MHT CET 2024 11th May Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

-0

If $\mathrm{f}(x)=\log _{x^2}(\log x)$, then at $x=\mathrm{e}, \mathrm{f}^{\prime}(x)$ has the value

$\frac{1}{\mathrm{e}^2}$

$\frac{1}{\mathrm{e}}$

$\mathrm{e}^2$

$\frac{1}{2 \mathrm{e}}$

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

-0

If $\mathrm{I}=\int_0^{\frac{\pi}{4}} \log (1+\tan x) \mathrm{d} x$, then value of $\mathrm{I}$ is

$\frac{\pi}{16} \log 2$

$\frac{\pi}{2} \log 2$

$\frac{\pi}{8} \log 2$

$\frac{\pi}{4} \log 2$

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

-0

The parametric equations of the circle $x^2+y^2-\mathrm{a} x-b y=0$ are

$x=\frac{\mathrm{a}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{2} \cos \theta, y=\frac{\mathrm{b}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{2} \sin \theta$

$x=\frac{-\mathrm{a}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \sin \theta, y=\frac{-\mathrm{b}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \cos \theta$

$x=\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{2}} \sin \theta, y=\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{2}} \cos \theta$

$x=\frac{\mathrm{a}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \cos \theta, y=\frac{\mathrm{b}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \sin \theta$

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

-0

The curve $y=a x^3+b x^2+c x+5$ touches the X - axis at $(-2,0)$ and cuts the Y -axis at a point Q where its gradient is 3 , then values of $a, b, c$ respectively, are

$3,-\frac{1}{2},-\frac{3}{4}$

$-\frac{3}{4},-\frac{1}{2}, 3$

$-\frac{1}{2},-\frac{3}{4}, 3$

$-\frac{1}{2}, 3,-\frac{3}{4}$

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