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| 351. |
If `int_(-1)^(3//2)|xsinpix|dx = (k)/(pi^(2))`, then the value of k is :A. `3pi+1`B. `2pi+1`C. `1`D. `4` |
| Answer» Correct Answer - A | |
| 352. |
`f(x)=int_0^xf(t)dt=x+int_x^1tf(t)dt ,t h e nt h ev a l u eoff(1)i s``1/2`(b)0 (c) 1(d) `-1/2`A. `(1)/(2)`B. 0C. 1D. `-(1)/(2)` |
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Answer» Correct Answer - A Given , `int_(0)^(x) f (t) = x + int_(x)^(1)t f (t) dt` On differentiating both sides W . R. t. x , we get `f(x)1 = 1x f (x) *1 rArr (1+x)f(x)=1` `rArr f(x)=(1)/(1 +x)rArr f (1) =(1)/( 1 + 1)=(1)/(2)` |
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| 353. |
The value of the integral `int_(0)^(1)(dx)/(x^(2)+2xcosalpha+1)`, where `0 lt alpha lt (pi)/(2)`, is equal to :A. `sin alpha`B. `alpha sin alpha`C. `(alpha)/(2sin alpha)`D. `(alpha)/(2) sin alpha` |
| Answer» Correct Answer - C | |
| 354. |
Find the value of `a(0 lt a lt 1)` for which the following definite integral is minimized. `int_(0)^(pi) |sinx-ax| dx` |
| Answer» Correct Answer - `a = (sqrt(2))/(pi) sin ((pi)/(sqrt(2)))` | |
| 355. |
Evaluate `int_(0)^(pi)(x dx)/(1+cos alpha sin x)`,where `0lt alpha lt pi`. |
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Answer» Correct Answer - `(pi alpha)/(sin alpha)` Let `I=int_(0)^(pi)(xdx)/(1+cos alpha sinx )`…………..1 `=int_(0)^(pi)((pi-x)dx)/(1+cos alpha (sin (pi-x)))` [using `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]` `:.I=int_(0)^(pi)((pi-x)dx)/(1+cos alpha sinx)`…………2 Adding 1 and 2 we get `2I=int_(0)^(pi)(x+pi-x)/(1+cos alpha sinx) dx` `=int_(0)^(pi)(pi)/(1+cos alpha sin x)dx` `:.I=(pi)/2 int_(0)^(pi)1/(1+cos alpha sin x)dx` `=(pi)/2 int_(0)^(pi)1/(1+cos alpha xx(2tanx//2)/(1+tan^(2)x//2) dx` `=(pi)/2 int_(0)^(pi)(sec^(2)x//2)/(1+tan^(2)x//2+2cos alpha tan x//2)` Put `tanx//2=t` or `1/2"sec^(2)x/2dx=dt` Also, when `xto 0, t to 0` and when `x to pi,t to oo` `:.I=(pi)/2 int_(0)^(oo) (2dt)/(t^(2)+(2cos alpha)t+1)` `=piint_(0)^(oo) (dt)/((t+cos alpha)^(2)+1-cos^(2)alpha` `=pi . 1/(sin alpha)[tan^(-1)((t+cos alpha)/(sin alpha))]_(0)^(oo)` `=(pi)/(sin alpha)[tan^(-1)oo-tan^(-)(cot alpha)]` `=(pi)/(sin alpha)[(pi)/2-tan^(-1)(cot alpha)]` `=(pi)/(sin alpha) [cot^(-1)(cot alpha)]` `=(pi alpha)/(sin alpha)` |
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| 356. |
`f(x)`is continuous function for all real values of `x`and satisfies`int_0^xf(t)dt=int_x^1t^2f(t)dt+(x^(16))/8+(x^6)/3+adot`Then the value of `a`is equal to:`-1/(24)`(b) `(17)/(168)`(c) `1/7`(d) `-(167)/(840)`A. `-1/24`B. `17/168`C. `1/7`D. `-167/840` |
