1
JEE Main 2022 (Online) 26th July Morning Shift
+4
-1

Let $$f(x) = \left\{ {\matrix{ {{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr } } \right.$$.

Then the set of all values of b, for which f(x) has maximum value at x = 1, is :

A
($$-$$6, $$-$$2)
B
(2, 6)
C
$$[ - 6, - 2) \cup (2,6]$$
D
$$\left[ {-\sqrt 6 , - 2} \right) \cup \left( {2,\sqrt 6 } \right]$$
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 $$\left( {{1 \over 2},{4 \over 3}} \right)$$
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
Calculus
EXAM MAP
Joint Entrance Examination