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| 251. |
Evaluate : (i) `int_(0)^(2pi) {sin(sinx)+sin(cosx)}dx`, (ii) `int_(0)^(pi) (dx)/(5+4cos2x)` (iii) `int_(0)^(pi//2) (2lnsinx-ln sin2x)dx` , (iv) `int_(0)^(oo) ln(x+(1)/(x)).(dx)/(1+x^(2))` |
| Answer» Correct Answer - (i) `0` , (ii) `pi/3` , (iii) `- (pi)/(2) ln 2` , (iv) `piln 2` | |
| 252. |
`int_(-pi/2)^(pi/2)(e^(|sinx|)cosx)/((1+e^(tanx))dxi se q u a lto``e+1`(b) `1-e``e-1`(d) none of theseA. `e+1`B. `2e`C. `e-1`D. `e-2` |
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Answer» Correct Answer - C `I=int_(-pi//2)^(pi//2) (e^(|sinx|)/cosx)/(1+e^(tanx))dx` ……….1 `=int_(-pi//2)^(pi//2) (e^(|sin(-x)|)cos(-x))/(1+e^(tan(-x)))dx` `:. I=int_(-pi//2)^(pi//2) (e^(tanx)e^(|sinx|)cosx)/(e^(tanx)+1)dx`………………2 Adding 1 and 2 we get `2I=int_(-pi//2)^(pi/2) e^(|sinx|)cosx dx` `=2int_(0)^(pi//2)e^(sinx)cosx dx` `=2[e^(sinx)]_(0)^(pi//2) =2(e-1)` `:. I=e-1` |
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| 253. |
Evaluate : (i) `int_(-pi//2)^(pi//2)sin^(2)xcos^(2)x(sinx+cosx)dx` , (ii) `int_(0)^(pi)xsin^(5)xdx` (iii) `int_(0)^(2)x^(3//2)sqrt(2-x)dx`, (iv) `int_(0)^(2pi)x(sin^(2)xcos^(2)x)` |
| Answer» Correct Answer - (i) `4/15` , (ii) `pi/2`, (iv) `(pi^(2))/(4)` | |
| 254. |
Let `f(x)` be a function satisying `f(x) = f((100)/(x)) AA x ge 0`. If `int_(1)^(10) (f(x))/(x) dx = 5` then find the value of `int_(1)^(100)(f(x))/(x) dx`. |
| Answer» Correct Answer - 10 | |
| 255. |
Evalaute `int_(pi//6)^(pi//3)(sqrt((sinx))dx)/(sqrt((sinx))+sqrt((cosx)))` |
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Answer» We have `I=int_(pi//6)^(pi//3)(sqrt((sinx))dx)/(sqrt((sinx))+sqrt((cosx)))`…………..1 or `I=int_(pi//6)^(pi/3)(sqrt((cosx))dx)/(sqrt((cosx))+sqrt((sinx)))` (Replacing `x` by `(pi)/2-x)`…………..2 Adding 1 and 2 we get `2I=int_(pi//6)^(pi//3)(sqrt((sinx))+sqrt((cosx)))/(sqrt((cosx))+sqrt((sinx))) dx` `int_(pi//6)^(pi//3)dx=[x]_(pi//6)^(pi//3)=pi//3-pi//6=pi//6` Hence `I=pi//12` |
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| 256. |
Let `u = int_(pi//6)^(pi//2)` min. (`sqrt(3)sinx, cosx`) dx and `V = int_(-3)^(5)x^(2)sgn (x-1) dx`. If `V = lambdaU`, then find the value of `lambda. |
| Answer» Correct Answer - 64 | |
| 257. |
Statement-I : The value of the integral `int_(pi//6)^(pi//3)(dx)/(1+sqrt(tanx))` is equal to `pi//6`. Statement-I : `int_(a)^(b)f(x)dx = int_(a)^(b)f(a+b-x) dx`A. Statement-I is true, Statement-II is true, Statement-II is a correct explanation for Statement-I.B. Statement-I is true, Statement-II is true, Statement-II is not a correct explanation for Statement-I.C. Statement-I is true, Statement-II is false.D. Statement-I is false, Statement-II is true. |
| Answer» Correct Answer - D | |
| 258. |
`int_(lnpi-ln2)^(lnpi) (e^(-x))/(1-cos(2/3e^(x)))` dx is equal toA. `sqrt(3)`B. `-sqrt(3)`C. `(1)/(sqrt(3))`D. `-(1)/(sqrt(3))` |
| Answer» Correct Answer - A | |
| 259. |
If `int_(ln2)^(x)(dx)/(sqrt(e^(x)-1))= (pi)/(6)`, then x is equal toA. `4`B. `ln B`C. `ln 4`D. `ln 2` |
| Answer» Correct Answer - C | |
