1
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
The line $ y=m x $ bisects the area unclosed by lines $ x=0, y=0 $ and $ x=\frac{3}{2} $ and the curve $ y=1+4 x-x^{2} $. Then, the value of $ m $ is
A
$ \frac{13}{6} $
B
$ \frac{13}{2} $
C
$ \frac{13}{5} $
D
$ \frac{13}{7} $
2
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
The area enclosed by the curves $ y=x^{3} $ and $ y=\sqrt{x} $ is
A
$ \frac{5}{3} $ sq. units
B
$ \frac{5}{4} $ sq. units
C
$ \frac{5}{12} $ sq. unit
D
$ \frac{12}{5} $ sq. units then $ k $ is
3
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
$$\frac{79}{16}$$
4
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
$$(2+\sqrt{3})+\frac{1}{2}\left(e+e^{-1}\right)$$
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