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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.
| 51. |
Evaluate the following limits: `lim_(xrarr0)(tan3x)/(sin4x)` |
| Answer» Correct Answer - `3/4` | |
| 52. |
Evaluate the following limits: `lim_(xrarr0)(sinxcosx0)/(3x)` |
| Answer» Correct Answer - `1/3` | |
| 53. |
Evaluate the following limits: `lim_(xrarr0)(sinmx)/(tannx)` |
| Answer» Correct Answer - `m/n` | |
| 54. |
` lim_(x to 1) sqrt(1-cos 2(x-1))/(x-1)`A. exists and it equals `sqrt2`B. exists and it equals `-sqrt2`C. does not exist because ` x -1 to 0`D. does not exist because left hand limit is not equal to right hand limit |
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Answer» Correct Answer - C LHL `underset(x to 1^(-)) lim sqrt(1-cos 2(x-1))/(x-1)` ` = underset(x to 1^(-)) limsqrt(2 sin^(2)(x-1))/(x-1) = sqrt2 underset(x to 1^(-)) lim (|sin (x-1)|)/(x-1)` Put ` x = 1 - h, h gt 0, " for " x to 1^(-) , h to 0` ` = sqrt2 underset ( h to 0) lim (|sin (-h)|)/(-h) ` ` = sqrt2 underset( h to 0) lim (sin h)/(-h) = - sqrt2` Again, RHL `underset( x to 1^(+)) lim sqrt(1-cos 2 (x -1))/(x-1)` ` underset ( x to 1^(+)) lim sqrt2 (|sin ( x - 1)|)/(x - 1) ` Put ` x = 1 + h, h gt 0 ` For ` x to 1^(+), h to 0` `underset( h to 0) lim sqrt2 (| sin h |)/h = underset( h to 0) lim sqrt2 (sin h)/h = sqrt2 ` ` :. ` LHL `ne` RHL. Hence, `underset( x to 1) lim f (x) ` does not exist. |
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| 55. |
Evaluate the following limit: `(lim)_(x->1)(sqrt(3+x)-sqrt(5-x))/(x^2-1)` |
| Answer» Correct Answer - `1/4` | |
| 56. |
Evaluate the following limits: `lim((1-cos2x))/(3tan^(2)x)` |
| Answer» Correct Answer - `2/3` | |
| 57. |
Evaluate:`("lim")_(xvecpi/2)(1+cos2x)/((pi-2x)^2)` |
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Answer» Correct Answer - `1/2` Putting `(pi-2x)=t,` we have giving limit `=lim_(t to a)(1+cos(pi-t))/(t^(2))=2/4.lim_(t to0){(sin(t//2))/((t//2))}^(2)=((1)/(2)xx1^(2))=1/2.` |
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| 58. |
Evaluate the following limit: `(lim)_(x->pi/6)(cot^2x-3)/(cos e c x-2)` |
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Answer» Correct Answer - 4 Given limit `=lim_(xto(pi)/(6))(cosec^(2)x-4)/(cosec x-2)=lim_(xto(pi)/(6))(cosec x+2)=(2+2)=4.` |
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| 59. |
Evaluate the following limits: `lim_(xrarr0)((e^(3+x)-sinx-e^(3)))/(x)` |
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Answer» Correct Answer - `(e^(3)-1)` Given limit `=e^(3). lim_(xto0)((e^(x)-1)/(x))-lim_(xto0)(sinx)/(x)=(e^(3)xx1-1)=(e^(3)-1).` |
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| 60. |
Evaluate the following limits: `lim_(xrarr0)((e^(tan-x)-1))/(tanx)` |
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Answer» Correct Answer - `(2a sin a+a^(2)cosa)` Given limit `=lim_(hto0)((a^(2)+h^(2)+2ah)sin(a+h)-a^(2)sina)/(h)` `=a^(2).lim_(hto0)(sin(a+h)-sina)/(h)+lim_(hto0)h sin(a+h)+lim_(hto0)2a sin(a+h)` `=a^(2). lim_(hto0)(cos(a+(h)/(2)).sin((h)/(2)))/((h//2))+0+2asina` `=a^(2).lim_(xto0)cos(a+(h)/(2)). lim_(hto0)(sin(h//2))/((h//2))+2asina` `=(a^(2)xxcosaxx1)+2asina=(a^(2)cosa+2asina).` |
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| 61. |
Evaluate the following limits: `lim_(xrarr0)((e^(2+x)-e^(2))/(x))` |
