The numbers of possible continuous `f(x)`
defined in `[0,1]`
for which
`I_1=int_0^1f(x)dx=1,I_2=int_0^1xf(x)dx-a ,I_3=int_0^1x^2f(x)dx=a^2i s//a r e`
1 (b) `oo`
(c) 2 (d) 0
A. `1`
B. `oo`
C. `2`
D. `0`
1 Answers
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`IfI_1=int_0^1 2x ,I_2=int_0^1 2^x^3dx ,I_3=int_1^22^x^2dx ,I_4=int_1^2 2^x^3dx ,` then which of the following is/are ture? `I_1> I_2` (b) `I_2> I_2`
Correct Answer - A::D For `0lt xlt 1, x^(3)gtx^(3)` or `2^(x^(2))gt2^(x^(3))` or `int_(0)^(1)2^(x^(2))dxgt int_(0)^(1)2^(x^(3))dx` Hence `I_(1)gtI_(2)` Also for `1ltxlt2, x^(2)ltx^(3)` or `2^(x^(2))lt2^(x^(3))` or `int_(1)^(2)2^(x^(2))dxlt int_(1)^(2)2^(x^(3))dx` or `I_(3)ltI_(4)`
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`IfI_I=int_0^(pi//2)cos(sinx)dx ,I_2=int_0^(pi/2)sin(cosx)d ,a n dI_3=int_0^(pi/2)cosx dx ,` then find the order in which the values `I_1,I_2,I_3,` ex
Correct Answer - `I_(1)gtI_(3)gtI_(2)` We know that `cosx` is decreasing function in `(0,pi//2)`. Also `xgtsinx` for `xepsilon(0,pi//2)` Thus, `cosx lt cos (sinx)` Further, `xgtsinx` and `cosxepsilon(0,1)` for `xepsilon(0,pi//2)` `:.cosxgtsin(cosx)` Thus, `sin...
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If `I_1=int_0^pixf(sin^3x+cos^2x)dxa n d` `I_2=int_0^(pi/2)f(sin^3x+cos^2x)dx ,t h e nr e l a t eI_1a n dI_2`
Correct Answer - `I_(1)=piI_(2)` `I_(1)=int_(0)^(1)(pi-x)f(sin^(3)x+cos^(2)x)dx` `:. 2I=pi int_(0)^(pi)f(sin^(3)x+cos^(2)x) dx` `=2pi int_(0)^(pi//2) f(sin^(3)x+cos^(2)x)dx` or `I_(1)=pi int_(0)^(pi//2) f(sin^(3)x+cos^(2)x)dx=piI_(2)`
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`"I f"I_1=int_(-100)^(101)(dx)/((5+2x-2x^2)(1+e^(2-4x)))` `"and"I_2=int_(-100)^(101)(dx)/(5+2x-2x^2),t h e n(I_1)/(I_2)"i s"` 2 (b) `1/2` (c) 1 (d) `-
Correct Answer - B `I_(1)=int_(-100)^(101)(dx)/((52x-2x^(2))(1+e^(2-4x)))` `=int_(-100)^(101)(dx)/((5+2(1-x)-2(1-x)^(2)(1+e^(2-4(1-x))))` or `2I_(1)=int_(-100)^(101)(dx)/(5+2x-2x^(2))=I_(2)` or `(I_(1))/(I_(2))=1/2`
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`Iff(x)=(e^x)/(1+e^x),I_1=int_(f(-a))^(f(a))xg(x(1-x)dx ,a n d` `I_2=int_(f(-a))^(f(a))g(x(1-x))dx ,t h e nt h ev a l u eof(I_2)/(I_1)i s` `-1` (b) `-
Correct Answer - C `f(x)=(e^(x))/(1+e^(x))` `:.f(a)=(e^(a))/(1+e^(a))` and `f(-a)=(e^(-a))/(1+e^(-a))=(e^(-a))/(1+1/(e^(a)))=1/(1+e^(a))` `:. f(a)+f(-a)=(e^(a)+1)/(1+e^(a))=1` Let `f(-a)=alpha` or `f(a)=1-alpha` Now `I_(1)=int_(alpha)^(1-alpha)xg(x(1-x))dx` `=int_(alpha)^(1-alpha)(1-x)g((1-x)(1-(1-x))dx` `=int_(alpha)^(1-alpha) (1-x)g(x(1-x))dx` `:. 2I_(1)=int_(alpha)^(1-alpha) g(x(1-x))dx=I_(2)` or `(I_(2))/(I_(1))=2`
