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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.
| 201. |
Evaluate `int_(0)^(pi)(x)/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx`. |
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Answer» Let `I=int_(0)^(pi)(x)/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx`…….`(i)` Then, `I=int_(0)^(pi)((pi-x))/((a^(2)cos^(2)(pi-x)+b^(2)sin^(2)(pi-x)])dx` or `I=int_(0)^(pi)((pi-x))/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx`........`(ii)` Adding `(i)` and `(ii)`, we get `2I=int_(0)^(pi)((x+pi-x))/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx=piint_(0)^(pi)(dx)/((a^(2)cos^(2)x+b^(2)sin^(2)x))` `=2piint_(0)^(pi//2)(dx)/((a^(2)cos^(2)x+b^(2)sin^(2)x))=2piint_(0)^(pi//2)(sec^(2)x)/((a^(2)+b^(2)tan^(2)x))dx` [dividing num. and denom. by `cos^(2)x`] `=2piint_(0)^(oo)(dt)/((a^(2)+b^(2)t^(2)))`, where `tanx=t` `=(2pi)/(b^(2))int_(0)^(oo)(dt)/(((a^(2))/(b^(2))+t^(2)))=[(2pi)/(b^(2))*(b)/(a)tan^(-1)((bt)/(a))]_(0)^(oo)` `=(2pi)/(ab)[tan^(-1)(oo)-tan^(-1)(0)]=(2pi)/(ab)((pi)/(2)-0)=((2pi)/(ab)xx(pi)/(2))=(pi^(2))/(ab)`. `:.I=(pi^(2))/(2ab)impliesint_(0)^(pi)(x)/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx=(pi^(2))/(2ab)`. |
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| 202. |
Evaluate `int_(0)^(pi)(xsinx)/((1+cos^(2)x))dx`. |
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Answer» Let `I=int_(0)^(pi)(xsinx)/((1+cos^(2)x))dx`…….`(i)` Then, `I=int_(0)^(pi)((pi-x)sin(pi-x))/(1+cos^(2)(pi-x))dx` or, `I=int_(0)^(pi)((pi-x)sinx)/((1+cos^(2)x))dx`……`(ii)` Adding `(i)` and `(ii)` , we get `2I=pi int_(0)^(pi)(sinx)/((1+cos^(2)x))dx=-piint_(1)^(-1)(dt)/((1+t^(2)))`, were `cosx=t` `=pi int_(-1)^(1)(dt)/((1+t^(2)))=pi[tan^(-1)t]_(-1)^(1)` `=pi[tan^(-1)-tan^(-1)]=pi[(pi)/(4)-(-(pi)/(4))]=(pi^(2))/(2)`. `:.I=(pi^(2))/(4)`. |
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| 203. |
Evaluate : `(i) int_(0)^(pi//2)(cosx)/((cos (x)/(2)+sin (x)/(2)))dx` `(ii) int_(0)^(pi//2)(cosx)/((1+cosx+sinx))dx` |
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Answer» `(i) int_(0)^(pi//2)(cosx)/((cos.(x)/(2)+sin.(x)/(2)))dx=int_(0)^(pi//2)((cos^(2).(x)/(2)-sin^(2).(x)/(2)))/((cos.(x)/(2)+sin.(x)/(2)))dx` `=int_(0)^(pi//2)(2(cos.(x)/(2)-sin.(x)/(2)))/((cos.(x)/(2)+sin.(x)/(2))^(2))dx=2*int_(1)^(sqrt(2))(dt)/(t^(2))=[(-2)/(t)]_(1)^(sqrt(2))=sqrt(2)(sqrt(2)-1)`. [putting `cos.(x)/(2)+sin.(x)/(2)=t` and `(1)/(2)(cos.