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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.
| 301. |
The value of `lim_(xrarr0) (|x|)/(x)`, isA. 1B. `-1`C. 0D. none of these |
| Answer» Correct Answer - D | |
| 302. |
`lim_(n rarr oo) (n(n+1))/(n^(2))=`________. |
| Answer» Correct Answer - 1 | |
| 303. |
`lim_(x rarr oo) (x^(n)+a^(n))/(x^(n)-a^(n))=`________. |
| Answer» Correct Answer - 1 | |
| 304. |
Evaluate:`lim_(x rarr -1) (logx^(2)-log((1)/(x^(4)))+log3)/(log((x^(3))/(-3)))`.A. 1B. 0C. -1D. Does not exist |
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Answer» Correct Answer - C Put x = -1 in the given limit. |
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| 305. |
Evaluate: `lim_(x rarr 4) (3x-8 sqrt(x)+4)/(5x-9sqrt(x)-2)`.A. `(1)/(5)`B. `(4)/(11)`C. `(3)/(10)`D. 5 |
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Answer» Correct Answer - B Factorize the numerator and denominator and cancel the common factor which is in both numerator and denominator then substitute x = 4. |
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| 306. |
Evaluate `lim_(x rarr 2) [(x^(2) - 5x + 6)/(x^(2) - 3x + 2)]`. |
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Answer» When `x = 2, (4 -10 + 6)/(4 - 6 + 2)=(0)/(0)`, which is an indeterminate form. Now, `x^(2) - 5x + 6 = (x-3)(x-2)` `x^(2) - 3x + 2 = (x-1)(x-2)` `" "underset(x rarr 2)("lim")[(x^(2) - 5x + 6)/(x^(2) - 3x + 2)]=underset(x rarr 2)("lim")[((x-3) (x-2))/((x-1) (x-2))]` `" "underset(x rarr 2)("lim")[(x-3)/(x-1)]=[(2-3)/(2-1)]= -1`. |
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| 307. |
`lim_(xto pi//2) (sin(xcosx))/(cos(xsinx))` is equal toA. `a=3" and "b=9//2`B. `a=3" and "b=9//2`C. `a=-3" and "b=-9//2`D. `a=3" and "b=-9//2` |
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Answer» Correct Answer - B `L=underset(xto pi//2)lim(sin(xcosx))/(sin((pi)/(2)-xsinx))` `L=underset(xto (pi)/(2))lim(sin(xcosx))/((xcosx))(((pi)/(2)-xsinx))/(sin((pi)/(2)-xsinx))(xcosx)/(((pi)/(2)-xsinx))` `=1xx1underset(xto pi//2)lim(xcosx)/(((pi)/(2)-xsinx))` Put `x=pi//2+h." Then "` `L=underset(hto0)lim(((pi)/(2)+h)cos((pi)/(2)+h))/((pi)/(2)-((pi)/(2)+h)sin((pi)/(2)+h))` `=underset(hto0)lim(-((pi)/(2)+h)sin h)/((pi)/(2)(1-cos h)-hcos h)` `=underset(hto0)lim(-((pi)/(2)+h)((sin h)/(h)))/((pi)/(2)((1-cos h))/(h)-cos h)`(Divide `N^(r)` and `D^(r)` by `h`) `=(-((pi)/(2)+0)1)/(0-1)=(pi)/(2)` |
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| 308. |
Evaluate `lim_(xto(pi)/(6)) (2-sqrt(3)cosx-sinx)/((6x-pi)^(2)).` |
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Answer» `underset(xto(pi)/(6))lim(2-sqrt(3)cosx-sinx)/((6x-pi)^(2))" "`(0/0 from) `=underset(xto(pi)/(6))lim2xx(1-((sqrt(3))/(2)cosx+1/2sinx))/((6x-pi)^(2))` `=underset(xto(pi)/(6))lim2xx(1-cos(x-(pi)/(6)))/(36(x-(pi)/(6))^(2))` `=underset(xto(pi)/(6))lim(2sin^(2)(x/2-pi/12))/(18xx4(x/2-pi/12)^(2))` `=(1)/(36)(underset(xto(pi)/(6))lim(sin(x/2-pi/12))/((x/2-pi/12)))^(2)=(1)/(36)` |