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Answer» Correct Answer - D `int_(0)^(x)f(t)dt=int_(x)^(1)t^(2)f(t)dt+(x^(16))/8+(x^(6))/3+a`……………..1 For `x=1, int_(0)^(1)f(t)dt=01/8+1/3+a=11/24+a`……………2 Differentiating both sides of w.r.t`x` We get `f(x)=0-x^(2)f(x)+2x^(15)+2x^(5)` `impliesf(x)=(2(x^(15)+x^(4)))/(1+x^(2))` `implies2int_(0)^(1)(x^(15)+x^(5))/(1+x^(2))dx+11/24+a`(from a) `implies2int_(0)^(1)(x^(13)-x^(11)+x^(9)-x^(7)+x^(5))dx=11/24+a` `implies 2(1/14-1/12+1/10-1/8+1/6)=11/24+a` `implies a=-167/840` |
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| 357. |
For `x >0,l e tf(x)=int_1^x(logt)/(1+t)dtdot`Find the function `f(x)+f(1/x)`and find the value of `f(e)+f(1/e)dot` |
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Answer» Let `F(x)=(x)+f(1/x)` `=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(1//x)(logt)/(1+t)dt` In second integral let `t=1//y`. So `dt=-1/(y^(2))dy` `:.F(x)=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(x)(-logy)/(1+1/y)(-(dy)/(y^(2)))` `=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(x)(logy)/(y(1+y))dy` `=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(x)(logt)/(t(1+t))dt` `=int_(1)^(x)(logt)/t dt=1/2(logx)^(2)` `:.F(e)=1/2` |
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| 358. |
The value of integral `int_(a)^(b)(|x|)/(x)dx`, `a lt b` is :A. `b -a` if `a gt 0`B. `a-b` if `b lt 0`C. `b +a` if `a lt 0 lt b`D. `|b | - |a|` |
| Answer» Correct Answer - A::B::C::D | |
| 359. |
If `int_(0)^(x)[x]dx=int_(0)^(|x|)x dx,` (where [.] and {.} denotes the greatest integer and fractional part respectively), thenA. `x epsilon[0,1)`B. `{x}=1//2`C. `{x}=1//3`D. `xgt0` |
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Answer» Correct Answer - A::B `int_(0)^(x)[x]dx=int_(0)^(1)0dx+int_(1)^(2)1dx+int_(2)^(3)2dx+…………….` `+int_([x]-1)^([x])([x]-1)dx+int_([x])^(x)[x]dx` `=0+1+2+3+……….+([x]-1)+[x](x-[x])` `=(([x]-1)[x])/2+[x]{x}`…………..1 Now `int_(0)^([x]) dx=[(x^(2))/2]_(0)^([x])=([x]^(2))/2`..............2 Comparing 1 and 2 we get `([x]-1)[x])/2+[x]{x}=([x]^(2))/2` `:.[x]^(2)-[x]+2[x]{x}=[x]^(2)` `=[x]=0` or `{x}=1//2` |
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| 360. |
The value of the integral `int_(0)^(1)(dx)/(x^(2)+2x cos alpha +1),0ltalpha lt pi` isA. `sin alpha`B. `alpha sin alpha`C. `(alpha)/(sin alpha)`D. `(alpha)/2 sin alpha` |
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Answer» Correct Answer - C Given integral `=int_(0)^(1) (dx)/((x+cos alpha)^(2)+(1-cos^(2)alpha))` `=int_(0)^(1)(dx)/((x+cos alpha)^(2)+sin^(2) alpha)` `=1/(sin alpha)| "tan"^(-1)(x+cos alpha)/(sin alpha)|_(0)^(1)` `=1/(sin alpha) ["tan"^(-1)(1+cos alpha)/(sin alpha) -"tan"^(-1) (cos alpha)/(sin alpha)]` `=1/(sin alpha) [tan^(-1) "cot"(alpha)/2-"tan"^(-1)(cot alpha)]` `=1/( sin alpha) [tan^(-1)"tan"((pi)/2-(alpha)/2)-tan^(-1)"tan"((pi)/2-alpha)]` `=1/(sin alpha) [((pi)/2-(alpha)/2)-((pi)/2-alpha)]` `=(alpha)/(2 sin alpha)` |
