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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. |
Prove that `int_0^x[cot^(-1)x]dx ,w h e r e[dot]`denotes the greatest integer function. |
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Answer» Correct Answer - `pi+cot1+cot2` We have `int_(-pi//2)^(2pi)[cot^(-1)x]dx` We have that `cot^(-1)xepsilon[0,pi]` so `[cot^(-1)x]=0` for `cot^(-1)x epsilon(0,1)` or `xepsilon(cot1,oo)` `[cot^(-1)x]=1` for `cot^(-1)xepsilon[1,2)` or `xepsilon(cot2, cot1]` `[cot^(-1)x=2` for `cot^(-1)x epsilon[2,3)` or `x epsilon(cot3, cot2]` `[cot^(-1)x]=3` for `cot^(-1)xepsilon[3,pi)` or `x epsilon(-oo,cot3]` `:. int_(cot1)^(2pi) 0dx+int_(cot2)^(cot1) 1dx+int_(-pi//2)^(cot2) 2dx` `=cot1-cot2+2cot2+pi` `=cot1+cot2+pi` |
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| 302. |
The value of `int_(0)^(1) 4x^(3){(d^(2))/(dx^(2))(1-x^(2))^(5)}dx` is |
| Answer» Correct Answer - 2 | |
| 303. |
The integral `int_0^(1. 5)[x^2]dx ,w h e r e[dot]`denotoes the greatest integer function, equals ........... |
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Answer» Correct Answer - `(2-sqrt(2))` `int_(0)^(1.5)[x^(2)]dx=int_(0)^(1)0dx+int_(1)^(sqrt(2))1dx+int_(sqrt(2))^(1.5)2 dx` `=0+[x]_(1)^(sqrt(2))+2[x]_(sqrt(2))^(1.5) ` `=(sqrt(2)-1)+2(1.5-sqrt(2))` `sqrt(2)-1+3-2sqrt(2)` ` =2-sqrt(2)` |
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| 304. |
The value of `int_(-(pi)/(2))^(pi/2)(x^(2)cosx)/(1+e^(x))dx` is equal toA. `(pi^(2))/(4) - 2`B. `(pi^(2))/(4) +2`C. `pi^(2) - e^(pi//2)`D. `pi^(2)+e^(pi//2)` |
| Answer» Correct Answer - A | |
| 305. |
The option(s) with the values of a and L that satisfy the following equation is (are) `(int_(0)^(4pi)e^(1)(sin^(6)at+cos^(4)at)dt)/(int_(0)^(pi)e^(1)(sin^(6)at+cos^(4)at)dt) = L ?`A. `a = 2, L = (e^(4pi) - 1)/( e^(pi) - 1)`B. `a = 2, L = (e^(4pi) +1)/(e^(pi) + 1)`C. `a = 4, L = (e^(4pi) - 1)/(e^(pi) - 1)`D. `a = 4, L = (e^(4pi) + 1 )/(e^(pi)- 1)=0` |
| Answer» Correct Answer - AC | |
| 306. |
`P rov et h a t0 |
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Answer» Since `0lt(x^(7))/(root(3)((1+x^(8))))ltx^(7)AA0ltxlt1,` We have `int_(0)^(1)0 dx lt int_(0)^(1)(x^(7))/(root(3)((1+x^(8))))dxltint_(0)^(1)x^(7)dx` Hence `0ltint_(0)^(1)(x^(7)dx)/(root(3)((1+x^(8))))lt1/8` |
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| 307. |
`u=int_0^(pi/2)cos((2pi)/3sin^2x)dx` and `v=int_0^(pi/2) cos(pi/3 sinx) dx`A. 2u = vB. 2u = 3vC. u = yD. u = 2v |
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Answer» Correct Answer - A `u=int_(0)^(pi//2)cos((2pi)/(3)sin^(2)x)dx` `u=int_(0)^(pi//2)cos((2pi)/(3)cos^(2)x)dx` `rArr" "2u=int_(0)^(pi//2)[cos((2pi)/(3)sin^(2)x)+cos((2pi)/(3).cos^(2)x)]dx` `rArr" "2u=int_(0)^(pi//2)2cos.(pi)/(3).cos((pi)/(3)cos2x)dx` `rArr" "u=(1)/(2)int_(0)^(pi)cos((pi)/(3)cost)dt" [Put 2x = t]"` `=int_(0)^(pi//2)cos((pi)/(3)cost)dt` |
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| 308. |
Evaluate `(int_(0)^(pi)/2)(tanxdx)/(1+m^(2)tan^(2)x)` |