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| 309. |
Evaluate:\(\lim\limits_{x \to 5}\frac{e^x-e^5}{x-5}\) |
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Answer» \(\lim\limits_{x \to 5}\frac{e^x-e^5}{x-5}\) \(=\lim\limits_{x \to 0}\frac{e^{n+5}-e^5}{n}\) [Let x – 5 = n ⇒ as x - 5, n → 0] \(=\lim\limits_{x \to 0}\frac{e^5(e^n-1)}{n}\) \(=e^5\lim\limits_{x \to 0}\frac{e^n-1}{n}\) \(=e^5.1 \) \(=e^5\) |
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| 310. |
If \(\lim\limits_{\text x \to a}\cfrac{\text x^9-a^9}{\text x-a}=\lim\limits_{\text x \to 5}\)(4 + x) lim(x→a) (x9 - a9)/(x - a) = lim(x→5)(4 + x), find all possible values of a. |
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Answer» Given, \(\lim\limits_{\text x \to a}\cfrac{\text x^9-a^9}{\text x-a}=\lim\limits_{\text x \to 5} \)(4 + x) we need to find value of n So we will first find the limit and then equate it with \(\lim\limits_{\text x \to5}\)(4 + x) to get the value of n. We need to find the limit for : \(\lim\limits_{\text x \to a}\cfrac{\text x^9-a^9}{\text x-a}\) As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach. Let, Z = \(\lim\limits_{\text x \to a}\cfrac{\text x^9-a^9}{\text x-a}\) Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem. Formula to be used: \(\lim\limits_{\text x \to a}\cfrac{(\text x)^n-(a)^n}{\text x-a} \) = nan -1 As Z matches exactly with the form as described above so we don’t need to do any manipulations– Z = \(\lim\limits_{\text x \to a}\cfrac{\text x^9-a^9}{\text x-a}\) Use the formula: \(\lim\limits_{\text x \to a}\cfrac{(\text x)^n-(a)^n}{\text x-a} \) = nan -1 ∴ Z = 9(a)9–1 = 9a8 According to question Z = \(\lim\limits_{\text x \to5}\)(4 + x) = 4 + 5 = 9 ∴ 9(a)8 = 9 ⇒ a8 = 1 = 1 8 or (–1)8 Clearly on comparison we have – a = 1 or –1 |
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| 311. |
Among (i) `lim_(xtooo) sec^(-1)((x)/(sinx))" and "(ii) lim_(xtooo) sec^(-1)((sinx)/(x)).`A. `-(1)/(2sqrt(2))`B. `(1)/(2sqrt(2))`C. `(1)/(sqrt(2))`D. does not exist |
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Answer» Correct Answer - A `underset(xtooo)limsec^(-1)((x)/(sinx))=sec^(-1)((oo)/(sinoo))` `=sec^(-1)((oo)/("any value between "-1" to "1))` `=sec^(-1)(+-oo)=(pi)/(2)` `underset(xtooo)limsec^(-1)((sinx)/(x))=sec^(-1)((sinoo)/(oo))` `=sec^(-1)(("any value between "-1" to "1)/(oo))` `=sec^(-1)0` = not defined Hence, (i) exists but (ii) does not exist. |
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| 312. |
The value of `lim_(xrarroo) a^xsin ((b)/(a^x))` is `(agt1)`A. blog aB. a log bC. bD. none of these |
| Answer» Correct Answer - C | |
