InterviewSolution
Saved Bookmarks
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.
| 51. |
Evaluate `lim_(xto0) (1-cos(1-cosx))/(x^(4)).` |
|
Answer» `1-cos(1-cosx)=2sin^(2)((1-cosx)/(2))` `=2sin^(2)("sin"^(2)(x)/(2))` `:. underset(xto0)lim(1-cos(1-cosx))/(x^(4))=underset(xto0)lim(2sin^(2)("sin"^(2)(x)/(2)))/(x^(4))` `=underset(xto0)lim(2sin^(2)("sin"^(2)(x)/(2)))/(("sin"^(2)(x)/(2))^(2)).("sin"^(4)(x)/(2))/(((x)/(2))^(4).16)=1/8` |
|
| 52. |
Evaluate `lim_(xto0) (cos^(-1)((1-x^(2))/(1+x^(2))))/(sin^(-1)x).` |
|
Answer» Correct Answer - 2 `underset(xto0)lim(cos^(-1)((1-x^(2))/(1+x^(2))))/(sin^(-1)x)=underset(xto0)lim(2tan^(-1)x)/(sin^(-1)x)=2` |
|
| 53. |
Evaluate `lim_(xto0) (8)/(x^(8)){1-"cos"(x^(2))/(2)-"cos"(x^(2))/(4)+"cos"(x^(2))/(2)"cos"(x^(2))/(4)}.` |
|
Answer» Correct Answer - `1//32` `underset(xto0)lim(8(1-"cos"(x^(2))/(4))(1-"cos"(x^(2))/(2)))/(x^(8))` `=underset(xto0)lim(8xx2"sin"^(2)(x^(2))/(8)xx2"sin"^(2)(x^(2))/(4))/(x^(8))` `=underset(xto0)lim(32)/(64xx16).("sin"^(2)(x^(2))/(8))/(x^(4)/(64)).(2"sin"^(2)(x^(2))/(4))/(x^(4)/(16))` `=1/32` |
|
| 54. |
The value of `lim_(xrarr1)(log_5 5x)^(log_x5)` , isA. 1B. eC. -1D. none of these |
| Answer» Correct Answer - B | |
| 55. |
Differentiate the following with respect to x(i) y = 25, (ii) y = π/4 |
|
Answer» (i) dy/dx = 0 (ii) dy/dx = 0 |
|
| 56. |
Differentiate the following with respect to x(i) y = 5 cos α, α is a constant.(ii) y = x6 |
|
Answer» (i) dy/dx = 0 (ii) dy/dx = 6x5 |
|
| 57. |
The value of `lim_(xrarr0) (logx-1)/(x-e)`, isA. 1B. `(1)/(e)`C. eD. 0 |
|
Answer» Correct Answer - B We have, `lim_(xtoe) (logx-1)/(x-e)=lim_(xtoe)(logx-loge)/(x-e)=lim_(xtoe)(log .x/e)/(x-e)` `lim_(xtoe)(log (1+x/e-1))/(x-e)=lim_(xto e) (log{1+((x-e)/(e))})/(e((x-e)/e))=(1)/(e)` |
|
| 58. |
The value of `lim_(xrarr1) (logx)/(sin pi x)`, isA. `(1)/(pi)`B. `-pi`C. `pi`D. `-(1)/(pi)` |
|
Answer» Correct Answer - D We have, `lim_(xto1) (logx)/(sin pi x)=lim_(xto1) (log {1+(x-1)})/(sin (pi-pix))` ` =lim_(xto1) (log {1+(x-1)})/(sin pi(1-x))` ` =lim_(xto1) (log {1+x(x-1)})/(x-1)xx(x-1)/(sinpi(1-x))` ` =-(1)/(2)lim_(xto1) (log {1+x(x-1)})/(x-1)xx (pi(x-x))/(sin pi(1-x))= -(1)/(pi)` |
|
| 59. |
`lim_(xrarrpi//2)(a^(cotx)-a^cosx)/(cotx-cosx)a gt 0` is equal toA. `log_(2)((pi)/(2))`B. `log_e 2`C. `log_e a`D. `a` |
|
Answer» Correct Answer - C We have, `lim_(xto pi//2)(a^(cotx)-a^cosx)/(cotx-cosx)` ` lim_(xto pi//2) a^cosx ((a^cotx-cos x-1)/(cot x -cos x)) =a ^(cos pi//2)xxlog _(e) a= log_e a ` |
