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| 151. |
Compute the integrals:`int_0^oof(x^n+x^(-n))logx(dx)/(1+x^2)` |
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Answer» Let `I=int_(0)^(oo) fx^(n)+x^(-n)In x (dx)/(1+x^(2))` Let `t=1//x` or `x=1//t`. So `dx=-1/(t^(2))dt`. Also when `xto0,t to oo`, when `xto oo, t to 0`. Thus, `I=int_(0)^(oo) f(x^(n)+x^(-n)) In x (dx)/(1+x^(2))` `=int_(oo)^(0)f(t^(-n)+t^(n))In (1/t)(-(dt)/(t^(2)))/(1+1/(t^(2)))` `=-int_(0)^(oo) f(t^(n)+t^(-n))In (t)(dt)/(1+t^(2))` `=-I` or `2I=0` or `I=0` |
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| 152. |
Evaluate `int_(-1)^(1) log ((2-x)/( 2+x)) dx ` |
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Answer» Let `f(x)=log_(e)((2-x)/(2+x))dx rArr f(-x) = log_(e) ((2+x)/(2-x)) = - log_(e) ((2-x)/(2+x)) =-f(x)` i.e., `f(x)` is odd function `:. underset(-1)overset(1)intlog_(e)((2-x)/(2+x)) dx = 0` |
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| 153. |
Evaluate:`int_(pi//4)^(pi//4)log(sinx+cosx)dx` |
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Answer» Let `I=int_(-pi//4)^(pi//4)log{sqrt(2)sin(x+(pi)/4)}dx` Putting `x+(pi)/4=theta, dx=d theta`, we get `I=int_(0)^(pi//2)log(sqrt(2)sin theta)d theta` `=1/2int_(0)^(pi//2)log 2 d theta +int_(0)^(pi//2) log sin theta d theta` `=(1/4 pi log 2)-1/2 pi log 2` `-1/4 pi log 2` |
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| 154. |
Evaluate:`int_(-pi/2)^(pi/2)log{(a x^2+b x+c)/(a x^2-b x+c)(a+b)|sinx|}dx` |
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Answer» `I=int_(-pi//2)^(pi//2) log {(ax^(2)+bx+c)/(ax^(2)-bx+c)(a+b)|sinx|}dx` `=int_(-pi//2)^(pi//2) log((ax^(2)+bx+c)/(ax^(2)-bx+c))dx+int_(-pi//2)^(pi//2) log (a+b)dx` `+int_(-pi//2)^(pi//2) log|sinx|dx`……………1 `=I_(1)+I_(2)+I_(3)` Now let `f(x)=log((ax^(2)+bx+c)/(ax^(2)-bx+c))` or `f(-x)=log((ax^(2)-bx+c)/(ax^(2)+bx+c))` `=-f(x)` `:.I_(1)=int_(-pi//2)^(pi//2) f(x)dx=0` `I_(2)=log(a+b)[x]_(-pi//2)^(pi//2)` `=pi log(a+b)` `I_(3)=int_(-pi//2)^(pi//2) log|sinx|dx` `=2int_(0)^(pi//2)log|sinx|dx` `=2int_(0)^(pi//2)logsinx dx` `=2(-1/2pi log 2)` Hence, from 1 we have `I=0+pi log(a+b)-pi log 2` `=pi log{(a+b)//2}` |
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| 155. |
Evaluate:`int_(-1)^1(x^3+|x|+1)/(x^2+2|x|+1)dx` |
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Answer» Correct Answer - `2log_(e)2` `I=int_(-1)^(1)(x^(3)+|x|+1)/(x^(2)+2|x|+1)dx` `=int_(-1)^(1)(x^(3))/(x^(2)+2|x|+1)dx+int_(-1)^(1)(|x|+1)/(x^(2)+2|x|+1)dx` `=0+2int_(0)^(1)((|x|+1)/(|x|+1)^(2))dx=2int_(0)^(1)(dx)/(1+x)` `=2In(1+x)|_(0)^(1)=2In2` |
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| 156. |
Compute the integrals:`int_0^oof(x^n+x^(-n))logx(dx)/x` |