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Answer» Let `I=int_(0)^(pi//2)(tan xdx)/(1+m^(2)tan^(2)x)` `=int_(0)^(pi//2)((sinx)/(cosx))/(1+m^(2) . (sin^(2) x)/(cos^(2)x)) dx` `= int_(0)^(pi//2)(sinx cos x dx)/(1-sin^(2)x+m^(2)sin^(2)x) dx` `=int_(0)^(pi//2)(sinx cosx)/(1+(m^(2)-1)sin^(2)x)` Put `sin^(2)x=t` `implies2sin x cos x dx=dt` when `xto0`, then `t to 0` and `xto pi/2,` then `t to 1` `:. I=1/2 int_(0)^(1)(dt)/(1+(m^(2)-1)t)` `=1/2[1/(m^(2)-1)log(1+(m^(2)-1)t)]_(0)^(1)` `=1/(2(m^(2)-1))[log(1+(m^(2)-1))-log1]` `=(log m^(2))/(2(m^(2)-1))` `=(2log|m|)/(2(m^(2)-1))` `=(log|m|)/(m^(2)-1)` |
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| 309. |
Let `d/(dx) (F(x))= e^(sinx)/x, x>0`. If `int_1^4 2e^sin(x^2)/x dx = F(k)-F(1)`, then possible value of k is: |
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Answer» `I=int_(1)^(4)(2e^(sinx^(2)))/xdx=F(k)-F(1)=[F(x)]_(1)^(k)` Now put `x^(2)=t` `:.2xdx=dt` `:.I=int_(1)^(16)(e^(sint))/tdt=[F(t)]_(1)^(16)` `:.I=F(16)-F(1)` `=:.k=16` |
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| 310. |
Evaluate `int_(1)^(e^(6))[(logx)/3]dx,` where [.] denotes the greatest integer function. |
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Answer» Correct Answer - `(e^(6)-e^(3))` When `1ltxlte^(3),[(logx)/3]=0` and when `e^(3)ltxlte^(6),[(logx)/3]=1` `:. int_(1)^(e^(6))[(logx)/3]dx=int_(1)^(e^(3))[(logx)/3]dx+int_(e^(3))^(e^(6))[(logx)/3]dx` `=int_(1)^(e^(3)) 0dx+int_(e^(3))^(e^(6))1dx=(e^(6)-e^(3))` |
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| 311. |
If`alpha=int_0^1(e^9x+3tan^((-1)x))((12+9x^2)/(1+x^2))dxw h e r etan^(-1)`takes only principal values, then the value of `((log)_e|1+alpha|-(3pi)/4)i s` |
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Answer» Correct Answer - 9 Here , `alpha = int_(0)^(1)e^((9x+3tan^(-1)x))((12+9x^(2))/(1+x^(2)))dx` Put `9x+3 tan^(-1) x=t` `rArr (9+(3)/(1+x^(2)))dx = dt` `:. alpha= int _(0)^(9+3pi//4)e^(t)dt = [e^(t)] _(0)^(9+3pi//4)=e^(9+3pi//4)-1` `rArr log _(e)|1+alpha|=9+(3pi)/(4)` ` rArr log _(e) | alpha+1|-(3pi)/(4)=9` |
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| 312. |
Evaluate:`int_(-pi/4)^(npi-pi/4)|sinx+cosx|dx` |
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Answer» `I=int_(-pi//4)^(npi-pi//4)|sinx+cosx|dx` `=int_(-pi//4)^(npi-(pi)/4) sqrt(2)|sin(x+pi//4)|dx` (Multiplying and dividing by `sqrt(2)`) `=nint_(0)^(pi)sqrt(2)|sin(x+pi//4)|dx` [As `|sin(x+pi//4)|` is periodic with period `pi`] `=sqrt(2)n[int_(0)^(3pi//4)sin(x+pi//4)dx+int_(3pi//4)^(pi)-sin(x+pi//4)dx]` `=2sqrt(2)n` |
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| 313. |
`int_0^x[sint]dt ,w h e r ex in (2npi,(2n+1)pi),n in N ,a n d[dot]`denotes the greatest integer function is equal to`-npi`(b) `-(n+1)pi``2npi`(d) `-(2n+1)pi`A. `4n-cosx`B. `4n-sinx`C. `4n+1-cosx`D. `4n-1-cosx` |
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Answer» Correct Answer - C `int_(0)^(x)|sint|dt=int_(0)^(2npi)|sint|dt+int_(2npi)^(x)|sint|dt` `=2nint_(0)^(pi) |sint|dt+int_(2npi)^(x)sin tdt` (as `x` lies in either first or second quadrant) `=2n(-cost)_(0)^(pi)+(-cost)_(2npi)^(x)=4n-cosx+1` |
