19say, the eqn in (y) is f(y).....now check the values of f(y) at each of the given intervals in the options,,,,the one in which f(y) gives opposite sign gives you the location of the roots !
let f(x)=x^{3}+x^{2}+100x+7sinx
then the eq
\frac{1}{y-f(1)}+ \frac{2}{y-f(2)}+ \frac{3}{y-f(3)}=0 has
1)exactly one root lying in (f(1),f(2))
2)booth roots lying in (f(2),f(3))
3)exactly one root lying in (-∞,f(1))
4)exactly one root lying in (f(3),∞)
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4 Answers
19
11Ans) f(x) = x3 + x2 + 100x + 7 sin x
f ' (x) = 3x2 + 2x + 100 + 7 cos x
= 3x2 + 2x + 93 + 7 (1 + cos x)
= 3 { (x + 1/3) 2 } + (278/3) + 7(1+cos x) > 0
Therefore, f(x) is increasing function
Thus f(1) < f(2) < f(3)
Now, let f(1) = a, f(2) = b, f(3) = c
therefore, a<b<c
Now, the given eqn is 1y-a + 2y-b + 3y-c = 0
or (y-b) (y-c) + 2 (y-c) (y-a) + 3 (y-a) (y-b) = 0
Now, let g(y) = (y-b) (y-c) + 2 (y-c) (y-a) + 3 (y-a) (y-b)
Therefore, g(a) > 0, g(b) < 0 and g(c) > 0
Hence one real root b/w f(1) and f(2) and other b/w f(2) and f (3) ..................[ Rolle's theorm]
11Alternately for this MCQ, we may have a different approach....
G(y)=\frac{1}{y-f(1)}+\frac{2}{y-f(2)}+\frac{3}{y-f(3)}
Now
\int G(y)dy=ln\left\{y-f(1) \right\}\left\{y-f(2) \right\}^2\left\{y-f(3) \right\}^3=0
That gives
y-f(1)=\frac{1}{\left\{y-f(2) \right\}^2.\left\{y-f(3) \right\}^3}
Now when y is between (-∞,f(1)), l.h.s decreases, but r.h.s increases, as y→f(1). Thus no solutions....
Similarly for the interval (f(3),∞).......Double root not possible since G'(y) ≠0.....
Only chances of solutions are when y lies in the intervals defined by options 1 and 2.