|
| 60. |
`lim_(xrarr oo) (1+(2)/(x))^x` equalsA. eB. `oo`C. `e^2`D. `(1)/(e)` |
|
Answer» Correct Answer - C We have, `lim_(xto oo) (1+(2)/(x))^(x) =e x ^(lim_(xtooo)(2)/(x)xxx)=e^2` |
|
| 61. |
If `l =lim_(xrarr0) (tanx^(n))/((tanx)^m)`, where `m,n in N`, thenA. `l=1 " for all " m,n in N`B. `l={(1" if "n gt m,,),(0" if "n lt m,,):}`C. `l={(1" if "n = m,,),(0" if "n gt m,,):}`D. `l=0 " for all " m,n in N` |
|
Answer» Correct Answer - C We have, ` l=lim_(xto0) (tanx^n)/((tanx)^m)` ` rArr l=lim_(xto0) ((tanx^n)/(x^n))xx(x^n)/(x^m)xx((x)/(tanx))^m` ` rArr l=lim_(xto0) 1xx x^(n-m)xx1` `rArr l=lim_(xto0) x^(n-m)rArr l={(0,if,ngtm,),(1,if,n=m,),(oo,if,n=m,):}` |
|
| 62. |
`lim_(xrarr-1)((x^4+x^2+x+1)/(x^2-x+1))^((1-cos(x+1))/((x+1)^2))` is equal toA. 1B. `sqrt(2//3)`C. `sqrt(3//2)`D. `e^(1//2)` |
|
Answer» Correct Answer - B We observe that `(x^4+x^2+x+1)/(x^2-x+1)to(2)/(3)as xto-1` But, ` (1-cos(x+1))/(x+1)^2to (0)/(0) as xto -1`. ` therefore lim_(xto-1)((x^4+x^2+x+1)/(x^2-x+1))^((1-cos(x+1))/((x+1)^2))` ` =((2)/(3))^(lim_(xto-1)((1-cos(x+1))/(x+1)^2))=((2)/(3))^(lim_(xto-1))(2sin^(2).(x+1)/(2))/(4((x+1)/2)^(2))=((2)/(3))^(1//2)` |
|
| 63. |
If `lim_(x->oo)((1+a^3)+8e^(1/ x))/(1+(1-b^3)e^(1/ x))=2,` then there existsA. `a=1,b=(-3)^(1//3)`B. `a=1,b=3^(1//3)`C. `a=-1,b=-(3)^(1//3)`D. none of these |
|
Answer» Correct Answer - A We have, `lim_(xto0)((1+a^3)+8e^(1//x))/(1+(1-b^3)e^(1//x))=2 [(oo)/(oo)"form"]` `rArr lim_(xto0) ((1+a^3)e^(-1//x)+8)/(e^(-1//x)+(1-b^3))=2` `rArr (0+8)/(0+(-b^3))=2 rArr 1-b^3=4rArr b^3=-3rArr b=(-3)^(1//3)` Again, `lim_(xto0) ((1+a^3)+8e^(1//x))/(1+(1-b^3)e^(1//x))=2` `rArrlim_(xto0) ((1+a^3)+8e^(1//x))/(1+4e^(1//x))=2` `rArr 1+a^3=2` ` rArr a=1` |
|
| 64. |
The largets value of non negative integer for which `lim_(x->1){(-a x+sin(x-1)+a]1-sqrt(x))/(x+sin(x-1)-1)}^((1-x)/(1-sqrt(x)))=1/4`A. 2B. 3C. 4D. 5 |
|
Answer» Correct Answer - A We have , `lim_(alphato1){(-ax+sin(x-1)+a)/(x+sin(x-1)-1)}^((1-x)/(1-sqrt(x)))=(1)/(4)` `lim_(alphato1){((sin(x-1))/(x-1)-a)/(1+(sin(x-1))/(x-1))}^(1+sqrt(x))=(1)/(4)` `rArr ((1-a)/(2))^2=(1)/(4)` `rArr a^2=2a=0rArra=0,2`. |
|
| 65. |
Let `f: R->[0,oo)` be such that `lim_(x->5) f(x)` exists and `lim_(x->5) ((f(x))^2-9)/sqrt(|x-5|)=0` then `lim_(x->5)f(x)=`A. 1B. 2C. 3D. 0 |
|
Answer» Correct Answer - C `lim_(xto5)((f(x))^2-9)/(sqrt(|x-5|))=0` ` rArr lim_(xto5)(f(x))^2-9=0` ` rArr l^2-9=0, "where" lim_(xto5)f(x)=l` `rArr l= +-3` `rArr l=3 [because f(x)ge 0 " for all " x in R ]` `rArr lim_(xto5)f(x)=3` |
