Q1: Give an example of a monomial and a binomial having degrees of 82 and 99, respectively.
Sol: An example of a monomial having a degree of 82 = x^{82}
An example of a binomial having a degree of 99 = x^{99} + x
Q2: Find the value of the polynomial 5x – 4x^{2} + 3 at x = 2 and x = –1.
Sol: Let the polynomial be f(x) = 5x – 4x^{2} + 3
Now, for x = 2,
f(2) = 5(2) – 4(2)^{2} + 3
⇒ f(2) = 10 – 16 + 3 = –3
Or, the value of the polynomial 5x – 4x^{2} + 3 at x = 2 is 3.
Similarly, for x = –1,
f(–1) = 5(–1) – 4(–1)^{2} + 3
⇒ f(–1) = –5 –4 + 3 = 6
The value of the polynomial 5x – 4x^{2} + 3 at x = 1 is 6.
Q3: Find the value of x^{3} + y^{3} + z^{3} – 3xyz if x^{2} + y^{2} + z^{2} = 83 and x + y + z = 15
Sol: Consider the equation x + y + z = 15
From algebraic identities, we know that (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)
So,
(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + xz)
From the question, x^{2} + y^{2} + z^{2} = 83 and x + y + z = 15
So,
15^{2} = 83 + 2(xy + yz + xz)
⇒ 225 – 83 = 2(xy + yz + xz)
Or, xy + yz + xz = 142/2 = 71
Using algebraic identity a^{3} + b^{3} + c^{3} – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca), x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – (xy + yz + xz))
Now,
x + y + z = 15, x^{2} + y^{2} + z^{2} = 83 and xy + yz + xz = 71
So, x^{3} + y^{3} + z^{3} – 3xyz = 15(83 – 71)
⇒ x^{3} + y^{3} + z^{3} – 3xyz = 15 × 12
Or, x^{3} + y^{3} + z^{3} – 3xyz = 180
Q4: If (x – 1/x) = 4, then evaluate (x^{2} + 1/x^{2}) and (x^{4} + 1/x^{4}).
Sol: Given, (x – 1/x) = 4
Squaring both sides we get,
(x – 1/x)^{2} = 16
⇒ x^{2} – 2.x.1/x + 1/x^{2} = 16
⇒ x^{2} – 2 + 1/x^{2} = 16
⇒ x^{2} + 1/x^{2} = 16 + 2 = 18
∴ (x^{2} + 1/x^{2}) = 18 ….(i)
Again, squaring both sides of (i), we get
(x^{2} + 1/x^{2})^{2} = 324
⇒ x^{4} + 2.x^{2}.1/x^{2} + 1/x^{4} = 324
⇒ x^{4} + 2 + 1/x^{4} = 324
⇒ x^{4} + 1/x^{4} = 324 – 2 = 322
∴ (x^{4} + 1/x^{4}) = 322.
Q5: Check whether (7 + 3x) is a factor of (3x^{3} + 7x).
Sol: Let p(x) = 3x^{3} + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.
By the remainder theorem, we know that when p(x) is divided by g(x) then the remainder is p(–7/3).
Now, p(–7/3) = 3(–7/3)^{3} + 7(–7/3) = –490/9 ≠ 0.
∴ g(x) is not a factor of p(x).
Q6: Compute the value of 9x^{2} + 4y^{2} if xy = 6 and 3x + 2y = 12.
Sol: Consider the equation 3x + 2y = 12
Now, square both sides:
(3x + 2y)^{2} = 12^{2}
⇒ 9x^{2} + 12xy + 4y^{2} = 144
⇒ 9x^{2} + 4y^{2} = 144 – 12xy
From the questions, xy = 6
So,
9x^{2} + 4y^{2} = 144 – 72
Thus, the value of 9x^{2} + 4y^{2} = 72
Q7: Calculate the perimeter of a rectangle whose area is 25x^{2} – 35x + 12.
Sol: Given,
Area of rectangle = 25x^{2} – 35x + 12
We know, area of rectangle = length × breadth
So, by factoring 25x^{2} – 35x + 12, the length and breadth can be obtained.
25x^{2} – 35x + 12 = 25x^{2} – 15x – 20x + 12
⇒ 25x^{2} – 35x + 12 = 5x(5x – 3) – 4(5x – 3)
⇒ 25x^{2} – 35x + 12 = (5x – 3)(5x – 4)
So, the length and breadth are (5x – 3)(5x – 4).
Now, perimeter = 2(length + breadth)
So, perimeter of the rectangle = 2[(5x – 3)+(5x – 4)]
= 2(5x – 3 + 5x – 4) = 2(10x – 7) = 20x – 14
So, the perimeter = 20x – 14
Q8: If a + b + c = 15 and a^{2} + b^{2} + c^{2} = 83, find the value of a^{3} + b^{3} + c^{3} – 3abc.
Sol: We know that,
a^{3} + b^{3} + c^{3} – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca) ….(i)
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca ….(ii)
Given, a + b + c = 15 and a^{2} + b^{2} + c^{2} = 83
From (ii), we have
152 = 83 + 2(ab + bc + ca)
⇒ 225 – 83 = 2(ab + bc + ca)
⇒ 142/2 = ab + bc + ca
⇒ ab + bc + ca = 71
Now, (i) can be written as
a^{3} + b^{3} + c^{3} – 3abc = (a + b + c)[(a^{2} + b^{2} + c^{2}) – (ab + bc + ca)]
a^{3} + b^{3} + c^{3} – 3abc = 15 × [83 – 71] = 15 × 12 = 180.
Q9: Find the values of a and b so that (2x^{3} + ax^{2} + x + b) has (x + 2) and (2x – 1) as factors.
Sol: Let p(x) = 2x^{3} + ax^{2} + x + b. Then, p( –2) = and p(½) = 0.
p(2) = 2(2)^{3} + a(2)^{2} + 2 + b = 0
⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i)
p(½) = 2(½)^{3} + a(½)^{2} + (½) + b = 0
⇒ a + 4b = –3 ….(ii)
On solving (i) and (ii), we get a = 5 and b = –2.
Hence, a = 5 and b = –2.
Q10: Factorise x^{2} + 1/x^{2} + 2 – 2x – 2/x.
Sol: x^{2} + 1/x^{2} + 2 – 2x – 2/x = (x^{2} + 1/x^{2} + 2) – 2(x + 1/x)
= (x + 1/x)^{2} – 2(x + 1/x)
= (x + 1/x)(x + 1/x – 2).
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