| 260. |
The value of `int_(sqrt(ln2))^(sqrt(ln3)) (xsinx^2)/(sinx^2+sin(ln6-x^2)) dx` isA. `(1)/(4)"log"(3)/(2)`B. `(1)/(2)"log"(3)/(2)`C. `"log"(3)/(2)`D. `(1)/(6)"log"(3)/(2)` |
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Answer» Correct Answer - A Put `x^(2)=trArrx dx=dt//2` `:.I=int_(log2)^(log3)(sint*(dt)/(2))/(sint+sin(log6-t))` . . . (i) Using , `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx` `=(1)/(2)int_(log2)^(log3)(sin(log2+log3-t))/(sin(log2+log3-t)+sin(log6-(log2+log3-t)))` `=(1)/(2)int_(log2)^(log3)(sin(log6-t))/(sin(log6-t)+sin(t))dt` `:. I=int_(log2)^(log3)(sin(log6-t))/(sin(log6-t)+sint)dt` . . . (ii) On adding Eqs . (i) and (ii) , we get `2I=(1)/(2)int_(log2)^(log3)(sint+sin(log6-t))/(sin(log6-t)+sint)dt` `rArr2I=(1)/(2)(t)_(log2)^(log3)=(1)/(2)(log3-log-2)` `:.I=(1)/(4)log((3)/(2))` |
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| 261. |
`int_0^1log(sqrt(1-x)+sqrt(1+x))dx` equals: |
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Answer» Correct Answer - `(1)/(2)log1-(1)/(2)+(pi)/(4)` Let `I = int _(0)^(1) log (sqrt(1-x)+sqrt(1+x))dx` Put `x= cos 2 theta` `rArr dx =- 2 sin 2 theta d theta` ` :. =- 2 int _(pi//4)^(0) log [ sqrt(1-cos 2 theta)+sqrt(1+cos 2 theta)] (sin 2 theta) d theta` `=- int _(pi//4)^(0) log [ (sqrt(2) sin theta+ cos theta)]sin 2 theta d theta` ` =- 2 int _(pi//4)^(0) [ (log sqrt(2))sin 2 theta+ log (sin theta + cos theta ) * sin 2 theta ] d theta` `=- 2 log sqrt(2)[(-cos 2 theta )/(2)] _(pi//4)^(0)- 2 int _(pi//4)^(0) log underset(I) ((sin theta + cos theta))underset(II) (* sin 2 theta d theta)` ` = log sqrt(2)-2 [ - {log (sin theta + cos theta) * (cos 2 theta)/(2)}_(pi//4)^(0)- int_(pi//4)^(0) (( cos theta - sin theta)/(cos theta sin theta)xx(- cos 2 theta)/(2)) d theta]` `= log (sqrt(2)) -2 [ 0 + (1)/(2) int _(pi//4)^(0) ( cos theta - sin theta) ^(2)d theta]` `= (1)/(2) log 2 - int _(pi//4)^(0) (1- sin theta) d theta` `= (1)/(2) log 2 - [ theta + (cos 2 theta)/(2)]_(pi//4)^(0)` ` =(1)/(2) log 2 - ((1)/(2) - (pi)/(4))=(1)/(2)+(pi)/(4)` |
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| 262. |
The range of the function `f(x)=int_(-1)^(1)(sinxdt)/(1+2tcosx+t^(2))` isA. `[-(pi)/2,(pi)/2]`B. `[0,pi]`C. `{0,pi}`D. `{-(pi)/2,(pi)/2}` |
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Answer» Correct Answer - D We have `f(x)=int_(-1)^(1)(sinx dt)/(sin^(2)x+(t-cosx)^(2))` `=(sinx)/(sinx)tan^(-1)((t-cosx)/(sinx))|_(-1)^(1)` `=tan^(-1)((1-cosx)/(sinx))-tan^(-1)((-1-cosx)/(sinx))` `=tan^(-1) (tanx//2) +tan^(-1) (cotx//2)` Now we know that `tan^(-1)x+tan^(-1)1/x={((pi)/2,xgt0),(-(pi)/2,xlt0):}` or `tan^(-1)("tan"x/2)+tan^(-1)(1/("tan"x/2))={((pi)/2, "tan"(pi)/2gt0),(-(pi)/2,"tan"x/2lt0):}` Hence range of `f(x)` is `{-(pi)/2,(pi)/2}` |
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| 263. |
If `int_0^oosinx/xdx=pi/2`, then `int_0^oosin^3x/xdx` is equal toA. `pi//2`B. `pi//4`C. `pi//6`D. `3pi//2` |
