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PHYSICS MADE EASY (KOTA)
JEE MAIN 2025 4th APRIL SHIFT -1 & 2 SOLVED PAPER; PERMITIVITY OF FREE SPACE; GAUSS LAW APPLICATION;

... #MATHEMATICAL FUNCTIONS, #EXPONENTIAL AND LOGARITHMIC FUNCTIONS, #FLOOR AND CELLING FUNCTIONS, ...

6:21
JEE MAIN 2025 4th APRIL SHIFT -1 & 2 SOLVED PAPER; PERMITIVITY OF FREE SPACE; GAUSS LAW APPLICATION;

2 views

55 minutes ago

Shivang Maths Academy JEE
Q) If 1/(2×3×4)+1/(3×4×5)+1/(4×5×6) +…..+1/(100×101×102)=𝑘/101, then 34 k #jeemains #class12 #jee

JEE MAINS 2022 ( 29 July – SHIFT 1 ) Q) If 1/(2×3×4)+1/(3×4×5)+1/(4×5×6) +…..+1/(100×101×102)=𝑘/101, then 34 k is equal to ...

2:49
Q) If 1/(2×3×4)+1/(3×4×5)+1/(4×5×6) +…..+1/(100×101×102)=𝑘/101, then 34 k #jeemains #class12 #jee

71 views

57 minutes ago

Shivang Maths Academy JEE
Q) If 𝜃∈[−7𝜋/6,4𝜋/3], then the number of solutions of√3 cosec^2 𝜃−2(√3−1)cosec𝜃 −4=0, is

JEE MAIN 2025 ( 2 April – Shift 2 ) jee advanced trigonometric question jee advanced maths solution jee advanced 2025 paper ...

3:01
Q) If 𝜃∈[−7𝜋/6,4𝜋/3], then the number of solutions of√3 cosec^2 𝜃−2(√3−1)cosec𝜃 −4=0, is

0 views

1 hour ago