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| 361. |
`Ifx=int_c^(sint)sin^(-1)z dz ,y=int_k^(sqrt(t))(sinz^2)/zdz ,t h e n(dy)/(dx)i se q u a lto``(tant)/(2t)`(b) `(tant)/(t^2)``t/(2t^2)`(d) `(t a tt^2)/(2t^2)`A. `(tant)/(2t)`B. `(tant)/(t^(2))`C. `(tan t)/(2t^(2))`D. `(tan t^(2))/(2t^(2))` |
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Answer» Correct Answer - C `(dx)/(dt)=sin^(-1)(sint)cost=tcost` and `(dy)/(dt)=(sint)/(sqrt(t)) . 1/(2sqrt(t))= (sin t)/(2t)` or `(dy)/(dx)= (sint)/(2t . t cost)=(tant)/(2t^(2))` |
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| 362. |
Let `f`be a continuous function on `[a ,b]dot`Prove that there exists a number `x in [a , b]`such that`int_a^xf(t)dx=int_x^bf(t)dtdot` |
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Answer» Let `g(x)=int_(a)^(x)f(t)dt-int_(x)^(b)f(t)dt, xepsilon[a,b]` We have `g(a)=int_(a)^(b)f(t)dt` and `g(b)=int_(a)^(b)f(t)dt` `:.g(a).g(b)=-[int_(a)^(b)f(t)dt]^(2)le0` clearly `g(x)` is continuous in `[a,b]` an `g(a).g(b)le0` It implies that `g(x)` will become zero at least once in `[a,b]`. Hence `int_(a)^(x)f(t)dt=int_(x)^(b)f(t)dt` for all least one value of `xepsilon[a,b]` |
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| 363. |
Evaluate: `int_(-1)^(3//2)|xsinpix|dx` |
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Answer» `|x sin pi x|={((-x)(-sinpix),"if",-1lexlt0),(xsinpix,"if",0lexle1),(x(-sinpix),"if",1ltxle3//2):}` `:.int_(-1)^(3//2)|x sin pi x|dx` `=int_(-1)^(0)sin pix dx +int_(-1)^(1) x sin dx +int_(1)^(3//2)(-x sin pi x)dx` `=int_(-1)^(1)x sin pi x dx - int_(1)^(3//2), x sin pi x dx` `=2int_(0)^(1)x sin x dx-int_(1)^(3//2)xsin pix dx` `=2[{x((-1)/(pi))cospix}_(0)^(1)-int_(0)^(1)1((-1)/(pi))cospix dx]` `-{x((-1)/(pi))cospix}_(1)^(3//2)+int_(1)^(3//2)1((-1)/(pi))cosx dx` `=(2/(pi))+(2/(pi^(2)))[sin pix]_(0)^(1)+{3/((2pi))}"cos"3/2pi+(1/(pi))` `-1/(pi^(2))[sinpix]_(1)^(3//2)` `=(2//pi)+0+0+(1//pi)+(1//pi^(2))` `=(3pi+1)//pi^(2)` |
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| 364. |
Which of the following statement(s) is/are TRUE?A. If function `y=f(x)` is continuous at `x=c` such that `f(c)!=0`, then `f(x)f(c)gt0AA x epsilon(c-h,c+h)`, where `h` is sufficiently small positive quantity.B. `lim_(n to oo) 1/n (n (1+1/n)(1+2/n)……(1+n/n))=1+2In2`.C. Let `f` be a continuous and non-negative function defined on`[a,b]` If `int_(a)^(b)f(x)dx=0`, then `f(x)=AA x epsilon [a,b]`D. Let `f` be continuous function defined on `[a,b]` such that `int_(a)^(b)f(x)dx=0`.Then there exists at least one `c epsilon(a,b)` for which `f(c)=0` |
