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0∫1 tan-1 (1 – x+ x2) dx ...
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y=(x-[x])[x] 0∫3 y.dx = ??? ...
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10∫19 (sin x / 1+x8 ) dx < A. 10 -10 B. 10 -9 C. 10 -8 D. 10 -7 ...
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lim { (1/n)((2n+1)(2n+2)2n+3)............(2n+n)^1/n))} n-->∞ ...
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0∫∞ dx / (x2+a2)(x2+b2) do we need to apply partial fraction here ?? ...
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*Image* ...
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(2^x)=(x^3)+1 ...
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(1) f:R-->R f((x-y)2)= (f(x))2 -2xf(y) +y2 then f(x)= ...
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limit x->pi/2 \frac{x-\pi /2}{cosx} i know by l'hopitol rule we can do.... this is my attempt without using l'hopitol rule but cosx=( {x-\pi /2} )(x-other root)..so on..... so \lim_{x-->\frac{\pi }{2}}( x-\frac{\pi }{2} ...
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*Image* ...
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*Image* ...
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Draw the rough sketch of of portions of curve x2=4[√x]y and y2=4[√y]x that lie within the square the square {(x,y) | 1≤x≤4, 1≤y<4}. Hence find the area that encloses the two curves and the line x+y=3 ...
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0∫1 x (1-x)/1+x) = ?? A. 1 B. 1- π/4 C. 1+π/4 D. none of these ...
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0∫π/3 cos x/(3+4 sin x)dx = k log(3 + 2√3/3).. the value of k = A. 1/2 B. 1/3 C. 1/4 D. 1/5 ...
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Nr= 2x^1/2 +3 x^1/3 + 5 x^1/5 Dr=(5x-2)^1/2 +(3x-2)^1/3 lt---->∞ Nr/Dr plz do without l'hospitol rule........ and plz tell me the logic behind l' hopitol ...
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1) Solve *Image* 2) Solve *Image* ...
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f(x)=\int_{0}^{\pi}\sin (x-t)\sin (2t-a)\ dt with respect to x... ...
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Find the area of the region bounded by the curves y = logex, y = sin4Ï€x and x = 0. ...
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Show that the curve (x/a)2n + (y/b)2n=2 touches the straight line x/a+y/b=2 at (a,b) no matter what the value of n be. ...
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(y3-2x2y)dx + (2xy2 - x3)dy =0, then the value of xy√(y2-x2) is 1. y2 +x 2. xy2 3. any constant 4. none of these ...
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Use the substitution y2=a-x to reduce the equation y3 dy/dx +x+y2=0 to homogeneous form and hence solve it. ...
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Find the value(s) of the parameter 'a' (a>0) for each of which the area of the figure bounded by the straight line, y = a2-ax/1+a4 and the parabola y = x2+2ax+3a2/1+a4 is the greatest. ...
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Find the no of solutions of |x2 - 6|x| +8|= 1/4 |sin|x|| 1) 4 2) 6 3) 8 4) 10 ...
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find the differential equation of the following y=c1sin2x+c2cos2x+c3 ...
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Find the curve y = f(x) where f(x)≥0, f(0)= 0, bounding a curfilinear trapezoid with the base {0,x] whose area is proportional to (n+1)th power of f(x). It is known that f(1) = 1. ...
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A point P moves inside the triangle formed by A (0,0) , B(1,1/ 3 ) and C(2,0) such that min { PA,PB,PC} =1 then the area bounded by the curve traced by P is a. 3 3 +3pi/2 b. 3 3 -3pi/2 c. 3 -pi/2 d. 3 +pi/2 NOTE→Please give ...
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Evaluate the following:- 1) *Image* 2) *Image* ...
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Ques) Find / Evaluate *Image* ...
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find the Domain and Range(of course give explanation) these are also last yrs question 1)f(x)=limn->∞cos(n! 2Πx) 2)g(x)=limn->∞cosn( 2Πx) 3)h(x)=limn->∞{cos(nx) + 2}1/n ...
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Prove the following (a2<1) a) \int _{0} ^ { \pi } ln(1-2a \cos \theta +a^2) d \theta=\begin{Bmatrix} 0 & a<1\\ 2\pi lna & a>1 \end{Bmatrix} b) \int _{0} ^ { \pi } ln(1-2a \cos \theta +a^2)\cos n \theta d \theta= - \f ...