(x)/(2)-sin(x)/(2))dx=dt`, also, `x=0impliest=1` and `x=(pi//2)impliest=sqrt(2)]` `(ii)int_(0)^(pi//2)(cosx)/((1+cosx+sinx))dx = int_(0)^(pi//2)(cosx)/((1+cosx)+sinx)dx` `=int_(0)^(pi//2)(cos^(2)(x//2)-sin^(2)(x//2))/([2cos^(2)(x//2)+2sin(x//2)cos(x//2)])dx` `=(1)/(2)int_(0)^(pi//2)(1-tan^(2)(x//2))/(1+tan(x//2))dx` [dividing num. and denom. by `cos^(2)(x//2)`] `=(1)/(2)int_(0)^(pi//2)[1-tan(x//2)]dx=(1)/(2)int_(0)^(pi//2)dx-(1)/(2)int_(0)^(pi//2)(sin(x//2))/(cos(x//2))dx` `=(1)/(2)*[x]_(0)^(pi//2)+[logcos(x//2)]_(0)^(pi//2)` `=(pi)/(4)+logcos.(pi)/(4)=(pi)/(4)+log((1)/(sqrt(2)))=((pi)/(4)-(1)/(2)log2)`. |
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| 204. |
Evaluate `int_(0)^(pi)(x)/((1+sinx))dx`. |
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Answer» Let `I=int_(0)^(pi)(x)/((1+sinx))dx`…..`(i)` Then, `I=int_(0)^(pi)((pi-x))/(1+xsin(pi-x))dx=int_(0)^(pi)((pi-x))/((1+sinx))dx`…….`(ii)` Adding `(i)` and `(ii)`, we get `2I=pi int_(0)^(pi)(dx)/((1+sinx))=pi*int_(0)^(pi)(1)/((1+sinx))xx((1-sinx))/((1-sinx))dx` or `2I=pi int_(0)^(pi)((1-sinx)/(cos^(2)x))dx=pi*[int_(0)^(pi)sec^(2)xdx-int_(0)^(pi)secxtanxdx]` `=pi*{[tanx]_(0)^(pi)-[secx]_(0)^(pi)}=2pi` `:.I=pi`, i.e, `int_(0)^(pi)(x)/((1+sinx))dx=pi`. |
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| 205. |
`int_(0)^(1)sin^(-1)sqrt(x)dx` |
| Answer» Correct Answer - Put `x=sin^(2)t`. | |
| 206. |
`int_(1//pi)^(2//pi)(sin(1//x))/(x^(2))dx=?`A. `1`B. `(1)/(2)`C. `(3)/(2)`D. none of these |
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Answer» Put `(1)/(x)=t` and `(1)/(x^(2))dx=-dt`. `[x=(1)/(pi)impliest=pi]` and `[x=(2)/(pi)impliest=(pi)/(2)]` `:. I=-int_(pi)^(pi//2)sintdt=int_(pi//2)^(pi)sintdt=[-cost]_(pi//2)^(pi)=1`. |
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| 207. |
`int_(1)^(2)|x^(2)-3x+2|dx=?`A. `(-1)/(6)`B. `(1)/(6)`C. `(1)/(3)`D. `(2)/(3)` |
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Answer» `(x^(2)-3x+2)=(x-1)(x-2)` `1 le x lt 2 implies (x^(2)-3x+2) le 0` `implies |x^(2)-3x+2|=-(x^(2)-3x+2)` `:. I=int_(1)^(2)-(x^(2)-3x+1)dx=int_(1)^(2)(-x^(2)+3x-2)dx` `=[(-x^(3))/(3)+(3x^(2))/(2)-2x]_(1)^(2)=((-2)/(3)+(5)/(6))=(1)/(6)`. |
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| 208. |
Let `f (x) = x – [x],` for every real number x, where `[x]` is the greatest integer less than or equal to x. Then, evaluate `int_-1^1 f(x) dx.`A. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - A `int_(-1)^(1)(x-[x])dx=int_(-1)^(0)(x+1)dx+int_(0)^(1)(x-0)dx` `=[(x^(2))/(2)+x]_(-1)^(0)+[(x^(2))/(2)]_(0)^(1)=(1)/(2)+(1)/(2)=1` |
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| 209. |
`int_0^(pi/4)sin^(- 1)sqrt(x/(a+x))dx` |
| Answer» Put `x=a tan^(2)theta`. | |
| 210. |
Prove:`int_0^(pi//2) log|tanx| dx=0` |
