(((2npi)/(sqrt(1+n^(2))+n)))((2pi)/(sqrt(1+n^(2))+n))}` `underset(ntooo)lim(2npi)/(n(sqrt(1+(1)/(n^(2)))+1))` `=(2pi)/(2)=pi` |
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| 432. |
Evaluate: `lim_(n rarr oo) (n^(2)(1+2+3+4+......+n))/(n^(4)+4n^(2))`. |
| Answer» Correct Answer - `(1)/(2)` | |
| 433. |
If `lim_(xto1) (2-x+a[x-1]+b[1+x])` exists, then a and b can take the values (where `[.]` denotes the greatest integer function)A. is always equal to -1B. is always equal to +1C. does not exist None of theseD. |
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Answer» Correct Answer - B::C Since the greatest integer function is discountinuous at integral values of x, for a given limit to exist both left-and right-hand limits must be equal. `L.H.L.=underset(xto1^(-))lim(2-x+a[x-1]+b[1+x])` `=2-1+a(-1)+b(1)=1-a+b` `R.H.L.=underset(xto1^(+))lim(2-x+a[x-1]+b[1+x])` `=2-1+a(0)+b(2)=1+2b` On comparing, we have `-a=b`. |
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| 434. |
`lim_(xto1) (xsin(x-[x]))/(x-1)`, where `[.]` denotes the greatest integer function, is equal toA. `(2)/(pi-1)`B. `(pi+1)/(2)`C. `(2)/(pi+1)`D. `(2(pi+1))/(pi-1)` |
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Answer» Correct Answer - C `underset(xto1)lim(xsin(x-[x]))/(x-1)` Now,`L.H.L.=underset(hto0)lim((1-h)sin(1-h-[1-h]))/((1-h)-1)` `=underset(hto0)lim((1-h)sin(1-h))/(-h)=-oo` `R.H.L.=underset(hto0)lim((1+h)sin(1+h-[1+h]))/((1+h)-1)=underset(hto0)lim((1+h)sinh)/(h)=1` Hence, the limit does not exist. |
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| 435. |
If `lim_(xto1) (1+ax+bx^(2))^((e)/((x-1)))=e^(3)`, then the value of bc is _________.A. `a=2`B. `a=1`C. `L=(1)/(64)`D. L=(1)/(32)` |
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Answer» Correct Answer - `(3)` `underset(xto1)lim(1+as+bx^(2))^((e)/(x-1))=e^(3)` or `e^(underset(xto1)lim(1+as+bx^(2)-1)^((e)/(x-1)))=e^(3)` or `e^(underset(xto1)lim(c(ax+bx^(2)))/(x-1))=e^(3)` or `underset(xto1)lim(c(ax+bx^(2)))/(x-1)=3` or `underset(hto0)lim(c"("a(1+h)+b(1+h)^(2)")")/(1+h-1)=3` or`underset(hto0)lim((ca+b)+(ac+2b)h+bh^(2))/(h)=3` or`ca+b=0 " and "ac+2b=3` or `b=3" and "ac=-3` Also, the form must be `1^(oo)` for which a+b=0, i.e.,a=-3" and "c=1`. |
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| 436. |
`lim_(xto1) [cosec(pix)/(2)]^(1//(1-x))` (where `[.]` represents the greatest integer function) is equal toA. (i) exists, (ii) does not existB. (i) does not exist, (ii) existsC. both (i) and (ii) existD. neither (i) nor (ii) exists |
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Answer» Correct Answer - B `cosec(pix)/(2)to1^(+)" when "xto1` or`[cosec(pix)/(2)]=1 `:." ""limit "=1` |
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| 437. |
Let `lim_(xto1) (x^(a)-ax+a-1)/((x-1)^(2))=f(a)`. Then the value of `f(4)` is_________.A. `+-(pi)/(4)`B. `+-(pi)/(3)`C. `+-(pi)/(6)`D. `+-(pi)/(2)` |
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Answer» Correct Answer - `(6)` Put `x=1+h`. Then, `f(a)=underset(hto0)lim((1+h)^(a)-a(1+h)+a-1)/(h^(2))` `=underset(hto0)lim((1+ah+(a(a-1))/(2!)h^(2)+...)-a-ah+a-1)/(h^(2))` `:.f(a)=(a(a-1))/(2)` `:.f(4)=6` |
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| 438. |
Let `m` and `n` be two positive integers greater than 1. If `lim_(ato0) (e^(cos(alpha^(n)))-e)/(alpha^(m))=-(e)/(2),`then the value of `(m)/(n)` is ___________. |