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Answer» Correct Answer - 1 Given limit `=lim_(xto0)((e^(3)-1)-(e^(2x)-1))/(x)` `=lim_(xto0)((e^(3x)-1))/(x)lim_(xto0)((e^(2x)-1)/(x))` `=lim_(3xto0){3((e^(3x)-1)/(3x))}-lim_(2xto0){2((e^(2x)-1)/(2x))}` `=(3xx1)-(2xx1)=(3-2)=1.` |
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| 62. |
Evaluate the following limits: `lim_(xrarr0)((e^(x)-x-1)/(x))` |
| Answer» Given limit `=lim_(xto0){((e^(x)-1)/(x))-1}=lim_(xto0)((e^(x)-1)/(x))-1=(1-1)=0.` | |
| 63. |
Evaluate: `(i) lim_(xrarr0)((tanx-sinx)/sin^(3)x)" "(ii)lim_(xrarr0)((tanx-sinx))/(x^(3))` |
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Answer» `(i) lim_(xto0)((tanx-sinx))/(sin^(3)x)=lim_(xto0)((sinx-sinxcosx)/(cosxsin^(3)x))` `=lim_(xto1)(sinx(1-cosx))/(cosxsin^(3)x)=lim_(xto0)((1-cosx))/(cosxsin^(2)x)` `=lim_(xto0)((1-cosx))/(cosx(1-cos^(2)x))` `=lim_(xto0)(1)/(cosx(1+cosx))=1/2" "["putting"x=0].` (ii) `lim_(xto0)((tanx-sinx))/(x^(3))=lim_(xto0)((sinx-sinxcosx)/(x^(3)cosx))` `=lim_(xto0)(sinx(1-cosx))/(x^(3)cosx)=lim_(xto0){(sinx)/(x).(2sin^(2)(x//2))/(x^(2)),(1)/(cosx)}` `=lim_(xto0){(sinx)/(x).(2sin^(2)(x//2))/((x//2)^(2)).1/4/(1)/(cosx)}` `1/2.lim_(xto0)(sinx)/(x).lim_(xto0){(sin(x//2))/((x//2))}^(2).lim_(xto0)(1)/(cosx)` `((1)/(2)1xx1xx1^(2)xx1)=1/2.` |
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| 64. |
Evaluate the following limits: `lim_(xrarr0)((a^(x)-a^(-x))/(x))` |
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Answer» Correct Answer - `2loga` Given limit `=lim_(xto0){((a^(2x)-1)/(2x))xx(2)/(a^(x))}=[(loga)xx2/1]=2loga.` |
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| 65. |
Evaluate the following limits: `lim_(xrarra)(cosx-cosa)/(cotx-cota)` |
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Answer» Correct Answer - `sin^(3)a` Given limit `=lim_(xtoa)(cosx-cosa)/(cosxsina-sinxcosa).sinxsina` `=lim_(xtoa)(-2sin((x+a)/(2))sin((x-a)/(2)))/(sin(a-x)).sinxsina` `=lim_(xto0)(-2sin((x+a)/(2))sin((x-a)/(2)))/(-2sin((x-a)/(2))cos((x-a)/(2))).sinxsina` `=lim_(xto0)sin((x+a)/(2))/(cos((x-a)/(2))).sinxsina=sin^(3)a.` |
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| 66. |
Evaluate the following limits: `lim_(xrarr0)((a^(x)-b^(x))/(x))` |
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Answer» Correct Answer - `(log a-logb)` Given limit `=lim_(xto0){((a^(x)-1)-(b^(x)-1))/(x)}=lim_(xto0)((a^(x)-1)/(x))-lim_(xto0)((b^(x)-1)/(x))=(loga-logb).` |
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| 67. |
`lim_(x rarr pi/4)(tan^3x-tanx)/(cos(x+pi/4)` |
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Answer» Correct Answer - `-4` Given limit `=lim_(xto(pi)/(4))(tanx(tanx+1)(sinx-cosx))/(cosx.cos(x+(pi)/(4)))` `=lim_(xto(pi)/(4))(tanx(tanx+1)(sinx-cosx))/(cosx.cos(x+(pi)/(4)))` `=-lim_(xto(pi)/(4))(tanx(tanx+1)(cosx-sinx))/(cosx. cos(x+(pi)/(4)))` `=-sqrt2lim_(xto(pi)/(4))(tanx(tanx+1)((1)/(sqrt2)cosx-(1)/(sqrt2)sinx))/(cosx.cos(x+(pi)/(4)))` `=-sqrt2lim_(xto(pi)/(4))(tanx(tanx+1).cos(x+(pi)/(4)))/(cosx. cos(x+(pi)/(4)))=-sqrt2lim_(xto(pi)/(4))(tanx(tanx+1))/(cosx)` `=-sqrt2.(1xx(1+1))/(((1)/(sqrt2)))=-4.` |
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| 68. |
`lim_(x->pi/2) (tan 2x)/(x-pi/2)` |
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Answer» Correct Answer - 2 Put `x-(pi)/(2)=h. Then, xto(pi)/(2)implies(x-(pi)/(2))to0implieshto0.` `thereforelim_(xto(pi)/(2))(tan2x)/((x-(pi)/(2)))=lim_(hto0)(tan2(h+(pi)/(2)))/(h)=2lim_(hto0)(tan(2h+pi))/(2h)` `=2 lim_(hto0)(tan2h)/(2h)=(2xx1)=2.` |