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`IfI_1=int_0^(pi/2)(cos^2x)/(1+cos^2x)dx ,I_2=int_0^(pi/2)(sin^2x)/(1+sin^2x)dx` `I_3=int_0^(pi/2)(1+2cos^2xsin^2x)/(4+2cos^2xsin^2x)dx ,t h e n` `I_1
Correct Answer - C `I_(1)int_(0)^(pi//2)(cos^(2)x)/(1+cos^(2)x)dx` `=int_(0)^(pi//2)(cos^(2)(pi//2-x))/(1+cos^(2)(pi//2-x))dx` `=int_(0)^(pi//2) (sin^(2)x)/(1+sin^(2)x)dx=I_(2)` Also `I_(1)+I_(2)=int_(0)^(pi//2)((sin^(2)x)/(1+sin^(2)x)+(cos^(2)x)/(1+cos^(2)x))dx` `=int_(0)^(pi//2)(sin^(2)x+sin^(2)xcos^(2)x+cos^(2)x+sin^(2)xcos^(2)x)/(1+sin^(2)x+cos^(2)x+sin^(2)xcos^(2)x)` `=int(1+2sin^(2)xcos^(2)x)/(2+sin^(2)x cos^(2)x)dx=2I_(3)` `:. 2I_(1)=2I_(3)` or `I_(1)=I_(3)` or `I_(1)=I_(2)=I_(3)`
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If `I_1=int_0^1(e^x)/(1+x) dx`aand `I_2=int_0^1 x^2/(e^(x^3)(2-x^3)) dx` then `I_1/I_2` is
Correct Answer - C `I_(1)=int_(0)^(1)(e^(x)dx)/(1+x), I_(2)=int_(0)^(1)(x^(2)dx)/(e^(x^(2))(2-x^(3)))` In `I_(2)` put `1-x^(3)=t` `:. I_(2)=1/3 int_(1)^(0) (-dt)/(e^(1-t)(1+t))=1/(3e)int_(0)^(1)(e^(t)dt)/(1+t)=1/(3e)I_(1)` `:. (I_(1))/(I_(2))=3e`
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If `I_n=int_0^pi e^x(sinx)^n dx ,` then `(I_3)/(I_1)` is equal to
Correct Answer - A `I_(3)=int_(0)^(pi)e^(x)(sinx)^(3)dx` `=e^(x)(sinx)^(3)|_(0)^(pi)-3int_(0)^(pi)(sinx)^(2)cosx e^(x)dx` `=0-3(sinx)^(2)cosx e^(x)|_(0)^(pi)+3int_(0)^(pi)(2sinx cos x cosx)` `-sin x sin^(2)x)e^(x)dx` `=0+6int_(0)^(pi)sin x cos^(2) xe^(2) dx-3 int_(0)^(pi) sin^(3) xe^(x) dx` `=6int_(0)^(pi) sinx(1-sin^(2)x)e^(x)dx-3int_(0)^(pi)sin^(3)xe^(x)dx` `=6int_(0)^(pi) sinxe^(x)dx-9int_(0)^(pi) sin^(3)x e^(x)dx=6I_(1)-9I_(3)` or...
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`IfI_n=int_0^1(dx)/((1+x^2)^n),w h e r en in N ,` which of the following statements hold good? `2nI_(n+1)=2^(-n)+(2n-1)I_n` `I_2=pi/8+1/4` (c) `I_2=pi
Correct Answer - A::B::D `I_(n)=int_(0)^(1)(dx)/((1+x^(2))^(n))=int_(0)^(1)(1+x^(2))^(-n)dx` `=x/((1+x^(2))^(n))|_(0)^(1)-int_(0)^(1)(-n)(1+x^(2))^(-n-1)2x.xdx` `=1/(2^(n))+2nint_(0)^(1)(x^(2)dx)/((1+x^(2))^(n+1))` `=1/(2^(n))+2n int_(0)^(1)(1+x^(2)-1)/((1+x^(2))^(n+1))dx` `=1/(2^(n))+2nI_(n)-2nI_(n+1)` or `2nI_(n+1)=2^(-n)+(2n-1)I_(n)` or `2I_(2)=1/2+I_(1)=1/2+tan^(-1)x|_(0)^(1)` or `I_(2)=1/4+(pi)/8` Also `4I_(3)=2^(-2)+3I_(2)=1/4+3(1/4+(pi)/8)`.so `I_(3)=(3pi)/32`
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`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`
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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