|
| 66. |
`lim_(xrarr0)(e^(1//x)-1)/(e^(1//x)+1)`,isA. -1B. 1C. 0D. non-existent |
|
Answer» Correct Answer - D Let `f(x)(e^1//x-1)/(e^(1//x)+1)`.Then, (LHL of `f(x)` at `x=0`) ` =lim_(xto0^-)f(x) =lim_(hto0) f(0-h)=lim_(hto0) (e^1//x-1)/(e^(-1//x)+1)` `=lim_(hto0) (((1)/(e^(1//h))-1)/((1)/(e^(1//h))+1))=-1[e^(1//h)to oo rArr (1)/(e^(1//h))to0]` and, ` [RHL of ` f(x)` at `x=0`] `=lim_(xto0^+) f(x) =lim_(hto0) f(0+h)=lim_(hto0) (e^(1//h)-1)/(e^(1//h)+1)` `lim_(hto0)(((1)/(e^(1//h))-1)/(1+e^(1//h)))["Dividing Nr and by "e^(1//h)]` Clearly, `lim_(xto0^-) f(x)ne lim_(xto0^+)f(x)`. Hence, `lim_(xto0) f(x)` does not exist. |
|
| 67. |
Let `f(x)=[x]=` Greatest integer less than or equal to x and k be an integer. Then, which one of the following in not correct?A. `lim_(xtok^-)f(x)=k-1`B. `lim _(xtok)f(x)=k`C. `lim _(xtok)f(x)"exists"`D. `lim_(xtok)f(x)` does not exist |
|
Answer» Correct Answer - C We have, `f(x)=[x]` `therefore ("LHS at" x =k)` `=lim_(xtok^-)f(x)=lim_(hto0)=lim_(hto0)f(k-h)=lim_(hto0)[h-k]` `=lim_(hto0)k-1=k-1[because k-1ltk-hltk-hlt ktherefore [k-h]=k-1]` `therefore (RHL "at" x=k)` `=lim_(xto0)f(x)=lim_(hto0)f(k-h)=lim_(hto0)[k+h]` `=lim_(hto0)k=k[because klt k+hlt k+1therefore [k+h]=k]` Clearly, `lim_(xt0k^+)f(x) ne lim_(xtok^+)f(x). So, lim_(xtok)f(x)` does not exist. |
|
| 68. |
Evaluate: \(\lim\limits_{x \to 0}\frac{sin\,nx}{x}\) |
|
Answer» \(\lim\limits_{x \to 0}\frac{sin\,nx}{x}\) \(=\lim\limits_{x \to 0}n.\frac{sin\,nx}{x}\) \(=n\lim\limits_{x \to 0}(\frac{sin\,nx}{x})\) = n .1 = n |
|
| 69. |
The left hand limit of the function f(x)=4A. 1B. -1C. 0D. non-existent |
|
Answer» Correct Answer - B `("LHL of" f(x) " at " x=4)` `-lim_(xto4^-)f(x)=lim_(hto0)f(4-h)=lim_(hto0)(|4-h-4|)/(4-h-4)` `=lim_(hto0)(|-h|)/(-h)=lim_(hto0)(h)/(-h)=lim_(hto0)-1=-1`. To evalate RHL of `f(x) at x=a` i.e.,`lim_(xtoa^+)f(x)` we may use the following algorithm: |
|
| 70. |
The right hand limit of the funtion f(x) = 4A. 1B. -1C. 0D. non-existent |
|
Answer» Correct Answer - A We have, (RHL of `f(x)at x=4`) `lim_(xto4^+)f(x)=lim_(hto0)f(4+h)=lim_(hto0)(|4+h-4|)/(4+h-4)` `lim_(hto0)(|h|)/(h)-lim_(hto0)(h)/(h)=lim_(hto0)1=1`. |
|
| 71. |
Let `f(theta)=(1)/(tan^9 theta){(1+tantheta)^10+(2+tantheta)^10+....+(20+tantheta)^10}-20tan theta` The left hand limit of `f(theta)" as "theta to (pi)/(2)`,isA. `1900`B. `2000`C. `2100`D. `2200` |
|