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Answer» Correct Answer - A::C::D (1) The expression `f(x)f(c)AAx epsilon(c-h,c+h)` where `hto0^(+)` is equivalent to `lim_(xtoc)f(x)f(c)`which is equal to `f(c))^(2)` because `f(x)` is continuous. There `f(x)f(c)gt0AA(c-h,c+h)` where `hto0^(+)`. (2) We have `I=lim_(nto oo) 1n In[(1+1/n)(1+2/n)........(1+n/n)]` `=lim_(nto oo) 1/n In prod_(k=1)^(n)(1+k/n)` `=lim_(nto oo) 1/n sum_(k=1)^(n)In(1+k/n)` `=int_(0)^(1)log_(e)(1+x)dx` `=int_(1)^(2)Inx dx=[(Inx-1)]_(1)^(2)=-1+2In2` ltbgt (3) Given `f(x)gt0` or `int_(a)^(b)f(x)dxge0`. But given `int_(a)^(b)f(x)dx=0`. So, this can be true only when `f(x)=0` (4) `int_(a)^(b)f(x)dx+0` i.e. `y=f(x)` cuts x-axis at least once. |
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| 365. |
If `f(y)=e^y,g(y)=y,y>0, and F(t)=int_0^t f(t-y)g(y) dy`, thenA. `F(t)=e^(t)-(1+t)`B. `F(t)=te^(t)`C. `F(t)=te^(-t)`D. `F(t)=1-e^(t)(1+t)` |
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Answer» Correct Answer - A We have `f(y)=e^(y),g(y)=y,ygt0` `F(t)=int_(0)^(t)f(t-y)g(y)dy` `=int_(0)^(t)e^(t-y) y dy` `=e^(t)int_(0)^(t) e^(-y) y dy` `=e^(t) ([-ye^(-y)]_(0)^(t)+int_(0)^(t) e^(-y) dy)` `=e^(t) (-te^(-t) -[e^(-y)]_(0)^(t))` `=e^(t) (-te^(-t)-e^(-1) +1)` `=e^(t)-(1+t)` |
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| 366. |
If `f(n)=int_(0)^(2015)(e^(x))/(1+x^(n))dx`, then find the value of `lim_(nto oo)f(n)` |
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Answer» `lim_(nto oo) int_(0)^(2015) (e^(x))/(1+x^(n))dx` `=int_(0)^(1) lim_(n to oo)((e^(x))/(1+x^(n)))dx+int_(1)^(2015)lim_(nto oo) ((e^(x))/(1+x^(n)))dx` `=int_(0)^(1)e^(x)dx+0=e-1` |
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| 367. |
`Ifx=int_0^y(dt)/(sqrt(1+9t^2))a n d(d^2y)/(dx^2)=a y ,t h e nfin da` |
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Answer» `x=int_(0)^(y)(dt)/(sqrt(1+9t^(2)))` Differentiating w.r.t `y`, we get `(dx)/(dy)=1/(sqrt(1+9y^(2)))` or `(dy)/(dx)=sqrt(1+9y^(2))` or `d/(dx)((dy)/(dx))=d/(dy)(sqrt(1+9y^(2)))(dy)/(dx)` or `(d^(2)y)/(dx^(2))=(18y)/(2sqrt(1+9y^(2)))sqrt(1+9y^(2))=9y` or `a=9` |
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| 368. |
`f(x)=int_1^x(tan^(-1)(t))/t dtAAx in R^+,t h e nfin dt h ev a l u eof``f(e^2)-f(1/(e^2))` |
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Answer» Correct Answer - `pi` `f(x)=int_(1)^(x)(tan^(-1)(t))/t dt` `:.f(1/x)=int_(1)^(1//x)(tan^(-1)(t))/tdt` Put `=t=1//u` `:.dt=-(du)/(u^(2))` `:.f(1//x)=int_(1)^(x)("tan"^(-1)(1/u))/(1/u)(-1/(u^(2)))du` `=-int_(1)^(x)("tan"^(-1)(1/u))/u du` `=-int_(1)^(x)("cot"^(-1)(u))/u du` `=-int_(1)^(x)(cot^(-1)(t))/t dt` Now` f(x)-f(1//x)=int_(1)^(x)(tan^(-1)t+cot^(-1)t)/t dt` `=int_(1)^(x) (pi)/2xx1/t dt` `=(pi)/2 log (x)` `:. f(e^(2))-f(1//e^(2))=(pi)/2log_(e)e^(2)=pi` |
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| 369. |