| 313. |
Evaluate `lim_(x rarr 3) [(x^(5)-243)/(x^(2)-9)]`. |
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Answer» `underset(x rarr 3)("lim")[(x^(5)-243)/(x^(2)-9)]=underset(x rarr 3)("lim")[(x^(5)-3^(5))/(x^(2)-3^(2))]` `" "=(5)/(2)3^(5-2)" "(therefore underset(x rarr a)("lim")[(x^(m)-a^(m))/(x^(n)-a^(n))]=(m)/(n)a^(m-n))` `" "=(5)/(2)*3^(3)=(135)/(2)`. |
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| 314. |
Evaluate: `lim_(d rarr 3) (d^(3)-27)/(d-3)`.A. 3B. 9C. 27D. 6 |
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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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| 315. |
`lim_(x rarr oo) {((3x)/(sqrt(x^(2)+5x-6)+2x)}=`________.A. -1B. 1C. 0D. `oo` |
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Answer» Correct Answer - B Numerator and denominator are of the same degree. `therefore "Limiting value of the given function"` `= ("Coefficient of x in numerator")/("Coefficient of x in denominator")` |
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| 316. |
`lim_(x rarr 1) (x^(3)-6x^(2)+11x-6)/(x^(2)-5x+4)=`______.A. `(2)/(3)`B. `-(5)/(3)`C. `(5)/(3)`D. `(-2)/(3)` |
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Answer» Correct Answer - D Factorize the numerator and the denominator, and cancel the common factor, then substitute x = 1. |
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| 317. |
Evaluate: `lim_(x rarr 2) [(2x^(2)-9x+10)/(5x^(2)-5x-10)]`. |
| Answer» Correct Answer - `(-1)/(15)` | |
| 318. |
Evaluate: `lim_(x rarr a) (sqrt(x+a)-sqrt(2a))/(x-a)`. |
| Answer» Correct Answer - `(1)/(2 sqrt(2a)` | |
| 319. |
Evaluate: `lim_(xto(5pi)/(4)) [sinx+cosx]`, [.] denotes the greatest integer function. |
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Answer» We have, `L=underset(xto(5pi)/(4))lim[sinx+cosx]=underset(xto(5pi)/(4))lim[sqrt(2)sin(x+pi/4)]` Now, `underset(xto(5pi)/(4)).[sqrt(2)sin(x+(pi)/(4))]=[sqrt(2)sin((3pi^(+))/(2))]=[-sqrt(2)]=-2` And `underset(xto(5pi^(+))/(4))lim[sqrt(2)sin(x+(pi)/(4))]=[sqrt(2)sin((3pi^(-))/2)=[-sqrt(2)]=-2` So, `L=-2` |
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| 320. |
The value of `lim_(xrarr3)(3^x-x^3)/(x^x-3^3)`, isA. `(log3-1)/(log3+1)`B. `(log 3+1)/(log 3-1)`C. 1D. none of these |
| Answer» Correct Answer - A | |
| 321. |
Evaluate the following limits:\(\lim\limits_{y\longrightarrow-3}[\frac {y^5+243}{y^3+27}]\)lim[y5 + 243/ y3 + 27], (y ∈-3) |
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Answer» = \(\lim\limits_{y\longrightarrow-3} \frac {y^5+243}{y^3+27}\) = \(\lim\limits_{y\longrightarrow-3} \frac {\frac {y^5+243}{y+3}}{\frac {y^3+27}{y+3}}\) ....[∴y→-3, y\(\neq\)-3, ∴ y+3 \(\neq\)0] = \(\frac{\lim\limits_{y\longrightarrow-3}[\frac {y^5-(-3)^5}{y-(-3)}]}{\lim\limits_{y\longrightarrow-3}[\frac {y^3-(-3)^3}{y-(-3)}]}\) = \(\frac {5(-3)^4}{3(-3)^2}\) = \(\frac {5}{3}\times9\) = 15 |
