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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.
| 201. |
`lim_(x rarr 0) (sqrt(8-3x)+sqrt(8+4x))/(sqrt(2-3x))=`_______.A. 5B. 3C. 2D. 4 |
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Answer» Correct Answer - D Substitute x = 0. |
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| 202. |
`lim_(x rarr 3) (sqrt(x+3)+sqrt(x+6))/(sqrt(x+1)-2)=`_______.A. 1B. 7C. 14D. None of these |
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Answer» Correct Answer - D Rationalize the denominator. |
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| 203. |
`lim_(x rarr 3)sqrt(9-x^(2))=`_______. |
| Answer» Correct Answer - does not exist | |
| 204. |
`lim_(x rarr oo) (2x+4)/(x-2)=`______.A. 1B. 0C. 2D. 6 |
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Answer» Correct Answer - C Divide both numerator and denominator by x, then us if `x rarr oo "then" (1)/(x) rarr 0`. |
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| 205. |
`lim_(x rarr 0) (x)/(sqrt(3+x)-sqrt(3-x))=`_______.A. `sqrt(3)`B. `sqrt(-3)`C. `2 sqrt(3)`D. `-2 sqrt(3)` |
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Answer» Correct Answer - A Rationalize the denominator and cancel common factor and then substitute x = 0. |
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| 206. |
`lim_(x rarr -2) (x+2)/(sqrt(2x+8)-sqrt(2-x))=`______.A. `(1)/(3)`B. `(2)/(3)`C. 1D. `(4)/(3)` |
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Answer» Correct Answer - D Rationalize the denominator and cancel the common factor and then substitute x = -2. |
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| 207. |
Evaluate: `lim_(x rarr 20) (sqrt(x+5)+5)/(sqrt(x+5)-5)`.A. 1B. 2C. 4D. `oo` |
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Answer» Correct Answer - D Substitute x = 20. |
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| 208. |
`lim_(x rarr 5^(+)) (|x-5|)/(x-5)=`______.A. 1B. -1C. 0D. Cannot be determined |
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Answer» Correct Answer - A When `x gt 5, |x-5|=x-5`. |
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| 209. |
`lim_(x rarr 1) (4-sqrt(15+x))/(1-x)=`______.A. `(1)/(6)`B. `(1)/(8)`C. `(1)/(10)`D. `(1)/(12)` |
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Answer» Correct Answer - B Rationalize the numerator and cancel the common factor and then substitute x = 1 |
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| 210. |
Evaluate: `lim_(x rarr 1) (2x^(2)+4x+4)/(2x-1)`.A. 1B. 10C. 20D. 5 |
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Answer» Correct Answer - B Substitute x = 1 in the given expression. |
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| 211. |
`lim_(x rarr 3)(x-3)/(sqrt(x+6)-sqrt(2x+3))=`_____.A. -5B. -6C. 9D. 6 |
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Answer» Correct Answer - B Rationalize the denominator and remove the common factor. Now substitute x = 3. |
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| 212. |
`lim_(x rarr 0) (sqrt(1+x+x^(2)+x^(3))-1)/(x)=`_____.A. 1B. 2C. `(1)/(2)`D. `(1)/(4)` |