Show that`int_a^b(|x|)/x dx=|b|=|a|dot` |
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Answer» Case I: If `0lealtb`, then `|x|//x=1` `:.I=int_(a)^(b)1dx=b-a=|b|-|a|` Case II: If `altble0`, then `|x|=-x` `:. I=int_(a)^(b)(-x)/xdx=int_(a)^(b)(-1)dx` `=[-x]_(a)^(b)=b-(-a)=|b|-|a|` Case III: If `alt0ltb` then `|x|=-x` when `altxlt0` and `|x|=x`, when `altxltb` `:.I=int_(a)^(|x|)/xdx=int_(a)^(0)(|x|)/x dx+int_(0)^(b)(|x|)/x dx` `=int_(a)^(0)(-x)/x dx+int_(0)^(b) x/x dx` `=int_(a)^(0)(-1)dx+int_(0)^(b) 1 dx` `=[-x]_(a)^(0)+[x]_(0)^(b)=a+b=b-(-a)=|b|-|a|` Hence, in all the cases `I=int_(a)^(b)(|x|)/x dx=|b|-|a|`. |
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| 370. |
`int_0^oo(dx)/([x+sqrt(x^2+1)]^3)i se q u a lto``3/8`(b) `1/8`(c) `-3/8`(d) none of theseA. `3/8`B. `1/8`C. `-3/8`D. none of these |
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Answer» Correct Answer - A Putting `x=tan theta` we get , `int_(0)^(pi//2) (dx)/([x+sqrt(x^(2)+1]^(3)))=int_(0)^(pi/2) (sec^(2) theta d theta)/((tan theta +sec theta)^(3))` `=int_(0)^(pi//2) (cos theta)/((1+sin theta)^(3))d theta` `=[-1/(2(1+sin theta)^(2))]_(0)^(pi//2)=-1/8+1/2=3/8` |
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| 371. |
`int_0^x{int_0^uf(t)dx}d ui se q u a lto``int_0^x(x-u)f(u)d u``int_0^x uf(x-u)d u``xint_0^xf(u)d u`(d) `xint_0^x uf(u-x)d u`A. `int_(0)^(x)(x-u)f(u)du`B. `int_(0)^(x) uf(x-u)du`C. `x int_(0)^(x)f(u)du`D. `x int_(0)^(x)uf(u-x)du` |
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Answer» Correct Answer - A::B L.H.S `=int_(0)^(x) {int_(0)^(u)f(t)dt}du` Integrating by parts choose 1 as the second function. Then, L.H.S`={uint_(0)^(u)f(t)dt}_(0)^(x)-int_(0)^(x)f(u)u du` `=x int_(0)^(x)f(t)dt-int_(0)^(x)f(u)u du` `=x int_(0)^(x)f(u)du-int_(0)^(x)f(u) udu-int_(0)^(x)f(u)(x-u)du` `=R.H.S`. |
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| 372. |
`Iff(x)=x+int_0^1t(x+t)f(t)dt ,``t h e nfin dt h ev a l u eoft h ed efin i t ein t egr a lint_0^1f(x)dxdot` |
| Answer» Correct Answer - 65 | |
| 373. |
`L e tA=int_0^oo(logx)/(1+x^3)dxdotT h e nfin dt h ev a l u eofint_0^oo(xlogx)/(1+x^3)dx`in terms of `Adot` |
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Answer» `B=int_(0)^(oo)(x log x)/(1+x^(3))dx` `=int_(0)^(oo) ((x+1)logx-log)/(1+x^(3))dx` `=int_(0)^(oo) (logx)/(x^(2)-x+1)dx-A` let `I=int_(0)^(oo) (logx)/(x^(2)-x+1) dx` Put `x=1/t` `:. I=int_(oo)^(0) ("log"1/t)/(-1/(t^(2))-1/t+1)((-dt)/(t^(2)))` `=-int_(0)^(oo) (logt)/(t^(2)-t+1)dt` `=-I` or `2I=0` or `I=0` `:.B=-A` |
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| 374. |
`int_0^oo(pi/(1+pi^2x^2)-1/(1+x^2))logxdxi se q u a lto``-pi/21npi`(b) 0`pi/21n2`(d) none of theseA. `-(pi)/2 In pi`B. `0`C. `(pi)/2 In 2`D. none of these |