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| 322. |
If `a gt 0` and `lim_(xrarr oo) {sqrt(x^2+x+1)-(ax+b)}=0`, then `(a,b)` lies on the line.A. `x-y+3=0`B. `3x+4y-5=0`C. `x+6y+2=0`D. `x+2y+3=0` |
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Answer» Correct Answer - B We have, `lim_(xto oo) {sqrt(x^2+x+1)-(ax+b)}=0` `lim_(xtooo)(x^2+x+1-(ax+b)^2)/(sqrt(x^2+x+1)ax+b)=0` `rArrlim_(xtooo)(x^2(1-a^2)+x(1-2ab)+(1-b^2))/(sqrt(x^2+x+1)(ax+b))=0` ` rArr 1-a^2=0,1-2ab=0` ` rArr a=1 and b=(1)/(2)` Clearly, `(a,b)=(1,(1)/(2))` lies on the line `3x+4y-5=0`. |
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| 323. |
If `l=lim_(xto-2) (tanpix)/(x+2)+lim_(xtooo) ( (1+1)/(x^2)^2)`, then which one of the following is not correct?A. `lgt 3`B. `lgt4`C. `l lt 4`D. l is a transcendental number |
| Answer» Correct Answer - C | |
| 324. |
The value of `lim_(xtooo) (cot^-1(x^-alog_ax))/(sec^-1(x^-alog_ax)),agt1`, is equal toA. 1B. 0C. `(pi)/(2)`D. does not exist |
| Answer» Correct Answer - A | |
| 325. |
Evaluate `lim_(x rarr oo) (3x^(2)+4x+5)/(4x^(2)+7)`. |
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Answer» Put `x = (1)/(y), "when" x rarr oo, y rarr 0` `therefore underset(x rarr oo)("lim")(3x^(2)+4x+5)/(4x^(2)+7)=underset(y rarr 0)("lim")(3*(1)/(y^(2))+4*(1)/(y)+5)/(4*(1)/(y^(2))+7)` `" "=underset(y rarr 0)("lim")(3+4y+5y^(2))/(4+7y^(2))=(3+0+0)/(4+7(0))=(3)/(4)`. |
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| 326. |
`lim_(x rarr oo) ((4x+5)(2x-1))/((27x^(2)+1))=`_______.A. `(4)/(27)`B. `(8)/(27)`C. `(2)/(27)`D. `(6)/(27)` |
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Answer» Correct Answer - B Divide both the numerator and the denominator by `x^(2)` and then use when `x rarr oo rArr (1)/(x) rarr 0`. |
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| 327. |
`lim_(x rarr oo)(4x-3)/((2x+3))=`_____. |
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Answer» Correct Answer - D Divide both the numerator and the denominator by x and apply the concept hat, as `x rarr oo, (1)/(x) rarr 0`. |
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| 328. |
Evaluate: `lim_(x rarr 3)(x^(5)-243)/(x-3)`. |
| Answer» Correct Answer - 405 | |
| 329. |
Evaluate: `lim_(x rarr a) (x^(14)-a^(14))/(x^(-7)-a^(-7))`. |
| Answer» Correct Answer - `-2a^(21)` | |
| 330. |
`lim_(xto-1) (1)/(sqrt(|x|-{-x}))` (where `{x}` denotes the fractional part of x) is equal toA. 16B. 24C. 32D. 8 |
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Answer» Correct Answer - A L.H.L.`=underset(xto-1^(-))lim(1)/(sqrt(|x|-{-x}))=underset(xto-1^(-))lim(1)/(sqrt(-x-(x+2)))` `=underset(xto-1^(-))lim(1)/(sqrt(-2x-2))=oo` R.H.L.`=underset(xto-1^(+))lim(1)/(sqrt(|x|-{-x}))=underset(xto-1^(-))lim(1)/(sqrt(-x-(x+1)))` `=underset(xto-1^(-))lim(1)/(sqrt(-2x-1))=1` Hence, the limit does not exist. |