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| 314. |
The following integral `int_(pi/4)^(pi/2)(2cosecx)^(17)` dx is equal toA. `underset(0)overset(log(1+sqrt(2)))int2(e^(u)+e^(-u))^(16)du`B. `underset(0)overset(log(1+sqrt(2)))int(e^(u)+e^(-u))^(17)du`C. `underset(0)overset(log(1+sqrt(2)))int(e^(u)+e^(-u))^(17)du`D. `underset(0)overset(log(1+sqrt(2)))int2(e^(u)+e^(-u))^(16)du` |
| Answer» Correct Answer - A | |
| 315. |
The value of `lim_(xrarr0)(1)/(x^(3))int_(x)^(0) (tln(1+t))/(t^(4)+4) dt` is |
| Answer» Correct Answer - B | |
| 316. |
The total number of distinct `x in (0,1]` for which `int_(0)^(x)(t^(2))/(1+t^(4)) dt = 2x-1` is |
| Answer» Correct Answer - 1 | |
| 317. |
If the value of the definite integral `int_0^1^(2007)C_7x^(2000)dot(1-x)^7dx`is equal to `1/k ,w h e r ek in N ,`then the value of `k/(26)`is____ |
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Answer» Correct Answer - 208 Let `I=int_(0)^(1).^(207)C_(7).ubrace(x^(200))_(II).ubrace((1-x)^(7))_(I)dx` `=.^(207)C_(7)[(1-x)^(7).(x^(201))/201|_(0)^(1)+7/201int_(0)^(1)(1-x)^(6).x^(201)dx]` `=.^(207)C_(7) . 1/207 int_(0)^(1)(1-x)^(6) . x^(201)dx` Integrating by parts againsix times more we get `I=.^(207)C_(7).(7!)/(201.202.203.204.205.206.207)int_(0)^(1)x^(207)dx` `=((207)!)/(7!(200)!) . (7!)/(201.202......207) . 1/208` `=((207)!)/((207)!7!) . (7!)/208=1/208=1/k` or `k=208` |
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| 318. |
The value of the integral `int_3^6 sqrtx/(sqrt(9-x)+sqrtx)dx` is |
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Answer» Correct Answer - `3//2` `I=int_(3)^(6) (sqrt(x))/(sqrt(9-x)+sqrt(x)) dx`………….1 `:. I=int_(3)^(6)(sqrt(9-x))/(sqrt(x)+sqrt(9-x))dx`…………2 Adding 1 and 2. `2I=int_(3)^(6)1.dx=[x]_(3)^(6)=6-3=3` Hence `I=3/2` |
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| 319. |
If the value of the definite integral `int_0^1(sin^(-1)sqrt(x))/(x^2-x+1)dx`is `(pi^2)/(sqrt(n))`(where `n in N),`then the value of `n/(27)`is |
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Answer» Correct Answer - 108 `I=int_(0)^(1)(sin^(-1)sqrt(x))/(x^(2)-x+1)dx`……………1 `I=int_(0)^(1)(sin^(-1)sqrt(1-x))/(x^(2)-x+1)dx=int_(0)^(1)(cos^(-1)sqrt(x))/(x^(2)-x+1)dx`…………..2 on adding 1 and 2 we get `2I=int_(0)^(1)(sin^(-1)sqrt(x)cos^(-1)sqrt(x))/(x^(2)-x+1) dx` `=(pi)/2 int_(0)^()(dx)/(x^(2)-x+1)dx` ` =(pi)/2 int_(0)^(1)(dx)((x-1/2)^(2)+((sqrt(3))/2)^(2))dx` `:.2I=(pi)/2 1/(((sqrt(3))/2))[tan^(-1)((2x-1)/(sqrt(3)))]_(0)^(1)=(pi)^(2)/(3sqrt(3))` Hence `I=(pi^(2))/(6sqrt(3))=(pi^(2))/(sqrt(108))=(pi^(2))/(sqrt(n))` |
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| 320. |
Find the mistake of the following evaluation of the integral`I=int_0^pi(dx)/(1+2sin^2x)``I=int_0^pi(dx)/(cos^2x+3sin^2x)``=int_0^pi(sec^2x dx)/(1+3tan^2x)=1/(sqrt(3))[tan^(-1)(sqrt(3)tanx)]pi0=0` |
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Answer» Here the anti derivative `1/(sqrt(3))[tan^(-1)(sqrt(3)tanx)]=F(x)` is discontinuous at `x=pi//2` in the interval `[0,pi]`. Since `F((pi^(+))/2)=lim_(hto0)F((pi)/2+h)` `=lim_(hto0)(1/(sqrt(3)))tan^(-1){sqrt(3)"tan"(1/2pi+h)}` `=lim_(hto0)(1//sqrt(3))"tan"^(-1){-sqrt(3)coth}` `=(1/(sqrt(3))) tan^(-1)(-oo)=-pi//(2sqrt(3))` and `F(1/2pi-0)=pi//(2sqrt(3))!=F(1/2pi+0)` the second fundamental theorem of integral calculus is not applicable. |