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| 69. |
`lim_(x->pi/6)(2-sqrt(3)cosx-sinx)/((6x-pi)^2)` |
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Answer» Correct Answer - `1/36` Given limit `=lim_(xto(pi)/(6))(2-2((sqrt3)/(2)cosx+(1)/(2)sinx))/((6x-pi))` `=lim_(xto(pi)/(6))(2[-1(cos""(pi)/(6))])/((6x-pi)^(2))=lim_((x)/(2)to(pi)/(12))(2xx2sin^(2)((x)/(2)-(pi)/(12)))/(144((x)/(2)-(pi)/(12))^(2))` `=(4)/(144)xxlim_((x)/(2)to(pi)/12){(sin((x)/(2)-(pi)/(12)))/(((x)/(2)-(pi)/(12)))}^(2)=((1)/(36)xx1^(2))=1/36.` |
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| 70. |
Evaluate `lim_(x to pi)((1+cosx)/(tan^(2)x)).` |
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Answer» `lim_(x to pi)((1+cosx)/(tan^(2)x))=lim_(x to pi)(cos^(2)x(1+cos x))/(sin^(2)x)` `=lim_(x to pi)(cos^(2)c(1+cosx))/((1-cos^(2)x))` `=lim_(x to pi)(cos^(2)x)/((1-cosx))=1/2" "["putting x"=pi.].` |
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| 71. |
Evaluate `lim_(xrarr0)((1-cosxsqrt(cos2x)))/(x^(2)).` |
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Answer» `lim_(xto0)((1-cosxsqrt(coss)))/(x^(2))` `=lim_(xto0){((1-cosxsqrt(cos2x)))/(x^(2))xx((1+cosxsqrt(cos2x)))/((1+cosxsqrt(cos2x)))}` `=lim{((1-cos^(2)xcos2x))/(x^(2))xx(1)/((1+cosxsqrt(cos2x)))}` `=lim_(xto0)((1-cos^(2)xcos2x))/(x^(2))xxlim_(xto0)(1)/((1+cosxsqrt(cos2x)))` `lim_(xto0){(1-cos^(2)x(1-2sin^(2)x))/(x^(2))xx(1)/((1+1sqrt1))` `((1-cos^(2)x+2sin^(2)xcos^(2)x))/(x^(2))xx1/2` `=1/2xxlim_(xto0)((sin^(2)x+2sin^(2)xcos^(2)x))/(x^(2))xx1/2xxlim_(xto0)(sin^(2)(1+2cos^(2)x))/(x^(2))` `=1/2xxlim_(xto0)((sinx)/(x))^(2)xxlim_(xto0)(1+2cos^(2)x)` `1/2xx1^(2)xx(1+2xx1^(2))=3/2.` |
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| 72. |
If ` f(x)={{:(|x|+1,xlt0),(0,x=0) ,(|x|-1,xgt0):}` for what value (s) of a does `underset(xrarra)"lim"f(x)` exists? |
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Answer» Case 1. When `a lt 0.` In this case, we have `underset(xtoa^(+))(lim)f(x)=underset(hto0)(lim)f(a+h)=underset(hto0)(lim){|a+h|+1}=|a|+1.` `underset(xtoa^(+))(lim)f(x)underset(hto0)(lim)f(a-h)=underset(hto0)(lim){|a-h|+1}=|a|+1.` `therefore underset(xtoa^(+))limf(x)=underset(xtoa)limf(x)=|a|+1.` So, in this case, `underset (xtoa)lim` f(x)n exists. Case 2. When `a lt 0.` In this case, we have `underset (xtoa^(+))limf(x)=underset(xto0)lim(a+h)=underset(hto0)lim{|a-h|-1}=|a|-1.` `underset(xtoa^(+))limf(x)=underset(hto0)limf(a-h)=underset(hto0)lim{|a-h|-1}=|a|-1.` `thereforeunderset(xtoa^(+))limf(x)=underset(xto0)limf(x)=|a|-1.` So, in this case, `unederset(xtoa)lim f(x)` exisets. Case 3. When `a lt 0.` In this case, we have `underset(xto0^(-))limf(x)=underset(hto0)limf(a+h)=underset(hto0)lim{|a-h|-1}=|a|-1.` `underset(xto0)limf(x)=underset(hto0)limf(0-h)=underset(hto0)lim{|-h|+1}=1.` Thus, Thus ` underset(xto0^(+))limf(x)neunderset(xto0^(-))limf(x).` Hence,` underset(xto0)limf(x)` does not exists. Thus, `underset(xtoa)limf(x)` exists only when `a ne 0.` |
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| 73. |
Evaluate `lim_(xrarr0){(sqrt(1+x)-sqrt(1-x))/(x)}.` |
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Answer» We have, `lim_(xto0){(sqrt(1+x)-sqrt(1-x))/(x)}=lim_(xto0){((sqrt(1+x)-sqrt(1-x)))/(x).((sqrt(1+x)+sqrt(1-x)))/((sqrt(1+x)+sqrt(1-x)))}` `=lim_(xto0)({(1+x)-(1-x)})/(x(sqrt(1+x)+sqrt(1-x)))` `=lim_(xto0)(2x)/(x(sqrt(1+x)+sqrt(1-x)))` `=lim_(xto0)(2)/((sqrt(1+x)+sqrt(1-x)))=1" "["putting"x=0].` |