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| 331. |
`lim_(xrarroo) {(e^(x)+pi^(x))^((1)/(x))}`= (where {.} denotes the fractional part of x ) is equal toA. `pi-e`B. `pi-3`C. `e-2`D. `3-e` |
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Answer» Correct Answer - B `underset(xrarroo)(lim)(e^(x)+pi^(x))^(1//x)=piunderset(xrarroo)(lim)(((e)/(pi))^(x)+1)^(1//x)=pi` Now, `{pi}=pi-3` `therefore" "underset(xrarroo)(lim){(x^(x)+pi^(x))^(1)/(x))}=pi-3` |
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| 332. |
If `f(a)=lim_(xto2)(sin^(x)a+cos^(x)a)^((1)/((x-2)))" for "ain[0,(pi)/(2)],` thenA. `-np`B. `np`C. `n^(2)p`D. `np^(2)` |
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Answer» Correct Answer - A::B::C `f(a)=underset(xto2)lim(sin^(x)alpha+cos^(x)alpha)^((1)/((x-2)))" "(1^(oo)" form")` `={{:(e^(underset(xto2)lim(sin^(x)alpha+cos^(x)alpha-1)/(x-2))","alphain(0","(pi)/(2))),(1", "alpha=0", "(pi)/(2)):}` Now `e^(underset(xto2)lim(sin^(x)alpha+cos^(x)alpha-1)/(x-2))=e^(underset(xto2)lim(sin^(x)alpha+cos^(x)alpha-sin^(2)alpha-cos^(2)alpha)/(x-2))` `=e^(underset(xto2)lim(sin^(2)alpha(sin^(x-2)a-1)+cos^(2)alpha(cos^(x-2)alpha-1))/(x-2))` `=e^(sin^(2)alphalog_(e)sinalpha+cos^(2)alphalog_(e)cosalpha)` `=e^(log_(e)(sinalpha)^(sin^(2)alpha)+log_(e)(cosalpha)^(cos^(2)alpha))` `=e^(log_(e)(sinalpha)^(sin^(2)alpha)(cosalpha)^(cos^(2)alpha))` `=(sinalpha)^(sin^(2)alpha)(cosalpha)^(cos^(2)alpha)` `:.f(x)={{:((cosalpha)^(cos^(2)alpha).(sinalpha)^(sin^(2)alpha)", "alphain(0", "(pi)/(2))),(1", "alpha=0", "(pi)/(2)):}` |
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| 333. |
`lim_(x to 0) (sin(x^(2)))/("ln"(cos(2x^(2)-x)))` is equal toA. `e^(a)`B. `-a`C. `e^(1-a)`D. `e^(1+a)` |
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Answer» Correct Answer - B `underset(xto0)lim(sin(x^(2)))/("ln"(cos(2x^(2)-x)))` `=underset(xto0)lim(sin(x^(2)))/(log(1-2sin^(2)((2x^(2)-x)/(2))))` `=underset(xto0)lim(sin(x^(2))x^(2))/((x^(2)log(1-2sin^(2)((2x^(2)-x)/(2))))/(-2sin^(2)((2x^(2)-x)/(2)))[-2sin^(2)((2x^(2)-x)/(2))])` `=underset(xto0)lim-(x^(2))/((2sin^(2)((2x^(2)-x)/(2)))/(((2x^(2)-x)/(2))^(2))((2x^(2)-x)/(2))^(2))` `=underset(xto0)lim-(2x^(2))/((2x^(2)-x)^(2))=underset(xto0)lim-(2)/((2x-1)^(2))=-2` |
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| 334. |
`lim_(xrarr(pi)/(2)) ((1-sinx)(8x^(3)-pi^(3)))/(pi-2x)^(4)`A. `-(pi^(2))/(16)`B. `(3pi^(2))/(16)`C. `(pi^(2))/(16)`D. `-(3pi^(2))/(16)` |