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Answer» Correct Answer - C Rationalize the numerator and cancel the common factor x, then substitute x = 0. |
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| 213. |
`lim_(x rarr b) (sqrt(x)-sqrt(b))/(x-b)=`_____.A. `(1)/(2b)`B. `(sqrt(2))/(b)`C. `(1)/(2sqrt(b))`D. `(1)/(sqrt(2)b)` |
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Answer» Correct Answer - C Use the formula, `underset(x rarr a)("lim")(x^(n)-a^(n))/(x-a)=na^(n-1)`. |
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| 214. |
Evaluate: `lim_(x rarr 2)(f(x)-f(2))/(x-2), "where" f(x)=x^(2)-4x`A. -1B. 2C. 0D. 4 |
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Answer» Correct Answer - C Substitute then factorize numerator f(x), then cancel the common factor in both the numerator and the denominator, then substitut x = 2. |
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| 215. |
Evaluate: `lim_(x rarr 9) (2x-7sqrt(x)+3)/(3x-11sqrt(x)+6)`.A. `(3)/(4)`B. `(5)/(3)`C. `(5)/(7)`D. `(3)/(7)` |
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Answer» Correct Answer - C Factorize numerator and denominator, eliminate the common factor then substitute x = 9. |
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| 216. |
Evaluate `lim_(x to oo) (log_(3)3x)^(log_(x)3).`A. l exists but m does notB. m exists but l does notC. both l and m existD. neither l nor m exists |
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Answer» Correct Answer - `e` `underset(xtooo)lim(log_(3)3x)^(log_(x)3)=underset(xto1)lim(log_(3)+log_(3)x)^((1)/(log_(3)x))` `=underset(xto1)lim(1+log_(3)x)^((1)/(log_(3)x))" "(1^(oo)"form")` =e |
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| 217. |
The value of `lim_(xtooo) ((20^(x)-1)/(19(5^(x))))^(1//x)` is ________.A. 3B. 1C. `(2)/(3)`D. `(3)/(2)` |
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Answer» Correct Answer - `(4)` We have, `y=underset(xtooo)lim((20^(x)-1)/(19(5^(x))))^((1)/(x))" "(oo^(0)" form")` `:." "log_(e)y=log_(e)underset(xtooo)lim((20^(x)-1)/(19(5^(x))))^((1)/(x))` `=underset(xtooo)limlog_(e)((20^(x)-1)/(19(5^(x))))^((1)/(x))` `=underset(xtooo)lim(1)/(x)log_(e)((20^(x)-1)/(19(5^(x))))` `=underset(xtooo)lim(log_(e)(20^(x)-1)-log_(e)19-log_(e)5^(x))/(x)` `=underset(xtooo)lim((20^(x)xxlog_(e)20)/(20^(x)-1)-0-(5^(x)log_(e)5)/(5^(x)))/(1)` `=underset(xtooo)lim(((log_(e)20)/(1))/(1-(1)/(20^(x)))-log_(e)5)` `=log_(e)20-log_(e)5=log_(e)4` |
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| 218. |
Evaluate `lim_(xto0) (sqrt(2)-sqrt(1+cosx))/(sin^(2)x).` |
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Answer» Correct Answer - `1//4sqrt(2)` We have `underset(xto0)lim(sqrt(2)-sqrt(1+cosx))/(sin^(2)x)` `underset(xto0)lim(2-(1+cosx))/(sin^(2)x)xx(1)/(sqrt(2)+sqrt(1+cosx))` `=underset(xto0)lim(1-cosx)/((1+cosx)(1-cosx))xxunderset(xto0)lim(1)/(sqrt(2)+sqrt(1+cosx))` `=underset(xto0)lim(1)/((1+cosx))xx(1)/(2sqrt(2))` `=(1)/(4sqrt(2))` |
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| 219. |
If `L=-lim_(ntooo) (2xx3^(2)xx2^(3)xx3^(4)...xx2^(n-1)xx3^(n))^((1)/((n^(2)+1)))`, then the value of `L^(4)` is _____________.A. `-(1)/(4)`B. `(1)/(2)`C. 1D. 2 |
