1
JEE Main 2023 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

Among the statements :

$$(\mathrm{S} 1)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow(\mathrm{p} \Rightarrow \mathrm{r})$$

$$(\mathrm{S} 2)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow((\mathrm{p} \Rightarrow \mathrm{r}) \vee(\mathrm{q} \Rightarrow \mathrm{r}))$$

A
only (S1) is a tautology
B
neither (S1) nor (S2) is a tautology
C
both (S1) and (S2) are tautologies
D
only (S2) is a tautology
2
JEE Main 2023 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

If $${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$$, then $${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$$ is equal to :

A
$${{51} \over {144}}$$
B
$${{49} \over {138}}$$
C
$${{50} \over {141}}$$
D
$${{52} \over {147}}$$
3
JEE Main 2023 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

A straight line cuts off the intercepts $$\mathrm{OA}=\mathrm{a}$$ and $$\mathrm{OB}=\mathrm{b}$$ on the positive directions of $$x$$-axis and $$y$$ axis respectively. If the perpendicular from origin $$O$$ to this line makes an angle of $$\frac{\pi}{6}$$ with positive direction of $$y$$-axis and the area of $$\triangle \mathrm{OAB}$$ is $$\frac{98}{3} \sqrt{3}$$, then $$\mathrm{a}^{2}-\mathrm{b}^{2}$$ is equal to :

A
$$\frac{392}{3}$$
B
98
C
196
D
$$\frac{196}{3}$$
4
JEE Main 2023 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let the solution curve $$y=y(x)$$ of the differential equation

$$ \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\} \text { pass through the origin. Then } y(1) \text { is equal to : } $$

A
$$\exp \left(\frac{1-\pi}{4 \sqrt{2}}\right)$$
B
$$\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)$$
C
$$\exp \left(\frac{4+\pi}{4 \sqrt{2}}\right)$$
D
$$\exp \left(\frac{\pi-4}{4 \sqrt{2}}\right)$$
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