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Answer» Let `I=int_(0)^(pi//2)log(tanx)dx`…….`(i)` Then, `I=int_(0)^(pi//2)log[tan((pi)/(2)-x)]dx` `[:.int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]` or `I=int_(0)^(pi//2)log(cotx)dx=int_(0)^(pi//2)log((1)/(tanx))dx=-int_(0)^(pi//2)logtanxdx=-I`. `:.I=-I` or `2I=0` or `I=0`. Hence, `int_(0)^(pi//2)log(tanx)dx=0`. |
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| 211. |
`int_(0)^(pi)(dx)/((1+sinx))=?`A. `(1)/(2)`B. `1`C. `2`D. `0` |
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Answer» `I=int_(0)^(pi)((1-sinx))/((1-sin^(2)x))dx=int_(0)^(pi)((1-sinx))/(cos^(2)x)dx` `=int_(0)^(pi)[sec^(2)x-secxtanx]dx=[tanx-secx]_(0)^(pi)=2`. |
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| 212. |
Let `[x]` denote the greatest integer less than or equal to `x`. Then, `int_(1)^(-1)[x]dx=?`A. `-1`B. `0`C. `(1)/(2)`D. `2` |
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Answer» `{-1 le x lt 0implies[x]=-1}` and `{0 le x lt 1implies[x]=0}` `:. I=int_(-1)^(0)[x]dx+int_(0)^(1)[x]dx=int_(-1)^(0)(-1)dx+int_(0)^(1)0*dx` `=[-x]_(-1)^(0)=0-[-(-1)]=-1`. |
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| 213. |
`int_0 ^(pi/4) sqrt(1+ sin 2x) dx`A. `0`B. `1`C. `2`D. `sqrt(2)` |
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Answer» `I=int_(0)^(pi//4)sqrt(cos^(2)x+sin^(2)x+2sinxcosx)dx=int_(0)^(pi//4)(cosx+sinx)dx` `=[sinx-cosx]_(0)^(pi//4)=1`. |
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| 214. |
`int_(0)^(pi//2)(sqrt(cosx))/((sqrt(cosx)+sqrt(sinx)))dx=?` |
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Answer» Let `I=int_(0)^(pi//2)(sqrt(cosx))/((sqrt(sinx)+sqrt(cosx)))dx`………..`(i)` Using the result `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx` in `(i)`, we get `I=int_(0)^(pi//2)(sqrt(cos[(pi//2)-x]))/(sqrt(sin[(pi//2)-x])+sqrt(cos[(pi//2)-x]))dx` or `I=int_(0)^(pi//2)(sqrt(sinx))/(sqrt(cosx)+sqrt(sinx))dx=int_(0)^(pi//2)(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx`.........`(ii)` Adding `(i)` and `(ii)` we get `2I=int_(0)^(pi//2)(sqrt(cosx))/((sqrt(sinx)+sqrt(cosx)))dx+int_(0)^(pi//2)(sqrt(sinx))/((sqrt(sinx)+sqrt(cosx)))dx` `=int_(0)^(pi//2)((sqrt(sinx)+sqrt(cosx)))/((sqrt(sinx)+sqrt(cosx)))dx=int_(0)^(pi//2)dx=[x]_(0)^(pi//2)=(pi)/(2)`. `:.I=(pi)/(4)`, i.e., `int_(0)^(pi//2)(sqrt(cosx))/((sqrt(sinx)+sqrt(cosx)))dx=(pi)/(4)`. |
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| 215. |
The value of `int_(1)^(4) |x-3|dx` is equal toA. 2B. `(5)/(2)`C. `(1)/(2)`D. `(3)/(2)` |
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Answer» Correct Answer - B ` int _(1)^(4) | x- 3| dx = int _(1)^(3) (3-x) dx + int _(3)^(4) (x-3) dx` ` = [ 3x - (x^(2))/2 ] _(1)^(3) + [ (x^(2))/2 - 3x ] _(3)^(4)` ` = (9- 9/2) - (3 - 1/2 ) + (16/2 - 12 ) - ( 9/2 - 9) = 5/2 ` |