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Answer» Correct Answer - B `sin^(3)x=3/4 sinxx =1/4 sinx 3x` `:. int_(0)^(oo) (sin^(3)x)/x dx=3/4 int_(0)^(oo) (sinx)/x dx-1/4 int_(0)^(oo) (sin 3x)/x dx` `=3/4 int_(0)^(oo) (sinx)/x dx-1/4 int_(o)^(oo) (sinu)/u du (u=3x)` `=3/4 (pi)/2-1/4 (pi)/2=(pi)/4` |
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| 264. |
Find `f(x)` if it satisfies the relation `f(x) = e^(x) + int_(0)^(1) (x+ye^(x))f(y) dy`. |
| Answer» Correct Answer - `(-3e^(x))/(2(e-1)) - 3x` | |
| 265. |
`int_- 1^1(e^(-1/ x))/(x^2(1+e^(-2/ x)))dx` is equal to :A. `(pi)/2=2tan^(-1)e`B. `(pi)/2-2cot^(-1)e`C. `2tan^(-1)e`D. `pi-2tan^(-1)e` |
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Answer» Correct Answer - D `I=int_(-1)^(1)(e^(-1//x))/(x^(2)(1+e^(-2//x)))dx=2 int_(0)^(1)(e^(-1//x))/(x^(2)(1+e^(-2//x)))dx` Put `e^(-1/x)=t` `implies 1/(x^(2))e^(-1/x)dx=dt` `:. I=2 int_(0)^(1//2) (dt)/(1+t^(2))` `=2[tan^(-1)t]_(0)^(1//e)` `=2"tan"^(-1)1/e` `=2[(pi)/2-tan^(-1)e]` `=pi-2 tan^(-1) e` |
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| 266. |
Evaluate the definiteintegrals`int0pi/4(sinx+cosx)/(9+16sin2x)dx`A. `1/20 log3`B. `1/40 log3`C. `1/20 log 6`D. `10 log 3` |
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Answer» Correct Answer - A `I=int_(0)^(pi//4)(sinx+cosx)/(25-16(sinx+cosx)^(2)) dx` Let `sinx-cosx=t` `:.I=int_(-1)^(0)(dt)/(25-16t^(2))` `=1/16int_(-1)^(0)(dt)/((5/4)^(2)-t^(2))` `=1/16 . 1/2 . 5/4 log [|(5/4+t)/(5/4-t)|]_(-1)^(0)` `=1/40[log1-"log"1/9]` `=(log 9)/40` `=1/20 log 3` |
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| 267. |
The value of `int_(0)^(pi//2)ln|tanx+cotx|` dx is equal to :A. `pi ln2`B. `-pi ln 2`C. `(pi)/(2) ln 2`D. `-(pi)/(2)ln 2` |
| Answer» Correct Answer - A | |
| 268. |
Complete the area of the region bounded by the parabolas `y^(2)+8x=16` and `y^(2)-24x=48`. |
| Answer» Correct Answer - `32sqrt((2)/(3))` sq. units | |
| 269. |
Consider a line `l: 2x -sqrt(3)y = 0` and a parameterized `C :x = tant, y=1/cost(0 |
| Answer» Correct Answer - 55 | |
| 270. |
The smaller area bounded by `x^2/16+y^2/9=1` and the line `3x+4y=12` is |
| Answer» Correct Answer - (i) `3(pi-2)` , (ii) `1/8` | |
| 271. |
If the area bounded by `f(x)=sqrt(tan x), y=f(c), x=0 and x=a, 0ltcltalt(pi)/(2)` is minimum then find the value of c. |
| Answer» Correct Answer - `4` sq. units | |
| 272. |
Find the area bounded by `y=sin^(-1)x ,y=cos^(-1)x ,a n dt h ex-a xi sdot`A. `underset(0)overset(1//sqrt(2))int(sin^(1)x)dx +underset(1//sqrt(2))overset(1)int(cos^(1)x)dx`B. `underset(pi//4)overset(pi//2)int(siny-cosy)dy`C. `underset(0)overset(pi//4)int(cosy-siny)dy`D. `(sqrt(2)-1)`sq.unit |
| Answer» Correct Answer - A::B::C::D | |
| 273. |
Find the area bounded by `y = [-0. 01 x^(4) - 0.02 x^(2)]`, (where `[*]` G.I.F.) and curve `3x^(2)+4y^(2) = 12`, which lies below `y = - 1`. |
| Answer» Correct Answer - `2sqrt(3) sin^(-1) sqrt((2)/(3)) - (2sqrt(2))/(sqrt(3))` | |
| 274. |
`P(2,2), Q(-2,2) R(-2,-2) & S(2,-2)` are vertices of a square. A parabola passes through `P. S &` its vertex liesn on x-axis. If this parabola bisect the area of the square PQRS, then vertex of the parabolaA. `(-2,0)`B. `(0,0)`C. `(-(3)/(2),0)`D. `(-1,0)` |