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Answer» Correct Answer - A `int_(0)^(oo) ((pi)/(1+pi^(2)x^(2))-1/(1+x^(2)))log x dx` `=int_(0)^(oo) ("log"(y/(pi))dy)/(1+y^(2))-int_(0)^(oo) (logx)/(1+x^(2))dx` `=-int_(0)^(oo) (log pi)/(1+y^(2))dy=-(pi)/2 In pi` |
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| 375. |
Evaluate`int_a^b(dx)/(sqrt(x)),w h e r ea , b > 0.` |
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Answer» `I=int_(a)^(b)(dx)/(sqrt(x))` where `a,bgt0` `=h lim_(nto oo) [1/(sqrt(a))+1/(sqrt(a+h))+1/(sqrt(a+2h))+…………..+1/(sqrt(a+(n-1)h))]` We know that `sqrt(r)+sqrt(r-h) lt 2sqrt(r)ltsqrt(r+h)+sqrt(r)` (for sufficiently small `hgt0`). Thus `1/(sqrt(r+h)+sqrt(r))lt1/(2sqrt(r)+sqrt(r))` or `(sqrt(r+h)-sqrt(r))/hlt1/(2sqrt(r))lt(sqrt(r)-sqrt(r-h))/h` Let put `r=a,a+h,a+2h,.........,a+(n-1)h` `:.(sqrt(a+h)-sqrt(a))/hlt 1/(2sqrt(a)) lt (sqrt(a)-sqrt(a-h))/h` `(sqrt(a+2h)-sqrt(a+h))/h lt 1/(2sqrt(a+h)) lt (sqrt(a+h)-sqrt(a))/h` `(sqrt(a+3h)-sqrt(a+2h))/h lt 1/(2sqrt(a+2h)) lt (sqrt(a+2h)-sqrt(a+h))/h` `(sqrt(a+nh)-sqrt(a+(n-1)h))/h lt 1/(2sqrt(a+(n-1)h))` `lt(sqrt(a+(n-1)h)-sqrt(a+(n-2)h))/h` Adding we get `(sqrt(a+nh)-sqrt(a))/h lt sum_(r=0)^(n-1)1/(2sqrt(a+rh)) lt (sqrt(a+(n-1)h)-sqrt(a-h))/h` or `2(sqrt(a+b-a)-sqrt(a))lth sum_(r=0)^(n-1)1/(sqrt(a+rh))` `lt 2(sqrt(a+b-a-h)-sqrt(a-h))` (Put `nh=b-a`) or `lim_(hto0) 2(sqrt(a+b-a)-sqrt(a))lt lim_(hto0)h sum_(r=0)^(n-1)1/(sqrt(a+rh))` `lt lim_(hto0)2(sqrt(a+b-ah)-sqrt(a-h))` or `2(sqrt(b)-sqrt(a)) lt lim_(hto0) h sum_(r=0)^(n-1)1/(sqrt(a+rh))lt 2(sqrt(b)-sqrt(a))` or `(sqrt(b)-sqrt(a))lt int_(a)^(b)1/(sqrt(x)) dx lt 2(sqrt(b)-sqrt(a))` or `int_(a)^(b)1/(sqrt(x)) dx=2(sqrt(b)-sqrt(a))` |
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| 376. |
`IfA_n=int_0^(pi/2)(sin(2n-1)x)/(sinx)dx ,b_n=int_0^(pi/2)((sinn x)/(sinx))^2dxforn in N ,`Then`A_(n+1)=A_n`(b) `B_(n+1)=B_n``A_(n+1)-A_n=B_(n+1)`(d) `B_(n+1)-B_n=A_(n+1)`A. `A_(n+1)=A_(n)`B. `B_(n+1)=B_(n)`C. `A_(n+1)-A_(n)=B_(n+1)`D. `B_(n+1)-B_(n)=A_(n+1)` |
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Answer» Correct Answer - A::D `A_(n+1)-A_(n)=int_(0)^(pi//2)(sin(2n+1)xsin(2n-1)x)/(sinx) dx` `=int_(0)^(pi//2)2 cos 2nx dx=0` or `A_(n+1)=A_(n)` `B_(n+1)-B_(n)=int_(0)^(pi//2)(sin^(2)(n+1)x-sin^(2)nx)/(sin^(2)x)dx` `=int_(0)^(pi//2) (sin(2n+1)x)/(sinx)dx=A_(n+1)` |
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| 377. |
The value of `int_0^1e^x^(2-x)dx`is`1`(c) `> e^(-1/4)`(d) `A. `lt1`B. `gt1`C. `gte^(-1/4)`D. `lte^(-1/4)` |
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Answer» Correct Answer - A::C `int_(0)^(1)e^(x^(2)-x)dx` For `x epsilon(0,1),x^(2)-x epsilon(-1//4,0)` `:.e^(-1//4)lt e^(x^(2)-x)lte^(0)` or `e^(-1/4)lt int_(0)^(1)e^(x^(2)-x)dxlt1` |