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| 216. |
`int_(0)^(pi//2)(sqrt(sinx)cosx)^(3)dx=?`A. `(2)/(9)`B. `(2)/(15)`C. `(8)/(45)`D. `(5)/(2)` |
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Answer» `I=int_(0)^(pi//2)(sinx)^(3//2)cos^(2)xcosxdx=int_(0)^(1)t^(3//2)(1-t^(2))dt`, where `sinx=t` `int_(0)^(1)(t^(3//2)-t^(7//2))dt=[(2)/(5)t^(5//2)-(2)/(9)t^(9//2)]_(0)^(1)=((2)/(5)-(2)/(9))=(8)/(45)`. |
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| 217. |
`int_(0)^(pi//2)(cosx)/((1+sinx)(2+sinx))dx` is equal toA. `log.(4)/(3)`B. `log.(1)/(3)`C. `log.(3)/(4)`D. None of these |
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Answer» Correct Answer - A Let ` l = int_(0)^(pi//2) (cos x)/((1+sin x) (2+ sin x) )dx` , Put `sin x = t rArr cos x dx = dt` ` :. l = int_(0)^(1) (dt)/((1+t)(2+t) )= int_(0)^(1) 1/((1+t))dt - int _(0)^(1) 1/((2+t))dt ` ` = [ log (1+t) - log (2+t) ]_(0)^(t)` ` = (log 2 - log 3) - (0 - log 2) = log. 4/3` |
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| 218. |
`int_(0)^(pi//2)(sinx)/((sinx+cosx))dx=?`A. `pi`B. `(pi)/(2)`C. `0`D. `(pi)/(4)` |
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Answer» `I=int_(0)^(pi//2)(sinx)/((sinx+cosx))dx` ………`(i)` `I=int_(0)^(pi//2)(sin((pi)/(2)-x))/({sin((pi)/(2)-x)+cos((pi)/(2)-x)})dx=int_(0)^(pi//2)(cosx)/((sinx+cosx))dx` ………`(ii)` `:.2I=int_(0)^(pi//2)intdx=[x]_(0)^(pi//2)=(pi)/(2)impliesI=(pi)/(4)`. |
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| 219. |
Property 8: If f(x) is a continuous function defined on `[-a; a]` then `int_(-a) ^a f(x) dx = int_0 ^a {f(x) + f(-x)} dx`A. `2int_(0)^(a){f(x)+f(-x)}dx`B. `2int_(0)^(a){f(x)-f(-x)}dx`C. `int_(0)^(a){f(x)+f(-x)}dx`D. none of these |
| Answer» We have `int_(-a)^(a)f(x)dx=int_(0)^(a){f(x)+f(-x)}dx`. | |
| 220. |
` int_(0)^(1) (dx)/(e^(x) +e^(-x)) dx ` is equal toA. `pi//4`B. `tan^(-1)e-(pi)/(4)`C. `tan^(-1)e`D. `(pi)/(4)tan^(-1)e` |
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Answer» Correct Answer - B Let ` int _(0)^(1) (dx)/(e^(x)+e^(-x))=int _(0)^(1) (e^(x))/(e^(2x)+1) dx` Put ` t = e^(x) ` ` rArr dt = e^(x) dx` Also at `x = 0 , t = e^(0) = 1 ` at ` x = 1, t = e^(1) = e` ` l = int _(1)^(e) (dt)/(t^(2)+1) = [ tan ^(-1) t] _(1)^(e)` ` tan ^(-1) e - tan ^(-1) (1)` ` = tan ^(-1) e - pi/4` |
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| 221. |
`int _(0)^(pi//2) (cosx)/(1+sinx)dx` is equal toA. log 2B. 2 log 2C. `(log 2)^(2)`D. `1/2 log2` |
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Answer» Correct Answer - A Let ` I = int _(0)^(pi//2)(cosx)/(1+sinx)dx` ` = [ log (1+sin x)]_(0)^(pi//2)= log 2- log 1 = log 2` |