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| 74. |
Evaluate `lim_(xrarra)((x^(m)-a^(m))/(x^(n)-a^(n))).` |
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Answer» `lim_(xtoa)((x^(m)-a^(m))/(x^(n)-a^(n)))=lim_(xtoa){((x^(m)-a^(m))/(x-a))div((x^(n)-a^(n))/(x-a))}` `=lim_(xtoa)((x^(m)-a^(m))/(x-a))divlim_(xtoa)((x^(n)-a^(2))/(x-a))` `=(ma^(m-1))(na^(n-1))` `=(ma^(m-1))/(na^(n-1))=m/na^(m-n).` |
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| 75. |
Evaluate `lim_(x-> 0) (tan3x-2x)/(3x- sin^2 x)` |
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Answer» `lim_(xto0)((tan3x-2x)/(3xsin^(2)x))` `=lim_(xto0)(((tan3x)/(3x)-2/3))/((1-(1)/(3).(sinx)/(x).sinx))" "["dividing num. and denom. by 3x"]` `=((lim_(xto0)(tan3x)/(3x)-2/3))/({1-1/3lim_(xto0)(sinx)/(x).lim_(xto0)sinx})=((1-(2)/(3)))/(1-((1)/(3)xx1xx0))=(((1)/(3)))/(1)=1/3.` |
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| 76. |
Evalute `lim_(xrarr0)((e^(3x)-1)/(x)).` |
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Answer» `lim_(xto1)((e^(3x)-1)/(x))=lim_(3xto0){((e^(3x)-1)/(3x))xx3}" "[because(xto0)implies(3xto0)]` `=3xxlim_(yto0)((e^(y)-1)/(y))where y=3x` `=(3xx1)=3" "[becauselim_(xto0)((e^(y)-1)/(y))=1].` |
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| 77. |
Evaluate `lim_(xrarr0)((sqrt(1+3x)-sqrt(1-3x)))/(x).` |
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Answer» `lim_(xto0)((sqrt(1+3x)-sqrt(1-3x)))/(x)` `=lim_(xto0){((sqrt(1+3x)-sqrt(1-3x)))/(x)xx((sqrt(1+3x)+sqrt(1-3x)))/((sqrt(1+3x)+sqrt(1-3x)))}` `=lim_(xto0)({1+3x)-(1-3x))/(x(sqrt(1+3x)+sqrt1-3x))=lim_(xto0)(6x)/((sqrt(1+3x)+sqrt(1-3x)))` `=lim_(xto0)(6)/((sqrt(1+3x)+sqrt(1-3x)))=(6)/((sqrt2+sqrt1))=6/3=3.` |
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| 78. |
Let `f(x)={x}=` greater integer less than or eqal to x. For any integer k, show that `lim_(xtok) f(x)` does not exist. |
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Answer» We have `underset(xtok^(+))limf(x)=underset(hto0)limf(k+h)=underset(hto0)lim[k+h]=underset(hto0)limk=k{because[k+h]=k}` ` and underset(xtok^(-))limf(x)=underset(hto0)lim(k-h)=underset(hto0)lim[k-h]=underset(hto0)lim(k-1)=(k-1)[because[k-h]=k-1]` Thus, `underset(xtok)limf(x)neunderset(xtok^(-))lim f(x).` Hence,`underset(xtok)limf(x)` does not exist. |
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| 79. |
Find the values of a and b so that the function `f(x)={{:(x+asqrt2 sin x", "0 le x le pi//4),(2x cot x + b", "pi//4 le x le pi//2),(a cot 2x - b sin x", "pi//2 lt x le pi):}` is continuous for ` 0 le x le pi`. |
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Answer» Correct Answer - `a=pi/6, b = (-pi)/12` Since, f(x) is continuous for ` 0 le x le pi` `:." " RHL("at" x = pi/4)= LHL ("at" x = pi/4)` ` rArr" " ( 2* pi/4 cot. Pi/4 + b) = (pi/4 + asqrt2 * sin. Pi/4) ` `rArr" " pi/2 + b = pi/4 + a rArr a - b = pi/4` .....(i) Also, RHL ` ("at" x = pi/2) = LHL ("at" x = pi/2)` ` rArr (a cot (2pi)/2 - b sin. pi/2) = ( 2* pi/2* cot. pi/2 + b) ` `rArr" " -a-b=b` `rArr" " a+ 2b = 0` ....(ii) On solving Eqs. (i) and (ii) , we get ` a = pi/6 and b =- pi/12 ` |
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| 80. |