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Answer» Correct Answer - `(2)` `mge2" and "nge2` `underset(alphato0)lim(e^(cos(alpha^(n)))-e)/(alpha^(m))` `=exxunderset(alphato0)lim(e^(cos(alpha^(n))-1)-1)/(cos(alpha^(n))-1)xx((cos(alpha^(n))-1)/((alpha^(n))^(2)))(alpha^(2n))/(alpha^(m))` `=exxunderset(alphato0)lim((e^(cos(alpha^(n))-1)-1)/(cos(alpha^(n))-1))xxunderset(alphato0)lim((cos(alpha^(n))-1)/((alpha^(2n))))xxunderset(alphato0)limalpha^(2n-m)` `=exx1xxunderset(alphato0)lim(-2"sin"^(2)(alpha^(n))/(2))/(alpha^(2n))xxunderset(alphato0)limalpha^(2n-m)` `=exx1xx(-(1)/(2))xxunderset(alphato0)limalpha^(2n-m)` Now, `underset(alphato0)limalpha^(2n-m)` must be equal to 1. i.e., `2n-m=0` or `(m)/(n)=2` |
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| 439. |
Evaluate `lim_(xtooo)[(x^(4)sin((1)/(x))+x^(2))/((1+|x|^(3)))].` |
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Answer» `underset(xtooo)lim[(x^(4)sin((1)/(x))+x^(2))/((1-x^(3)))]" "`(as x is negative) `=underset(xtooo)lim[(xsin((1)/(x))+1/x)/(1/x^(3)-1)]` `underset(xtooo)lim[(sin((1)/(x))/((1)/x)+(1)/(x))/((1)/(x^(3))-1)]` `=(1+0)/(0-1)=-1` |
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| 440. |
Evaluate `lim_(x to oo) x("tan"^(-1)(x+1)/(x+4)-(pi)/(4)).` |
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Answer» We have `underset(xtooo)limx("tan"^(-1)(x+1)/(x+4)-(pi)/(4))` `=underset(xtooo)limx("tan"^(-1)(x+1)/(x+4)-tan^(-1)1)` `=underset(xtooo)limxtan^(-1)(((x+1)/(x+4)-1)/(1+(x+1)/(x+4)))` `=underset(xtooo)limxtan^(-1)((-3)/(2x+5))` `=underset(xtooo)lim{(tan^(-1)((-3)/(2x+5)))/((-3)/(2x+5))}((-3)/(2x+5))` `=underset(xtooo)lim{(tan^(-1)((-3)/(2x+5)))/((-3)/(2x+5))}underset(xtooo)lim((-3)/(2x+5))` `=1xxunderset(xtooo)lim((-3)/(2+(5)/(x)))=1xx((-3)/(2))=-(3)/(2)` |
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| 441. |
The largest value of the non-negative integer a for which `lim_(xto1) {(-ax+sin(x-1)+a)/(x+sin(x-1)a)}^((1-x)/(1-sqrt(x)))=(1)/(4)` is ___________. |
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Answer» Correct Answer - `(2)` `underset(xto1)lim{(-ax+sin(x-1)+a)/(x+sin(x-1)-1)}^((1-x)/(1-sqrt(x)))=(1)/(4)` `implies" "underset(xto1)lim{((sin(x-1))/((x-1))-a)/((sin(x-1))/((x-1))+1)}^(1+sqrt(x))=(1)/(4)` `implies" "((1-a)/(2))^(1)=(1)/(4)` `implies" "a=0,a=2` |
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| 442. |
Evaluate `lim_(xto0)(tanx-sinx)/(x^(3)).` |
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Answer» We have `underset(xto0)lim(tanx-sinx)/(x^(3))" "`(0/0 form) `=underset(xto0)lim((sinx-sinxcosx)/(x^(3)cosx))` `=underset(xto0)lim{(sinx(1-cosx))/(x^(3)cosx)}=underset(xto0)lim{(sinx)/(x)(1-cosx)/(x^(2))(1)/(cosx)}` `={underset(xto0)lim(sinx)/(x)}{underset(xto0)lim(2"sin"^(2)(x)/(2))/(((x)/(2))xx4)}underset(xto0)lim(1)/(cosx)` `={underset(xto0)lim(sinx)/(x)}1/2{underset(xto0)lim((sin(x/2))/((x)/2))^(2)}underset(xto0)lim(1)/(cosx)` `=1xx1/2(1)^(2)xx1/2=1/2` |
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| 443. |
If `lim_(xrarr0)(x^3)/(sqrt(a+x)(bx-sinx))=1,agt0`, then `a+b` is equal toA. 36B. 37C. 38D. 40 |
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Answer» Correct Answer - B `underset(xrarr0)(lim)(x^(3))/(sqrt(a+x)(bx-sinx))=1` `rArr" "underset(xrarr0)(lim)(x^(3))/(sqrta(bx-sinx))=1` `rArr" "underset(hrarr0)(lim)(3x^(2))/(sqrta(b-cos x))=1` `therefore" "b=1` `therefore" "l=underset(xrarr0)(lim)(3x^(2))/(sqrta(1-cosx))` `" "=underset(xrarr0)(lim)(3x^(2))/(sqrta 2sin^(2).(x)/(2))=(6)/(sqrta)` `therefore" "a=26` Thus `a+b=37` |