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| 222. |
`int_(0)^(pi//2)(sinx-cosx)log(sinx+cosx)dx=0` |
| Answer» Apply `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx` to get `I=-I`. | |
| 223. |
The value of `overset(pi)underset(-pi)int(cos^(2)x)/(1+a^(2))dx,a gt 0`, isA. `2pi`B. `(pi)/(a)`C. `(pi)/(2)`D. `api` |
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Answer» Correct Answer - C Let ` l = int _(-pi)^(pi) (cos^(2)x)/(1+a^(x)) dx, a gt 0 " "` …(i) ` rArr l = int _(-pi)^(pi) (cos^(2)x)/(1+a^(-x))dx " "` …(ii) On adding Eqs. (i) and (ii), we get ` 2l = int _(-pi)^(pi) cos^(2) x dx = int _(-pi)^(pi) ((cos 2x+1))/2 dx ` ` rArr 2l = 1/2 [ ((sin2x)/2+x)]_(-pi)^(pi) = 1/2 (pi+pi) rArr l = pi/2 ` |
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| 224. |
` int _(0)^(1) (dx)/(1+x+x^(2))` is equal toA. `(pi)/(sqrt3)`B. `(pi)/(2sqrt3)`C. `(2pi)/(3sqrt3)`D. `(pi)/(3sqrt3)` |
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Answer» Correct Answer - D `int_(0)^(1) (dx)/(1+x+x^(2)) = int_(0)^(1) (dx)/(x^(2) + x + 1/4 - 1/4 +1 )` ` = int _(0)^(1) (dx)/((x+1/2)^(2)+ ((sqrt(3))/2)^(2))` , ` = 2/(sqrt(3)) [ tan^(-1). ((x+1/2))/(sqrt(3)/2)]_(0)^(1)` ` = 2/(sqrt(3)) [ tan^(-1) ((3/2)/(sqrt(3)/2))-tan^(-1) ((1/2)/(sqrt(3)/2))]` ` = 2/(sqrt(3)) [ tan^(-1) sqrt(3) - tan ^(-1). 1/(sqrt(3))]` ` =2/(sqrt(3)) (pi/3 - pi/6) = (2pi)/(6sqrt(3)) = pi/(3sqrt(3))` |
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| 225. |
` int _(0)^(2pi) (sin x + |sin x|)dx` is equal to |
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Answer» Correct Answer - B ` int _(0)^(2pi) ( sin x + | sin x| ) dx ` ` int _(0)^(pi) ( sin x + sin x) dx + int _(pi)^(2pi) (sin x - sin x) dx ` ` = int _(0)^(pi) 2 sin x dx + int _(pi)^(2pi) 0 dx = 2 [ - cos x] _(0)^(pi) +0` ` = int _(0)^(pi) 2 sin x dx + int _(pi)^(2pi) 0 dx = 2 [ - cos x ] _(0)^(pi) + 0 ` ` = - 2 ( cos pi - cos 0) = -2 (-1-1) = 4` |
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| 226. |
` int _(pi//4)^(pi//2) "cosec"^(2)dx` is equal toA. `-1`B. 1C. 0D. `(1)/(2)` |
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Answer» Correct Answer - B ` int _(pi//4)^(pi//2) "cosec"^(2) x dx = [ -cotx ] _(pi//4)^(pi//2) = ( -cot. pi/2 + cot .pi/2 ) = 1 ` |
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| 227. |
` int _(0)^(pi//4) log ((sin x+ cos x)/(cosx))dx ` is equal toA. `(pi)/(8)log2`B. `(pi)/(4)log2`C. `log2`D. `(pi)/(2)log2` |
