Integral Calculus & Area · JEE Main previous year questions

2018 9 questions

1. Q122018 · 16 Apr 2018 · Shift 1

   If $\int \frac{\tan x}{1+\tan x+{\tan }^{2}x}dx=x-\frac{K}{\sqrt{A}}{\tan }^{-1}\left(\frac{K\tan x+1}{\sqrt{A}}\right)+C$, (C is a constant of integration) then the ordered pair (K, A) is equal to :

   1. A.(2, 1)
   2. B.$(-2,3)$
   3. C.(2, 3)
   4. D.$(-2,1)$
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   Answer: C

2. Q132018 · 16 Apr 2018 · Shift 1

   If $f(x)={\int }_0^{x}t(\sin x-\sin t)dt$ then :

   1. A.$f^{\prime \prime \prime }(x)+f^{\prime \prime }(x)=\sin x$
   2. B.$f^{\prime \prime \prime }(x)+f^{\prime \prime }(x)-f^{\prime }(x)=\cos x$
   3. C.$f^{\prime \prime \prime }(x)+f^{\prime }(x)=\cos x-2x\sin x$
   4. D.$f^{\prime \prime \prime }(x)-f^{\prime \prime }(x)=\cos x-2x\sin x$
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   Answer: C

3. Q142018 · 16 Apr 2018 · Shift 1

   If the area of the region bounded by the curves, $y=x^2,y=\frac{1}{x}$ and the lines $y=0$ and $x=t$ ($t>1$) is 1 sq. unit, then t is equal to :

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

4. Q132018 · 15 Apr 2018 · Shift 1

   If $f\left(\frac{x-4}{x+2}\right)=2x+1$, $(x\in \mathbf{R}-\{1,-2\})$, then $\int f(x)dx$ is equal to : (where C is a constant of integration)

   1. A.$12{\log }_e|1-x|+3x+\text{C}$
   2. B.$-12{\log }_e|1-x|-3x+\text{C}$
   3. C.$12{\log }_e|1-x|-3x+\text{C}$
   4. D.$-12{\log }_e|1-x|+3x+\text{C}$
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   Answer: B

5. Q142018 · 15 Apr 2018 · Shift 1

   The area (in sq. units) of the region $\{x\in \mathbf{R}:x\ge 0,y\ge 0,y\ge x-2\text{ and }y\le \sqrt{x}\}$, is :

   1. A.$\frac{13}{3}$
   2. B.$\frac{8}{3}$
   3. C.$\frac{10}{3}$
   4. D.$\frac{5}{3}$
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   Answer: C

6. Q152018 · 15 Apr 2018 · Shift 1

   The value of the integral ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{4}x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right)dx$ is :

   1. A.0
   2. B.$\frac{3}{4}$
   3. C.$\frac{3}{8}\pi$
   4. D.$\frac{3}{16}\pi$
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   Answer: C

7. Q112018 · 8 Apr 2018 · Shift 1

   The value of ${\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2^x}dx$ is

   1. A.$\frac{\pi }{4}$
   2. B.$\frac{\pi }{8}$
   3. C.$\frac{\pi }{2}$
   4. D.$4\pi$
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   Answer: A

8. Q132018 · 8 Apr 2018 · Shift 1

   The integral $$\int \frac{{\sin }^{2}x{\cos }^{2}x}{{\left({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x\right)}^2}dx$$ is equal to

   1. A.$\frac{-1}{1+{\cot }^{3}x}+C$
   2. B.$\frac{1}{3(1+{\tan }^{3}x)}+C$
   3. C.$\frac{-1}{3(1+{\tan }^{3}x)}+C$
   4. D.$\frac{1}{1+{\cot }^{3}x}+C$
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   Answer: C

9. Q142018 · 8 Apr 2018 · Shift 1

   Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $\sin x\frac{dy}{dx}+y\cos x=4x,x\in (0,\pi )$. If $y\left(\frac{\pi }{2}\right)=0$, then $y\left(\frac{\pi }{6}\right)$ is equal to :

   1. A.$-\frac{4}{9}{\pi }^2$
   2. B.$\frac{4}{9\sqrt{3}}{\pi }^2$
   3. C.$-\frac{8}{9\sqrt{3}}{\pi }^2$
   4. D.$-\frac{8}{9}{\pi }^2$
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   Answer: D