Differential Equations · JEE Main previous year questions

2019 7 questions

1. Q142019 · 12 Apr 2019 · Shift 2

   Let $z\in \mathbb{C}$ with $\text{Im}(z)=10$ and it satisfies $\frac{2z-n}{2z+n}=2i-1$ for some natural number n. Then :

   1. A.$n=20$ and $\text{Re}(z)=-10$
   2. B.$n=40$ and $\text{Re}(z)=10$
   3. C.$n=40$ and $\text{Re}(z)=-10$
   4. D.$n=20$ and $\text{Re}(z)=10$
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   Answer: C

2. Q52019 · 10 Apr 2019 · Shift 1

   If y = y(x) is the solution of the differential equation $\frac{dy}{dx}=(\tan x-y){\sec }^{2}x$, $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, such that y(0) = 0, then $y\left(-\frac{\pi }{4}\right)$ is equal to :

   1. A.$\frac{1}{2}-e$
   2. B.$e-2$
   3. C.$2+\frac{1}{e}$
   4. D.$\frac{1}{e}-2$
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   Answer: B

3. Q162019 · 10 Apr 2019 · Shift 2

   Let y = y(x) be the solution of the differential equation, $\frac{dy}{dx}+y\tan x=2x+x^2\tan x$, $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, such that y(0) = 1. Then :

   1. A.$y\left(\frac{\pi }{4}\right)-y\left(-\frac{\pi }{4}\right)=\sqrt{2}$
   2. B.$y^{\prime }\left(\frac{\pi }{4}\right)-y^{\prime }\left(-\frac{\pi }{4}\right)=\pi -\sqrt{2}$
   3. C.$y\left(\frac{\pi }{4}\right)+y\left(-\frac{\pi }{4}\right)=\frac{{\pi }^2}{2}+2$
   4. D.$y^{\prime }\left(\frac{\pi }{4}\right)+y^{\prime }\left(-\frac{\pi }{4}\right)=-\sqrt{2}$
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   Answer: B

4. Q112019 · 9 Apr 2019 · Shift 1

   The solution of the differential equation $x\frac{dy}{dx}+2y=x^2(x\ne 0)$ with y(1) = 1, is :

   1. A.$y=\frac{4}{5}{x}^3+\frac{1}{5x^2}$
   2. B.$y=\frac{3}{4}{x}^2+\frac{1}{4x^2}$
   3. C.$y=\frac{x^2}{4}+\frac{3}{4x^2}$
   4. D.$y=\frac{x^3}{5}+\frac{1}{5x^2}$
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   Answer: C

5. Q92019 · 12 Jan 2019 · Shift 1

   Let $y=y(x)$ be the solution of the differential equation, $x\frac{dy}{dx}+y=x{\log }_{e}x$, $(x>1)$. If $2y(2)={\log }_{e}4-1$, then $y(e)$ is equal to :

   1. A.$-\frac{e}{2}$
   2. B.$-\frac{e^2}{2}$
   3. C.$\frac{e^2}{4}$
   4. D.$\frac{e}{4}$
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   Answer: D

6. Q12019 · 11 Jan 2019 · Shift 2

   The solution of the differential equation, $$\frac{dy}{dx}=(x-y{)}^2,\text{ when }y(1)=1,\text{ is :}$$

   1. A.$-{\log }_e\left|\frac{1+x-y}{1-x+y}\right|=x+y-2$
   2. B.${\log }_e\left|\frac{2-x}{2-y}\right|=x-y$
   3. C.${\log }_e\left|\frac{2-y}{2-x}\right|=2(y-1)$
   4. D.$-{\log }_e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$
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   Answer: D

7. Q132019 · 9 Jan 2019 · Shift 1

   If $y=y(x)$ is the solution of the differential equation, $$x\frac{dy}{dx}+2y=x^2,\text{ satisfying }y(1)=1,\text{ then }y\left(\frac{1}{2}\right)\text{ is equal to :}$$

   1. A.$\frac{7}{64}$
   2. B.$\frac{49}{16}$
   3. C.$\frac{1}{4}$
   4. D.$\frac{13}{16}$
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   Answer: B