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Answer» Correct Answer - A Let ` l = int _(0)^(pi//4) log ((sin x + cos x)/(cos x)) dx` ` l =- int _(0)^(pi//4) log (1+tan x ) dx" " ` ` = int _(0)^(pi//4) log [ 1 + tan (pi/4 -x) ] dx` ` = int _(0)^(pi//4) log (1+ (tan pi//4-tan x)/(1+tan pi //4 tanx))dx` ` = int _(0)^(pi//4) log (1+ (1-tanx)/(1+tan x)) dx` ` rArr l = int _(0)^(pi//4) log (2/(1+tan x))dx" "` ...(ii) On adding Eqs. (i) and (ii) , we get ` 2l = int _(0)^(pi//4) log 2 dx = log 2 (pi/4)` ` rArr l = pi/8 log 2 ` |
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| 228. |
The value of the integral `overset(pi//4)underset(0)int sin^(-4)x dx`, isA. `-(8)/(3)`B. `(3)/(2)`C. `(8)/(3)`D. None of these |
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Answer» Correct Answer - A Let ` l = int_(-pi//4)^(pi//4) sin^(-4) x dx = int _(-pi//4)^(pi//4) x^(4) dx` Put ` cot x = t rArr - 1^(2) x dx = dt` ` :. l = int_(-1)^(1) (1+t^(2)) dt = -2 int_(0)^(1) (1+ t^(2))dt` ` = - 2 [ t+(t^(3))/3]_(0)^(1) = - 2 [ 1+1/3] = - 8/3` |
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| 229. |
If `int _(a)^(b) x^(3) dx = 0 ` and if ` int _(a)^(b) x^(2) dx = 2/3 ,` then the values of a and b respectively areA. 1,1B. `-1,-1`C. `1,-1`D. `-1,1` |
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Answer» Correct Answer - D Given , ` int _(a)^(b) x^(3) dx = 0 and int _(a)^(b) x^(2) dx = 2/3 ` If we take `a=-1` and b= 1 , then it will satisfy the given integration . |
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| 230. |
`int _(-1//2)^(1//2) cos x log ((1+x)/(1-x)) dx = k log 2 , ` then k equals |
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Answer» Correct Answer - A `cosx log((1+x)/(1-x))dx` is an odd function . `thereforeint_(-1//2)^(1//2)cosx log((1+x)/(1-x))dx=0` Then k = 0 |
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| 231. |
Let `f : (-1,1) to R ` be continous function , if ` int _(0)^(sinx) f(t) dt = (sqrt(3))/2 x , "then " f ((sqrt(3))/2 )` is equal toA. `(1)/(2)`B. `(sqrt3)/(2)`C. `sqrt((3)/(2))`D. `sqrt3` |
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Answer» Correct Answer - D Given , ` int _(0)^(sin x) f(t) dt = (sqrt(3))/2 x` On differentiating w.r.t x , we get `f(sin) cos x = (sqrt(3))/2 ` [ Using Leibnitz theorem ] If ` sin x = (sqrt(3))/2 rArr x = pi/3 ` ` :. f (sin pi/3) cos.pi/3 = (sqrt(3))/2 rArr f (sqrt(3)/2) . 1/2 = (sqrt(3))/2 ` ` rArr f((sqrt(3))/2) = sqrt(3)` |
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| 232. |
If `f(t)`is an odd function, then `varphi(x)=int_a^xf(t)dx`is an even function.A. an odd functionB. an even functionC. neither even nor oddD. 0 |
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Answer» Correct Answer - B if f(t) is an odd function , then `int_(0)^(x)f(t)` dt is even function . |
