Page 30 - mathsvol1ch1to3ans

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2n 2 −n
3 9 3
= 27.
3 3n
3n
3
4
2n
3 (3 )(3 −n ) = 3 (3 )
n + 4 = 3n + 3
1
Hence n =
2
5. Find the radius of the spherical tank whose volume is 32π/3 units.
4 32π
3
3
Solution:Since Volume (V ) = ( )πr = we have r = 8. Hence radius r = 2 units.
3 3
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6. Solve (1 − x) 1/4 + (15 + x) 1/4 = 2.
1
4
Solution: Let u = (1 − x) 4 and hence x = 1 − u .The given equation can be rewritten as
1
4 ( )
u + (16 − u ) 4 = 2
1
4 ( )
(16 − u ) 4 = 2 − u
Raising to fourth power
16 − u 4 = (2 − u) 4

2
16 − u 4 = 16 − 32u + 24u + 32u = 0
3
4
2
2u + 8u − 24u + 32u = 0
2
2u(u − 2)(u − 2u + 8) = 0

1
( )
2
u − 2u + 8 has no solution in R. From other factors, u = 0, u = 2. (1 − x) 4 = 0 ⇒ x = 1.
1
( )
(1 − x) 4 = 2 ⇒ x = −15. Hence the values of x are 1, −15.
√
7. Solve (x + 1) 1/3 = x − 3.
2
Solution:Squaring both sides, we have (x + 1) 3 = x − 3. Raising to the power of 3 we have
3
2
2
2
3
(x + 1) = (x − 3) . The expression now becomes, x + 2x + 1 = x − 9x + 27x − 27.
2
3
Hence we have x − 10x + 25x − 28 = 0 a 0 = 28, a n = 1The dividers of a 0 :
1, 2, 4, 7, 14, 28, The dividers of a n : 1
1, 2, 4, 7, 14, 28
Therefore, check the following rational numbers : ±
1
7
is a root of the expression, so factor out x − 7
1
3
2
x − 10x + 25x − 28
Compute to get the rest of the equation :
x − 7
2
2
x − 3x + 4 = (x − 7) (x − 3x + 4) √
7 + 6
8. Simplify by rationalising the denominator. √ .
3 − 2
Solution:Multiplying both numerator and denominator by the conjugate of the denominator, we
√
√
√
√
√
(7 + 6)(3 + 2) 21 + 3 6 + 7 2 + 2 3
have =
9 − 2 7