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Answer» Here limits (reciprocal) and type of functions (reciprocal terms are present i.e. `x` and `1//x`) suggest that we must substitute `1//t` for `x`. Let `t=1//x` or `x=1//t`. So `dx=-1/(t^(2))dt`. Also when `xto 0, t to oo` when `xto oo, t to 0`. Thus, `I=int_(0)^(oo) f(x^(n)+x^(-n)) In x(dx)/x` `=int_(oo)^(0)f(t^(-n)+t^(n))In(1/t)(-(dt)/(t^(2)))/(1/t)` `=-int_(0)^(oo) ft^(n)+t^(-n)In(t)(dt)/t` `=-1` or `2I=0` or `I=0` |
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| 157. |
Prove that `int_(0)^(x)e^(xt)e^(-t^(2))dt=e^(x^(2)//4)int_(0)^(x)e^(-t^(2)//4)dt`. |
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Answer» Let `I=int_(0)^(x)e^(xt)e^(-t^(2))dt` `=e^(x^(2)//4) int_(0)^(x)e^(-x^(2)//4)e^(xt)e^(-t^(2))dt` `=e^(x^(2)//4)int_(0)^(x)e^(-(x^(2)//4-tx+t^(2)))dt` `=e^(x^(2)//4)int_(0)^(x)e^(-x//2-t^(2))dt` The result clearly suggests that we have to substitute `y//2` for `x//2-t`. Then `dt=-dy//2`. Also when `t=0,y=x`, and when `t=x, y=-x`. Thus, `I=e^(x^(2)/4)int_(x)^(-x)e^(-y^(2)//4)(-dy//2)` `=(e^(x^(2)//4))/2int_(-x)^(x)e^(-y^(2)//4)dy` `(e^(x^(2)//4))/2 2int_(0)^(x)e^(-y^(2)//4)dy` [`e^(y^(2)//4)` is an even function] `=e^(x^(2)//4)int_(0)^(x)e^(-t^(2)//4)dt` |
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| 158. |
Evaluate the following: `int_(-1//2)^(1//2)cos x "log" (1-x)/(1+x)dx` |
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Answer» Correct Answer - `0` Since `cosx "log" (1-x)/(1+x)` is an odd function of `x`, `int_(-1//2)^(1//2)cos x "log"(1-x)/(1+x)dx=0` |
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| 159. |
Prove that `int_(0)^(2a)f(x)dx=int_(a)^(a)[f(a-x)+f(a+x)]dx` |
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Answer» `I=int_(0)^(a)[f(a-x)+f(a-x)]dx` `=int_(0)^(a)f(a-x)dx+int_(0)^(a)f(a+x)dx` `=int_(0)^(a)f(a-(a-x))dx+ubrace(int_(a)^(2a)f(t)dt)_(x+a=t)` `-int_(0)^(a)f(x)dx+ubrace(int_(a)^(2a)f(x)dx)_("Change of dummy variable")` `=int_(0)^(2a)f(x)dx` |
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| 160. |
Evaluate the following: `int_(-pi)^(pi)(1-x^(2))xsinx cos^(2)x dx` |
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Answer» Correct Answer - `0` Value `=0` since `(1-x^(2)) sinx cos^(2)x` is an odd function of `x`. |
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| 161. |
Find the value of`int_0^1{(sin^(-1)x)//x}dx` |
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Answer» Correct Answer - `(pi)/2 log_(e)2` `int_(0)^(1){(sin^(-1)x)//x}dx` `[(sin^(-1))(logx)]_(0)^(1)-int_(0)^(1)1/(sqrt(1-x^(2)))logxdx` `=0-lim_(xto0)(sin^(-1)x)/x(x log x)` `-int_(0)^(pi//2)1/(sqrt(1-sin^(2)theta))(log sin theta) cos theta d theta` `=-lim_(xto0) x log x -int_(0)^(pi//2) log sin theta d theta` `=(pi)/2 log 2` |
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| 162. |
`int_(-(3pi)/2)^(-pi/2) {(pi+x)^3+cos^2(x+3pi)}dx` is equal to (A) `pi/4-1` (B) `pi^4/32` (C) `pi^4/32+pi/2` (D) `pi/2` |