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| 233. |
Let `F (x) = f(x) + f ((1)/(x)),` where `f (x) = int _(1) ^(x ) (log t)/(1+t) dt.` Then F (e) equalsA. `1//2`B. 0C. 1D. 2 |
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Answer» Correct Answer - A `F(e)=f(e)+f((1)/(e))` `rArrF(e)=int_(1)^(e)(logt)/(1+t)dt+int_(1)^(1//e)(logt)/(1+t)dt[becausef(x)=int_(1)^(x)(logt)/(1+t)dt]` On putting t `=(1)/(t)` in second integration , we get `F(e)=int_(1)^(e)(logt)/(1+t)dt+int_(1)^(e)(logt)/(t(1+t))` `=int_(1)^(e)(logt)/(t)dt=[((logt)^(2))/(2)]_(1)^(e)` `=(1)/(2)[(loge)^(2)-(log1)^(2)]=(1)/(2)` |
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| 234. |
If `kint_(0)^(1)xf(3x)dx=int_(0)^(3)tf(t)dt`, then the value of k isA. 9B. 3C. `1/9`D. `1/3` |
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Answer» Correct Answer - A Let `l=kint_(0)^(1)xf(3x)dx` Put 3x=t `rArrdx=(dt)/(3)` `thereforel=kint_(0)^(3)(t)/(3)*f(t)*(dt)/(3)=(k)/(9)int_(0)^(3)t*f(t)dt` Now `(k)/(9)int_(0)^(3)t*f(t)dt=int_(0)^(3)t*f(t)dt` [ Given] `rArr(k)/(9)=1rArrk=9` |
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| 235. |
Evaluate :`int_0^1sqrt(x(1-x))dx` |
| Answer» `I=intsqrt({((1)/(2))^(2)-(x-(1)/(2))^(2)})dx`. | |
| 236. |
`int_1^3(dx)/(x^2(x+1))` |
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Answer» Let `(1)/(x^(2)(x+1))=(A)/(x)+(B)/(x^(2))+(C )/(x+1)`. Then `Ax(x+1)+B(x+1)+Cx^(2)-=1`. This, gives `A=-1`, `B=1`, `C=1`. |
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| 237. |
`int_(0)^(1)(logx)/(sqrt(1-x^(2)))dx=-(pi)/(2)(log2)` |
| Answer» Correct Answer - Put `x= sin theta` | |
| 238. |
`int_(0)^(1)((sin^(-1)x)/(x))dx=(pi)/(2)(log2)` |
| Answer» Put `x=sin theta` and integrate by parts. | |
| 239. |
`int_(0)^(pi)log(1+cosx)dx=-pi(log2)` |
| Answer» `2I=int_(0)^(pi)log(1-cos^(2)x)dx=2int_(0)^(pi)log(sinx)dx=4int_(0)^(pi//2)log(sinx)dx`. | |
| 240. |
`int_1^2 (5x^2)/(x^2+4x+3) dx` |
| Answer» Let `(5x^(2))/((x^(2)+4x+3))=5+(A)/((x+1))+(B)/((x+3))`. | |
| 241. |
`int_(0)^(pi//2)log(tanx+cotx)dx=pi(log2)` |
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Answer» `I=int_(0)^(pi//2)log((sinx)/(cosx)+(cosx)/(sinx))dx=int_(0)^(pi//2)log((1)/(sinxcosx))dx` `=-[int_(0)^(pi//2)log(sinx)dx+int_(0)^(pi//2)log(cosx)dx]`. |
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| 242. |
`int_(pi//3)^(pi//4)(tanx+cotx)^(2)dx` |
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Answer» `I=int(tan^(2)x+cot^(2)x+2)dx=int{(sec^(2)x-1)+(cosec^(2)x-1)+2}dx` `=int(sec^(2)x+cosec^(2)x)dx`. |
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| 243. |
`int_(1)^(2)(dx)/((x+1)sqrt(x^(2)-1))` |
| Answer» Put `(x+1)=(1)/(t)`. | |