| Answer» Correct Answer - D | |
| 275. |
The area bounded by the curve `x = a cos^3t,, y = a sin^3t, ` is :A. `(3pia^(2))/(8)`B. `(3pia^(2))/(16)`C. `(3pia^(2))/(16)`D. `(3pia^(2))/(32)` |
| Answer» Correct Answer - A | |
| 276. |
The area bounded by the curve `f(x)=x+sinx`and its inverse function between the ordinates `x=0a n dx=2pi`is`4pis qdotu n i t s`(b) `8pis qdotu n i t s``4s qdotu n i t s`(d) `8s qdotu n i t s`A. `4 pi`B. `8 pi`C. `4`D. 8 |
| Answer» Correct Answer - D | |
| 277. |
Consider the curve `C: y = sin 2x - sqrt(3)|sinx|, C` cuts the x-axis at `(a,0) , x in (-pi, pi)`. `A_(1)` : The area bounded by the curve C and the positive x-axis between the origin and the line `x = a`. `A_(2)` : The area bounded by the curve C and the negative x-axis between the line x = a and the origin. Prove that `A_(1) + A_(2) +8A_(1) A_(2) = 4`. |
| Answer» Correct Answer - 3 | |
| 278. |
Let `f(x) = (1-x)^(2) sin^(2)x+ x^(2)` for all `x in IR` and let `g(x) = int_(1)^(x)((2(t-1))/(t+1)-lnt) f(t)` dt for all `x in (1,oo)`. Consider the statements : P : There exists some `x in IR` such that `f(x) + 2x = 2 (1+x^(2))` Q : There exist some `x in IR` such that `2f(x) + 1 = 2x(1+x)` ThenA. both P and Q are trueB. P is true and Q is falseC. P is false and Q is trueD. both P and Q are false |
| Answer» Correct Answer - C | |
| 279. |
Let `f(x) = (1-x)^(2) sin^(2)x+ x^(2)` for all `x in IR` and let `g(x) = int_(1)^(x)((2(t-1))/(t+1)-lnt) f(t)` dt for all `x in (1,oo)`. Which of the following is true ?A. g is increasing on `(1,oo)`B. g is decreasing on `(1,oo)`C. g is increasing on `(1,2)` and decreasing on `(2,oo)`D. g is increasing on `(1,2)` and decreasing on `(2,oo)` |
| Answer» Correct Answer - B | |
| 280. |
Let m,n be two positive real numbers and define `f(n)=int_(0)^(oo)x^(n-1)e^(-x)dx` and `g(m,n)=int_(0)^(1)x^(m-1)(1-m)^(n-1)dx`. It is known that f(n) for n gt 0 is finite and g(m, n) = g(n, m) for m, n gt 0. `int_(0)^(1)x^(m)(log_(e).(1)/(x))dx=`A. `(f(n+1))/((m+1)^(n))`B. `(f(n))/((m+1)^(n+1))`C. `(f(n+1))/((m+1)^(n+1))`D. `g(m+1),n+1)` |
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Answer» Correct Answer - C Putting `log_(e)..(1)/(x)=t` `rArr" "x=e^(-t)` `rArr" "int_(0)^(1)x^(m)(log_(e)(1)/(x))^(n)dx` `" "=int_(oo)^(0)e^(-mt)t^(n)(-e^(t))dt` `" "=int_(0)^(oo)t^(n)e^(-(m+1))dt` `" "=(1)/((m+1)^(n+1))int_(0)^(oo)t^(n)e^(-y)dy" (putting (m + 1) t = y)"` `" "=(f(n+1))/((m+1)^(n+1))` |
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| 281. |
Let m,n be two positive real numbers and define `f(n)=int_(0)^(oo)x^(n-1)e^(-x)dx` and `g(m,n)=int_(0)^(1)x^(m-1)(1-m)^(n-1)dx`. It is known that f(n) for n gt 0 is finite and g(m, n) = g(n, m) for m, n gt 0. `int_(0)^(oo)(x^(m-1))/((1+x)^(m+n))dx=`A. g(m,n)B. `g(m-1,n)`C. `g(m-1,n-1)`D. `g(m,n-1)` |
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Answer» Correct Answer - A `g(m,n)=int_(0)^(1)x^(m-1)(1-x)^(n-1)dt` Put `x=(1)/(1+y)` `rArr" "g(m,n)=int_(oo)^(0)(1)/((1+y)^(m-1))(1-(1)/(1+y))^(n-1)(-(1)/((1+y)^(2)))dy` `" "=int_(0)^(oo)(y^(n-1))/((1+y)^(m+n))dy` `" "=int_(0)^(oo)(x^(n-1))/((1+x)^(m+n))dx` |