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| 321. |
`int_(2)^(0){x^(3)+3x^(2)+3x+3+(x+1)cos(x+1)} dx` is equal toA. `-4`B. `0`C. `4`D. `6` |
| Answer» Correct Answer - C | |
| 322. |
Let `f"[0,oo)rarr R` be a continuous and stricity increasing function such that `f^(3) (x) =int_(0)^(x) tf^(2)(t) dt, x ge0`. The area enclosed by `y = f(x)` , the x-axis and the ordinate at `x = 3` is ` "_______"`A. `3/2`B. `5/2`C. `7/2`D. `1/2` |
| Answer» Correct Answer - A | |
| 323. |
The value of`int_0^1 4x^3{(d^2)/(dx^2)(1-x^2)^5}dxi s` |
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Answer» Correct Answer - 2 PLAN Integration by parts `intf(x)g(x)dx = f (x) int g (x) dx - int((d)/(dx)[f(x)] intg (x) dx ) dx` Given , `I= int _(0)^(1) 4x^(3) (d^(2))/(dx^(2))(1-x^(2))^(5)dx` ` = [ 4x^(3)(d)/(dx) (1-x^(2))^(5)]_(0)^(1) - int _(0)^(1)12 x^(2) (d)/(dx) (1- x^(2))^(5)dx` `=[4x^(3)xx5(1-x^(2))^(4)(-2x)]_(0)^(1)` `-12 [ [x^(2)(1-x^(2))^(5)]_(0)^(1)- int 2x(1-x^(2))^(5)dx]` `=0-0- 12 (0-0)+12 int_(0)^(1)2x (1- x^(2))^(5)dx` `=12xx[-((1-x^(2))^(6))/(6)]_(0)^(1)=12[0+ (1)/(6)]=2` |
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| 324. |
Let g (x) ` int_(0)^(x) f(t) dy=t` , where f is such that ` (1)/(2) le f (t) le 1` for ` tin [0,1] and 0 le f (t) le (1)/(2) ` for ` t in [1 ,2]`. then . g (2) satis fies the inequalityA. `-(3)/(2)leg (2) lt(1)/(2)`B. ` 0leg (2) lt2`C. `(3)/(2) lt g (2) le 5//2`D. `2lt g (2) lt 4` |
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Answer» Correct Answer - B Given , g (x) `= int_(0)^(x)f(t) dt` ` rArr g (2) = int_(0)^(2) f (t) dt = int_(0)^(1)f(t)dt + int_(1)^(2)f(t)dt` Now , `(1)/(2) le f (t) le 1 " for" t in [0,1]` We get `int_(0)^(1)(1)/(2)dt leint_(0)^(1)f(t)dt le int_(0)^(1)1 dt` `rArr (1)/(2)leint_(0)^(1)f (t) dt le 1` . . . (i) Again , `0le f (t) le (1)/(2) " for " t in [1,2]` . . . (ii) `rArr int_(1)^(2)0 dt le int_(1)^(2) f (t) dtle int _(1)^(2) dt` `rArr 0 le int_(1)^(2) f (t) dt le (1)/(2)` From Eqs . (i) and (ii) , we get `(1)/(2)le int_(0)^(1)f (t) dt + int_(1)^(2) f (t) dt le (3)/(2)` ` rArr (1)/(2) le g (2) le (3)/(2)` ` rArr 0leg (2) lt 2` |
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| 325. |
`L e tf(x)=x^3=(3x^2)/2+x+1/4`Then the value of `(int_(1/4)^(3/4)f(f(x))dx)^(-1)`os____ |
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Answer» Correct Answer - 4 Given `f(x)=x^(3)-(3x^(2))/2+x+1/4=1/4(4x^(3)(4x^(3)-6x^(2)+4x+1)` `=1/4(4x^(3)-6x^(2)+4x-1+2)` `=1/4[x^(4)-(1-x)^(4)]+2/4` `:.f(1-x)=1/4[(1-x)^(4)-x^(4)]+2/4` `:.f(x)+f(1-x)=2/4+2/4=1` Replacing `x` by `f(x)`,...................1 we have `f[f(x)]+f[1-f(x)]=1`...............2 Now, `I=int_(1//4)^(3//4)f(f(x))dx`.............3 Also `I=int_(1//4)^(3//4)f(f(1-x))dx=int_(1//4)^(3//4) f(1-f(x))dx` [using 1 ] ...........4 Adding 3 and 4 we get `2I=int_(1//4)^(3//4)[f(f(x))+f(1-f(x))]dx=int_(1//4)^(3//4)dx=1/2` or `I=1/4` `:.I^(-1)=4` |