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Answer» Correct Answer - `(6)` Clearly, n is even. Then, `underset(ntooo)lim"("2^(1+3+5+...+n//2" terms ").3^(2+4+6+...+n//2" terms")")"^((1)/((n^(2)+1)))` `=underset(ntooo)lim(2^((n^(2))/(4)).3^((n(n+2))/(4)))^((1)/((n^(2)+1)))` `=2^(underset(ntooo)lim(1)/(4(1+(1)/(n^(2))))).3^(underset(ntooo)lim((1+(2)/(n)))/(4(1+(1)/(n^(2)))))` `=2^((1)/(4))3^((1)/(4))=(6)^((1)/(4))` |
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| 220. |
The value of `lim_(ntooo) [root(3)((n+1)^(2))-root(3)((n-1)^(2))]` is __________.A. Equals `(1)/(sqrt(2))`B. Does not existC. Equals `sqrt(2)`D. Equals -sqrt(2)` |
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Answer» Correct Answer - `(0)` `underset(ntooo)lim[root(3)((n+1)^(2))-root(3)((n-1)^(2))]` `=underset(ntooo)limn^(2//3)[(1+(1)/(n))^(2//3)-(1-(1)/(n))^(2//3)]` `=underset(ntooo)limn^(2//3)[(1+(2)/(3).(1)/(n)+((2)/(3)((2)/(3)-1))/(2!)(1)/(n^(2))...)-(1-(2)/(3).(1)/(n)+((2)/(3)((2)/(3)-1))/(2!)(1)/(n^(2))...)]` `=underset(ntooo)limn^(2//3)[(4)/(3).(1)/(n)+(8)/(81).(1)/(n^(3))+...]` `=underset(ntooo)lim[(4)/(3).(1)/(n^(1//3))+(8)/(81).(1)/(n^(7//3))+...]=0` |
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| 221. |
If `f(x)=|x-1|-[x]`, where `[x]` is the greatest integer less than or equal to x, thenA. `underset(xto0)lim[f(x)]=0`B. `underset(xto0)lim[f(x)]=1`C. `underset(xto0)lim[(f(x))/(x)]` does not existD. `underset(xto0)lim[(f(x))/(x)]` exists |
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Answer» Correct Answer - A::D `f(1+0)=underset(hto0)lim(|1+h-1|-[1+h])=underset(hto0)lim(h-1)=-1` `f(1-0)=underset(hto0)lim(|1-h-1|-[1-h])=underset(hto0)lim(h-0)=0` |
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| 222. |
Evaluate `lim_(x to oo) (sinx^(0))/(x).` |
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Answer» Correct Answer - `pi//180` `underset(xto0)lim(sinx^(@))/(x)=underset(xto0)lim("sin"(pix)/(180))/(x)=(pi)/(180)" "{becausex^(@)=(pix)/(180)" radian"}` |
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| 223. |
Evaluate `lim_(xtosqrt(10)) (sqrt(7+2x)-(sqrt(5)+sqrt(2)))/(x^(2)-10).` |
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Answer» Correct Answer - `(1)/(2sqrt(10)"("sqrt(5)+sqrt(2)")")` Rationalizing Nr., we get `=underset(xtosqrt(10))lim(7+2x-(sqrt(5)+sqrt(2))^(2))/(x^(2)-10).(1)/(sqrt(7+2x)+(sqrt(5)+sqrt(2)))` `= (1)/(sqrt(7+2sqrt(5))+(sqrt(5)+sqrt(2)))underset(xtosqrt(10))lim(2(x-sqrt(10)))/((x-sqrt(10))(x+sqrt(10)))` `= (1)/((sqrt(5)+sqrt(2))+(sqrt(5)+sqrt(2))).(1)/(sqrt(10))` `=(1)/(2sqrt(10)(sqrt(5)+sqrt(2)))` |
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| 224. |
If `lim_(ntooo) (an-(1+n^(2))/(1+n))=b`, where a is a finite number, thenA. `f(0)=1`B. `f((pi)/(2))=1`C. `f(a)=(cosa)^(cos^(2)a).(sina)^(sin^(2)a)" if "ain(0,(pi)/(2))`D. `f(a)=((sina)^(sin^(2)a))/((cosa)^(cos^(2)a))" if "ain(0,(pi)/(2))` |
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Answer» Correct Answer - A::C Limit`=underset(ntooo)lim(an(1+n)-(1+n^(2)))/(1+n)=underset(ntooo)lim((a-1)n^(2)+an-1)/(n+1)` If `a-1=0," limit "=underset(ntooo)lim(an-1)/(n+1)=a=b` `:. a=b=1` |
