Javascript is required
1
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Let 'a' be a real number such that the function f(x) = ax2 + 6x $$-$$ 15, x $$\in$$ R is increasing in $$\left( { - \infty ,{3 \over 4}} \right)$$ and decreasing in $$\left( {{3 \over 4},\infty } \right)$$. Then the function g(x) = ax2 $$-$$ 6x + 15, x$$\in$$R has a :
A
local maximum at x = $$-$$ $${{3 \over 4}}$$
B
local minimum at x = $$-$$$${{3 \over 4}}$$
C
local maximum at x = $${{3 \over 4}}$$
D
local minimum at x = $${{3 \over 4}}$$
2
JEE Main 2021 (Online) 16th March Evening Shift
MCQ (Single Correct Answer)
+4
-1
The maximum value of

$$f(x) = \left| {\matrix{ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|,x \in R$$ is :
A
$$\sqrt 5 $$
B
$${3 \over 4}$$
C
5
D
$$\sqrt 7 $$
3
JEE Main 2021 (Online) 26th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = $${\pi \over 2}$$ in the first quadrant. Then,
A
$${A_1}:{A_2} = 1:\sqrt 2 $$ and $${A_1} + {A_2} = 1$$
B
$${A_1} = {A_2}$$ and $${A_1} + {A_2} = \sqrt 2 $$
C
$$2{A_1} = {A_2}$$ and $${A_1} + {A_2} = 1 + \sqrt 2 $$
D
$${A_1}:{A_2} = 1:2$$ and $${A_1} + {A_2} = 1$$
4
JEE Main 2021 (Online) 26th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Let slope of the tangent line to a curve at any point P(x, y) be given by $${{x{y^2} + y} \over x}$$. If the curve intersects the line x + 2y = 4 at x = $$-$$2, then the value of y, for which the point (3, y) lies on the curve, is :
A
$$ - {{18} \over {19}}$$
B
$$ - {{4} \over {3}}$$
C
$${{18} \over {35}}$$
D
$$ - {{18} \over {11}}$$
JEE Main Subjects
© 2023 ExamGOAL