Content Curator | Updated On - May 12, 2024
TS EAMCET 2023 Question Paper May 14 Shift 2 PDF is available here for download. TS EAMCET 2023 Question Paper consists of 160 questions carrying 1 mark each. TS EAMCET 2023 Question Paper May 14 Shift 2 PDF for MPC includes three subjects, Physics, Chemistry and Mathematics. The Physics and Chemistry section of the paper includes 40 questions each while the Mathematics section includes a total of 80 questions.
TS EAMCET 2023 Question Paper May 14 Shift 2 PDF
| TS EAMCET 2023 Question Paper PDF | TS EAMCET 2023 Answer Key PDF | TS EAMCET 2023 Solution PDF |
|---|---|---|
| Download PDF | Download PDF | Download PDF |
Also Check:
TS EAMCET Previous Year Question Papers
| TS EAMCET 2022 Question Paper | TS EAMCET 2021 Question Paper |
| TS EAMCET 2020 Question Paper | TS EAMCET 2019 Question Paper |
| TS EAMCET 2018 Question Paper | TS EAMCET 2017 Question Paper |
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TS EAMCET Questions
1. The orthocenter of the triangle whose sides are given by x + y + 10 = 0, x - y - 2 = 0 and 2x + y - 7 = 0 is
The orthocenter of the triangle whose sides are given by x + y + 10 = 0, x - y - 2 = 0 and 2x + y - 7 = 0 is
(-4, -3)
(-4, -6)
(4,6)
(3,6)
2. The velocity of a particle having a magnitude of 10 ms-1 in the direction of 60° with positive X-axis is
The velocity of a particle having a magnitude of 10 ms-1 in the direction of 60° with positive X-axis is
5i - 5√3j
5√3i - 5j
5√3i + 5j
5i + 5√3j
3. If \(\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1}+ \frac{Bx+C}{2x^2+3}\), then A - B + C=
If \(\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1}+ \frac{Bx+C}{2x^2+3}\), then A - B + C=
2
1
3
6
4. If order and degree of the differential equation corresponding to the family of curves y2 = 4a(x+a)(a is parameter) are m and n respectively, then m+n2 =
If order and degree of the differential equation corresponding to the family of curves y2 = 4a(x+a)(a is parameter) are m and n respectively, then m+n2 =
3
4
5
2
5. If 2i - j + 3k, -12i - j - 3k, -i + 2j -4k and λi + 2j - k are the position vectors of four coplanar points, then λ =
If 2i - j + 3k, -12i - j - 3k, -i + 2j -4k and λi + 2j - k are the position vectors of four coplanar points, then λ =
9
-2
8
6
6. If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]
If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]\(\frac{-π}{4}\)
\(\frac{π}{4}\)
\(\frac{3π}{4}\)
\(\frac{-3π}{4}\)
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