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| 225. |
Find the values of a and b in order that `lim_(xto0) (x(1+acosx)-bsinx)/(x^(3))=1.` |
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Answer» Correct Answer - `a=(-5)/(2),b=(-3)/(2)` `underset(xto0)lim(x(1+acosx)-bsinx)/(x^(3))=1" "`(0/0 form) `impliesunderset(xto0)lim(x+axcosx-bsinx)/(x^(3))=1` `impliesunderset(xto0)lim(x+ax(1-(x^(2))/(2!)+(x^(4))/(4!)+...)-b(x-(x^(3))/(3!)+(x^(5))/(5!)+...))/(x^(3))=1` `impliesunderset(xto0)lim((1+a-b)x+(-(a)/(2)+(b)/(6))x^(3))/(x^(3))=1` `implies1+a-b=0" "(1)` and `-a/2+b/6=1" "(2)` Solving equations (1) and (2), we get `a=-(5)/(2),b=-(3)/(2).` |
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| 226. |
If `m,n inN,lim_(xto0) (sinx^(n))/((sinx)^(m))` isA. `(p)/(2)`B. `(pi)/(2sqrt(2))`C. `(pi)/(sqrt(2))`D. `sqrt(2)pi` |
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Answer» Correct Answer - A::B::C `L=underset(xto0)lim(sinx^(n))/((sinx)^(m))=underset(xto0)lim((sinx^(n))/(x^(n))x^(n))/(((sinx)^(m))/(x^(m))x^(m))=underset(xto0)limx^(n-m)` If `n=m`, then `L=("a very small value near to zero")^("exactly zero")=1` If `ngtm`, then `L=("a very small value near to zero")^("Positive integer")=0` If `nltm`, then `L=("a very small value near to zero")^("negative integer")=oo` |
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| 227. |
If `f(x)={(xsin,((1)/(x)),xne0),(0,,x=0):}` Then, `lim_(xrarr0) f(x)`A. is equal to 1B. is equal to`-1`C. is equal to 0D. does not exist |
| Answer» Correct Answer - C | |
| 228. |
Find the value of a and b if `lim_(x->0)(x(1+a c o s x)-bsinx)/(x^3)=1`A. `(5)/(3),(3)/(2)`B. `(5)/(2),(3)/(2)`C. `-(5)/(2),(3)/(2)`D. none of these |
| Answer» Correct Answer - C | |
| 229. |
If `|f(x)|lex^(2),` then prove that `lim_(xto0) (f(x))/(x)=0.` |
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Answer» We have `|f(x)|lex^(2)` `:." "|(f(x))/(x)|le|x|` `implies" "underset(xto0)lim|(f(x))/(x)|leunderset(xto0)lim|x|` `implies" "|underset(xto0)lim(f(x))/(x)|le0` `implies" "|underset(xto0)lim(f(x))/(x)|=0` `implies" "underset(xto0)lim(f(x))/(x)=0` |
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| 230. |
Let `f(x)={{:(x^(n)sin(1//x^(2))","xne0),(0", "x=0):},(ninI).` ThenA. `(p)/(2)`B. `(pi)/(2sqrt(2))`C. `(pi)/(sqrt(2))`D. `sqrt(2)pi` |
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Answer» Correct Answer - A::B For `ngt0, underset(xto0)limx^(n)sin(1//x^(2))=0xx(" any value between-1 to 1")=0` For `nlt0,underset(xto0)limx^(n)sin(1//x^(2))=ooxx(" any value between-1 to 1")=+-oo` |
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| 231. |
If `lim_(x->a)(f(x)/(g(x)))` exists, thenA. both `lim_(xto a) f(x)` and `lim_(xtoa) g(x)` must existB. `lim_(xtoa) f(x)` need not exist but `lim_(xtoa) g(x)` existsC. neither `lim_(xtoa) f(x)` nor `lim_(xtoa)g(x)` many existD. `lim_(xto a)f(x)` exists but `lim_(xto a)g(x)` need not exist |
| Answer» Correct Answer - C | |
| 232. |
If `f(x)={{:(sinx","" "xnenpi", "ninI),(2","" ""otherwise"):}` and `g(x)={{:(x^(2)+1","" "xne0", "2),(4","" "x=0),(5","" "x=2):}` then find `lim_(xto0) g{f(x)}`. |