Answer» Correct Answer - C Let `xtan theta`. Then, `f(tan^-1x)=((1+x)^10+(2x+x)^10+.....+ (20+x)^10-20x^10)/(x^9)` and `xtooo "as" theta to (pi^-)/(2)` `lim_(theta to(pi^-)/(2))f(theta) =lim_(xtooo) f(tan^-1x)` ` =lim_(xt0 oo) ((1+x)^10+(2+x)^10+....+(20+x)^10-20x^10)/(x^9)` ` =lim_(xto oo) (sum_(r=1)^(20)(r+x)^10-x^10)/(x^9)` ` =lim_(xtooo) sum_(r=1)^(20) (.^10C_(0) r^10+.^10C_(1) r^9+.....+.^10C_(9))/ (x^9)` ` =sum_(r=1)^(20)lim_(xtooo) {(.^10C_(0) r^10)/(x^9)+(.^10C_(1) r^9)/(r^8)+.....+.^10C_(9) r}` ` =sum_(r=1)^(20).^10C_(0)r=10 (sum_(r=1)^(20)r)=10xx(20xx21)/(2)=2100` |
|
| 72. |
Find the value of \(\lim\limits_{x \to 0}\frac{e^x-1}{x}\) |
|
Answer» \(\lim\limits_{x \to 0}\frac{e^x-1}{x}\) \(=\lim\limits_{x \to 0}\frac{[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....]-1}{x}\) \(=\lim\limits_{x \to 0}\frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...}{x}\) \(=\lim\limits_{x \to 0}[\frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...}{x}]\) = 1. |
|
| 73. |
Find the left hand limit and right hand limit of the greatest integer function f(x) = [x] = greatest integer less than or equal to x at x = k, where k is an integer. Also, show that \(\lim\limits_{x \to k}\) f(x) does not exit. |
|
Answer» We have− f(x) = [x] L.H.L (of x = k) \(=\lim\limits_{x \to k-}f(x)\) \(=\lim\limits_{h \to 0}f(k-h)\) \(=\lim\limits_{h \to 0}[k-h]\) \(=\lim\limits_{h \to 0}(k-1)\) [∵ k − 1 < k − h < k ⇒ [k − h] = k − 1] = k-1 R.H.L (of x = k) = \(\lim\limits_{h\to 0}f(k+h)\) \(=\lim\limits_{h\to 0}(k+h)\) \(=\lim\limits_{h\to 0}k\) [∵ k < k + h < k + 1 ⇒ [k + h] = k] = k Here, L.H.L ≠ R.H.L ∴ \(\lim\limits_{x \to k}f(x)\) does not exit. |
|
| 74. |
\(\lim\limits_{x \to a}\frac{\sqrt x+\sqrt a}{x+a}=?\) |
|
Answer» \(\lim\limits_{x \to a}\frac{\sqrt x+\sqrt a}{x+a}\) \(=\frac{\sqrt a+\sqrt a}{a+a}\) \(=\frac{2\sqrt a}{2a}\) \(=\frac{\sqrt a}{a}\) \(=\frac{1}{\sqrt a}\) |
|
| 75. |
The value of `lim_(x->a) (sinx/sina)^(1/(x-a))=`A. `e^sin a `B. `e^tan a `C. `e^cot a `D. 1 |
| Answer» Correct Answer - C | |
| 76. |
Evaluate:- \(\lim\limits_{h \to a}\frac{(x+2)^{3/2}-(a+2)^{3/2}}{x-a}\) |
|
Answer» Let x + 2 = y, then y → a + 2 as x → a ∴ \(\lim\limits_{h \to a}\frac{(x+2)^{3/2}-(a+2)^{3/2}}{x-a}\) \(=\lim\limits_{h \to a}\frac{y^{3/2}-(a+2)^{3/2}}{y-(a+2)}\) [∵ x + 2 = y ⇒ x = y − 2] \(=\frac{3}{2}(a+2)^{\frac{3}{2}-1}\)-1 \(=\frac{3}{2}(a+2)^\frac{1}{2}\) [∵ \(\lim\limits_{h \to a}\frac{x^n-a^n}{x-a}=na^{n-1}\) ] |
|
| 77. |
Evaluate :\(\lim\limits_{x \to \frac{\pi}{6}}\frac{cot^2x-3}{cosec\,x-2}\) |