| 244. |
`int_(0)^(pi//2)(1)/((1+sqrt(cotx)))dx=?`A. `0`B. `(pi)/(4)`C. `(pi)/(2)`D. `pi` |
| Answer» `I=int_(0)^(pi//2)(sqrt(sinx))/((sqrt(sinx)+sqrt(cosx)))dx`. | |
| 245. |
`int_2^3(2-x)/(sqrt(5x-6-x^2))` |
| Answer» Let `(2-x)=A*(d)/(dx)(5x-6-x^(2))+B`. | |
| 246. |
` int_(0)^(lambda) (ydy)/(sqrt(y+lambda))` is equal toA. `(2)/(3)(2-sqrt2)lambda sqrt(lambda)`B. `(2)/(3)(2+sqrt2)lambda sqrt(lambda)`C. `(1)/(3)(2-sqrt2)lambdasqrt(lambda)`D. `(1)/(3)(2+sqrt2)lambdasqrt(lambda)` |
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Answer» Correct Answer - A Let ` l = int_(0)^(lambda) (ydy)/(sqrt(y+lambda))` , ` = int_(0)^(lambda) [ sqrt(y+lambda) - lambda/(sqrt(y+lambda))]dy` , ` = [ (y+lambda)^(3//2)/(3//2)]_(0)^(lambda) - [ (lambda sqrt(y+lambda))/(1//2) ] _(0)^(lambda)` ` = 2/3 [ (2 lambda )^(3//2) - lambda ^(3//2) ] - 2lambda[(2 lambda )^(1//2) - (lambda)^(1//2)] ` ` = 2lambda sqrt(lambda) [ (2 sqrt(2)-1)/3 - (sqrt(2)-1) ] = 2/3 lambda sqrt(lambda) (2 - sqrt(2))` , |
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| 247. |
`int_(0)^(pi//2)(dx)/((1+sqrt(tanx)))=(pi)/(4)` |
| Answer» `I=int_(0)^(pi//2)sqrt((sinx)/(cosx))/((1+sqrt((sinx)/(cosx))))dx=int_(0)^(pi//2)(sqrt(sinx))/((sqrt(sinx)+sqrt(cosx)))dx` | |
| 248. |
`int_(8)^(15)(dx)/((x-3)sqrt(x+1))` is equal toA. `(1)/(2)log.(5)/(3)`B. `(1)/(3)log.(5)/(3)`C. `(1)/(5)log.(3)/(5)`D. `(1)/(2)log.(3)/(5)` |
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Answer» Correct Answer - A Put `x +1 = t^(2) rArr dx = 2t dt ` At x= 8, t=3 and x=15 , t =4 ` :. l = int_(3)^(4) (2t dt)/((t^(2)-1-3)t)` , ` = int _(3) ^(4) (2dt)/((t^(2)-4))= 2. 1/4 [log. (t-2)/(t+2)]_(3)^(4)` ` = 1/2 [ log. 1/3 - log. 1/5] =1/2 log. 5/3` |
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| 249. |
`int_(0)^(pi//2)(sqrt(tanx)+sqrt(cotx)) dx` |
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Answer» `I=int_(0)^(pi//2)(sqrt(2)(sinx+cosx))/(sqrt(2sinxcosx))dx=sqrt(2)*int_(0)^(pi//2)((sinx+cosx))/(sqrt(1-(1-sin2x)))dx` `=sqrt(2)*int_(0)^(pi//2)((sinx+cosx))/(sqrt(1-(sinx-cosx)^(2)))dx`. Put `(sinx-cosx)=t`. |
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| 250. |
If `f(x) = tanx-tan ^(3) x + tan^(5) x - tan ^(7) x + ... infty` for `olt x lt pi/4 , "than" int_(0)^(pi//4) f (x) dx=`A. 1B. 0C. `1/4`D. `1/2` |
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Answer» Correct Answer - C `f(x)=tanx-tan^(3)x+tan^(5)x-...infty` [ it is a Gp] `rArrf(x)=(tanx)/(1+tan^(2)x)=(tanx)/(sec^(2)x)=(sin2x)/(2)` `thereforeint_(0)^(pi//4)f(x)dx=int_(0)^(pi//4)(sin2x)/(2)dx=[-(cos2x)/(4)]_(0)^(pi//4)=(1)/(4)` |
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