9th Maths 2.2
NCERT Class 9th solution of Exercise 2.1
NCERT Class 9th solution of Exercise 2.3
NCERT Class 9th solution of Exercise 2.4
NCERT Class 9th solution of Exercise 2.5
Exercise 2.2
Q1. Find the value of the polynomial `5x - 4x^2 + 3` at
i) `x = 0` ii) `x = - 1` iii) `x = 2`
Sol. :
i) Let `p(x) = 5x - 4x^2 + 3`
`p(0) = 5(0) - 4(0)^2 + 3`
`p(0) = 0 - 0 + 3`
`p(0) = 3`
Answer :
`p(0) = 3`
ii) Let `p(x) = 5x - 4x^2 + 3`
`p(-1) = 5(-1) - 4(-1)^2 + 3`
`p(-1) = - 5 - 4 + 3`
`p(-1) = - 9 + 3`
`p(-1) = - 6`
Answer :
`p(-1) = - 6`
iii) Let `p(x) = 5x - 4x^2 + 3`
`p(2) = 5(2) - 4(2)^2 + 3`
`p(2) = 10 - 16 + 3`
`p(2) = 13 - 16`
`p(2) = - 3`
Answer :
`p(2) = - 3`
Q2. Find `p(0),` `p(1)` and `p(2)` for eac of the following polynomials:
i) `p(y) = y^2 - y + 1` ii) `p(t) = 2 + t + 2t^2 - t^3` iii) `p(x) = x^3` iv) `p(x) = (x-1)(x+1)`
Sol. :
i) `p(y) = y^2 - y + 1` ii) `p(t) = 2 + t + 2t^2 - t^3` iii) `p(x) = x^3` iv) `p(x) = (x-1)(x+1)`
Sol. :
i) `p(y) = y^2 - y + 1`
`p(0) = (0)^2 - (0) + 1`
`p(0) = 0 - 0 + 1`
`p(0) = 1`
`p(1) = (1)^2 - (1) + 1`
`p(1) = 1 - 1 +1`
`p(1) = 0 + 1`
`p(1) = 1`
`p(2) = (2)^2 - (2) + 1`
`p(2) = 4 - 2 + 1`
`p(2) = 2 + 1`
`p(2) = 3`
Answer :
`= 1, 1, 3`
ii) `p(t) = 2 + t + 2t^2 - t^3`
`p(0) = 2 + (0) + 2(0)^2 - (0)^3`
`p(0) = 2 + 0 + 0 - 0`
`p(0) = 2`
`p(1) = 2 + (1) + 2(1)^2 - (1)^3`
`p(1) = 2 + 1 + 2 - 1`
`p(1) = 4`
`p(2) = 2 + (2) + 2(2)^2 - (2)^3`
`p(2) = 2 + 2 + 8 - 8`
`p(2) = 4`
Answer :
`= 2, 4, 4`.
iii) `p(x) = x^3`
`p(0) = (0)^3`
`p(0) = 0`
`p(1) = (1)^3`
`p(1) = 1`
`p(2) = (2)^3`
`p(2) = 8`
Answer :
`= 0, 1, 8.`
iv) `p(x) = (x-1)(x+1)`
`p(0) = (0-1)(0+1)`
`p(0) = (-1)(1)`
`p(0) = - 1`
`p(1) = (1-1)(1+1)`
`p(1) = 0(2)`
`p(1) = 0`
`p(2) = (2-1)(2+1)`
`p(2) = (1)(3)`
`p(2) = 3`
Answer :
` = - 1, 0, 3.`
Q3. Verify whether the following are zeros of the polynomial indicated against them:
i) `p(x) = 3x + 1, x = -frac{1}{3}` ii) `p(x) = 5x - π, x = frac{4}{5}`
iii) `p(x) = x^2 - 1, x = 1, - 1`
iv) `p(x) = (x+l)(x-2), x = - 1, 2`
v) `p(x) = x^2, x = 0`. vi) `p(x) = lx + m, x = -frac{m}{l}`
vii) `p(x) = 3x^2 - 1, x = -frac{1}{sqrt3}, frac{2}{sqrt3}`
viii) `p(x) = 2x + 1, x = frac{1}{2}`
Sol. :
i) `p(x) = 3x + 1`
`p(-frac{1}{3}) = 3(-frac{1}{3}) + 1`
`p(-frac{1}{3}) = - 1 + 1`
`p(-frac{1}{3}) = 0`
Answer :
`-frac{1}{3}` is a zero.