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Answer» Correct Answer - `pi//2` Let `I=int_(-3pi//2)^(-pi//2)[(x+pi)^(3)+cos^(2)(x+3pi)]dx` Put `x=pi=t`, so that `dx=dt` when `x=-(3pi)/2, t=-(pi)/2` When `x=-(pi)/2,t=(pi)/2` `:.I=int_(-pi//2)^(pi//2) [t^(3)+cos^(2)(t+2pi)]dt` `=int_(-pi//2)^(pi//2) [t^(3)+cos^(2)t]dt` `=0+2int_(0)^(pi//2) cos^(2) t dt =int_(0)^(pi//2)(1+cos2t)dt` `=(pi)/2+0=(pi)/2` |
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| 163. |
show that the sum of the two integrals `int_(-4)^(-5) e^((x+5)^2)dx+3int_(1/3)^(2/3) e^(9(x-2/3)^2)dx` is zero |
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Answer» `I=int_(-4)^(-5)e^((x+5)^(2))dx+3int_(1//3)^(2//3)e^(9(x-2/3)^(2))dx` `=int_(-4)^(-5)e^((x+5)^(2))dx+3int_(1//3)^(2//3)e^((3x-2)^(2))dx` `=I_(1)+I_(2)` Note that in both `I_(1)` and `I_(2)`, function has same form i.e. `e^(t^(2))`. Also `e^(t^(2))` is no integrable. Now in `I_(1)` let `x+t=y` and in `I_(2)`. Let `3x-2=-t`. Then `I=int_(0)^(0)e^(y^(2))dy+int_(1)^(0)(-dt)=0`. |
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| 164. |
`int_(pi)^((3pi)/2)(sin^(4)x+cos^(4)x)dx` |
| Answer» Correct Answer - `(3pi)/(8)` | |
| 165. |
Evaluate:`int_(-pi)^(3pi)log(s e ctheta-t a ntheta)dtheta` |
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Answer» Let `I=int_(-pi)^(3pi)log(sec theta-tan theta)d theta`…………..1 Using the property `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`, we get `I=int_(-pi)^(3pi)log[sec (2pi-theta)-tan(2pi-theta)]d theta` `int_(-pi)^(3pi) log[sec theta+tan theta]d theta`.................2 Adding 1 and 2 we get `2I=int_(-pi)^(3pi)log[(sec theta-tan theta )(sec theta+tan theta)]d theta` `=int_(-pi)^(3pi) log(1)d theta=int_(-pi)^(3pi)0, d theta=0` or `I=0` |
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| 166. |
Find the value of `ln(int_(0)^(1) e^(t^(2)+t)(2t^(2)+t+1)dt)` |
| Answer» Correct Answer - 2 | |
| 167. |
Evaluate:`int_(-pi)^pi(xsinx dx)/(e^x+1)` |
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Answer» Let `I=pi_(-pi)^(pi)(x sin x dx)/(e^(x)+1)`……………..1 Replacing `x` by `0-x` or `r-x`, we get `I=int_(-pi)^(pi)((-x)sin(-x)dx)/(e^(-x)+1)=int_(-pi)^(pi)(e^(x)xsin xdx)/(e^(x)+1)`…………….2 Adding 1 and 2 we get `2I=int_(-pi)^(pi) x sin x dx` or `I=int_(0)^(pi)xsinxdx` `int_(0)^(pi)(pi-x)sin(pi-x)dx=int_(0)^(pi)x sin x dx-I` or `I=pi` |
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| 168. |
Evaluate: `int_0^pi log(1+cosx)dx` |
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Answer» Correct Answer - `-pilog_(e)2` `I=int_(0)^(pi)log(1+cosx)dx=int_(0)^(pi)log(2"cos"^(2)x/2)dx` `=int_(0)^(pi)(log2+2"log cos"x/2)dx=pi log 2+2int_(0)^(pi)"log cos"x/2 dx` `=pi log 2+2xx2int_(0)^(pi//2) log cos t dt, ` where `t=x/2` and `dx=2dt` `pi log 2=4xx(-(pi)/2 log 2)` `=-pi log 2` |
