1
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1 Let $$f(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}$$.

Consider

$$(\mathrm{S} 1): f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$$

$$(\mathrm{S} 2): \int\limits_{-2}^{2} f(x) \mathrm{d} x=12$$

Then,

A
both (S1) and (S2) are correct
B
both (S1) and (S2) are wrong
C
only (S1) is correct
D
only (S2) is correct
2
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1 The area of the region enclosed by $$y \leq 4 x^{2}, x^{2} \leq 9 y$$ and $$y \leq 4$$, is equal to :

A
$$\frac{40}{3}$$
B
$$\frac{56}{3}$$
C
$$\frac{112}{3}$$
D
$$\frac{80}{3}$$
3
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1 $$\int\limits_{0}^{2}\left(\left|2 x^{2}-3 x\right|+\left[x-\frac{1}{2}\right]\right) \mathrm{d} x$$, where [t] is the greatest integer function, is equal to :

A
$$\frac{7}{6}$$
B
$$\frac{19}{12}$$
C
$$\frac{31}{12}$$
D
$$\frac{3}{2}$$
4
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1 Consider a curve $$y=y(x)$$ in the first quadrant as shown in the figure. Let the area $$\mathrm{A}_{1}$$ is twice the area $$\mathrm{A}_{2}$$. Then the normal to the curve perpendicular to the line $$2 x-12 y=15$$ does NOT pass through the point. A
(6, 21)
B
(8, 9)
C
(10, $$-$$4)
D
(12, $$-$$15)
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