1
JEE Main 2023 (Online) 25th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$\mathrm{z_1=2+3i}$$ and $$\mathrm{z_2=3+4i}$$. The set $$\mathrm{S = \left\{ {z \in \mathbb{C}:{{\left| {z - {z_1}} \right|}^2} - {{\left| {z - {z_2}} \right|}^2} = {{\left| {{z_1} - {z_2}} \right|}^2}} \right\}}$$ represents a

A
hyperbola with the length of the transverse axis 7
B
hyperbola with eccentricity 2
C
straight line with the sum of its intercepts on the coordinate axes equals $$-18$$
D
straight line with the sum of its intercepts on the coordinate axes equals $$14$$
2
JEE Main 2023 (Online) 25th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three non zero vectors such that $$\overrightarrow b $$ . $$\overrightarrow c $$ = 0 and $$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = {{\overrightarrow b - \overrightarrow c } \over 2}$$. If $$\overrightarrow d $$ be a vector such that $$\overrightarrow b \,.\,\overrightarrow d = \overrightarrow a \,.\,\overrightarrow b $$, then $$(\overrightarrow a \times \overrightarrow b )\,.\,(\overrightarrow c \times \overrightarrow d )$$ is equal to

A
$$\frac{1}{2}$$
B
$$-\frac{1}{4}$$
C
$$\frac{1}{4}$$
D
$$\frac{3}{4}$$
3
JEE Main 2023 (Online) 25th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let S$$_1$$ and S$$_2$$ be respectively the sets of all $$a \in \mathbb{R} - \{ 0\} $$ for which the system of linear equations

$$ax + 2ay - 3az = 1$$

$$(2a + 1)x + (2a + 3)y + (a + 1)z = 2$$

$$(3a + 5)x + (a + 5)y + (a + 2)z = 3$$

has unique solution and infinitely many solutions. Then

A
$$\mathrm{n({S_1}) = 2}$$ and S$$_2$$ is an infinite set
B
$$\mathrm{{S_1} = \Phi} $$ and $$\mathrm{{S_2} = \mathbb{R} - \{ 0\}}$$
C
$$\mathrm{{S_1} = \mathbb{R} - \{ 0\}}$$ and $$\mathrm{{S_2} = \Phi} $$
D
S$$_1$$ is an infinite set and n(S$$_2$$) = 2
4
JEE Main 2023 (Online) 25th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$y = y(x)$$ be the solution curve of the differential equation $${{dy} \over {dx}} = {y \over x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x > 0,y(1) = 3$$. Then $${{{y^2}(x)} \over 9}$$ is equal to :

A
$${{{x^2}} \over {5 - 2{x^3}(2 + {{\log }_e}{x^3})}}$$
B
$${{{x^2}} \over {3{x^3}(1 + {{\log }_e}{x^2}) - 2}}$$
C
$${{{x^2}} \over {7 - 3{x^3}(2 + {{\log }_e}{x^2})}}$$
D
$${{{x^2}} \over {2{x^3}(2 + {{\log }_e}{x^3}) - 3}}$$
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