01Easy38×since 2002Q3624If y=y(x)y = y(x)y=y(x) is the solution of the differential equation 2x2dydx−2xy+3y2=02{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 02x2dxdy−2xy+3y2=0 such that y(e)=e3y(e) = {e \over 3}y(e)=3e, then y(1) is equal to :A13{1 \over 3}31B23{2 \over 3}32C32{3 \over 2}23D3Check answerSkip
02Hard38×since 2002Q3625Let g:(0,∞)→Rg:(0,\infty ) \to Rg:(0,∞)→R be a differentiable function such that ∫(x(cosx−sinx)ex+1+g(x)(ex+1−xex)(ex+1)2)dx=x g(x)ex+1+c\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c}∫(ex+1x(cosx−sinx)+(ex+1)2g(x)(ex+1−xex))dx=ex+1xg(x)+c, for all x > 0, where c is an arbitrary constant. Then :Ag is decreasing in (0,π4)\left( {0,{\pi \over 4}} \right)(0,4π)Bg' is increasing in (0,π4)\left( {0,{\pi \over 4}} \right)(0,4π)Cg + g' is increasing in (0,π2)\left( {0,{\pi \over 2}} \right)(0,2π)Dg −-− g' is increasing in (0,π2)\left( {0,{\pi \over 2}} \right)(0,2π)Check answerSkip
03Medium38×since 2002Q3626If the solution curve y=y(x)y = y(x)y=y(x) of the differential equation y2dx+(x2−xy+y2)dy=0{y^2}dx + ({x^2} - xy + {y^2})dy = 0y2dx+(x2−xy+y2)dy=0, which passes through the point (1, 1) and intersects the line y=3xy = \sqrt 3 xy=3x at the point (α,3α)(\alpha ,\sqrt 3 \alpha )(α,3α), then value of loge(3α){\log _e}(\sqrt 3 \alpha )loge(3α) is equal to :Aπ3{\pi \over 3}3πBπ2{\pi \over 2}2πCπ12{\pi \over 12}12πDπ6{\pi \over 6}6πCheck answerSkip
04Easy38×since 2002Q3646Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation secx dy+{2(1−x)tanx+x(2−x)}dx=0\sec x \mathrm{~d} y+\{2(1-x) \tan x+x(2-x)\} \mathrm{d} x=0secx dy+{2(1−x)tanx+x(2−x)}dx=0 such that y(0)=2y(0)=2y(0)=2. Then y(2)y(2)y(2) is equal to:A2{sin(2)+1}2\{\sin (2)+1\}2{sin(2)+1}B2C1D2{1−sin(2)}2\{1-\sin (2)\}2{1−sin(2)}Check answerSkip
05Medium38×since 2002Q3627The general solution of the differential equation (x−y2)dx+y(5x+y2)dy=0\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0(x−y2)dx+y(5x+y2)dy=0 is :A(y2+x)4=C∣(y2+2x)3∣\left(y^{2}+x\right)^{4}=\mathrm{C}\left|\left(y^{2}+2 x\right)^{3}\right|(y2+x)4=C(y2+2x)3B(y2+2x)4=C∣(y2+x)3∣\left(y^{2}+2 x\right)^{4}=C\left|\left(y^{2}+x\right)^{3}\right|(y2+2x)4=C(y2+x)3C∣(y2+x)3∣=C(2y2+x)4\left|\left(y^{2}+x\right)^{3}\right|=\mathrm{C}\left(2 y^{2}+x\right)^{4}(y2+x)3=C(2y2+x)4D∣(y2+2x)3∣=C(2y2+x)4\left|\left(y^{2}+2 x\right)^{3}\right|=C\left(2 y^{2}+x\right)^{4}(y2+2x)3=C(2y2+x)4Check answerSkip
06Medium38×since 2002Q3628Let the solution curve of the differential equation x dy=(x2+y2+y)dx,x>0x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0x dy=(x2+y2+y)dx,x>0, intersect the line x=1x=1x=1 at y=0y=0y=0 and the line x=2x=2x=2 at y=αy=\alphay=α. Then the value of α\alphaα is :A12\frac{1}{2}21B32\frac{3}{2}23C−-−32\frac{3}{2}23D52\frac{5}{2}25Check answerSkip
07Medium38×since 2002Q3629If y=y(x),x∈(0,π/2)y=y(x), x \in(0, \pi / 2)y=y(x),x∈(0,π/2) be the solution curve of the differential equation (sin22x)dydx+(8sin22x+2sin4x)y=2e−4x(2sin2x+cos2x)\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)(sin22x)dxdy+(8sin22x+2sin4x)y=2e−4x(2sin2x+cos2x), with y(π/4)=e−πy(\pi / 4)=\mathrm{e}^{-\pi}y(π/4)=e−π, then y(π/6)y(\pi / 6)y(π/6) is equal to :A23e−2π/3\frac{2}{\sqrt{3}} e^{-2 \pi / 3}32e−2π/3B23e2π/3\frac{2}{\sqrt{3}} \mathrm{e}^{2 \pi / 3}32e2π/3C13e−2π/3\frac{1}{\sqrt{3}} e^{-2 \pi / 3}31e−2π/3D13e2π/3\frac{1}{\sqrt{3}} e^{2 \pi / 3}31e2π/3Check answerSkip
08Medium38×since 2002Q3630If the solution curve of the differential equation dydx=x+y−2x−y\frac{d y}{d x}=\frac{x+y-2}{x-y}dxdy=x−yx+y−2 passes through the points (2,1)(2,1)(2,1) and (k+1,2),k>0(\mathrm{k}+1,2), \mathrm{k}>0(k+1,2),k>0, thenA2tan−1(1k)=loge(k2+1)2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)2tan−1(k1)=loge(k2+1)Btan−1(1k)=loge(k2+1)\tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)tan−1(k1)=loge(k2+1)C2tan−1(1k+1)=loge(k2+2k+2)2 \tan ^{-1}\left(\frac{1}{k+1}\right)=\log _{e}\left(k^{2}+2 k+2\right)2tan−1(k+11)=loge(k2+2k+2)D2tan−1(1k)=loge(k2+1k2)2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(\frac{k^{2}+1}{k^{2}}\right)2tan−1(k1)=loge(k2k2+1)Check answerSkip
09Medium38×since 2002Q3631Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation dydx+(2x2+11x+13x3+6x2+11x+6)y=(x+3)x+1,x>−1\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1dxdy+(x3+6x2+11x+62x2+11x+13)y=x+1(x+3),x>−1, which passes through the point (0,1)(0,1)(0,1). Then y(1)y(1)y(1) is equal to :A12\frac{1}{2}21B32\frac{3}{2}23C52\frac{5}{2}25D72\frac{7}{2}27Check answerSkip
10Hard38×since 2002Q3632Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (3y2−5x2)y dx+2x(x2−y2)dy=0\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0(3y2−5x2)y dx+2x(x2−y2)dy=0 such that y(1)=1y(1)=1y(1)=1. Then ∣(y(2))3−12y(2)∣\left|(y(2))^{3}-12 y(2)\right|(y(2))3−12y(2) is equal to :A64B16216 \sqrt{2}162C32D32232 \sqrt{2}322Check answerSkip