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| 169. |
If `int_(1)^(0) (x)/(x+1+e^(x))dx` is equal to `-lnk` then find the value of `k`. |
| Answer» Correct Answer - 2 | |
| 170. |
Evaluate : `3int_(0)^(pi)(a^(2)sin^(2)x+b^(2)cos^(2)x)/(a^(4)sin^(2)x+b^(4)cos^(2)x)dx`, where `a^(2)+b^(2)=(3pi)/(4), a^(2) ne b^(2)` and `ab ne 0` |
| Answer» Correct Answer - 8 | |
| 171. |
`int_(pi//4)^(3pi//4)(dx)/(1+cosx)` is equal toA. 2B. `-2`C. `(1)/(2)`D. `-(1)/(2)` |
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Answer» Correct Answer - A Let `I=int_(pi//4)^(3pi//4)(dx)/(1+cosx)` . . . (i) `rArrI=int_(pi//4)^(3pi//4)(dx)/(1+cos(pi-x))` `I=int_(pi//4)^(3pi//4)(dx)/(1+cosx)` . . . (ii) On adding Eqs . (i) and (ii) , we get `2I=int_(pi//4)^(3pi//4)((1)/(1+cosx)+(1)/(1-cosx))`dx `rArr2I=int_(pi//4)^(3pi//4)((2)/(1-cos^(2)x))`dx `rArrI=int_(pi//4)^(3pi//4)"cosec"^(2)xdx=[-cotx]_(pi//4)^(3pi//4)` `=[-cot(3pi)/(4)+cot(pi)/(4)]=-(-1)+1=2` |
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| 172. |
Evaluate: `int_0^(pi/2) sin^2x/(sinx+cosx)dx` |
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Answer» Let `I=int(sin^(2)xdx)/(sinx+cosx)`………………….1 `:.I=int_(0)^(pi//2)(sin^(2)(1/2pi-x)dx)/("sin"(1/2pi-x)+"cos"(1/2pi-x))` `=int_(0)^(pi//2)(cos^(2)xdx)/(cosx+sinx)` ………..2 Now adding 1 and 2 we get `2I=int_(0)^(pi//2)((sin^(2)x+cos^(2)x)dx)/(sinx+cosx)` or `I=1/2int_(0)^(pi//2)(dx)/(sinx+cosx)` `=1/(2sqrt(2))int_(0)^(pi//2)(dx)/(sin(x+1/4pi))` `=(1/(2sqrt(2)))int_(0)^(pi//2)cosec(x+1/4)dx` `=(1/(2sqrt(2)))[log{cosec(x+1/4pi)-cot(x+1/4pi)}]_(0)^(pi//2)` `=(1//2sqrt(2))[log{cosec(1/2pi+1/4pi)-cot(1/2pi+1/4pi)}` `-log{cosec(1/4pi)-cot(1/4pi)}]` `=(1//2sqrt(2)[log{sec(1/4pi)+tan(1/4pi)}-log(sqrt(2)-1)]` `=(1//2sqrt(2))[log(sqrt(2)+1)-log(sqrt(2)-1)]` `=(1/(2sqrt(2)))log((sqrt(2)+1)/(sqrt(2)-1))` |
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| 173. |
If `f(x)=(sinx)/xAAx in (0,pi],`prove that`pi/2int_0^(pi/2)f(x)f(pi/2-x)dx=int_0^pif(x)dx` |
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Answer» `int_(0)^(pi)f(x)dx=int_(0)^(pi)(sin x)/x dx` Let `I=int_(0)^(pi//2)f(x)f((pi)/2-x)dx` `=int_(0)^(pi//2)(sinx)/x ("sin"((pi)/2-x))/((pi)/2-x)dx` `=int_(0)^(pi//2)(2sinx cosx)/(x(pi-2x)) dx` `=int_(0)^(pi//2)(sin2x)/(x(pi-2x))dx` Let `2x=t`. So `dt=2dx` `:. I=int_(0)^(pi)(sint)/(t/2(pi-t)^(2))=int_(0)^(pi)(sint)/(t(pi-t))dt` `=1/(pi) int_(0)^(pi) sin t(1/t+1/(pi-t))dt` `=1/(pi) int_(0)^(pi)(sint)/t dt+1/(pi) int_(0)^(pi)(sint)/((pi-t))dt` `=1/(pi) int_(0)^(pi) (sint)/t dt+1/(pi) int_(0)^(pi)(sin(pi-t))/(pi-(pi-t))dt` `=1/(pi) int_(0)^(pi)(sint)/t dt+1/(pi) int_(0)^(pi) (sint)/t dt=2/(pi) (sint)/t dt` or`(pi)/2 int_(0)^(pi//2)f(x)f(pi/2-x)dx=int_(0)^(pi)(sinx)/x dx` |
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| 174. |
If `[x]`denotes the greatest integer less than or equal to `x ,`then find the value of the integral `int_0^2x^2[x]dxdot` |