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| 444. |
Evaluate `lim_(xto0) (e^(x)+e^(-x)-2)/(x^(2))` |
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Answer» Correct Answer - 1 `underset(xto0)lim(e^(x)+e^(-x)-2)/(x^(2))" "`(form 0/0) `=underset(xto0)lim(1)/(e^(x))((e^(x)-1)/(x))^(2)` =1 |
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| 445. |
`lim_(xto0) ((1-3^(x)-4^(x)+12^(x)))/(sqrt((2cosx+7))-3)` |
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Answer» Correct Answer - `-12ln2xxln3` `underset(xto0)lim((1-3^(x)-4^(x)+12^(x)))/(sqrt((2cosx+7))-3)` `=underset(xto0)lim((3^(x)-1)(4^(x)-1))/(sqrt((2cosx+7))-3)` `=underset(xto0)lim((3^(x)-1)(4^(x)-1)(sqrt(2cosx+7))+3)/((2cosx+7-9))` `=underset(xto0)lim(((3^(x)-1))/(x)xx((4^(x)-1))/(x)(sqrt((2cosx+7))+3))/((-2(1-cosx))/(x^(2)))` `=(("1n "3)("1n "4)6)/(-2xx(1)/(2))=-6" 1n "3xx" 1n "4` `=-12" 1n "2xx" 1n "3` |
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| 446. |
`lim_(xto0) ((1+tanx)/(1+sinx))^(cosecx)` is equal toA. `underset(xto0)limf(x)` exists for `ngt0`B. `underset(xto0)limf(x)` does not exists for `nlt0`C. `underset(xto0)limf(x)` does not exists for any value of nD. `underset(xto0)limf(x)` exists for any value of n |
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Answer» Correct Answer - C `underset(xto0)lim((1+tanx)/(1+sinx))^(cosecx)=underset(xto0)lim((1+tanx)^((1)/(sinx)))/((1+sinx)^((1)/(sinx)))` `=(underset(xto0)lim((1+tanx)^((1)/(tanx)))^((1)/(cosx)))/((1+sinx)^((1)/(sinx)))` `=(e^((1)/(cos0)))/(e)=1` |
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| 447. |
If `lim_(xto0) (p sin2x+(1-cos2x))/(x+tanx)=1`, then the value of `p` is___________. |
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Answer» Correct Answer - `(1)` `underset(xto0)lim(p sin2x+(1-cos2x))/(x+tanx)=1` `implies" "underset(xto0)lim(2p((sin2x)/(2x))+((1-cos2x)/(x)))/(1+(tanx)/(x))=1` `implies" "(2p+0)/(1+1)=1` `implies" "p=1` |
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| 448. |
Let `f(x) = ((1 - x(1+ |1-x | )) /(|1-x|)) cos(1/(1-x))` for `x!=1` |
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Answer» Correct Answer - C::D `f(1^(+))=underset(hto0)lim(1-(1+h)(1+h))/(h)"cos"(1)/(h)` `=underset(hto0)lim(-h^(2)-2h)/(h)"cos"(1)/(h)=underset(hto0)lim(-h-2)"cos"(1)/(h)` Thus, `underset(xto1^(+))limf(x)` does nto exist. `f(1^(-))=underset(hto0)lim(1-(1-h))/(h)"cos"(1)/(h)` `=underset(hto0)lim(-1(1-h^(2)))/(h)"cos"(1)/(h)``=underset(hto0)lim(h^(2))/(h)"cos"(1)/(h)=underset(hto0)lim h"cos"(1)/(h)=0` |
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| 449. |
Evaluate `lim_(xto0) (a^(tanx)-a^(sinx))/(tanx-sinx),agt0` |
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Answer» Correct Answer - `ln a` `underset(xto0)lima^(sinx)xx((a^(tanx-sinx)-1))/((tanx-sinx))=a^(0)" 1n "a=" 1n "a` |
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| 450. |
Let`L=lim_(xto0) (a-sqrt(a^(2)-x^(2))-(x^(2))/(4))/(x^(4)),agt0`. If L is finite, then |
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Answer» Correct Answer - A::C `L=underset(xto0)lim(a-sqrt(a^(2)-x^(2))-(x^(2))/(4))/(x^(4))` `=underset(xto0)lim((1)/(x^(2)(a+sqrt(a^(2)-x^(2))))-(1)/(4x^(2)))` `=underset(xto0)lim((4-a)-sqrt(a^(2)-x^(2)))/(4x^(2)(a+sqrt(a^(2)-x^(2))))` Numerator` to" if "a=2` and then `L=(1)/(64)`. |
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