If `.^(20)C_(1)+(2^(2)) .^(20)C_(2) + (3^(2)).^(20)C_(3) +......+(20^(2)).^(20)C_(20)=A(2^(beta))` , then the ordered pair `(A, beta)` is equal toA. `(420 , 19)`B. `(420 , 18)`C. `(380, 18)`D. ` (380, 19)` |
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Answer» Correct Answer - B We Know, `(1+x)^(n) = .^(n)C_(0) + .^(n)C_(1)x + .^(n)C_(2) x^(2) +... +.^(n)C_(n) x^(n) ` On differentiating both sides w. r. t. X, we get `n(1+x)^(n-1)=.^(n)C_(1) + 2 .^(n)C_(2) x + ... + n .^(n)C_(n) x^(n-1)` On multiplying both sides by x, we get `n x(1+x)^(n-1)=.^(n)C_(1)x+2^(n) C_(2)x^(2)+...+n .^(n)C_(n) x^(n) ` Again on differentiating both sides w.r.t. x, we get `n[(1+x)^(n-1) +(n-1) x (1+x)^(n-2)]=.^(n)C_(1) + 2^(2) .^(n)C_(2)x+...+ n^(2) .^(n)C_(n) x^(n-1) ` Now putting x = 1 in both sides, we get `.^(n)C_(1) +(2^(2)).^(n)C_(2)+(3^(2)) .^(n)C_(3) + ...+(n^(2)).^(n)C_(n) = n(2^(n-1)+(n-1) 2^(n-2))` For n = 20, we get `.^(20)C_(1) +(2^(2)) .^(20)C_(2)+(3^(2)) .^(20)C_(3) + ...+ (20)^(2) .^(20)C_(20) ` ` = 20 (2^(19)+(19) 2^(18))` ` = 20 (2+19) 2^(18) = 420 (2^(18))` ` = A(2^(B))` (given) On comparing, we get (A,B) = (420, 18) |
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| 81. |
The derivative of an even function is always an odd function. |
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Answer» Correct Answer - True It is always true that differential of even function is and odd function. |
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| 82. |
Let `f(x) = {x-1)^2 sin(1/(x-1)-|x|`,if `x!=1` and -1, if `x=1` 1valued function. Then, the set of pointsf, where `f(x)` is not differentiable, is .... . |
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Answer» Correct Answer - x = 0 Given, `f(x) = {{:((x-1)^(2) sin. 1/((x-1))-|x|", if "x ne 1),(" -1, if " x = 1):}` As, `f(x) = {{:((x-1)^(2)sin. 1/(x-1) -x", "0le x-{1}),((x-1)^(2)sin 1/((x-1))+x", " x lt 0),(" -1 , " x= 1):}` Here, f(x) in not differentiable at x = 0 due to |x|. Thus, f(x) is not differentiable at x = 0 . |
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| 83. |
A discontinuous function `y = f(x)` satisfying `x^2 + y^2= 4` is given by `f(x) = ....` |
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Answer» Correct Answer - `f(x)=sqrt(4-x^(2))` Given, ` x^(2) + y^(2) = 4 rArr y = sqrt(4-x^(2)) ` or `" " f(x) - sqrt( 4- x^(2)) ` |
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| 84. |
Q. f={`(x+a if x=0)` `g(x)={(x+1 if x |
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Answer» Correct Answer - `g{f(x)}={{:(x+a+1", if "xlt -a),((x+a-1)^(2)", if "ale x lt c),(x^(2) +b", if "0le xle 1),((x-2)^(2)+b", if "xgt1):}` a=1, b = 0 gof is differentiable at x = 0 gof `(x) = {{:(f(x) +1", if " f(x) lt 0),({f(x)-1}^(2)+b", if " f(x) ge 0):}` ` ={{:(x+a+1", if " x lt -a),((x+a-1)^(2)+b", if " -a le x lt 0),((|x-1|-1)^(2)+b", if " c ge 0):}` As gof (x) is continuous at x =- a gof `(-a) = "gof "(-a^(+)) = " gof " (-a^(-)) ` `rArr" " 1+b = 1 + b = 1 rArr b = 0 ` Also, gof (x) is continuous at x = 0 `rArr" " "gof " (0) =" gof " (0^(+)) = " gof " (0^(-))` ` rArr" " b = b = (a-1)^(2) + b rArr a = 1` `"Hence, "gof(x) = {{:(x+2", if " x lt -1),(x^(2)", if " x lt -1),((|x-1|-1)^(2)", if " x ge 0):}` In the neighbourhood of x = 0, gof `(x) = x^(2)`, which is differentiable at x = 0. |
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| 85. |
