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1
JEE Main 2021 (Online) 20th July Morning Shift
+4
-1
Let $$A = [{a_{ij}}]$$ be a 3 $$\times$$ 3 matrix, where $${a_{ij}} = \left\{ {\matrix{ 1 & , & {if\,i = j} \cr { - x} & , & {if\,\left| {i - j} \right| = 1} \cr {2x + 1} & , & {otherwise.} \cr } } \right.$$

Let a function f : R $$\to$$ R be defined as f(x) = det(A). Then the sum of maximum and minimum values of f on R is equal to:
A
$$- {{20} \over {27}}$$
B
$${{88} \over {27}}$$
C
$${{20} \over {27}}$$
D
$$- {{88} \over {27}}$$
2
JEE Main 2021 (Online) 25th February Morning Slot
+4
-1
If Rolle's theorem holds for the function $$f(x) = {x^3} - a{x^2} + bx - 4$$, $$x \in [1,2]$$ with $$f'\left( {{4 \over 3}} \right) = 0$$, then ordered pair (a, b) is equal to :
A
($$-$$5, $$-$$8)
B
(5, $$-$$8)
C
($$-$$5, 8)
D
(5, 8)
3
JEE Main 2021 (Online) 24th February Evening Slot
+4
-1 English
Hindi
Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) $$\ne$$ 0 for all x $$\in$$ R. If $$\left| {\matrix{ {f(x)} & {f'(x)} \cr {f'(x)} & {f''(x)} \cr } } \right|$$ = 0, for all x$$\in$$R, then the value of f(1) lies in the interval :
A
(0, 3)
B
(9, 12)
C
(3, 6)
D
(6, 9)
4
JEE Main 2020 (Online) 5th September Evening Slot
+4
-1
The derivative of
$${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$$ with
respect to $${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$$ at x = $${1 \over 2}$$ is :
A
$${{2\sqrt 3 } \over 3}$$
B
$${{2\sqrt 3 } \over 5}$$
C
$${{\sqrt 3 } \over {10}}$$
D
$${{\sqrt 3 } \over {12}}$$
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