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Answer» Correct Answer - 1 `underset(xto0^(+))limg{f(x)}=g(f(0^(+)))=g((sin0^(+)))=g(0^(+))=(0)^(2)+1=1` `underset(xto0^(-))limg{f(x)}=g(f(0^(-)))=g((sin0^(-)))=g(0^(-))=(0)^(2)+1=1` |
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| 233. |
If `f(x)={{:((x-|x|)/(x)","xne0),(2", "x=0):},`show that `lim_(xto0) f(x)` does not exist. |
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Answer» L.H.L of `f(x)` at `x=0` is `underset(xto0)limf(x)=underset(hto0)limf(0-h)=underset(hto0)lim(-h-|-h)/((-h))` `=underset(hto0)lim(-h-h)/(-h)=underset(hto0)lim(-2h)/(-h)=underset(hto0)lim2=2` R.H.L of `f(x)` at `x=0` is `underset(hto0)limf(x)=underset(hto0)limf(0+h)=underset(hto0)lim(h-|h)/((h))` `underset(hto0)lim(h-h)/(h)=underset(hto0)lim0/h=underset(hto0)lim0=0` Clearly, `underset(xto0^(-))limf(x)neunderset(xto0^(+))limf(x)` So, `underset(xto0^(-))limf(x)` does not exist. |
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| 234. |
Evaluate `lim_(x to 0) (tan(sgn(x)))/(sgn(x))` if exists. |
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Answer» Correct Answer - `tan1` `underset(xto0^(+))lim(tan(sgn(x)))/(sgn(x))=(tan1)/(1)=tan1` `underset(xto0^(-))lim(tan(sgn(x)))/(sgn(x))=tan(-1)/((-1))=tan1` `:." "underset(xto0)lim(tan(sgn(x)))/(sgn(x))=tan1` |
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| 235. |
Show that `lim_(xto0) (e^(1//x)-1)/(e^(1//x)+1)` does not exist. |
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Answer» Let `f(x)=(e^(1//x)-1)/(e^(1//x)+1)` L.H.L of `f(x) " at "x=0` is `underset(xto0^(-))limf(x)=underset(hto0)lim(0-h)=underset(hto0)lim(e^(-1//h)-1)/(e^(-1//h)+1)` `underset(hto0^(-))lim(((1)/(e^(1//h))-1)/((1)/(e^(1//h))+1))=-1` `[becausehto0implies1/htoooimpliese^(1//h)toooimplies(1)/(e^(1//h))to0]` R.H.L. of `f(x)` at `x=0` is `underset(xto0)limf(x)=underset(hto0)limf(0+h)=underset(hto0)lim(e^(1//h)-1)/(e^(1//h)+1)` `=underset(hto0)lim((1-(1)/(e^(1//h)))/(1+(1)/(e^(h))))" "`[Dividing Nr and Dr by `e^(1//h`)] `=(1-0)/(1+0)=1` Clearly, `underset(xto0^(-))limf(x)neunderset(xto0^(+))limf(x )` Hence, `underset(xto0)limf(x)` does not exist. |
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| 236. |
Consider the following graph of the function y=f(x). Which of the following is//are correct? `(a) lim_(xto1) f(x)` does not exist. `(b) lim_(xto2)f(x)` does not exist. `(c) lim_(xto3) f(x)=3.` `(d)lim_(xto1.99) f(x)`exists. |
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Answer» Correct Answer - (a), ( c ), (d) `a. underset(xto1^(+))limf(x)=3" and "underset(xto1^(-))limf(x)=2." Thus,"underset(xto1)limf(x)` does not exist. `b. underset(xto2^(+))limf(x)=underset(xto2^(-))limf(x)=3." Thus,"underset(xto2)limf(x)` exist. `c. underset(xto3^(+))limf(x)=underset(xto3^(-))limf(x)=3." Thus,"underset(xto3)limf(x)` exist. `d. underset(xto1.99^(+))limf(x)=underset(xto1.99^(-))limf(x)=3." Thus,"underset(xto1.99)limf(x)` exist. |
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| 237. |
Evaluate `lim_(xto0) (3x+|x|)/(7x-5|x|).` |