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Answer» Correct Answer - D `underset(xrarr(pi)/(2))(lim)((1-sinx)(8x^(3)-pi^(3))cosx)/(pi-2x)^(4)` `=underset(xrarr(pi)/(2))(lim)((1-cos((pi)/(2)-x))(2x-pi)(4x^(2)+pi^(2)+2pix)sin((pi)/(2)-x))/(16((pi)/(2)-x)^(4))` `underset(hrarr0)(lim)((1-cos h)(-2h)sin h)/(16h^(2))underset(xrarr(pi)/(2))(lim)(4x^(2)+pi^(2)+2pix)` `=underset(hrarr0)(lim)((1-cos h)(-2h sin h))/(16h^(4))underset(xrarr(pi)/(2))(lim)(4x^(2)+pi^(2)+2pix)` `=-underset(hrarr0)(lim)((1-cos h))/(8h^(4))(4((pi)/(2))^(2)+pi^(2)+2pi((pi)/(2)))` `=-(3pi^(2))underset(hrarr0)(lim)(2sin^(2).(h)/(2))/(32(h^(2))/(4))` `=-(3pi^(2))/(16)` |
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| 335. |
The value of `lim_(x->0) [x^2/(sin x tan x)]` (Wherer `[*]` denotes greatest integer function) is |
| Answer» Correct Answer - A | |
| 336. |
The value of `lim_(xrarroo) (5^(x+1)-7^(x+1))/(5^x-7^x)`,isA. 5B. `-5`C. `7`D. `-7` |
| Answer» Correct Answer - C | |
| 337. |
Statement-1 : `lim_(xrarr oo) ((x+1)^10+(x+2)^10+.....+(x+100)^(10)10)/(x^10+9^10)` Statement -2 : If `f(x)and g(x) ` are polynomials of same degree, then `lim_(xrarroo) (f(x))/(g(x))=("Coefficient of leading term in" f(x))/("Coefficient of leading term in" g(x))`A. Statement -1 is true, Statement-2 is true,, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for statement -1.C. Statement-1 is true, Statement-2 is False.D. Statement-1 is False, Statement-2 is true. |
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Answer» Correct Answer - A Clearly, statement-2 is true (See Theory on page 34.9) . Now, `lim_(xto oo) ((x+1)^10+(x+2)^10+.....+(x+100)^(10))/(x^10+9^10)` `=lim_(xtooo) (100x^100 +.^10C_1(1+2+.....+100)x^9+.....+(1^10+2^10+......100^10))/(x^10+9^10)` `=(100)/(1)=100` [Using statement -2] |
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| 338. |
Given `lim_(x to 0)(f(x))/(x^(2))=2`, where `[.]` denotes the greatest integer function, thenA. `(1)/(3)`B. `(1)/(4)`C. `(1)/(2)`D. `(2)/(3)` |
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Answer» Correct Answer - A::C Since `x^(2)gt0` and limit equals `2,f(x)` must be a positive quantity. Also, since `underset(xto0)lim(f(x))/(x^(2))=2`. Denominator `to` zero and limit is finite. Therefore, `f(x)` must be approaching zero or `underset(xto0)lim[f(x)]=0^(+)`. Hence, `underset(xto0)lim[(f(x))/(x)]=0^(+)`. `underset(xto0^(+))lim[(f(x))/(x)]=underset(xto0^(+))lim[x(f(x))/(x^(2))]=0` and `underset(xto0^(-))lim[(f(x))/(x)]=underset(xto0^(-))lim[x(f(x))/(x^(2))]=-1` Hence, `underset(xto0)lim[(f(x))/(x)]` does not exist. |
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| 339. |
The value of `lim_(xtooo) {(1)/(3)+(2)/(21)+(3)/(91)+...+(n)/(n^4+n^2+1)}`, isA. 1B. `1//2`C. `1//3`D. none of these |
| Answer» Correct Answer - B | |
| 340. |
The value of `lim_(xrarr-2) (x^2-x-6)^2/(x+2)^2`, isA. 6B. 25C. 9D. 16 |