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| 326. |
`int_(0)^(pi)[cotx]dx,` where [.] denotes the greatest integer function, is equal toA. `(pi)/2`B. `1`C. `-1`D. `-(pi)/2` |
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Answer» Correct Answer - D Let `I=int_(0)^(pi)[cotx]dx` …………..i `=int_(0)^(pi)[cot(pi-x)]dx=int_(0)^(pi)[-cotx]dx`………….ii Adding i and ii we get `2I=int_(0)^(pi)[cotx] dx+int_(0)^(pi)[-cotx]dx=int_(0)^(pi)(-1)dx` [since `[x]+[-x]` is equal to `-1` if `x !inZ` and is equal to 0 if `x epsilonZ`] `=[-x]_(0)^(pi)=-pi` `:.I=-(pi)/2` |
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| 327. |
Evaluate : (i) `int_(0)^(4)[x]^(2) dx` (where `[*]` denotes greatest integer function) (ii) `int_(0)^(pi)sqrt(1+sin2x)dx` , (iii) `int_(0)^(2)f(x)dx` where `f(x)-[{:(2x+1, 0lexlt1),(3x^(2),1lexle2):}` (iv) `int_(0)^(4)|x^(2)+4x+3|dx` , (v) `int_(0)^(oo)[cot^(1)x]dx` (where `[*]` denotes greatest integer function) (vi) `int_(-5)^(5)|x+2|dx` , (vii) `int_(-1)^(1)[cos^(-1)x]dx` (where `[*]` denotes greatest integer function) |
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Answer» Correct Answer - (i) `5-sqrt(2) - sqrt(3)` , (ii) `2sqrt(2)`, (iii) `9` , (iv) `4` , (v) `cot 1` , (vi) `29` (vii) `cos 1+cos2 + cos 3 + 0` |
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| 328. |
If `int_(0)^(11) (11^(x))/(11^([x]))dx = k/(log11)`, (where [] denotes greatest integer function) then value of k isA. `11`B. `101`C. `110`D. `121` |
| Answer» Correct Answer - C | |
| 329. |
The value of `int_(e)^(pi^(2))[log_(pi)x]d(log_(e)x)` (where [.] denotes greatest integer function) isA. `2log_(e)pi`B. `log_(e)pi`C. 1D. 0 |
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Answer» Correct Answer - B Let `log_(e)x=t` `I=int_(1)^(log_(e)pi^(2))[log_(pi)e^(t)]dt` `rArr" "I=int_(1)^(log_(e)pi^(2))[tlog_(pi)e]dt=int_(1)^(log_(e)pi)0dt+int_(log_(e)pi)^(2log_(e)pi)1dt=log_(e)pi` |
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| 330. |
Given a function `f:[0,4]toR` is differentiable ,then prove that for some `alpha,beta epsilon(0,2), int_(0)^(4)f(t)dt=2alphaf(alpha^(2))+2betaf(beta^(2))`. |
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Answer» `I=int_(0)^(4)f(t)dt,` put `t=x^(2)` `:.I=2int_(0)^(2)xf(x^(2))dx` using LMVT, we have `(int_(0)^(2)2xf(x^(2))dx-int_(0)^(0)2xf(x^(2))dx)/(2-0)=2yf(y^(2))` for somd `yepsilon(0,2)` `impliesint_(0)^(2)2xf(x^(2))dx=2.2yf(y^(2))=2{(2 alpha f(alpha^(2))+2betaf(beta^(2)))/2}` [where `0lt beta lt y lt alpha lt 2`. and using intermediate Vlue Theorem] `implies int_(0)^(2)2xf(x^(2))dx=2alpha f(alpha^(2))+2betaf(beta^(2))` |
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| 331. |
Given that for each `a epsilon(0,1),lim_(hto 0^(+)) int_(h)^(1-h)t^(-a)(1-t)^(a-1)dt` exists. Let this limit be `g(a)`. In addition it is given the function `g(a)` is differentiable on`(0,1)`. The value of `g(1/2)` isA. `pi/2`B. `pi`C. `-(pi)/(2)`D. `0` |
| Answer» Correct Answer - D | |
| 332. |