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Answer» Correct Answer - Limit does not exist. `underset(xto0^(-))lim(3x+|x|)/(7x-5|x|)=underset(xto0^(-))lim(3x-x)/(7x-5x)=1/6` and `underset(xto0^(+))lim(3x+|x|)/(7x-5|x|)=underset(xto0)lim(3x+x)/(7x-5x)=2` Hence, the limit does not exist. |
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| 238. |
If `f(x)={{:(x","" "xlt0),(1","" "x=0),(x^(2)","" "xgt0):}," then find " lim_(xto0) f(x)"` if exists. |
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Answer» We have `underset(xto0^(-))limf(x)=underset(xto0)limx=0` and `underset(xto0^(+))limf(x)=underset(xto0)limx^(2)=2` Hence, `underset(xto0)limf(x)` is equal `" to "0.` |
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| 239. |
Evaluate `lim_(xto2) (x^(2)-5x+6)/(x^(2)-4).` |
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Answer» When`x=2,` the expression `(x^(2)-5x+6)/(x^(2)-4)` assumes the indeterminate form `0/0.` Here, (x-2) is a common factor in numerator and denominator. Factorizing the numerator and denominator, we have `underset(xto2)lim(x^(2)-5x+6)/(x^(2)-4)=underset(xto2)lim((x-2)(x-3))/((x+2)(x-2))` `=underset(xto2)lim(x-3)/(x+2)=(2-3)/(2+2)=-(1)/(4)` |
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| 240. |
If `3-((x^(2))/(12))lef(x)le3+((x^(3))/(9))` in the neighborhood of x=0, then find the value of `lim_(xto0) f(x).` |
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Answer» According to question, `underset(xto0)lim(3-(x^(2))/(12))leunderset(xto0)limf(x)leunderset(xto0)lim(3+(x^(3))/(9))` or `(3-0)leunderset(xto0)limf(x)le(3+0)` Hence, `underset(xto0)limf(x)=3" "`(Using sandwich theorem) |
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| 241. |
Using product rule,differentiate the following with respect to x:(i) y = (x3 + x2 + 1)sinx(ii) y = (x2 - 5x + 6)secx |
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Answer» (i) dy/dx = (x3 + x2 + 1)d/dx sinx + sinx d/dx(x3 + x2 + 1) = (x3 + x2 + 1)cosx + (3x2 + 2x)sinx (ii) dy/dx = (x2 - 5x + 6)d/dx(secx) + sec. d/dx(x2 - 5x + 6) = (x2 - 5x + 6)secxtanx + secx.(2x - 5) |
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| 242. |
Differentiate the following with respect to x(i) y = x-5(ii) y = 5x7/2(iii) y = 8.x5/2x-5/3(iv) y = (2 + x)2(v) y = 3/x5 |
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Answer» (i) dy/dx = -5x-6 (ii) dy/dx = 5 . 7/2x7/2 - 1 = 35/2x5/2 (iii) y = 8 . x(5/2 - 5/3) = 8x5/6 (iv) y = (2 + x)2 = 4 + 4x + x2 dy/dx = d/dx(4 + 4x + x2) = 0 + 4(1) + 2x = 4 + 2x (v) y = 3x-5 dy/dx = 3(-5x-6) = - 15x-6 |
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| 243. |
Using quotient rule, differentiable with respect to x(i) y = (secx + 1)/(secx - 1)(ii) y = (1 - tanx)/(1 + tanx)(iii) y = (secx - 1)/(secx + 1) |
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Answer» (i) dy/dx = ((secx - 1)secxtanx - (secx + 1)secxtanx)/(secx - 1)2 = (-2secxtanx)/(secx - 1)2 (ii) dy/dx ((1 + tanx)(-sec2x) - (1 - tanx)secx2x)/(1 + tanx)2 = -2sec2x/(1 + tanx)2 (iii) dy/dx = ((secx + 1)secxtanx - (secx - 1)secxtanx)/(secx + 1)2 = 2secxtanx/(secx + 1)2 |
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| 244. |
Evaluate `lim_(xto0)(3^(2x)-2^(3x))/(x).` |