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| 282. |
If `U_n=int_0^pi(1-cosnx)/(1-cosx)dx ,`where `n`is positive integer or zero, then show that `U_(n+2)+U_n=2U_(n+1)dot`Hence, deduce that`int_0^(pi/2)(sin^2ntheta)/(sin^2theta)=1/2npidot`A. `pi//2`B. `pi`C. `npi//2`D. `npi` |
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Answer» Correct Answer - C `U_(n+2)-U_(n+1)=int_(0)^(pi)((1-cos(n+2)x)-(1-cos(n+1)x))/(1-cosx)dx` `=int_(0)^(pi)(cos(n+1)x-cos(n+2)x)/(1-cosx)` `=int_(0)^(x)(2sin(n+3/2)x . "sin"x/2)/(2sin^(2)x//2) dx` `implies U_(n+2)-U_(n+1)=int_(0)^(pi)("sin"(n+3/2)x)/("sin"x/2)dx`.................1 `impliesU_(n+1)-U_(n)=int_(0)^(pi)("sin"(n+1/2)x)/("sin"x/2)dx`.............2 From 1 and 2 we get `(U_(n+2)-U_(n-1))-(U_(n+1)-U_(n))` `=int_(0)^(pi)(sin(n+3/2)x-sin(n+1/2)x)/("sin"x/2)dx` `implies U_(n+2)+U_(n)-2U_(n+1)` `=int(2cos(n+1)x.sinx//2)/(sinx//2) dx` `=2int_(0)^(pi)cos(n+1)x dx` `=2((sin(n+1)x)/(n+1))-(0)^(pi)=0` `impliesU_(n+2)+U_(n)=2U_(n+1)` `implies U_(n),U_(n+1),U_(n+2)` are in A.P. `U_(0)=int_(0)^(pi)(1-1)/(1-cosx)dx=0` `U_(1)=int_(0)^(pi)(1-cosx)/(1-cosx) dx=pi` `U_(1)=U_(0)=pi` (common difference) `:.U_(n)=U_(0)+npi=npi` Now, `I_(n)=int_(0)^(pi//2) (sin^(2) n theta)/(sin^(2) theta) d theta` `=int_(0)^(pi//2) (sin^(2) n theta)/(sin^(2)theta) d theta` `=int_(0)^(pi//2) (1-cos2 n theta)/(1-cos 2theta) d theta=1/2 int_(0)^(pi)(1-cosn x)/(1-cosx) dx` `impliesI_(n)=1/2npi` |
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| 283. |
Let `I = int_(2)^(oo)((2x)/(x^(2)+1)- (1)/(2x+1))` dx & I is a finite real number, thenA. `lambda = 1/2`B. `lambda = 1`C. `I = 1/2 ln (5/2)`D. `I = 1/4 ln (5/4)` |
| Answer» Correct Answer - A::D | |
| 284. |
Find the equation of tangent to`y=int_(x^2)^(x^3)(dt)/(sqrt(1+t^2))a tx=1.` |
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Answer» Correct Answer - `sqrt(2)y=x=1` `y|_(x=1)=0,(dy)/(dx)=1/(sqrt(1+x^(5))) 3x^(2)-1/(sqrt(1+x^(4)))2x` or `(dy)/(dx)|_(x=1)=1/(sqrt(2))` Thus, required equation is `ysqrt(2)=x-1`. |
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| 285. |
If `I_(n)=int_(0)^(pi/2) sin^(x)x dx`, then show that `I_(n)=((n-1)n)I_(n-2)`. Hence prove that `I_(n)={(((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))………(1/2)(pi)/2,"if",n"is even"),(((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))………(2/3)1,"if",n"is odd"):}` |
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Answer» `I_(n)=int_(0)^((pi)/2 sin^(n)x dx` `=int_(0)^((pi)/2) sin^(n-1)x sin x dx` `=[-sin^(n-1)x cosx]_(0)^((pi)/2)+int_(0)^((pi)/2)(n-1)sin^(n-2)x cos^(2)x dx` `=(n-1)int_(0)^((pi)/2)sin^(n-2)x(1-sin^(2)x)dx` `=(n-1)int_(0)^((pi)2/2)sin^(n-2)x x-(n-1)int_(0)^((pi)/2)sin^(n)x dx` or `I_(n)+(n-1)I_(n)=(n-1)I_(n-2)` or `I_(n)=((n-1)/n)I_(n-2)` `=((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))...........I_(0)` or `I_(1)` Accordingly if `n` is even or odd, `I_(0)=(pi)/2,I_(1)=1` Hence `I_(n)= {(((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))………(1/2)(pi)/2,"if",n"is even"),(((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))………(2/3)1,"if",n"is odd"):}` |