Let ` f(x) = {{:({1+|sin x|}^(a//|sin x|)", " pi/6 lt x lt 0),(" b, " x = 0 ),(e^(tan 2x//tan 3x) ", "0ltx ltpi/6):}` Determine a and b such that f(x) is continous at x = 0. |
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Answer» Correct Answer - `a=2/3,b=e^(2//3)` ` f(x) = {{:({1+|sin x|}^(a//|sin x|) " , "pi//6 lt x lt 0),(" , " x = 0),(e^(tan 2x//tan 3x)" , "0 lt x lt pi //6):}` Since, f(x) is continuous at x = 0 . `:. ` RHL (at x = 0) = LHL (at x = 0) = f(0) ` rArr underset( h to 0) lim e^(tan 2h//tan 3h) = underset( h to 0) lim { 1+ | sin h| } ^(a//| sin h|) = b` ` rArr" " e^(23) = e^(a) = b` ` :." " a = 2//3` ` and" " b = e^(2//3) ` |
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| 86. |
Let `f(x)={1+x ,0 lt x leq 2 and 3-x ,2 lt x leq 3` Determine the composite function `g(x) = f(f(x))` & hence find the point of discontinuity of g. if any. |
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Answer» Given, `f(x) = {{:(1+x", "0 le x le 2),(3-x", " 2 lt x le 3):}` `:." "fof(x) = f[f(x)]={{:(1+f(x)", " 0 le f(x) le 2),(3-f(x)", "2ltf(x)le3):}` `= fof = {{:(1+f(x)" ,"0le f(x) le 1),(1+f(x)", "1ltf(x)le2),(3-f(x)", "2ltf(x) le 3):}={{:(1+(3-x)", "2lt xle 3),(1+(1+x)", " 0 le x le 1 ),(3-(1+x)", " 1 lt x le 2 ):}` `rArr (fof) (x) = {{:(4-x", " 2 lt x le 3),(2+x", " 0 le x le 1),(2-x", " 1 lt x le 2 ):}` Now, RHL (at x = 2) = 2 and LHL (at x = 2) = 0 Also, RHL (at x = 1) = 1 and LHL (at x =1) = 3 Therefore, f(x) is discontinuous at x = 1,2 ` :. ` f[f(x)] is discontinuous at x = {1, 2}. |
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| 87. |
Let ` f(x) = 15 -|x-10|, x in R`. Then, the set of all values of x, at which the function, ` g(x) = f(f(x))` is not differentiable, isA. `{5, 10, 15, 20}`B. ` {5, 10, 15}`C. ` { 10}`D. `{10, 15}` |
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Answer» Correct Answer - B Given function is `f(x) = 15 - |x-10|, x in R and g(x) = f(f(x))` `= f(15-|x-10|)` ` = 15-|15-|x-10|-10|` ` = 15-|5-|x - 10||` `{{:(15-|5-(x-10)|" , "x ge 10),(15-|5+(x-10)|" , " x lt 10):}` ` ={{:(15-|15-x|" , " x ge 10),(15-|x-5|" , " x lt 10):}` `={{:(15+(x-5)= 10 +x" , "x lt 5),(15-(x-5)=20-x" , "5le x lt 10),( 15+(x-15) = x" , " 10 le x lt 15),(15-(x-15) = 30-x", " x ge 15):}` From the above definition it is clear that g(x) is not differentiable at x = 5, 10, 15. |
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| 88. |
Let ` f(x+y) = f(x) + f(y)` for all x and y. If the function f (x) is continuous at x = 0 , then show that f (x) is continuous at all x. |
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Answer» Since, f(x) is continuous at x = 0. `rArr" " underset( x to 0) lim f(x) = f(0) ` `rArr" "f(0^(+)) = f(0^(-))= f (0) = 0` ….(i) To show, continuous at x = k RHL = ` underset( h to 0) lim f(k+h) = underset( h - 0) lim [f(k)+f(h)] = f(k) + f(0^(+))` ` " " f(k) + f(0) ` LHL = ` underset( h to 0) lim f(k - h) = underset( h to 0) lim [f(k) + f(-h)]` ` " " = f(k) + f(0^(-)) = f(k) + f(0) ` ` :." " underset( x to k) lim f(x) = f(k) ` ` rArr f(x) ` is continuous for all ` x in R`. |
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| 89. |
For every pair of continuous functions `f,g:[0,1]->R` such that `max{f(x):x in [0,1]}= max{g(x):x in[0,1]}` then which are the correct statementsA. `[f(c )]^(2) + 3f(c ) = [g(c )]^(2) + 3g(c )` for some ` c in[0, 1]`B. `[f(c )]^(2)+f(c ) = [g(c )]^(2) + 3g(c )` for some `c in [0, 1]`C. `[f(c )]^(2) + 3f(c ) = [g(c )]^(2) + g(c )` for some ` c in [0, 1]`D. `[f(c )]^(2) = [g(c )]^(2) ` for some `c in [0, 1]` |