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Answer» Correct Answer - B We have, `lim_(xto-2) (x^2-x-6)^2/(x+2)^2` `=lim_(xto-2)((x-3)(x+2)^2)/((x+2)^2)=lim_(xto-2)(x-3)^2=(-2-3)^2=25` |
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| 341. |
Evaluate: `lim_(x rarr oo) (x^(5)+3x^(4)-4x^(3)-3x^(2)+2x+1)/(2x^(5)+4x^(2)-9x+16)`. |
| Answer» Correct Answer - `(1)/(2)` | |
| 342. |
`lim _(xrarr oo) (sqrt(x^2+x+1)-sqrt(x^2+1))=`A. `-(1)/(2)`B. `(1)/(2)`C. `1`D. `-1` |
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Answer» Correct Answer - B Here , the expression assumes the form `oo-oo` as `xto oo`.So, we reduce it to the rational form `(f(x))/(8(x))`. We have, `lim _(xto oo) sqrt(x^2+x+)-sqrt(x^2+1)` `lim _(xto oo) ({sqrt(x^2+x+1)-sqrt(x^2+1)}{sqrt(x^2+x+1)+sqrt(x^2+1)})/(sqrt(x^2+x+1)+sqrt(x^2+1))` `lim _(xto oo) (x^2+x+1-x^2-1)/(sqrt(x^2+x+1)+sqrt(x^2+1))=lim_(x to oo)(x)/(sqrt(x^2+x+1)+sqrt(x^2+1))` `=lim_(x to oo) (1)/(sqrt(1+(1)/(x)+(1)/(x^2))+sqrt(1+(1)/(x^2)))` `=(1)/(1+1)=(1)/(2)`. `[{:("Dividing " N^2,),(and D^r" by " x,):}]` |
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| 343. |
`lim_(x->pi/4)(int_2^(sec^2x)f(t)dt)/(x^2-(pi^2)/16)` equals :A. `(8)/(pi)f(2)`B. `(2)/(pi)f(2)`C. `(2)/(pi)f((1)/(2))`D. `4f(2)` |
| Answer» Correct Answer - A | |
| 344. |
`Let f(x)=(sin^(-1)(1-{x})xxcos^(-1)(1-{x}))/(sqrt(2{x})xx(1-{x}))`, where `{x}` denotes the fractional part of x. `R=lim_(xto0+) f(x)` is equal toA. `np(1-n)`B. `-np(1+n)`C. `n^(2)p`D. `np(1+n)` |
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Answer» Correct Answer - A We have `f(x)=(sin^(-1)(1-{x})cos^(-1)(1-{x}))/(sqrt(2{x})(1-{x}))` `:.underset(xto0^(+))limf(x)=underset(hto0)limf(0+h)` `=underset(hto0)lim(sin^(-1)(1-{0+h})cos^(-1)(1-{0+h}))/(sqrt(2{0+h})(1-{0+h}))` `=underset(hto0)lim(sin^(-1)(1-h)cos^(-1)(1-h))/(sqrt(2h)(1-h))` `=underset(hto0)lim(sin^(-1)(1-h))/((1-h))underset(hto0)lim(cos^(-1)(1-h))/(sqrt(2)h)` In second limit, put `1-h=costheta.` Then `underset(xto0^(+))limf(x)=underset(hto0)lim(sin^(-1)(1-h))/((1-h))underset(hto0)lim(cos^(-1)(costheta))/(sqrt(2(1-costheta)))` `=underset(hto0)lim(sin^(-1)(1-h))/((1-h))underset(thetato0)lim(theta)/(2sin(theta//2))(becausethetagt0)` `=sin^(-1)1xx1=pi//2` and `underset(xto0^(-))limf(x)=underset(hto0)limf(0-h)` `=underset(hto0)lim(sin^(-1)(1-{0-h})cos^(-1)(1-{0-h}))/(sqrt(2{0-h}")")(1-{0-h}))` `=underset(hto0)lim(sin^(-1)(1+h-1)cos^(-1)(1+h-1))/(sqrt(2(-h+1))(1+h-1))` `=underset(hto0)lim(sin^(-1)h)/(h)underset(hto0)lim(cos^(-1)h)/(sqrt(2(1-h)))` `=1(pi//2)/(sqrt(2))=(pi)/(2sqrt(2))` |
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| 345. |
`lim_(x to 0) {(1+x)^((2)/(x))}` (where `{.}` denotes the fractional part of x) is equal toA. 1B. `e`C. `e^(-1)`D. none of these |
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Answer» Correct Answer - A `(1+x)^(2//x)=(1+x)^(2//x)-[(1+x)^(2//x)]` Now, `underset(xto0)lim(1+x)^(2//x)=e^(2)` or`underset(xto0)lim{(1+x)^(2//x)}=e^(2)-[e^(2)]=e^(2)-7` |