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| 378. |
`T h ev a l u eofint_0^oo(dx)/(1+x^4)i s``s a m ea st h a tofint_0^oo(x^2+1dx)/(1+x)``pi/(2sqrt(2))``s a m ea st h a tofint_0^oo(x^2+1dx)/(1+x^4)`(d) `pi/(sqrt(2))`A. same as that of `int_(0)^(oo) (x^(2)+1dx)/(1+x^(4))`B. `(pi)/(2sqrt(2))`C. same as that of `int_(0)^(oo) (x^(2)dx)/(1+x^(4))`D. `(pi)/(sqrt(2))` |
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Answer» Correct Answer - B::C `I=int_(0)^(oo) (dx)/(1+x^(4))` ……………1 `=int_(0)^(oo) (x^(2)+1-x^(2))/(1+x^(4))dx` `=int_(0)^(oo)(x^(2))/(1+x^(4))dx+int_(0)^(oo)(1-x^(2))/(1+x^(4))dx=I_(1)+I_(2)` `I_(2)=int_(0)^(oo) (1/(x^(2))-1)/(1/(x^(2))+x^(2))dx` Put `x+1/x=y` `:.I_(2)=int_(oo)^(oo) (-1)/(y^(2)-2)dy=0` `:.I=int_(0)^(oo) (dx)/(1+x^(4))=int_(0)^(oo)(x^(2)dx)/(1+x^(4))` Adding 1 and 2 we get `2I=int_(0)^(oo) (1+x^(2)dx)/(1+x^(4))=int_(0)^(oo) (1/(x^(2))+1)/(1/(x^(2))+x^(2))dx` (Putting `x=1/x=`) `=int_(-oo)^(oo) (dy)/(y^(2)+2)` `=[1/(sqrt(2))"tan"^(-1)y/(sqrt(2))]_(-oo)^(oo) =(pi)/(sqrt(2))` or `(pi)/(2sqrt(2))` |
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| 379. |
Let `I_(1)=int_(0)^(oo)(x^(2)sqrtx)/((1+x)^(6))dx,I_(2)=int_(0)^(oo)(xsqrtx)/((1+x)^(6))dx`, thenA. `I_(1)=2I_(2)`B. `I_(2)=2I_(1)`C. `I_(1)=I_(2)`D. `I_(1)=I_(2)` |
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Answer» Correct Answer - D `I_(1)=int_(0)^(oo)(x^(2)sqrtx)/((1+x)^(6))dx` Let `x=(1)/(t)` `rArr" "I_(1)=int_(oo)^(0)((1)/(t^(2)sqrtt))/((1+(1)/(t))^(6))(-(1)/(t^(2))dt)` `rArr" "I_(1)=int_(0)^(oo)(tsqrtt)/((1+t)^(6))dt=I_(2)` |
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| 380. |
Let `S_n=sum_(k=0)^n n/(n^2+k n+k^2) and T_n=sum_(k=0)^(n-1)n/(n^2+k n+k^2)`,for `n=1,2,3,.......,` thenA. `S_(n)lt(pi)/(3sqrt(3))`B. `S_(n)gt(pi)/(3sqrt(3))`C. `T_(n) lt (pi)/(3sqrt(3))`D. `T_(n)gt(pi)/(3sqrt(3))` |
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Answer» Correct Answer - A::D `S_(n)lt lim_(nto oo) S_(n)=lim_(n to oo) sum_(k=1)^(n)1/n 1/(1+k//n+(k//n)^(2))` `=int_(0)^(1)(dx)/(1+x+x^(2))` `=int_(0)^(1)(dx)/((x+1/2)^(2)+3/4)` `=[2/(sqrt(3))"tan"^(-1)((x+1/2)/((sqrt(3))/2))]_(0)^(1)=(pi)/(3sqrt(3))` Now `T_(n)gt(pi)/(3sqrt(3))`as `h sum_(k=0)^(n-1)f(k//n)gt int_(0)^(1)f(x)dxgt h sum_(k=1)^(n)f(k//n)` |
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| 381. |
Let `S_n=sum_(k=0)^n n/(n^2+k n+k^2) and T_n=sum_(k=0)^(n-1)n/(n^2+k n+k^2)`,for `n=1,2,3,.......,` thenA. `S_(n)lt(pi)/(3sqrt(3))`B. `T_(n)lt(pi)/(3sqrt(3))`C.D. `T_(n)gt(pi)/(3sqrt(3))` |
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Answer» Correct Answer - A::D Given , `S_(n)=sum_(k=n)^(n)(n)/(n^(2)+kn+k^(2))` `=sum_(k=0)^(1)(1)/(n) *((1)/(1+(k)/(n)+(k^(2))/(n^(2))))ltunderset(n tooo)(lim)sum_(k=0)^(n)(1)/(n)((1)/(1+(k)/(n)+((k)/(n))^(2)))` `=int_(0)^(1)(1)/(1+x+x^(2))dx` `= [(2)/(sqrt(3))tan^(-1)((2)/(sqrt(3))(x+(1)/(2)))]_(0)^(1)` ` = (2)/(sqrt(3))*((pi)/(3)-(pi)/(6))=(pi)/(3sqrt(3))i.e S_(n)lt(pi)/(3sqrt(3))` Similarly , `T_(n)gt(pi)/(3sqrt(3))` |
