1
JEE Main 2022 (Online) 29th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let the solution curve of the differential equation

$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}} $$, $$y(1) = 3$$ be $$y = y(x)$$. Then y(2) is equal to:

A
15
B
11
C
13
D
17
2
JEE Main 2022 (Online) 29th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

If the mirror image of the point (2, 4, 7) in the plane 3x $$-$$ y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to :

A
54
B
50
C
$$-$$6
D
$$-$$42
3
JEE Main 2022 (Online) 29th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f:R \to R$$ be a function defined by :

$$f(x) = \left\{ {\matrix{ {\max \,\{ {t^3} - 3t\} \,t \le x} & ; & {x \le 2} \cr {{x^2} + 2x - 6} & ; & {2 < x < 3} \cr {[x - 3] + 9} & ; & {3 \le x \le 5} \cr {2x + 1} & ; & {x > 5} \cr } } \right.$$

where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and $$I = \int\limits_{ - 2}^2 {f(x)\,dx} $$. Then the ordered pair (m, I) is equal to :

A
$$\left( {3,\,{{27} \over 4}} \right)$$
B
$$\left( {3,\,{{23} \over 4}} \right)$$
C
$$\left( {4,\,{{27} \over 4}} \right)$$
D
$$\left( {4,\,{{23} \over 4}} \right)$$
4
JEE Main 2022 (Online) 29th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$$, $$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$$ and $$\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$$ where $$\alpha ,\,\beta \in R$$, be three vectors. If the projection of $$\overrightarrow a $$ on $$\overrightarrow c $$ is $${{10} \over 3}$$ and $$\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$$, then the value of $$\alpha + \beta $$ is equal to :

A
3
B
4
C
5
D
6
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