IIT JEE Main PYQ Differential Equation 4 April 2024 shift 1

[@sumant81]

Love the fact that you solved it like how a student would and what should one think .It would have been very easy to say x=i is a factor and start off .Well done .

[@sanjeevsir29077]

Thanks

[@abhijitjoshi9287]

Brilliantly solved. Discovering i and -i as factors was key to the solution.

[@sanjeevsir29077]

Thanks ji 😊

[@HarshitPandey-d3p]

beautiful question

[@sanjeevsir29077]

Yes ji 👌

[@Shubham-d8v]

good question with best solution 👍

[@sanjeevsir29077]

@@Shubham-d8v thanks ji

[@chuzzles]

3x² = 2x² + x² (x⁴ + 2x³ + 2x²) + (x² + 2x + 2) =(x² + 2x + 2)(x² +1)

[@notanigga829]

I tried the same but couldn't factor 😅

[@sanjeevsir29077]

Difficult to think 😊

[@sanjeevsir29077]

​@@notanigga829yes difficult to think 😊

[@Smitopam]

great thikinig

[@raghvendrasingh1289]

👍 yes sir aise hi x^5+x+1 aur x^5+x^4+1 ke factor karte hain x = w se zero milta hai isliye x^2+x+1 ek factor hoga.waise x^2 minus aur plus karke bhi kar sakte hain.a^3 - b^3 use karke ek aur hai x^4+12x-5 = x^4+12x+4-9 = (x^2+2)^2 - 4x^2 + 12x -9 = ( x^2+2)^2 - (2x - 3)^2 = (x^2+2x-1)(x^2 - 2x+5)

[@rishabhtiwari2325]

A tough question

[@amarendrasingh7327]

A good question not tough

[@sanjeevsir29077]

@@amarendrasingh7327 yes ji

[@rishabhtiwari2325]

@@amarendrasingh7327 mujhe factorization tricky laga, baaki partial fraction integration toh ez hai