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Answer» `int_(0)^(2)x^(2)[x]dx=int_(0)^(1)x^(2)[x]dx+int_(1)^(2)x^(2)[x]dx` `=int_(0)^(1)x^(2)(0)dx+int_(1)^(2)x^(2)(1)dx` `=0+int_(1)^(2)x^(2)dx=|(x^(3))/3|_(1)^(2)` `=(8-1)/3=7/3` |
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| 175. |
Let `g(x)=int_(0)^(x)f(t)dt`, where `f` is such that `1/2lef(t)le1`, for `tepsilon[0,1]` and `0lef(t)le1/2`, for `tepsilon[1,2]`. Then prove that `1/2leg(2)le3/2`. |
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Answer» `g(x)=int_(0)^(x)f(t)dt` `:.g(2)=int_(0)^(2)f(t)dt=int_(0)^(1)f(t)dt+int_(1)^(2)f(t)dt` Now `1/2lef(t)le1` for `tepsilon[0,1]` `impliesint_(0)^(1)1/2dtleint_(0)^(1)f(t)dtleint_(0)^(1)1 dt` `implies 1/2 le int_(0)^(1)f(t)dtle1`……………….1 Also `0lef(t)le1/2` for `tepsilon[1,2]` `implies int_(1)^(2)0dt le int_(1)^(2)f(t)le int_(1)^(2)1/2dt` `implies 0 le int_(1)^(2) f(t) dt le 1/2`...............2 Adding 1 and 2 we get. `1/2 le int_(0)^(1)f(t)+int_(1)^(2)f(t)dtle3/2` `implies1/2le int_(0)^(2)f(t)dtle3/2` `implies1/2 le g(2)le3/2` |
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| 176. |
Prove that`int_0^1tan^(-1)(1/(1-x+x^2))dx=2int_0^1tan^(-1)x dxdot`Hence or otherwise, evaluate the integral`int_0^1tan^(-1)(1-x+x^2)dx` |
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Answer» `int_(0)^(1)"tan"^(-1)1/(1-x+x^(2))dx=int_(0)^(1)"tan"^(-1)(x+(1-x))/(1-x(1-x))dx` `=int_(0)^(1)[tan^(-1)x+tan^(-1)(1-x)]dx` `=int_(0)^(1)tan^(-1)xdx+int_(0)^(1)tan^(-1)(1-x)dx` `=int_(0)^(1)tan^(-1)xdx+int_(0)^(1)tan^(-1)tan^(-1)[1-(1-x)]dx` `=2int_(0)^(1)tan^(-1)x dx`................1 Now, `I=int_(0)^(1)tan^(-1)(1-x+x^(2))dx` `=int_(0)^(1)cot^(-1)(1/(1-x+x^(2)))dx` `=int_(0)^(1)[(pi)/2-tan^(-1)(1/(1-x+x^(2)))]dx` `=(pi)/2-2int_(0)^(1)x dx`[from equation 1] `=(pi)/2-2{x tan^(-1) x-1/2 log(1+x^(2))}_(0)^(1)` `=log_(e)2` |
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| 177. |
`int_(0)^(1)e^(2x)e^(e^(x) ` dx =)A. `e^(e)(2e-1)`B. `e^(e)(e-1)`C. `e^(2e)(e-1)`D. none of these |
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Answer» Correct Answer - B `I=int_(0)^(1)e^(2x)e^(e^(x))dx` Let `e^(x)=1` `rArr" "I=int_(0)^(e)te^(t)dt=(te^(t)-e^(t))_(1)^(e)=e^(e)(e-1)` |
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| 178. |
The number of solution of the equation `int_(-2)^(1)|cos x|dx=0,0ltxlt(pi)/(2)`, is |
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Answer» Correct Answer - A `int_(-2)^(x)|cosx|dx=int_(-2)^(-pi//2)|cos x|dx+int_(-pi//2)^(x)|cos x|dx` `" "=int_(-2)^(-pi//2)-cosxdx+int_(-pi//2)^(x)cosxdx` `" "1-sin 2+sinx+1` `therefore" "int_(-2)^(x)|cos x|dx=0` `rArr" "-1sin2+sinx+1=0` `rArr" "sinx=sin2-2lt-1` `therefore" No solution in "(0,(pi)/(2)).` |
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| 179. |