Given that for each `a epsilon(0,1),lim_(hto 0^(+)) int_(h)^(1-h)f^(-a)(1-t)^(a-1)dt` exists. Let this limit be `g(a)`. In addition it is given the function `g(a)` is differentiable on`(0,1)`. The value of `g(1/2)` isA. `pi`B. `2pi`C. `(pi)/2`D. `(pi)/4` |
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Answer» Correct Answer - A `g(1/2)=lim_(kto 0^(+))int_(k)^(1+k)t^(-1//2)(1-t)^(-1//2)dt` `=int_(0)^(1)(dt)/(sqrt(t-t^(2)))=int_(e)^(1)(dt)/(sqrt(1/4-(t-1/2)^(2)))=sin^(-1)((t-1/2)/(1/2))|._(0)^(1)` `=sin^(-1)1-sin^(-1)(-1)=pi` |
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| 333. |
Given that for each `a epsilon(0,1),lim_(hto 0^(+)) int_(h)^(1-h)t^(-a)(1-t)^(a-1)dt` exists. Let this limit be `g(a)`. In addition it is given the function `g(a)` is differentiable on`(0,1)`. The value of `g(1/2)` isA. `pi`B. `2pi`C. `pi/2`D. `pi/4` |
| Answer» Correct Answer - A | |
| 334. |
Evaluate `int_(0)^(npi+t)(|cosx|+|sinx|)dx,` where `n epsilonN` and `t epsilon[0,pi//2]`. |
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Answer» Correct Answer - `4n+sint-cost+1` Let `I=int_(0)^(npi+1)(|cosx|+|sinx|)dx` `=int_(0)^(npi)(|cosx|+|sinx|)dx+int_(npi)^(npi+1)(|cosx|+|sinx|)dx` `=2nint_(0)^(pi//2) (|cosx|+|sinx|)dx+int_(0)^(1)(|cosx|+|sinx|)dx` `=2n int_(0)^(pi//2) (cosx+sinx)dx+int_(0)^(1)(cosx+sinx)dx` `=4n+sint-cost+1` |
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| 335. |
`f(x)` satisfies the relation `f(x)-lambdaint_0^(pi//2)sinx*costf(t)dt=sinx` If `lambda > 2` then `f(x)` decreases inA. `(0,pi)`B. `(pi/2,3pi//2)`C. `(-pi//2,pi//2)`D. none of these |
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Answer» Correct Answer - C `f(x)-lamda int_(0)^(pi//2)sinx cost f (t) dt=sinx` or `f(x)-lamda sinx int_(0)^(pi//2) cost f(t)dt=sinx` or `f(x)=Asinx=sinx` or `f(x)=(A+1)sinx` where `A=lamdaint_(0)^(pi//2) cost f(t)dt` or `A=lamda int_(0)^(pi//2) cos (A+1)sin dt` `=(lamda(A+1))/2int_(0)^(pi//2) sin 2tdt` `=(lamda(A+1))/2[(-cos 2t)/2]_(0)^(pi//2)` `=(lamda(A+1))/2` `:. A=(lamda)/(2-lamda)` `:.f(x)=((lamda)/(2-lamda)+1)sinx=(2/(2-lamda))sinx` `(2/(2-lamda))sinx=2` or `sinx=(2-lamda)` or `|2-lamda|le1` or `-1le lamda-2le 1` or `1 le lamda le 3` `int_(0)^(pi//2) f(x)dx=3` or `int_(0)^(pi//2) 2/(2-lamda) sinxdx=3` or `-[2/(2-lamda) cosx]_(0)^(pi//2) =3` or `2/(2-lamda)=3` |
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| 336. |
`int_(-1)^(1)xln(1+e^(x))dx=`A. `0`B. `ln(1+e)`C. ` ln(1+e)-1`D. `1//3` |
| Answer» Correct Answer - D | |
| 337. |
The value of definite integral `int_0^(pi^2/4) dx/(1+sin^2 sqrtx+ cos^2 sqrtx)=` isA. `pi ln 2`B. `(piln2)/(2)`C. `(piln2)/(4)`D. `2pi ln 2` |
| Answer» Correct Answer - B | |
| 338. |
Let `f(x) = {{:(1-x ,"If" 0 le x le 1),(0, "if" 1 lt x le 2),((2-x)^(2),"if"2 lt x le 3):}` and function `F(x) = int_(0)^(x)f(t) dt`. If number of points of discountinuity in [0,3] and non-differentiablity in (0,2) and `F(x)` are `alpha` and `beta` respectively , then `(alpha - beta)` is equal to . |
| Answer» Correct Answer - 2 | |
| 339. |
For`x >0,l e tf(x)=int_1^x((log)_e t)/(1+t)dtdot`Findthe function `f(x)+f(1/x)`andshow that `f(e)+f(1/e)=1/2dot` |