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| 286. |
If for nonzero `x ,af(x)+bf(1/x)=1/x-5,`where `a!=b`then `int_1^2f(x)dx=1____` |
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Answer» Correct Answer - `(1)/(a^(2)-b^(2))[a log 2-5a+(7)/(2)b]` Given , of ` (x)+bf (1//x)=(1)/(x)-5` . . . (i) Replacing x by `1//x` in Eq . (i) , we get af `(1//x) +bf (x)=x-5` .. . (ii) On multiplying Eq . (i) by a and Eqs . (ii) by b , we get `a^(2)f(x)+abf(1//x)=a((1)/(x)-5)`. . . (iii) `abf(1//x)+b^(2)f(x)=b(x-5)` ... (iv) On subtracting Eq . (iv) from Eq . (iii) , we get `(a^(2)-b^(2)) f(x)=(a)/(x)-bx-5a+5b` `rArr f(x)=(1)/((a^(2)-b^(2)))((a)/(x)-bx-5a+5b)` `rArr int_(1)^(2)f(x)dx=(1)/((a^(2)-b^(2)))int_(1)^(2)((a)/(x)-bx-5a+5b)dx` `=(1)/((a^(2)-b^(2)))[alog|x|-(b)/(2)x^(2)-5(a-b)x]_(1)^(2)` `=(1)/((a^(2)-b^(2)))[alog2-2b-10(a-b)-alog1+(b)/(2)+5(a-b)]` `=(1)/((a^(2)-b^(2)))[alog 2-5a+(7)/(2)b]` |
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| 287. |
For `x epsilonR`, and a continuous function `f` let `I_(1)=int_(sin^(2)t)^(1+cos^(2)t)xf{x(2-x)}dx` and `I_(2)=int_(sin^(2)t)^(1+cos^(2)t)f{x(2-x)}dx`. Then `(I_(1))/(I_(2))` isA. `-1`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - B `I_(1)=int_(sin^(2)t)^(1+cos^(2)t)xf(x(2-x))dx` `=int_(sin^(2)t)^(1+cos^(2)t)(2-x)f(x(2-x))dx=2I_(2)-I_(1)` or `2I_(1)=2I_(2)` or `(I_(1))/(I_(2))=1` |
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| 288. |
`(lim)_(nvecoo)n/(2^n)int_0^2x^n dxe q u a l s___` |
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Answer» Correct Answer - 2 `lim_(n to oo) n/(2^(n)) . (x^(n+1))/(n+1)]_(0)^(2)=lim_(n to oo) n/(2^(n)) . (2^(n+1))/(n+1)=lim_(n to 0)2/(1+(1//n)=2` |
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| 289. |
If `I=int_(0)^(3pi//4) ((1+x)sinx(1-x)cosx)dx`, then the value of `(sqrt(2)-1)I` is_______ |
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Answer» Correct Answer - 2 `I=int_(0)^(3pi//4) (sinx+cosx)dx+int_(0)^(3pi//4)ubrace(x)_(1)ubrace((sinx-cosx))_(II)dx` `=int_(0)^(3pi//4) (sinx+cosx)dx+ubrace(x(-cosx-sinx)|_(0)^(3pi//4))_("zero")` `+int_(0)^(3pi//4) (sinx+cosx)dt` `=2int_(0)^(3pi//4)(sinx+cosx)dx=2sqrt(2)+1` |
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| 290. |
Let `f: RvecR`be a continuous odd function, which vanishes exactly at one point and `f(1)=1/2dot`Suppose that`F(x)=int_(-1)^xf(t)dtfora l lx in [-1,2]a n dG(x)=int_(-1)^x t|f(f(t))|dtfora l lx in [-1,2]dotIf(lim)_(xvec1)(F(x))/(G(x))=1/(14),`Then the value of `f(1/2)`is |
| Answer» Correct Answer - 7 | |
| 291. |
Let `f(a,b) = int_(a)^(b)(x^(2)-4x+3)dx, (bgt 0)` thenA. `f(a,3)` is least when `a = 1`B. `f(a,b)` is an increasing function `AA b ge 4`C. `f(0,b)` is least for `b = 2`D. `min{f(a,b)} = -4/3` |
| Answer» Correct Answer - A::B::D | |
| 292. |
