1for the first one it's (d) , get the roots as -1 ± √5 4
Let -1 + √54 = m ........ (1)
and -1 - √54 = x ...........(2)
eliminate √5 from first equation , we get √5 = 4m +1
Put this value in (2) -1 - (4m + 1)4 = x
this gives -m -12 = x
1) If m is a root of the eqn. 4x2 + 2x - 1 = 0, then its other root is given by
(a) 4m3 - 3m (b) 4m3 + 3m (c) m - (1/2) (d) -m-(1/2)
2) if a,b,c,p,q,r are six complex numbers, such that p/a + q/b + r/c = 1 + i and a/p + b/q + c/r = 0,
where i = √(-1) then value of
p2/a2 + q2/b2 + c2/r2 is
(a) 0 (b) -1 (c) 2i (d) -2i
3) if log126 = x, log2454 = y, then the value of xy + (5x - 2y) + 4 is .................
share the solutions please....
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6 Answers
11for the second one take the square of the first equation
p2a2 + q2b2 + r2c2 + 2 ( pqab + qrbc + prac ) = 2i
now take out pqrabc common from the second term , the second sum becomes 0 , so ans is (c) 2i
3This is a nice problem.
x = (log6)/(log12) and y = (log54)/(log24).
So, xy = [(log6)(log54)]/[(log24)(log12)].
Also 5x = 5(log6)/(log12)
And 2y = 2(log54)/(log24).
Putting these values in
xy + 5x – 2y + 4
we get ,
[log6(log54) + 4(log12)(log24) + 5(log6)(log24) – 2(log54)(log12)]/[(log12)(log24)].
Or, [log54(log6-2log12) + log24(4log12 + 5log6)]/[(log12)(log24)]/
Or, [-log54(log24) + log24(log(12^4*6^5))]/[(log12)(log24)].
Or, [log(12^4*6^5/54)]/[(log12)].
Or, [log(12^6)]/[(log12)]
= 6 answer…
There may be other methods to do it. I used the base change formula for it.
1708hemang nice answer
\hspace{-16}$Here $\mathbf{x=\log_{12}6}$ and $\mathbf{y=\log_{24}54}$\\\\ Now We have to calculate $\mathbf{xy+5x-2y+4}$\\\\ $\mathbf{y(x-2)+5x+4}$\\\\ $\mathbf{y.\left(\log_{12}6-\log_{12}(12)^2\right)+5x+4}$\\\\ $\mathbf{y.\log_{12}\left(\frac{6}{144}\right)+5x+4}$\\\\ $\mathbf{y.\log_{12}\left(\frac{1}{24}\right)+5x+4}$\\\\ $\mathbf{-\log_{24}54.\log_{12}24+5.\log_{12}6+4}$\\\\ $\mathbf{-\log_{12}54+\log_{12}6^5+4}$\\\\ $\mathbf{=\ln_{12}\left(\frac{6.6.6.6.6}{54}\right)+4}$\\\\ $\mathbf{=\log_{12}(12)^2+4=2.\ln_{12}12+4=6}$