|
Answer» \(\lim\limits_{x \to\frac{\pi}{6}}\frac{cot^2x-3}{cosec\,x-2}\) \(=\lim\limits_{x \to\frac{\pi}{6}}\frac{cosec^2x-1-3}{cosec\,x-2}\) \(=\lim\limits_{x \to\frac{\pi}{6}}\frac{cosec^2x-4}{cosec\,x-2}\) \(=\lim\limits_{x \to\frac{\pi}{6}}\frac{(cosec\,x-2)(cosec\,x+2)}{(cosec\,x-2)}\) \(=\lim\limits_{x \to\frac{\pi}{6}}(cosec\,x+2)\) [x ≠ π/6] \(= cosec\frac{\pi}{6}+2\) = 2 + 2 = 4 |
|
| 78. |
`lim_(h -> 0) (sin(a+3h)-3sin(a+2h)+3sin(a+h)-sina)/h^3 =`A. `sin a `B. `-sina`C. `cos a `D. `-cos a ` |
| Answer» Correct Answer - D | |
| 79. |
The value of `lim_(h->0) ((a+h)^2sin(a+h)-a^2sina)/h=`A. `2asina+a^2cos a`B. ` 2a sin a -a^2 cos a`C. `2a cos a +a^2sin a`D. none of these |
| Answer» Correct Answer - A | |
| 80. |
If `lim_(xrarr-1)(sin(x^3+bx^2+cx +d))/((sqrt(2+x)-1){log_e(x+2)}^2)` exists and is equal to l, then `b+d+l` is equal toA. 5B. 6C. 7D. 4 |
|
Answer» Correct Answer - B It is given that `lim_(xto-1)(sin(x^3+bx^2+cx +d))/((sqrt(2+x)-1){log_e(x+2)}^2)=l` `rArrlim_(xto-1)(sin(x^3+bx^2+cx +d))/((x+1)^3{(log_e(1+(x+1)))/((x+1))}^2)xx (sqrt(2+x+1)=l` `rArrlim_(xto-1)(sin(x^3+bx^2+cx +d))/((x^3+bx^2+cx +d))xx(1)/({(log_e(1+(x+1)))/((x+1))}^2)xx(x^3+bx^2+cx +d)/((x+1)^3)xx(sqrt(2+x+1)=l` `rArrlim_(xto-1)1xx(1)/((1)^2)xx(x^3+bx^2+cx +d)/((x+1)^3)xx2=l` `rArrlim_(xto-1)(x^3+bx^2+cx +d)/((x+1)^3)xx(l)/(2)` `rArrb=3,c=3,d=1 and l=2` `therefore b+d+l=3+1+2=6` |
|
| 81. |
`lim_(xto1) (1+sinpi((3x)/(1+x^(2))))/(1+cospix)` is equal toA. `logn((2)/(3))`B. 0C. `n log n((2)/(3))`D. not defined |
|
Answer» Correct Answer - A `underset(xto1)lim(1+sinpi((3x)/(1+x^(2))))/(1+cospix)=underset(xto1)lim(1-cos"((3pi)/(2)-(3pix)/(1+x^(2))))/(1-cos(pi-pix))` `=underset(xto1)lim(2sin^(2)"((3pi)/(4)-(3pix)/(2(1+x^(2)))))/(2-sin^(2)((pi)/(2)-(pix)/(2)))` `=underset(xto1)lim(((3pi)/(4)-(3pix)/(2(1+x^(2))))/((pi)/(2)-(pix)/(2)))^(2)` `=underset(xto1)9(((1)/(2)-(x)/(1+x^(2)))/(1-x))^(2)` `=underset(xto1)lim9((x-1)/(2(1+x^(2))))^(2)=0` |
|
| 82. |
Let `f(x) = 1 /(sqrt( 18 - x^2)` The value of `Lt_(x -> 3) (f(x)-f(3)) / (x-3)` is |
| Answer» Correct Answer - D | |
| 83. |
Find the value of \(\lim\limits_{x\to-3}\sqrt[3]{2-2x}\) |
|
Answer» \(\lim\limits_{x\to-3}\sqrt[3]{2-2x}\) =\(\sqrt[3]{2-2(-3)}\) \(=\sqrt[3]{2+6}\) \(=\sqrt[3]{8}=2\) |
|
| 84. |
The value of `lim_(xrarr1)(log_2 2x)^(log_x5)`, isA. `5//2`B. `e^log_2 5`C. `log 5//log 2`D. `e^log ar 5^2` |
| Answer» Correct Answer - B | |
| 85. |
