1
JEE Main 2023 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right)} & {,\,x \ne 0} \cr 0 & {,\,x = 0} \cr } } \right.$$

Then at $$x=0$$

A
$$f$$ is continuous but $$f'$$ is not continuous
B
$$f$$ and $$f'$$ both are continuous
C
$$f$$ is continuous but not differentiable
D
$$f'$$ is continuous but not differentiable
2
JEE Main 2023 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

Let $$\overrightarrow u = \widehat i - \widehat j - 2\widehat k,\overrightarrow v = 2\widehat i + \widehat j - \widehat k,\overrightarrow v .\,\overrightarrow w = 2$$ and $$\overrightarrow v \times \overrightarrow w = \overrightarrow u + \lambda \overrightarrow v $$. Then $$\overrightarrow u .\,\overrightarrow w $$ is equal to :

A
$$ - {2 \over 3}$$
B
$${3 \over 2}$$
C
2
D
1
3
JEE Main 2023 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

Let $$\alpha$$ be a root of the equation $$(a - c){x^2} + (b - a)x + (c - b) = 0$$ where a, b, c are distinct real numbers such that the matrix $$\left[ {\matrix{ {{\alpha ^2}} & \alpha & 1 \cr 1 & 1 & 1 \cr a & b & c \cr } } \right]$$ is singular. Then, the value of $${{{{(a - c)}^2}} \over {(b - a)(c - b)}} + {{{{(b - a)}^2}} \over {(a - c)(c - b)}} + {{{{(c - b)}^2}} \over {(a - c)(b - a)}}$$ is

A
3
B
6
C
12
D
9
4
JEE Main 2023 (Online) 24th January Morning Shift
Numerical
+4
-1
Change Language

Let $$\lambda \in \mathbb{R}$$ and let the equation E be $$|x{|^2} - 2|x| + |\lambda - 3| = 0$$. Then the largest element in the set S = {$$x+\lambda:x$$ is an integer solution of E} is ______

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