1plsss. reply soon , i m just not getting my mistake.......
\begin{vmatrix} 3x &4x &-5 \\ 2x^{2}&9 &2x \\ x& 8 & 3x \end{vmatrix}=ax^{4}+ bx^{3}+cx^{2}+ dx+e,then,
the value of 6a+5b+4c+3d+2e is equal to.......
a.273 b.235 c.-15 d.-53
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10 Answers
13x(27x-16x)-4x(6x3-2x2)-5(16x2-9x)
=33x2-24x4+8x3+45x
=-24x4+8x3-47x2+45x
comparing we have a=-24
b=8
c=-47
d=45
e=0
6a= -144
5b=40
4c=-188
3d=135
2e=0
adding all we get -157 so either ur options are wrong or the question typed is wrong
62There is a much simpler method by derivatives...
can you guys see it?!
1but bhaiyya wont tht be too big.....i men three determinants and differntiaion ................so wont just opening be better?????
21No AKand, not at all
it wud be real short,
put x=0; get e
take deriv. put x=0, get d
divide by x^4 on both sides put x --> ∞ get a
similarly get b,c