`lim_(xrarr0) (sinx^n)/((sinx)^m), (mltn)` is equal toA. 1B. 0C. `n//m`D. none of these |
| Answer» Correct Answer - B | |
| 86. |
If `0lt xlty`, then `lim_(xrarroo) (y^n+x^n)^(1//n)` is equal toA. eB. xC. yD. none of these |
| Answer» Correct Answer - C | |
| 87. |
Let `S_(n)=1+2+3+...+n " and " P_(n)=(S_(2))/(S_(2)-1).(S_(3))/(S_(3)-1).(S_(4))/(S_(4)-1)...(S_(n))/(S_(n)-1)`, where `n inN,(nge2) Then ``underset(ntooo)limP_(n)`=__________.A. `(1)/(4)`B. `(1)/(24)`C. `(1)/(16)`D. `(1)/(8)` |
|
Answer» Correct Answer - `(3)` `S_(n)=(n(n+1))/(n)" and "S_(n)-1=((n+2)(n-1))/(2)` `:." "(S_(n))/(S_(n)-1)=(n(n+1))/(2).(2)/((n+2)(n-1))=((n)/(n-1))((n+1)/(n+2))` `:." "P_(n)=((2)/(1).(3)/(2).(4)/(3).(5)/(4)...(n)/(n-1))((3)/(4).(4)/(5).(5)/(6)...(n+1)/(n+2))` `=((n)/(1))((3)/(n+2))` `:." "underset(ntooo)limP_(n)=3` |
|
| 88. |
`lim_(xto0) (x^(4)(cot^(4)x-cot^(2)x+1))/((tan^(4)x-tan^(2)x+1))` is equal toA. -1B. 1C. 0D. none of these |
|
Answer» Correct Answer - A `(x^(4)(cot^(4)x-cot^(2)x+1))/((tan^(4)x-tan^(2)x+1))` `(x^(4)(1-tan^(2)x+tan^(4)x))/(tan^(4)x(tan^(4)x-tan^(2)x+1))=(x^(4))/(tan^(4)x),xne0` `:." "underset(xto0)lim(x^(4)(cot^(4)x-cot^(2)x+1))/((tan^(4)x-tan^(2)x+1))=underset(xto0)lim(x^(4))/(tan^(4x))=1` |
|
| 89. |
`lim_(xto1) (1-x^(2))/(sin2pix)` is equal to |
|
Answer» Correct Answer - B `underset(xto1)lim(1-x^(2))/(sin2pix)=-underset(xto1)lim(2pi(1-x)(1+x))/(2pisin(2pi-2pix))` `=underset(xto1)lim((2pi-2pix))/((2pi-2pix))(1+x)/(2pi)=(-1)/(pi)` |
|
| 90. |
Evaluate `lim_(yto0)(y^(2)+sin x)/(x^(2)+siny^(2)),` where `(x,y)to(0,0)` along the curve `x=y^(2)`. |
|
Answer» Correct Answer - 2 `underset(yto0)underset(xto0)lim(y^(2)+sin x)/(x^(2)+siny^(2))=underset(xto0)lim(x+sinx)/(x^(2)+sinx)=underset(xto0)lim(1+(sinx)/(x))/(x+(sinx)/(x))=2` |
|
| 91. |
`lim_(xto0) (1)/(x)cos^(1)((1-x^(2))/(1+x^2))` is equal toA. `f(x)f(y)`B. `f(x)+f(y)`C. `f(x)-f(y)`D. none of these |
|
Answer» Correct Answer - D We know that `cos^(-1)((1-x^(2))/(1+x^2))={{:(2tan^(-1)x", "xge0),(-2tan^(-1)x", "xle0):}` or `underset(xto0)lim(1)/(x)cos^(-1)((1-x^(2))/(1+x^2))=underset(xto0^(+))lim(2tan^(-1)x)/(x)=2` and`underset(xto0)lim(1)/(x)cos^(-1)((1-x^(2))/(1+x^2))=underset(xto0^(-))lim[-(2tan^(-1)x)/(x)]=-2` |
|
| 92. |
`lim_(yto0) ((x+y)sec(x+y)-xsecx)/(y)` is equal to |
|
Answer» Correct Answer - A `underset(yto0)lim{(x{sec(x+y)-secx})/(y)+sec(x+y)}` `=underset(yto0)lim[(x)/(y){(cosx-cos(x+y))/(cos(x+y)cosx)}]+underset(yto0)limsec(x+y)` `=underset(yto0)lim[(x2sin(x+(y)/(2))sin((y)/(2)))/(ycos(x+y)cosx)]+secx` `=underset(yto0)lim[(xsin(x+(y)/(2)))/(cos(x+y)cosx)xx(sin((y)/(2)))/((y)/(2))]+secx` `=xtanxsecx+secx` `=secx(xtanx+1)` |