Find the value of`int_(-pi/3)^(pi/3)(pi+4x^3)/(2-cos(|x|pi/3))dx` |
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Answer» Let `I=int_(-pi//3)^(pi//3) (pi+4x^(3))/(2-cos(|x|+(pi)/3))dx+int_(-pi//3)^(pi/3)(4x^(3))/(2-cos(|x|+(pi)/3))dx` The second integral becomes zero as integrand being an odd function of `x`. `:. I=2pi int_(0)^(pi//3)(dx)/(2-cos(x+(pi)/3))` Let `x+pi//3=y` or `dx=dy`. Also as `xto0,y to pi//3`, and as `x to pi//3, yto2pi//3`. `:. I=2pi int_(pi//3)^(2pi//3)(dy)/(2-cosy)` `=2pi int_(pi//3)^(2pi//3) (dy)/(2-(1-tan^(2)y//2)/(1+tan^(2)y//2))` `=2pi int_(pi//3)^(2pi//3)(1/2sec^(2)y//2)/(tan^(2)y//2+(1//sqrt(3))^(2))dy` `=(4pisqrt(3))/3[tan^(-1)(sqrt(3) tan y//2)]_(pi//3)^(2pi//3)` `=(4pi)/(sqrt(3))[tan^(-1)3-tan^(-1)1]` `=(4pi)/(sqrt(3))[tan^(-1)3-pi//4]` |
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| 293. |
Prove that`int_0^oo[n e^(-x)]dx=1n((n^n)/(n !)),w h e r en`is a natural number greater than 1 and [.] denotes the greatest integerfunction.. |
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Answer» `AAxe[0,oo),n e^(-x)epsilon(0,n]` If `0ltn e^(-x)1,xepsilon(In n,oo)`. If `1le n e^(-x)lt 2,xepsilon(In n//2, In n]` If `2le n e^(-x)lt3,xepsilon(In n//3, In n//2]` If `n-1le n e^(-x)ltn,xepsilon(0,In n/(n-1)]` `:. int_(0)^(oo) [n e^(-x)]dx=int_(0)^(In n/(n-1))(n-1)dx+int_(In n/(n-1))^(In n/(n-2))(n-2)dx` `+............+int_(In n/2)^(In n) 1dx+int_(In n)^(oo)0 dx` `=(n-1)(In n/(n-1))+(n-2)[In (n/(n-2))` `-In(n/(n-1))]+..............+1[In n-In-n/2]` `=In((n^(n))/(n!))`. |
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| 294. |
Evaluate `int_(0)^(1)(tx+1-x)^(n)dx`, where n is a positive integer and t is a parameter independent of x . Hence , show that |
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Answer» Correct Answer - `[(t^(n+1)-1)/((t-1)(n+1))]` Let ` I= int _(0)^(1) (tx =1-x)^(n) dx= int_(0)^(1){(t-1)x+1}^(n) dx` `=[(((t -1)x+1)^(n+1))/((n+1)(t-1))]_(0)^(1)=(1)/( n+1)((t^(n+1-1))/(t -1))` ` = (1)/(n+1)(1+t+t^(2)+ . . . + t^(n))` . . . (i) Again , `I= int_(0)^(1) ( tx +1-x)^(n) dx = int _(0)^(1) [ (1- x)+tx]^(n) dx` `=int _(0)^(1) [.^(n)C_(0)(1- x)^(n) + ^(n)C_(1)(1-x)^(n-1)(t x)` `=int_(0)^(1) [ sum_(r=0)^(n).^(n)C_(r)(1-x)^(n-r)( tx)^(r)]dx` ` = sum_(r =0)^(n) .^(n)C_(r)[int_(0)^(1)(1-x)^(n-r) .x^(r)dx]t^(r)` . . . (iii) From Eqs . (i) and (ii) , we get ` sum_(r =0)^(n) .^(n)C_(r)[int_(0)^(1)(1-x)^(n-r) .x^(r)dx]t^(r)=(1)/(n +1) (1+t+ . . . t^(n))` On equating coefficient of `t ^(k)` on both sides , we get ` .^(n)C_(k) [ int_(0)^(1) (1-x)^(n-k)*x^(k) dx ] =(1)/( n=1)` `rArr int_(0)^(1) (1- x)^(n-k)x^(k) dx = (1)/((n+1)^(n)C_(k)` |
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| 295. |
If `y = int_(4)^(4x^(2))t^(4)e^(4t)dt`, find `(d^(2)y)/(dx^(2))` |
| Answer» Correct Answer - `2048 e^(16x^(2))` | |
| 296. |