ii) `p(x) = 5x - π`
`p(frac{4}{5}) = 5(frac{4}{5}) - π`
`p(frac{4}{5}) = 4 - π ≠ 0`
Answer :
`frac{4}{5}` is not a zero.
iii) `p(x) = x^2 - 1`
`p(1) = (1)^2 - 1`
`p(1) = 1 - 1`
`p(1) = 0`
`p(-1) = (-1)^2 - 1`
`p(-1) = 1-1`
`p(-1) = 0`
Answer :
`1` and `-1` both are the zeros.
iv) `p(x) = (x+l)(x-2)`
`p(-1) = (-1+l)(-1-2)`
`p(-1) = (0)(-3)`
`p(-1) = 0`
`p(2) = (2+l)(2-2)`
`p(2) = 3(0)`
`p(2) = 0`
Answer :
`-1` and `2` both are the zeros.
v)`p(x) = x^2`
`p(0) = (0)^2`
`p(0) = 0`
Answer :
`0` is the zero.
vi)`p(x) = lx + m`
`p(-frac{m}{l}) = l(-frac{m}{l}) + m`
`p(-frac{m}{l}) = - m + m`
`p(-frac{m}{l}) = 0`
Answer:
`-frac{m}{l}` is a zero.
vii)`p(x) = 3x^2 - 1`
`p(-frac{1}{sqrt3}) = 3(-frac{1}{sqrt3})^2 - 1`
`p(-frac{1}{sqrt3}) = 1 - 1`
`p(-frac{1}{sqrt3}) = 0`
`p(frac{2}{sqrt3}) = 3(frac{2}{sqrt3})^2 - 1`
`p(frac{2}{sqrt3}) = 4 - 1`
`p(frac{2}{sqrt3}) = 3 ≠ 0`
Answer :
`-frac{1}{sqrt3}` is zero but `frac{2}{sqrt3}` is not zero.
viii) `p(x) = 2x + 1`
`p(frac{1}{2}) = 2(frac{1}{2}) + 1`
`p(frac{1}{2}) = 1 + 1`
`p(frac{1}{2}) = 2 ≠ 0`
Answer :
`frac{1}{2}` is not a zero.
Q4. Find the zero of the polynomial in each of the following cases :
i) `p(x) = x + 5` ii) `p(x) = x - 5`
iii) `p(x) = 2x + 5` iv) `p(x) = 3x - 2` v) `p(x) = 3x`
vi) `p(x) = ax : a ≠ 0` vii) `p(x) = cx + d, c ≠ 0, c and d` are real numbers.
Sol. :
Let `p(x) = x + 5 = 0`
`x = - 5`
Answer :
Required zero `- 5`.
ii)`p(x) = x - 5`
Let `p(x) = x - 5 = 0`
`x = 5`
Answer :
Required zero `5`.
iii) `p(x) = 2x + 5`
Let `p(x) = 2x + 5 = 0`
`x = -frac{5}{2}`
Answer :
Required zero `- frac{5}{2}`.
iv) `p(x) = 3x - 2`
Let `p(x) = 3x - 2 = 0`
`x =frac{2}{3}`
Answer :
Required zero `frac{2}{3}`
v) `p(x) = 3x`
Let `p(x) = 3x = 0`
`x = 0`
Answer :
Requried zero `0`
vi) `p(x) = ax`
Let `p(x) = ax = 0`
`x = 0`
Answer :
Requried zero `0`.
vii) `p(x) = cx + d`
Let `p(x) = cx + d = 0`
`x = -frac{d}{c}`
Answer :
Requried zero `-frac{d}{c}`.
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