Newest Relation Functions Questions

The answer is given as only transitive. Why? Can someone please explain ...

which one is greater: A=20104019 or B=2009200920112011 ...

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1.\int_{0}^{x}[1+t]^{3}dt 2.\int_{0}^{x}[x]dx plz help sir ...

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1) Let f:R to R be continuous function which satisfies f(x)=∫(0 to x)f(t)dt, then the value of f(ln5) is: a)0; b)2; c)4; d) 6. 2) Let I=∫(0 to 1) (sinx)/ x and J=∫(0 to 1) (cosx)/ x .Then which is true? a)I<2/3 and J ...

Let f(x) be a non constant twice differentiable function defined on (-∞,∞) such that f(x)=f(1-x) and f'(1/4)=0.Then a) f''(x) vanishes at least twice on [0,1] b) f'(1/2)=0 c) ∫(frm -1/2 to 1/2)f(x+1/2)sinxdx=0 d) ∫(fr ...

Integrate:: 4x3[d2/dx2(1-x2)5]dx from 0 to 1.Here d2/dx2 means the double derivative of the function w.r.t.x. This sum is meant for students only ,not for experts. ...

If x>0 then integrate (2x-[2x])2 from 0 to 100[x] where [.] means G.I.F. ...

1) Prove that ∫x3 2ax-x2 dx=7πa5/8. 2) Integrate: [ x4/(1-x4)] *cos-1(2x/(1+x2). Upper limit=1/ 3 ;Lower limit=-1/ 3 . 3) Integrate::: 2-x2/[(1+x) 1-x2 ].Upper limit=1;Lower limit=0. 4) Integrate:::: log(1+tanx) from 0 to ...

2^sinx+2^cosx>1 ...

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INTEGRATE: 1) (x3+3)/x6(x2+1). Upper limit=∞;Lower limit=0 2) xe-x/( 1-e-2x ). Upper limit=0;Lower limit=∞. 3) If f(x) be a function satisfying f'(x)=f(x) with f((0)=1 and g be the function satisfying f(x) + g(x)=x2,then ...

Integrate: 1) ( sinx )/( sinx + cosx ) 2) xln(x)/(x2-1)3/2. ...

Integrate: 1) (1+x-2/3)/(1+x) 2) cos2x /sinx 3)(x2 + n(n-1))/(xsinx+ncosx)2 ...

Integrate: 1) [(x-1) x4+2x3-x2+2x+1 ]/x2(x+1) 2) (x2-1)/(x3 2x4-2x2+1 ) ...

try this integration: \frac{x^{2}-2}{x^{3}\sqrt{x^{2}-1}} ...

INTEGRATE: emsin-1x ...

Integrate this: \frac{x^{3}sin^{-1}x}{\sqrt{1-x^{2}}} ...

PROVE THAT : iim x100ln(x)/extan-1(π/3 + sinx)=∞ x→∞ ...

The minimum value of mod of(sinA + cosA + tanA + secA + cotA + cosecA) is 2 2 -k.Find k ...

If f(x)=∫xa 1/f(x) dx (x>0) and ∫1a 1/f(x) dx= 2 ,then f(50) is: ...

If f(x) and φ(x) are continuous function in[0,4] satisfying f(x)=f(4-x),φ(x) + φ(4-x)=3 and ∫40f(x)dx=2,then the value of ∫40 f(x)φ(x)dx is: NOTE:The upper and lower limits are 4 and 0 ...

If each ai>0, then the shortest distance between the point(0,-3) and the curve y=1+a1x2 + a2x4 + ...........................................+anx2n is a) 1 b) 2 c) 3 d) 4 ...

If each ai>0, then the shortest distance between the point(0,-3) and the curve y=1+a1x2 + a2x4 + ...........................................+anx2n is a) 1 b) 2 c) 3 d) 4 ...

Given that f is areal valued differentiable function such that f(x)f'(x)<0,for all real x.It fpllows that a) f(x) is an increasing function b) f(x) is a decreasing fnction C) mod(f(x)) is an increasing function d) mod(f(x) ...

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π∫0 xcotxdx ...

π∫0 dx/1+cos2x ...