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Answer» Correct Answer - `[(1)/(2)(Inx)^(2)]` f(x) `=int_(1)^(x)(Int)/( 1+t)dt " for" x gt 0` Now , `f(1//x)=int_(1)^(1//x)(Int)/( 1+t)dt` Put `t=1 //u rArr dt = (-1//u^(2))du` `:. f (1//x)= int_(1)^(x)(In(1//u))/(1+1//u)*((-1))/(u^(2))du` Now , `f(x) + f((1)/(x))= int_(1)^(x) (Int)/((1 +t))dt + int_(1)^(x) (Int)/(t(1+t))dt` `=int_(1)^(x)((1+t)Int)/(t( 1+t))dt = int_(1)^(x)(Int)/(t) dt = (1)/(2)[(In t)^(2)]_(1)^(x)=(1)/(2) (In x)^(2)` Put x = e , `f(e) + f ((1)/(e))=(1)/(2)(In e)^(2)=(1)/(2)` Hence proved. |
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| 340. |
`L e tI_1=int_(-2)^2(x^6+3x^5+7x^4)/(x^4+2)dxa n d``I_2=int_(-3)^1(2(x+1)^2+11(x+1)+14)/((x+1)^4+2)dtdot`Then the value of `I_1+I_2i s`8 (b)`(200)/3`(c) `(100)/3`(d) `non eoft h e s e`A. `8`B. `200//3`C. `100//3`D. noe |
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Answer» Correct Answer - C In `I_(2)` put `x+1=t`. Then `I_(2)=int_(-2)^(2)(2t^(2)+11t+14)/(t^(4)+2)dt=int_(-2)^(2)(2x^(2)+11x+14)/(x^(4)+2)dx` `:. I_(1)+I_(2)=int_(-1)^(2)(x^(6)+3x^(5)+7x^(4)+2x^(2)+11x+14)/(x^(4)+2) dx` `int_(-2)^(2)((x^(2)+3x+7)(x^(4)+2)+5x)/(x^(4)+2)dx` `=int_(-2)^(2)(x^(2)+3x+7)dx+5int_(-2)^(2)x/(x^(4)+2)dx` `=2 int_(0)^(2)(x^(2)+7)dx=100/3` |
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| 341. |
`f`is an odd function, It is also known that `f(x)`is continuous for all values of `x`and is periodic with period 2. If `g(x)=int_0^xf(t)dt ,`then`g(x)i sod d`(b) `g(n)=0,n in N``g(2n)=0,n in N`(d) `g(x)`is non-periodicA. `g(x)` is oddB. `2(n)=0, n epsilonN`C. `g(2n)=0,n epsilonN`D. `g(x)` is non-periodic |
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Answer» Correct Answer - C `g(x)=int_(0)^(x)f(t)dt` `g(-x)=int_(0)^(-x)f(t)dt=-int_(0)^(x)f(-t)dt=int_(0)^(x)f(t)dt` as `f(-t)=-f(t)` or `g(-x)=g(x)` Thus `g(x)` is even Also `g(x+2)=int_(0)^(x+2)f(t)dt` `=int_(0)^(2)f(t)dt+int_(2)^(2+x)f(t)dt` `g(2)+int_(0)^(x)f(t+2)dt` `=g(2)+int_(0)^(x)f(t)dt` `=g(2)+g(x)` Now `g(2) =int_(0)^(2)f(t)dt int_(0)^(1)f(t)dt+int_(1)^(2)f(t)dt` `=int_(0)^(1)f(t)dt+int_(-1)^(0)f(t+2)dt` `=int_(0)^(1)f(t)dt+int_(-1)^(0)f(t)dt` `=int_(-1)^(1)f(t)dt=0` as `f(t)` is odd `g(2)=0implies g(x+2)=g(x)`. i.e. `g(x)` is periodic with period 2. `:. g(4)0` or `g(6)=0, g(2n)=0, n epsilon N`. |
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| 342. |
Let `T >0`be a fixed real number. Suppose `f`is continuous function such that for all `x in R ,f(x+T)=f(x)dot`If `I=int_0^Tf(x)dx ,`then the value of `int_3^(3+3T)f(2x)dx`is`3/2I`(b) `2I`(c) `3I`(d) `6I`A. `(3)/(2)I`B. IC. 3ID. 6I |
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Answer» Correct Answer - C `int_(3)^(3+3T)f(2x)dx " put "2x =yrArr dx = (1)/(2)dy` `:. (1)/(2)int_(6)^(6++T)f(y)dy = (6I)/(2)=3I` |
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| 343. |
`lim_(xto oo) (int_(0)^(x)tan^(-1)dt)/(sqrt(x^(2)+1))` is equal toA. `(pi)/2`B. `(pi)/4`C. 1D. `pi` |