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Answer» Correct Answer - A::D PLAN If a continous function has values of opposite sign inside an interval, then it has a root in that interval. `" " f,g : [0, 1] to R` We take two cases. Case I Let f and g attain their common maximum value at p. `rArr" " f(p) = g(p)`, where ` p in [0, 1]` Case II Let f and g attain their common maximum value at different points. ` rArr" " f(a) = M and g(b) = M` ` rArr" " f(a) - g(a) gt 0 and f(b) - g(b) lt 0` ` rArr f(c ) - g(c ) = 0 " for some " c in [0, 1]` as f and g are continuous functions. ` rArr f(c ) - g(c ) = 0 " for some " c in [0, 1]` for all cases. ....(i) Option `(a) rArr f^(2)(c ) - g^(2)(c ) + 3 [f(c ) - g (c ) ] = 0` which is true from Eq. (i) . Option ` (d) rArr f^(2) (c ) - g^(2) (c ) - g^(2) (c ) = 0` which is true from Eq. (i) Now, if we take ` f (x) = 1 and g(x) = 1, AA x in [0, 1]` Options (b) and (c ) does not hold. Hence, options (a) and (d) are correct. |
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| 90. |
Let m and n be two positive integers greater then 1. If ` lim_( a to 0) ((e^(cos(alpha^(n)))-e/alpha^(m)) =- (e/2) ` , then the value of `m/n` is |
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Answer» Correct Answer - `(2)` Given, `underset( alpha to 0) lim [(e^(cos(alpha^(n))-e))/alpha^(m)]=- e/2` `rArr underset( alphato 0)lim(e{e^(cos(alpha^(n))-1)})/(cos (alpha^(n))-1)*(cos(alpha^(n))-1)/alpha^(m) = (-e)/2` `rArr underset( alpha to 0) lim e{(e^(cos(alpha^(n))-1))/(cos(alpha^(n))-1)}*underset(alphato 0) lim (-2sin^(2) .alpha^(n)/2)/alpha^(m) =- e//2` `rArr e xx1 (-2) underset( alphato 0) lim (sin^(2).((alpha^(n))/2))/(alpha^(2n)/4 )* alpha^(2n)/(4 alpha^(m)) = (-e)/2` ` rArr e xx 1xx -2 xx1 underset( alphato 0) lim (alpha^(2n-m))/4 = (-e)/2 ` For this to be exists, 2n - m = 0 ` rArr" " m/n = 2` |
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| 91. |
Let `alpha, beta in R " be such that " lim_( x to 0) (x^(2)sin (beta x))/(ax - sin x) = 1."Then,"6(alpha+beta)"equals"` |
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Answer» Correct Answer - `(7)` Here, ` underset( x to 0) lim (x^(2) sin (beta x))/(ax - sin x ) = 1` ` rArr underset( x to 0) lim (x^(2)(beta x-((betax)^(3))/(3!) +((betax)^(5))/(5!) - ...))/(ax - (x - x^(3)/(3!) + x^(5)/(5!) - ...))= 1` `rArr underset(x to 0) lim (x^(3)(beta-(beta^(3) x^(2))/(3!) + (beta^(5)x^(4))/(5!) - ...))/((alpha-1)x+x^(3)/(3!) +x^(5)/(5!)-...)=1` Limit exists only, when `alpha - 1 = 0` `rArr" " alpha = 1` ....(i) `:. underset( x to 0) lim (x^(3) (beta-(beta^(3)x^(2))/(3!)+(beta^(5)x^(4))/(5!)-...))/(x^(3)(1/(3!)-x^(2)/(5!)-...))= 1 ` ` rArr" " 6beta=1` ...(ii) From Eqs. (i) and (ii) , we get ` 6(alpha + beta) = 6alpha + 6 brta` ` = 6 + 1 = 7` |
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| 92. |
`lim_(x to pi//2) (cot x - cos x)/((pi - 2x)^(3))` equalsA. `1/24`B. `1/16`C. `1/8`D. `1/4` |