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| 346. |
`lim_(xrarr pi/2) ((1-tanx//2)(1-sinx))/((1-tanx//2)(pi-2x)^3)`A. `oo`B. `(1)/(8)`C. `0`D. `(1)/(32)` |
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Answer» Correct Answer - D We have, ` lim_(xto pi//2)((1-tanx//2)/(1+tanx//2)){(1-sinx)/((pi-2x)^3)}` ` (1)/(64) lim_(xto pi//2) tan((pi)/(4)-(x)/(2))/(((pi)/(4)-(x)/(2)))([1-cos((pi)/(2)-x)])/(((pi)/(4)-(x)/(2))^2)=(1)/(64)xx1xx2=(1)/(32)` |
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| 347. |
If `f: R-> R` is defined by (where [.] is g.i.f) `f(x) [x-3] + |x-4|` for `x in R` then `lim_(x->3^-) f(x) =`A. -2B. -1C. 0D. 1 |
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Answer» Correct Answer - C We have, , `f(x)=(x)=[x-3]+|x-4|` `therefore lim_(xto3^-)f(x)=lim_(xto3^-) [x-3]+lim_(xto3^-) |x-4|` `rArr lim_(xto3^-) f(x) lim_(hto0) [3-h-3]+lim_(hto0) |3-h-4|` `rArr lim_(xto3^-) f(x)=lim_(hto0) [-h]+lim_(hto0) |-1-h|` `rArr lim_(xto3^-) f(x) lim_(hto0) -1 +lim_(hto0) (1+h)=-1+1=0` |
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| 348. |
Statement -1: If a and b are positive real numbers and `[.]` denotes the greatest integer function , thenA. Statement -1 is true, Statement-2 is true,, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for statement -1.C. Statement-1 is true, Statement-2 is False.D. Statement-1 is False, Statement-2 is true. |
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Answer» Correct Answer - A For any non-zero real number x, we have ` 0le {x}lt 1` `rArr 0le ({x})/(x) lt (1)/(x) " for all " x gt 0 rArr lim_(xto oo) ({x})/(x)=0` Hence, statement -2 is true. Now, `lim_(xto0^+)(x)/(a) [(b)/(a)]=lim_(xto0 ^+)(x)/(a)((b)/(x)-{(b)/(x)})=lim_(xto0^+) ((b)/(a)-(x)/(a){(b)/(x)})` ` rArr lim_(xto 0^+)(x)/(a) [(b)/(x)]=(b)/(a)-(b)/(a)lim_(xto0^+)(x)/(b) {(b)/(x)}=(b)/(a)-(b)/(a)lim_(xto0^+)({(b)/(x)})/((b)/(x))` ` rArr lim_(xto 0^+)(x)/(a) [(b)/(x)]=(b)/(a)-(b)/(a)lim_(xtooo) ({y})/(y)," where "y=(b)/(x)` ` rArr lim_(xto 0^+)(x)/(a) [(b)/(x)]=(b)/(a)-(b)/(a)xx0=(b)/(a)" "["Using statements -2"]` |
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| 349. |
The value `lim_(xrarr pi//2)(sinx)^(tanx)`, is |
| Answer» Correct Answer - B | |
| 350. |
`lim_(xrarr(pi)/(2)) (1-sinx)tanx=`A. `(pi)/(2)`B. 1C. 0D. `oo` |
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Answer» Correct Answer - C `underset(xrarrpi//2)(lim)(1-sinx)tanx` `" "=underset(xrarrpi//2)(lim)sinx(1-sinx)/(cosx)` `" "=underset(xrarrpi//2)(lim)((1-sin^(2)x))/(cosx(1+sinx))` `" "=underset(xrarrpi//2)(lim)(cos^(2)x)/(cosx(1+sinx))` `" "=underset(xrarrpi//2)(lim)(cosx)/(1+sinx)=0` |
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