For every function f (x) which is twice differentiable , these will be good approximation of `int_(a)^(b)f(x)dx=((b-a)/(2)){f(a)+f(b)}`, for more acutare results for ` cin(a,b),F( c) = (c-a)/(2)[f(a)-f( c)]+(b-c)/(2)[f(b)-f( c)]` When ` c= (a+b)/(2)` `int_(a)^(b)f(x)dx=(b-a)/(4){f(a)+f (b)+2 f ( c) }dx` Good approximation of `int_(0)^(pi//2)sinx dx , `isA. `pi//4`B. `pi(sqrt(2)+1)//4`C. `pi(sqrt(2)+1)//8`D. `(pi)/(8)` |
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Answer» Correct Answer - C `int_(0)^(pi//2)sin x dx =((pi)/(2)-0)/(4)[sin +sin ((pi)/(2))+2 sin((0+(pi)/(2))/(2))]` `=(pi)/(8)(1+sqrt(2))` |
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| 180. |
If`Isum_(k=1)^(98)int_k^(k+1)(k+1)/(x(x+1))dx ,t h e n:``I(log)_e 99``I >(log)_e 99`(d) `I |
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Answer» Correct Answer - B::D `I=sum_(k=1)^(98)int_(k)^(k+1)((k+1))/(x(x+1))dx` Clearly , `Igtsum_(k=1)^(98)int_(k)^(k+1)((k+1))/((x+1)^(2))dx` `rArrIgtsum_(k=1)^(98)(k+1)int_(k)^(k+1)(1)/((x+1))dx` `rArrIgtsum_(k=1)^(98)(-(k+1))[(1)/(k+2)-(1)/(k+21)]rArrIgtsum_(k=1)^(98)(1)/(k+2)` `rArrIgt (1)/(3)+ ...+(1)/(100)gt(98)/(100)rArrIgt(49)/(50)` Also , `Igtsum_(k=1)^(98)int_(k)^(k+1)(k+1)/(x(k+1))dx=sum_(k=1)^(98)[log_(e)(k+1)-log_(e)k]` `Igtlog_(e)99` |
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| 181. |
Consider the statements : P : There exists some x IR such that f(x) + 2x = 2(1+x2) Q : There exists some x IR such that 2f(x) +1 = 2x(1+x) Then (A) both P and Q are true (B) P is true and Q is false (C) P is false and Q is true (D) both P and Q are false.A. 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 |
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Answer» Correct Answer - C Here , `f(x) + 2x = (1 - x)^(2) * sin ^(2) x + x^(2)+2x` . . . (i) where , P :` f (x) = 2x = 2 (1+x)^(2)` . . . (ii) `:. 2 (1+x^(2))=(1-x)^(2) sin^(2) x+x^(2)+2x` `rArr (1-x)^(2)sin^(2)x = x^(2)-2x+2` `rArr(1-x)^(2)sin^(2)x = (1-x)+1` `rArr (1 -x)^(2) cos^(2)=-1` which is never possible . `:.` P is false. Again , let Q : h (x) = 2 f (x) + 1 - 2x ( 1+ x) where , h (0) = 2 f (0) = 1 - 0 =1 h (1) = f (1) + 1 4 =- 3 , as h (0) h (1) `lt` 0 `rArr` h (x) must have a solution. `:.` Q is true. |
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| 182. |
Find the area of the region formed by `x^2+y^2-6x-4y+12 le 0`. `y le x` and `x le 5/2`. |
| Answer» Correct Answer - `(pi/6+1/8-(sqrt(3))/(8))` sq. units | |
| 183. |
Find the area bounded by the curves `x = |y^(2)-1|` and `y = x- 5` |
| Answer» Correct Answer - `109/6` sq. units | |
| 184. |
Find the area bounded by the curves `4 y = |4-x^(2)|, y = 7 - |x|` |
| Answer» Correct Answer - `32` sq. units | |
| 185. |
Find the area bounded by the curves `y=-x^2+6x-5,y=-x^2+4x-3,`and the straight line `y=3x-15a n d`lying right to `x=1.` |