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| 382. |
Let `f`be a real-valued function satisfying `f(x)+f(x+4)=f(x+2)+f(x+6)`Prove that `int_x^(x+8)f(t)dt`is constant function. |
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Answer» Given that `f(x)+f(x+4)=f(x+2)+f(x+6)`…………1 Replacing `x` by `x+2` we get `f(x+2)+f(x+6)=f(x+4)+f(x+8)`…………………2 From equation 1 and 2 we get `f(x)=f(x+8)`……………3 or `int_(x)^(x+8)f(t)dt+int_(0)^(8)f(t)dt` Thus, `g` is a constant function. |
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| 383. |
f is a real valued function from R to R such that `f(x)+f(-x)=2`, then `int_(1-x)^(1+X)f^(-1)(t)dt=`A. `-1`B. 0C. 1D. none of these |
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Answer» Correct Answer - B Given `f(x)+f(-x)=2` Let f(x) = t `therefore" "f(-x)=2-t` `therefore" "f^(-1)(t)+f^(-1)(2-t)=0` `I=int_(1-x)^(1+x)f^(-1)(t)dt=int_(1-x)^(1+x)f^(-1)(2-t)dt` `therefore" "2I=int_(1-x)^(1+x)(f^(-1)(t)+f^(-)(2-t))dt` I = 0 |
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| 384. |
Let `f(x)=lim_( n to oo)(cosx)/(1+(tan^(-1)x)^(n))`. Then the value of `int_(o)^(oo)f(x)dx` is equal toA. `cos (tan 1)`B. `sin(tan1)`C. `tan(tan1)`D. none of these |
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Answer» Correct Answer - B `I=int_(0)^(tan1)f(x)dx+int_(tan1)^(oo)f(x)dx` `=int_(0)^(tan1)cosxdx+0` `=sin(tan1)` |
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| 385. |
`Lim_(hto0)(int_(a)^(x+h)ln^(2)tdt-int_a^(x)ln^(2)tdt)/(h)` equals to : |
| Answer» Correct Answer - B | |
| 386. |
If `L(m,n)=int_(0)^(1)t^(m)(1+t)^(n),dt`, then prove that `L(m,n)=(2^(n))/(m+1)-n/(m+1)L(m+1,n-1)` |
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Answer» Correct Answer - NA `L(m,n)=int_(0)^(1)t^(m)(1+t)^(n)dx` `=[(t^(m+1))/(m+1)(1+t)^(n)]_(0)^(1)-int_(0)^(1)n(1+t)^(n-1)(t^(m+1))/(m+1)dx` `=[(t^(m+1))/(m+1)(1+t^(n))]_(0)^(1)-n/(m+1)int_(0)^(1)t^(m+1)(1+t)^(n-1)dx` `(2^(n))/(m+1)-n/(m+1)L(m+1,n-1)` |
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| 387. |
If `I_(n)=int_(0)^(pi)x^(n)sinxdx`, then find the value of `I_(5)+20I_(3)`. |
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Answer» Correct Answer - `pi^(5)` `I_(m)=int_(0)^(pi)x^(m) sin x dx` `=[-x^(m)cosx]_(0)^(pi)+n int_(0)^(pi)x^(n-1)cos x dx` `=pi^(n)+n[x^(n-1)sinx ]_(0)^(pi)-n(n-1)int_(0)^(pi)x^(n-2)sin x dx` `implies I_(m)=pi^(n)+n.0-n(n-1)I_(n-2)` Put `n=5` `I_(5)=pi^(5)-20I_(3)` `I_(5)+20I_(3)=pi^(5)` |
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