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Answer» Correct Answer - A `lim_(xto oo) (int_(0)^(x)tan^(-1)tdt)/(sqrt(x^(2)+1))` (`(oo)/(oo)` form) `=lim_(xto oo) (tan^(-1))/(x/(sqrt(x^(2)+1)))` `=(pi)/2 lim_(xto oo) (sqrt(x^(2)+1))/x` `=(pi)/2 lim_(xto oo) sqrt(1+1/(x^(2)))=(pi)/2` |
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| 344. |
`Ifint_(sinx)^1t^2f(t)dt=1=1-s inx ,w h e r ex in (0,pi/2),`then find the value of `f(1/(sqrt(3)))dot`A. 3B. `sqrt(3)`C. `1//3`D. None of these |
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Answer» Correct Answer - A Since `int_(sinx)^(1)t ^(2)f(t)dt =1-sin x`, thus to find f(x). On differentiating both sides using Newton Leibnitz formula i . e. `(d)/(dx)int_(sinx)^(1)t^(2)f(t)dt =(d)/(dx)(1-sinx)` `rArr {1^(2)f(1)} *(0)- (sin^(2)x)* cosx=-cosx` `rArr f(sin x)=(1)/(sin^(2)x)` For `f ((1)/(sqrt(3)))` is obtained when ` sin x = 1//sqrt(3)` i .e `f((1)/(sqrt(3)))=(sqrt(3))^(2)=3` |
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| 345. |
If `[f((sqrt(3))/2)]`is [.] denotes the greatest integer function)4 (b) 5(c) 6 (d) `-7`A. `4`B. `5`C. `6`D. `-7` |
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Answer» Correct Answer - B `int_(cosx)^(1)t^(2)f(t)dt=1-cosx` Differentiating both sides w.r.t `x`, we get `d/(dx) int_(cosx)^(1)t^(2)f(t)dt=d/(dx)(1-cosx)` or `-cos^(2)x f(cosx)(-sinx)=sinx` or `cos^(2)xf(cosx)sinx=sinx` or `f(cosx)=1/(cos^(2)x)` Now `f((sqrt(3))/4)` is attained when `cosx=(sqrt(3))/4` `:.f((sqrt(3))/4)=16/3=5.33` `[f((sqrt(3))/4)]=5` |
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| 346. |
The value of `int_(0)^(1)({2x}-1)({3x}-1)dx`, (where {} denotes fractional part opf x) is equal to :A. `19/36`B. `19/144`C. `19/72`D. `19/18` |
| Answer» Correct Answer - C | |
| 347. |
Evaluate `int_(-3)^(5) e^({x})dx`, where `{.}` denotes the fractional part functions. |
| Answer» `underset(-3)overset(5)inte^((x))dx=(5-(-3))underset(0)overset(1)inte^({x})dx=8underset(0)overset(1)inte^(x)dx = 8(e^(x))_(0)^(1)=8(e-1)` | |
| 348. |
The value of `int_(-pi//2)^(pi//2)(dx)/([x] + [ sin x] +4)`, where [t] denotes the greatest integer less than or equal to t , isA. `1/12 (7pi - 5)`B. `3/10 (4pi - 3)`C. `3/20 (4pi - 3)`D. `1/12(7pi + 5)` |
| Answer» Correct Answer - C | |
| 349. |
`f:[0,1)toR` be a non increasing function the for `alpha epsilon(0,1)`A. `alpha int_(0)^(1)f(x) dx le int_(0)^(alpha)f(x)dx`B. `alpha int_(0)^(1)f(x)dx ge int_(0)^(alpha)f(x)dx`C. `alpha^(2) int_(0)^(1)f(x)dxleint_(0)^(alpha) f(x)dx`D. `sqrt(alpha) int_(0)^(1)f(x)dxgeint_(0)^(alpha)f(x)dx` |
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Answer» Correct Answer - A::C `int_(0)^(alpha) f(x)dx`………..1 Put `x=alphat` `:. int_(0)^(alpha) f(x)dx=alpha int_(0)^(1)f (alpha t)dt ge alpha int_(0)^(1)f(t)dt` as `alpha epsilon(0,1)` Also `alpha int_(0)^(1) f(t)dt ge alpha^(2) int_(0)^(1)f(t)dt` `:. int_(0)^(alpha)f(x)dxgealpha^(2)int_(0)^(1)f(x)dx` |
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| 350. |
`int_(0)^(pi) [cotx]dx`, where `[*]` denotes the greatest integer function, is equal to :A. `1`B. `-1`C. `-(pi)/(2)`D. `(pi)/(2)` |
| Answer» Correct Answer - C | |