Consider the following statement : `S_(1)` : The value of `int_(0)^(2pi)cos^(-1)(cosx)` dx is `pi^(2)` `S_(2)` : Area enclosed by the curve `|x-2| + |y+1| = 1` is equal to `3 sq`. Unit `S_(3)` : If `d/dx f(x) = g(x)` for `a le x le b`, then `int_(a)^(b)f(x)g(x)dx` equals to `([f(b)]^(2)-[f(a)]^(2))/(2)`. `S_(4)` : Area of the region `R = {(x,y), x^(2) le y le x}` is `1/6` State, in order, whether `S_(1), S_(2), S_(3), S_(4)` are true or falseA. `"TFTT"`B. `"TTTT"`C. `"FFFF"`D. `"TFTF"` |
| Answer» Correct Answer - A | |
| 297. |
A continuous real function `f`satisfies`f(2x)=3(f(x)AAx in RdotIfint_0^1f(x)dx=1,`then find the value of `int_1^2f(x)dx` |
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Answer» Correct Answer - 5 We have `f(2x)=3f(x)`……………1 and `int_(0)^(1)f(X)dx=1` ……………..2 From equatons 1 and 2 `1/3int_(0)^(1)f(2x)dx=1` Put `2x=t`. Then `1/6int_(0)^(2)f(t)dt=1` or `int_(0)^(2)f(t)dt=6` or `int_(0)^(1)f(t)dt+int_(1)^(2)f(t)dt=6` Hence `int_(1)^(2)f(t)dt=6-int_(0)^(1)f(t)dt=6-1=5` |
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| 298. |
If `int_a^b(f(x)-3x)dx=a^2-b^2`then the value of `f(pi/6)`is ___ |
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Answer» `int_(a)^(b)(f(x)-3x)dx=a^(2)-b^(2)` `impliesint_(a)^(b)f(x)dx-3/2[x^(2)]_(a)^(b)=a^(2)-b^(2)` `impliesint_(a)^(b)f(x)dx-3/2[b^(2)-a^(2)]=a^(2)-b^(2)` `int_(a)^(b)f(x)dx=(b^(2))/2-(a^(2))/2` `impliesf(x)=x` `=f((pi)/6)=(pi)/6` |
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| 299. |
Prove that`int_0^x[t]dt=([x]([x]-1))/2+[x](x-[x]),`where [.] denotes the greatest integer function. |
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Answer» Correct Answer - NA Let `x=n+fAAn epsilon` and `0leflt1` `:.[x]=n` `int_(0)^(x)[t]dt=int_(0)^(1)[t]dt+int_(1)^(2)[t]dt+int_(2)^(3)[t]dt+………..+int_(n)^(n+f)[t]dt` `=0+1int_(1)^(2)dt+2int_(2)^(3)dt+………….+n int_(n)^(n+f)dt` `=(2-1)+2(3-2)+……..+n(n+f-n)` `=1+2+3..........+(n-1)+nf` `=((n-1)n)/2+nf` `=([x]([x]-1))/2+[x](x-[x])` [from equation 1] |
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| 300. |
Let `f: RvecR`be a function defined by`f(x)={[x],xlt=2 0,x >2`where `[x]`is the greatest integer less than or equal to `xdot`If`I=int_(-1)^2(xf(x^2))/(2+f(x+1))dx ,t h e nt h ev a l u eof(4I-1)i s` |
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Answer» Here , ` f (x) = {{:([x]_(,),xle2),(0_(,),xgt2):}` `:. I= int_(-1)^(2) ( xf (x^(2)))/(2+f (x+1))dx` `=int _(-1)^(0)(x f(x^(2)))/(2+f (x+1))dx + int _(0)^(1) (xf (x^(2)))/(2+f(x=1))dx` `+int_(1)^(sqrt(2))(x f (x^(2)))/(2+f(x+1))dx + int _(sqrt(2))^(sqrt(3))(xf (x^(2)))/(2+f(x+1))dx+ int_(sqrt(3))^(2)(xf (x^(2)))/(2+f(x+1))dx` `=int_(-1)^(0)dx+int_(0)^(1) 0 dx + int_(1)^(sqrt(2))(x*1)/(2+0)dx+int_(sqrt(2))^(sqrt(3))0 dx + int_(sqrt(3))^(2)0 dx` `[{:(because-1ltxltrArr0 ltx^(2)lt 1 rArr [x^(2)]=0_(,),),(0 lt x lt 1 rArr 0 lt x^(2) lt 1 rArr [x^(2)] = 0 _(,),),(1 ltx lt sqrt(2)rArr{{:(1ltx^(2)lt2 rArr [x^(2)]=1,),(2ltx+1lt1+sqrt(2)rArrf(x+1)=0_(,),):}:},),(sqrt(2)ltxltsqrt(3)rArr 2 lt x^(2)lt3 rArr f(x^(2))=0_(,),),(andsqrt(3)lt xlt 2 rArr 3 lt x^(2)lt 4 rArr f (x^(2))=0,):}]` `rArr I = int_(1)^(sqrt(2))(x)/(2)dx = [(x^(2))/(4)]_(1)^(sqrt(2))=(1)/(4)(2-1)=(1)/(4)` `:. 4I =1 rArr 4 I -1=0` |
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