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Answer» Correct Answer - B `underset ( x to pi//2)lim(cot x - cos x ) /((pi-2x)^(3)) = underset( x to pi//2 ) lim 1/8 * ( cos x ( 1 - sin x ))/(sin x (pi/2 - x )^(3)) ` ` underset ( h to 0) lim 1/8*(cos(pi/2 - h)[1 - sin (pi/2 - h)])/(sin(pi/2 - h)(pi/2 - pi/2 + h)^(3))` ` = 1/8 underset ( h to 0 ) lim ( sin h ( 1 - cos h ) )/( cos h * h ^(3)) ` ` = 1/8 underset( h to 0 ) lim ( sin h (2 sin ^(2) h/2))/( cos h * h ^(3 )) ` ` = 1/4 underset( h to 0 ) lim ( sin h * sin^(2) ( h/2 ) )/( h ^(3) cos h ) ` ` 1/4 underset ( h to 0 ) lim (( sin h ) /h ) ( ( sin h/2)/( h/2))^(2) * 1/( cos h) * 1/4 = 1/4 xx 1/4 = 1/16` |
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| 93. |
`(d^2x)/(dy^2)`equals:(1) `((d^2y)/(dx^2))^(-1)`(2) `-((d^2y)/(dx^2))^(-1)((dy)/(dx))^(-3)`(3) `((d^2y)/(dx^2))^(-1)((dy)/(dx))^(-2)`(4) `-((d^2y)/(dx^2))^(-1)((dy)/(dx))^(-3)`A. `((d^(2)y)/(dx^(2)))^(-1)`B. `-((d^(2)y)/(dx^(2)))^(-1)((dy)/(dx))^(-3)`C. `((d^(2)y)/(dx^(2)))((dy)/(dx))^(-2)`D. `-((d^(2)y)/(dx^(2)))((dy)/(dx))^(-3)` |
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Answer» Correct Answer - D Since, `(dx)/(dy) = 1/(dy//dx) = ((dy)/(dx))^(-1) ` ` rArr" " d/(dy) ((dx)/(dy)) = d/(dx) ((dy)/(dx))^(-1) (dx)/(dy) ` ` rArr" " (d^(2)x)/(dy^(2)) =- ((d^(2)y)/(dx^(2)))((dy)/(dx))^(-2)((dx)/(dy)) =- ((d^(2)y)/(dx^(2))((dy)/(dx))^(-3) ` |
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| 94. |
Evaluate the following limits: `lim_(hto0)((sqrt(1+h)-1)/(h))` |
| Answer» Correct Answer - `(1)/(2)` | |
| 95. |
Express 0.001 neighbourhood of – 5 in modulus form. |
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Answer» 0.001 neighbourhood of – 5 in modulus form is expressed by |x – (- 5) | < 0.001 = |x + 5] < 0.001. |
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| 96. |
1.998 < x < 2.002 Express in modulus and neighbourhood form |
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Answer» Here, a – δ = 1.998; a + δ = 2.002 In Neighbourhood form : N (a, δ) |
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| 97. |
|x + 3| < 0.15 Express in interval and neighbourhood form |
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Answer» Here, a = – 3: δ = 0.15 In Interval form : (a – δ, a + δ) |x + 3| < 0.15 = (- 3 – 0.15, – 3 + 0.15) = (- 3.15, – 2.85) In Neighbourhood form : N (a, δ) |x + 3| < 0.15 = N(-3, 0.15) |
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| 98. |
1.95 < x < 2.05 Express in modulus and neighbourhood form |
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Answer» Here, a – δ = 1.95; a + δ = 2.05 In Modulus form: | x – a | < 8 In Neighbourhood form : N (a, δ) |
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| 99. |
3.8 < x < 4.8 Express in modulus and neighbourhood form |
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Answer» Here, a – δ = 3.8; a + δ = 4.8 By adding, 2a = 8.6 ∴ a = \(\frac{8.6}2\) = 4.3 Putting a = 4.3 in a + δ = 4.8, 4.3 + 8 = 4.8 ∴ δ = 4.8 – 4.3 = 0.5 In Modulus form : | x – a| < δ In Neighbourhood form: N (a, δ) |
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| 100. |
State multiplication working rule of limit. |
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Answer» f(x) and g (x) are two functions of real variable x and \(\lim\limits_{x\to a}\) f(x) = 1,\(\lim\limits_{x\to a}\) g (x) = m. If f (x) g ∙ (x) is the product of two functions, then multiplication working rule of limit is as follows: Thus, limit of product of two functions is equal to the product of limit of these two functions. |
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