|
| 93. |
Evaluate :\(\lim\limits_{x \to 1}\frac{x^{15}-1}{x^{10}-1}\) |
|
Answer» \(\lim\limits_{x \to 1}\frac{x^{15}-1}{x^{10}-1}\) = \(\lim\limits_{x \to 1}\frac{(x^5)^3-1^3}{(x^5)^2-1^2}\) ∵ a3 − b3 = (a − b) (a2 + b2 + ab) \(=\lim\limits_{x \to 1}\frac{(x^5-1)(x^{10}+x^5+1)}{(x^5-1)(x^5+1)}\) = \(\lim\limits_{x \to 1}\frac{x^{10}+x^5+1}{x^5+1}\) = \(\frac{1^{10}+1^{5}+1}{1^5+1}\) = \(\frac{3}{2}\) |
|
| 94. |
Evaluate the following limit : \(\lim\limits_{\text x \to 0} \) 9lim(x→0) |
|
Answer» Given the limit \(\Rightarrow\lim\limits_{\text x\to 0}\) 9 Always remember the limiting value of a constant (such as 4, 13, b, etc.) is the constant itself. So, the limiting value of constant 9 is itself, i.e., 9. |
|
| 95. |
`lim_(x rarr -1)(x^(5)+1)/(x+1)=`_____.A. 1B. -5C. 5D. None of these |
|
Answer» Correct Answer - C `underset(x rarr -1)("lim")(x^(5)+1)/(x+1)` `underset(x rarr -1)("lim")(x^(5)-(-1)^(5))/(x-(-1))=5(-1)^(5-1) = 5(-1)^(4)=5`. |
|
| 96. |
The scientists who put calculus in mathematical form are _______ and _______. |
| Answer» Correct Answer - Newton and Leibnitz | |
| 97. |
`lim_(x rarr a)x^(n) + ax^(n-1) +a^(2)x^(n-2) + .........+a^(n)=`________. |
| Answer» Correct Answer - `na^(n)` | |
| 98. |
The limiting position of a secant is _______. |
| Answer» Correct Answer - tangent | |
| 99. |
Evaluate : \(\lim\limits_{x \to \frac{\pi}{4}}\frac{tan^3x-tanx}{cos(x+\frac{\pi}{4})}\) |
|
Answer» \(\lim\limits_{x \to \frac{\pi}{4}}\frac{tan^3x-tanx}{cos(x+\frac{\pi}{4})}\) \(=\lim\limits_{x \to \frac{\pi}{4}}\frac{tanx(tan^2x-1)}{cos(x+\frac{\pi}{4})}\) \(=\lim\limits_{x \to \frac{\pi}{4}}tanx.\lim\limits_{x \to \frac{\pi}{4}}[\frac{-(1-tan^2\,x)}{cos(x+\frac{\pi}{4})}]\) \(=1\times[-\lim\limits_{x \to \frac{\pi}{4}}\frac{(1+tan\,x)(1-tan\,x)}{cos(x+\frac{\pi}{4})}]\) \(=\lim\limits_{x \to \frac{\pi}{4}}(1+tanx)\lim\limits_{x \to \frac{\pi}{4}}[\frac{cos\,x-sin\,x}{cos\,x.cos(x+\frac{\pi}{4})}]\) \(=-2\sqrt 2\times\) \(\lim\limits_{x \to \frac{\pi}{4}}\frac{cos(x+\frac{\pi}{4})}{cos\,x.cos(x+\frac{\pi}{4})}\) ∵ [cos x − sin x = \(\sqrt{2}\,cos\,(x+\frac{\pi}{4})]\) |
|
| 100. |
Differentiate the following with respect to x(i) y = (ax)m + bm(ii) y = x3 + 4x2 + 7x + 2 |
|
Answer» (i) y = amxm + bm dy/dx = ammxm-1 + 0 = ammxm-1 (ii) dy/dx = 3x2 + 4(2x) + 7 = 0 = 3x2 + 8x + 7 |
|