Page 22 - mathsvol1ch1to3ans

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2
Interval Sign of (x + 3) Sign of x − Sign of 2x + x − 15
2
(−∞, −3) − − +

5
−3, + − −
2

5
, ∞ + + +
2

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So the inequality is satisfied in −3, .
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2
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(ii) −x + 3x − 2 ≥ 0. can be re-written as x − 3x + 2 ≤ 0. On factorizing the polynomial we
get (x − 1)(x − 2) ≤ 0

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Interval Sign of (x − 1) Sign of (x − 2) Sign of x − 3x + 2 ≤ 0
(−∞, 1) − − +

(1, 2) + − −

(2, ∞) + + +

So the inequality is satisfied in [1, 2].
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2. Solve the inequality x + ax + a − 1 > 0.

Solution: We use completion of squares method to factorize the polynomial
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f(x) = x + ax + a − 1
2 2
a a
2
= x + ax + − + a − 1
4 4

a a
h i 2 2
= x + − − a + 1
2 4
a a
h i 2 h i 2
= x + − − 1
2 2
= (x + a − 1)(x + 1)
The inequality holds true only when,either x + a > 1, x > −1 (or) x + a < 1, x < −1.

Exercise - 2.11

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1. Find the zeros of the polynomial function f(x) = 4x − 25.
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2
Solution:f(a) = 4a − 25 = 0 ⇒ a = ± . Hence the zeros of the polynomial function f(x) =
2
5 5
− and + .
2 2
3
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2. If x = −2 is one root of x − x − 17x = 22, then find the other roots of equation.
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2
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Solution:x − x − 17x − 22 = (x + 2)(ax + bx + c).
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Comparing the coefficients of x , we have a = 1.Comparing constant terms, we have c = −11.