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Answer» We have `underset(xto0)lim(3^(2x)-2^(3x))/(x)" "`(0/0 from) `=underset(xto0)lim{((3^(2x)-1)/(x))-((2^(3x)-1)/(x))}` `=underset(xto0)lim((3^(2x)-1)/(2x).2)-underset(xto0)lim((2^(3x)-1)/(3x).3)` `=2log3-3log2=log9-log8=log((9)/(8))` |
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| 245. |
If `m, n in I_(0)` and `lim_(xto0) (tan2x-nsinx)/(x^(3))` = some integer, then find the value of n and also the value of limit. |
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Answer» `L=underset(xto0)lim(tan2x-nsinx)/(x^(3))` `underset(xto0)lim(sin2x-nsinxcos2x)/(x^(3)cos2x)` `=underset(xto0)lim("sin"x)/(x)((2cosx-ncos2x))/(x^(2))=(1)/(cos2x)` `=underset(xto0)lim((2cosx-ncos2x))/(x^(2))` Now, for `xto0`,`x^(2)to0`. Therefore, for `xto0, 2cosx-ncos2xto0.` So, n=2. For, n=2. `L=underset(xto0)lim((2cosx-ncos2x))/(x^(2))` `=4underset(xto0)lim("sin"(x)/(2)"sin"(3x)/(2))/(x^(2))` =3 |
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| 246. |
Evaluate `lim_(xto0) (tan2x-x)/(3x-sinx).` |
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Answer» Correct Answer - `1//2` `underset(xto0)lim(tan2x-x)/(3x-sinx)=underset(xto0)lim{((2tan2x)/(2x)-1)/(3-(sinx)/(x))}=1/2` |
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| 247. |
Evaluate `lim_(xto0) (e^(x)-e^(-x)-2x)/(x-sinx).` |
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Answer» Correct Answer - 2 `underset(xto0)lim(e^(x)-e^(-x)-2x)/(x-sinx)` `=underset(xto0)lim((1+x+(x^(2))/(2!)+(x^(3))/(3!)+...)-(1-x+(x^(2))/(2!)-(x^(3))/(3!)+...)-2x)/(x-(x-(x^(3))/(3!)+(x^(5))/(5!)......))` `=underset(xto0)lim(2.(x^(3))/(6)+2.(x^(5))/(5!)+...)/((x^(3))/(6)-(x^(5))/(5!)...)` `=underset(xto0)lim((1)/(3)+(1)/(60)x^(2)+...)/((1)/(6)-(1)/(120)x^(2)+...)=(1//3)/(1//6)=2` |
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| 248. |
Evaluate `lim_(xto0) {(sinx-x+(x^(3))/(6))/(x^(5))}.` |
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Answer» Correct Answer - `1//120` `underset(xto0)lim(sinx-x+(x^(3))/(6))/(x^(5))=underset(xto0)lim(x-(x^(3))/(3!)+(x^(5))/(5!)-(x^(7))/(7!)+* * *-x+(x^(3))/(6))/(x^(5))` `=underset(xto0)lim((1)/(5!)-(x^(2))/(7!)+...)=(1)/(120)` |
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| 249. |
Evaluate `lim_(xto0) (e^(x)-e^(xcosx))/(x+sinx).` |
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Answer» `underset(xto0)lim(e^(x)-e^(xcosx))/(x+sinx)=underset(xto0)lime^(x)(e^(x-cosx)-1)/(x+sinx)` `=e^(0)underset(xto0)lim(e^(x-cosx)-1)/(x-xcosx).underset(xto0)lim(x-xcosx)/(x+sinx)` `=1xx1xxunderset(xto0)lim(1-cosx)/(1+(sinx)/(x))` =`(1-1)/(1+1)` =0 |
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| 250. |
Evaluate `lim_(ntooo) {cos((x)/(2))cos((x)/(4))cos((x)/(8))...cos((x)/(2^(n)))}`. |
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Answer» We know that `costhetacos2thetacos4theta...cos2^(n-1)theta=(sin2^(n)theta)/(2^(n)sintheta)` `:." "cos((x)/(2))cos((x)/(4))cos((x)/(8))...cos((x)/(2^(n)))=(sinx)/(2^(n)"sin"((x)/(2^(n))))` `:." ""Given limits is "underset(ntooo)lim(sinx)/(2^(n)sin((x)/(2^(n))))=underset(ntooo)lim((sinx)/(x)((x)/(2^(n))))/(sin((x)/(2^(n))))` `=(sinx)/(x)` |
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