MHT CET 2025 19th April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2025 19th April Evening Shift

MCQ (Single Correct Answer)

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The lines $\frac{x-3}{1}=\frac{y-2}{1}=\frac{z-5}{-k}$ and $\frac{x-4}{\mathrm{k}}=\frac{y-3}{1}=\frac{\mathrm{z}-3}{2}$ are coplanar, hence $\mathrm{k}=$

1,2

$-2,3$

$-1,2$

$\frac{1}{2}, 1$

MHT CET 2025 19th April Evening Shift

MCQ (Single Correct Answer)

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$$ \int \frac{x \mathrm{~d} x}{(x-1)(x-2)}= $$

$\log \left(\frac{x-1}{x-2}\right)+\mathrm{c}$, where c is the constant of integration

$\quad \log \left(\frac{x-2}{(x-1)^2}\right)+\mathrm{c}$, where c is the constant of integration

$\log \left(\frac{x-2}{x-1}\right)+\mathrm{c}$, where c is the constant of integration

$\quad \log \left(\frac{(x-2)^2}{x-1}\right)+\mathrm{c}$, where c is the constant of integration

MHT CET 2025 19th April Evening Shift

MCQ (Single Correct Answer)

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Let $A$ and $B$ are independent events with $\mathrm{P}(\mathrm{B})=\frac{2}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{11}{20}$, then $\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right)$ is root of the equation

$4 x^2-7 x+3=0$

$4 x^2+7 x+3=0$

$4 x^2-3 x-7=0$

$6 x^2-5 x+1=0$

MHT CET 2025 19th April Evening Shift

MCQ (Single Correct Answer)

-0

The equation of tangent to the curve $y=\cos (x+y)$ where $-2 \pi \leq x \leq 2 \pi$ and which is parallel to the line $x+2 y=0$, is

$2 x+4 y+\pi=0$

$2 x+4 y-\pi=0$

$2 x+4 y-3 \pi=0$

$2 x-4 y+3 \pi=0$