| Answer» Correct Answer - `73/6` sq. units | |
| 186. |
The area of the figure bounded by right of the line `y = x + 1, y= cos x` and x-axis is :A. `1/2`B. `2/3`C. `5/6`D. `3/2` |
| Answer» Correct Answer - D | |
| 187. |
Let `I_(n) = int_(0)^(1)(1-x^(3))^(n)dx, (nin N)` thenA. `3n I_(n) = (3n-1)I_(n-1) AA in ge 2`B. `(3n-1)I_(n) = 3n I_(n-1) AA n ge 2`C. `(3n-1)I_(n)= (3n+1)I_(n-1) AA n ge 2`D. `(3n+1)I_(n)=3nI_(n-1) AA n ge 2` |
| Answer» Correct Answer - D | |
| 188. |
The area bounded by the x-axis and the curve `y = 4x - x^2 - 3` isA. `1/3`B. `2/3`C. `4/3`D. `8/3` |
| Answer» Correct Answer - C | |
| 189. |
The area bounded by `y = 2-|2-x| and y=3/|x|` isA. `(4+3ln3)/(2)`B. `(4-3ln3)/(2)`C. `3/2 + ln 3`D. `1/2+ln3` |
| Answer» Correct Answer - B | |
| 190. |
The area bounded by the curve `y^(2) = 4x` and the line `2x-3y+4=0` isA. `1/3`B. `2/3`C. `4/3`D. `5/3` |
| Answer» Correct Answer - A | |
| 191. |
Area of region bounded by `x=0, y=0, x=2, y=2, y=lnx` isA. `6-4ln2`B. `4 ln 2-2`C. `2 ln 2-4`D. `6 - 2 ln 2` |
| Answer» Correct Answer - A | |
| 192. |
If `int_(0)^(x^(2)(1+x))f(t)dt=x`, then the value of f(2) is.A. `1//2`B. `1//3`C. `1//4`D. `1//5` |
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Answer» Correct Answer - D Differentiating both sides w.r.t. x, then `f(x^(2)(1+x))xx(2x+3x^(2))=1` At `x=1rArr f(2)=(1)/(5)` |
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| 193. |
Let `f(x)=int_(0)^(x)(e^(t))/(t)dt(xgt0),` then `e^(-a)[f(x+1)-f(1+a)]=`A. `int_(0)^(x)=(e^(t))/((t+a))dt`B. `int_(1)^(x)(e^(t))/(t+a)dt`C. `e^(-a)int_(1+a)^(x+a)(e^(t))/(t)dt`D. `int_(0)^(x)(e^(t-a))/((t+a))dt` |
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Answer» Correct Answer - B::C `e^(-a)[f(x+a)-f(1+a)]` `" "=e^(-a)[int_(0)^(x+a)(e^(t).dt)/(t)-int_(0)^(1+a)(e^(t))/(t)dt]` `" "=e^(-a)[int_(0)^(x+a)(e^(t).dt)/(t)+int_(1+a)^(0)(e^(t))/(t)dt]` `" "=e^(-a)[int_(1+a)^(x+a)(d^t.dt)/(t)]` `" "=e^(-a)int_(1)^(x)(e^(y+a))/(y+a).dy" "("Put, t"=y+a, dt=dy)` `" "=int_(1)^(x)(e^(y).dy)/(t+a)` `" "=int_(1)^(x)(e^(t).dt)/(t+a)` |
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| 194. |
The value of `int_(-pi//2)^(pi//2)(sin^(2)x)/(1+2^(x))dx` isA. `pi//4`B. `pi//8`C. `pi//2`D. `4pi` |
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Answer» Correct Answer - A `I=int_(-pi//2)^(pi//2) (sin^(2)x)/(1+2^(x))dx`…………….i or `I=int_(-pi//2)^(pi//2) (sin^(2)(-x))/(1+2^(-x))dx` or `I=int_(-pi//2)^(pi//2) (2^(x).sin^(2)x)/(2^(x)+1)dx`……………ii Adding i and ii we get `2I=int_(-pi//2)^(pi//2) sin^(2)xdx=2int_(0)^(pi//2) sin^(2)xdx` `=int_(0)^(pi//2)(1-cos^(2)x)dx` `=[x-(sin2x)/2]_(0)^(pi//2)=(pi)/2` `:. I=(pi)/4` |
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| 195. |
The value of `int_0^1(x^4(1-x)^4)/(1+x^2) dx` isA. `22/7-pi`B. `2/105`C. `0`D. `71/15-(3pi)/2` |
