1
JEE Main 2022 (Online) 25th July Morning Shift
+4
-1

If the absolute maximum value of the function $$f(x)=\left(x^{2}-2 x+7\right) \mathrm{e}^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}$$ in the interval $$[-3,0]$$ is $$f(\alpha)$$, then :

A
$$\alpha=0$$
B
$$\alpha=-3$$
C
$$\alpha \in(-1,0)$$
D
$$\alpha \in(-3,-1]$$
2
JEE Main 2022 (Online) 25th July Morning Shift
+4
-1

The curve $$y(x)=a x^{3}+b x^{2}+c x+5$$ touches the $$x$$-axis at the point $$\mathrm{P}(-2,0)$$ and cuts the $$y$$-axis at the point $$Q$$, where $$y^{\prime}$$ is equal to 3 . Then the local maximum value of $$y(x)$$ is:

A
$$\frac{27}{4}$$
B
$$\frac{29}{4}$$
C
$$\frac{37}{4}$$
D
$$\frac{9}{2}$$
3
JEE Main 2022 (Online) 30th June Morning Shift
+4
-1

If xy4 attains maximum value at the point (x, y) on the line passing through the points (50 + $$\alpha$$, 0) and (0, 50 + $$\alpha$$), $$\alpha$$ > 0, then (x, y) also lies on the line :

A
y = 4x
B
x = 4y
C
y = 4x + $$\alpha$$
D
x = 4y $$-$$ $$\alpha$$
4
JEE Main 2022 (Online) 30th June Morning Shift
+4
-1

Let $$f(x) = 4{x^3} - 11{x^2} + 8x - 5,\,x \in R$$. Then f :

A
has a local minina at $$x = {1 \over 2}$$
B
has a local minima at $$x = {3 \over 4}$$
C
is increasing in $$\left( {{1 \over 2},{3 \over 4}} \right)$$
D
is decreasing in $$\left( {{1 \over 2},{4 \over 3}} \right)$$
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