🎓 JEE MAIN📅 Year: 2022📚 Mathematics🏷 Straight line
2
Let $\alpha_1, \alpha_2 ; (\alpha_1 < \alpha_2)$ be the values of $\alpha$ for the points $(\alpha, -3), (2, 0)$ and $(1, \alpha)$ to be collinear. Then the equation of the line, passing through $(\alpha_1, \alpha_2)$ and making an angle of $\frac{\pi}{3}$ with the positive direction of the x-axis, is :
Let the eccentricity of the ellipse ${x^2} + {a^2}{y^2} = 25{a^2}$ be b times the eccentricity of the hyperbola ${x^2} - {a^2}{y^2} = 5$, where a is the minimum distance between the curves y = ex and y = logex. Then ${a^2} + {1 \over {{b^2}}}$ is equal to :
🎓 JEE MAIN📅 Year: 2022📚 Mathematics🏷 Inverse Trigonometrical Function
3
$ \alpha = \tan\left(\frac{5\pi}{16} \sin\left(2\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)\right) $
$ \beta = \cos\left(\sin^{-1}\left(\frac{4}{5}\right) + \sec^{-1}\left(\frac{5}{3}\right)\right) $
where the inverse trigonometric functions take principal values.
Then, the equation whose roots are $ \alpha $ and $ \beta $ is :
🎓 JEE MAIN📅 Year: 2022📚 Mathematics🏷 Maxima and Minima
2
If the absolute maximum value of the function $f(x)=\left(x^{2}-2 x+7\right) \mathrm{e}^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}$ in the interval $[-3,0]$ is $f(\alpha)$, then :
🎓 JEE MAIN📅 Year: 2022📚 Mathematics🏷 Maxima and Minima
1
The curve $y(x)=a x^{3}+b x^{2}+c x+5$ touches the $x$-axis at the point $\mathrm{P}(-2,0)$ and cuts the $y$-axis at the point $Q$, where $y^{\prime}$ is equal to 3 . Then the local maximum value of $y(x)$ is:
For any real number $x$, let $[x]$ denote the largest integer less than equal to $x$. Let $f$ be a real valued function defined on the interval $[-10,10]$ by $f(x)=\left\{\begin{array}{l}x-[x], \text { if }[x] \text { is odd } \\ 1+[x]-x, \text { if }[x] \text { is even } .\end{array}\right.$Then the value of $\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x \,d x$ is :
The slope of the tangent to a curve $C: y=y(x)$ at any point $(x, y)$ on it is $\dfrac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}$.
If $C$ passes through the points $\left(0, \tfrac{1}{2}+\tfrac{\pi}{2\sqrt{2}}\right)$ and $\left(\alpha, \tfrac{1}{2}e^{2\alpha}\right)$, then $e^{\alpha}$ is equal to :