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Answer» Correct Answer - A `int_(0)^(1)(x^(6)-4x^(5)+5x^(4)-4x^(2)+4-4/(1+x^(2)))dx` `=[(x^(7))/7-(2x^(6))/3+x^(5)-(4x^(3))/3+4x]_(0)^(1)-pi` `=1/7-2/3+1-4/3+4-pi=22/7-pi` |
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| 196. |
The valued of `int_(sqrt(In2))^(sqrt(In3)) (x sinx^(2))/(sinx^(2)+sin(In 6-x^(2)))dx` isA. `1/4 In 3/2`B. `1/2 In 3/2`C. `In 3/2`D. `1/6 In 3/2` |
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Answer» Correct Answer - A Put `x^(2)=t` and `2xdx=dt` `:. I=1/2int_(In2)^(In3)(sint)/(sint+sin(In6-t)dt` `=1/2int_(In2)^(In3)(sin(In6-t))/(sin(In6-t)+sint)dt` `:.2I=1/2int_(In2)^(In3)1dt` or `I=1/4 In 3/2` |
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| 197. |
If `y=int_(x^(2))^(x^(3))1/(logt)dt(xgt0)`, then find `(dy)/(dx)` |
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Answer» `y=int_(x^(2))^(x^(3))1/(logt)dt` `:. (dy)/(dx)=1/(logx^(3)) d/(dx)(x^(3))-1/(logx^(2))d/(dx)(x^(2))` `=(3x^(2))/(3logx)-(2x)/(2logx)` `=(x^(2)-x)/(logx)` |
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| 198. |
If `A(x+y)=A(x)A(y) and A(0) ne 0 and B(x)=(A(x))/(1+(A(x))^(2))`, thenA. `int_(-2010)^(2010)B(x)dx=int_(0)^(2011)B(x)dx`B. `int_(-2010)^(2011)B(x)dx-int_(0)^(2010)B(x)dx=int_(0)^(2011)B(x)dx`C. `int_(-2010)^(2011)B(x)dx=0`D. `int_(-2010)^(2010)(2B(-x)-B(x))dx=2int_(0)^(2010)B(x)dx` |
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Answer» Correct Answer - B::D `A(x+y)=A(x)A(y)` `rArr" "A(0+0)=A(0)A(0)` `rArr" "A(0)=1` Put `y=-x,` we get `A(0)=A(x)A(-x)" (i)"` `B(-x)=(A(-x))/(1+(A(-x))^(2))` `=((1)/(A(x)))/(1+(1)/((A(x))^(2)))` `=(A(x))/(1+(A(x))^(2))` = B(x) Thus, B(x) is even. `int_(-2010)^(2011)B(x)dx=int_(-2010)^(2010)B(x)dx+int_(2010)^(2011)B(x)dx` `=2int_(0)^(2010)B(x)dx+int_(2010)^(2011)B(x)dx` `=int_(0)^(2010)B(x)dx+int_(0)^(2011)B(x)dx` |
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| 199. |
`int_(pi//4)^(3pi//4)(dx)/(1+cosx)` is equal toA. `-2`B. 2C. 4D. `-1` |
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Answer» Correct Answer - B Let `I=int_(pi//4)^(3pi//4)(dx)/(1+cosx)=int_(pi//4)^(3pi//4)(1-cosx)/(1-cos^(2)x)dx` `= int_(pi//4)^(3pi//4)(1-cosx)/(sin^(2)x)dx` `=int_(pi//4)^(3pi//4)("cosec"^(2)x-"cosec" x cot x)dx` `=[-cotx+"cosec"x]_(pi//4)^(3pi//4)` `[(1+sqrt(2))-(-1+sqrt(2))]=2` |
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| 200. |
`int_(0)^(1) sqrt((1-x)/(1+x)) dx=?`A. `(pi)/(2)+1`B. `(pi)/(2)-1`C. `-1`D. 1 |
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Answer» Correct Answer - B `I=int_(0)^(1)sqrt((1-x)/(1+x))dx=int_(0)^(1)(1-x)/(sqrt(1+x^(2)))` `=int_(0)^(1)(1)/(sqrt(1-x^(2)))dx-int_(0)^(1)(x)/(sqrt(1-x^(2)))dx` `=[sin^(-1)x]_(0)^(1)+int_(0)^(1)(t)/(t)dt` where , `t^(2)=1-x^(2)rArrtdt =-x dx]` `=(sin^(-1)-sin^(-1)0)+[t]_(1)^(0)=pi//2-1` |
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