1
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

The area of the region bounded by the parabola $$y=x^2+1$$ and lines $$y=x+1, y=0, x=\frac{1}{2}$$ and $$x=2$$ is

A
$$\frac{23}{6}$$
B
$$\frac{23}{16}$$
C
$$\frac{79}{24}$$
D
$$\frac{79}{16}$$
2
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

Let the functions $$f: R \rightarrow R$$ and $$g: R \rightarrow R$$ be defined by $$f(x)=e^{x-1}-e^{-|x-1|}$$ and $$g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$$. Then, the area of the region in the first quadrant bounded by the curves $$y=f(x), y=g(x)$ and $x=0$$ is.

A
$$(2-\sqrt{3})+\frac{1}{2}\left(e-e^{-1}\right)$$
B
$$(2+\sqrt{3})+\frac{1}{2}\left(e-e^{-1}\right)$$
C
$$(2-\sqrt{3})+\frac{1}{2}\left(e+e^{-1}\right)$$
D
$$(2+\sqrt{3})+\frac{1}{2}\left(e+e^{-1}\right)$$
3
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

What is the area enclosed by the parabola described by $${(y - 2)^2} = (x - 1)$$, its tangent line at the point (2, 3), and the X-axis?

A
3
B
6
C
9
D
12
4
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

The area enclosed by the curves $$y = \sin x + \cos x$$ and $$y = |\cos x - \sin x|$$ over the interval $$\left[ {0,{\pi \over 2}} \right]$$ is

A
$$4(\sqrt 2 - 1)$$
B
$$2\sqrt 2 (\sqrt 2 - 1)$$
C
$$2(\sqrt 2 + 1)$$
D
$$2\sqrt 2 (\sqrt 2 + 1)$$
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