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Correct option is  D) 1

Correct option is (D) 1

The value of \(2x^3-2x^2+1\) at x = 1

\(=2\times1^3-2\times1^2+1\)

= 2 - 2 + 1

= 1.

Correct option is  C) 7

Correct option is (C) 7

Degree of \(3x^2y^2z^3\) = 2+2+3 = 7.

Correct option is  D) – 4ab

Correct option is  C) (a + b)²

Correct option is (C) (a + b)²

\((a–b)^2+4ab=a^2+b^2-2ab+4ab\)

\(=a^2+b^2+2ab\)

= \((a+b)^2\).

A) (a + b)² – 2ab 

Correct option is  B) 0

Correct option is (B) 0

\((a-b)^2-(a-b)^2=0.\)

Correct option is  C) 4ab

D) 16x² – 25y²

Correct option is (D) 16x² – 25y²

(4x + 5y) (4x – 5y) = \((4x)^2-(5y)^2\)

= \(16x^2-25y^2\).

A) – 21 m² n + 14 mn³

Correct option is (A) –21 m²n + 14 mn³

\((3m-2n^2)\times(-7\,mn)\) \(=3m\times-7mn-2n^2\times-7mn\)

\(=(3\times-7)m^2n+(-2\times-7)mn^3\)

= \(–21\,m^2n+14\,mn^3\).

Correct option is  A) 9,74,000

Correct option is (A) 9,74,000

\(987^2–13^2=(987-13)(987+13)\)

= 974 \(\times\) 1000

= 9, 74, 000.

C) (a + b) (a – b)

Correct option is (C) (a + b) (a – b)

\(302\times298\) = (300+2) (300-2)

Let a = 300 & b = 2

\(\therefore\) Formula involved in the product of \(302\times298\) is (a+b) (a–b).

Correct option is  B) (a – b)² 

Correct option is (B) (a – b)²

\(196^2=(200-4)^2,\)

Let a = 200, b = 4.

Then formula use in the product of \(196^2\) is \((a-b)^2.\)

Correct option is  C) A&B

Correct option is (C) A&B

a(a + 1) = 6 \(\Rightarrow\) \(a^2+a-6=0\)

\(\Rightarrow\) \(a^2+3a-2a-6=0\)

\(\Rightarrow\) a(a+3) - 2(a+3) = 0

\(\Rightarrow\) (a-2) (a+3) = 0

\(\Rightarrow\) a - 2 = 0 or a+3 = 0

\(\Rightarrow\) a = 2 or a = -3

Correct option is  D) 0

Correct option is (D) 0

\(196^2–(194+2)^2\) \(=196^2–196^2=0\)

A) (a + b + c)² 

Correct option is (A) \((a+b+c)^2\)

\(a^2+b^2+c^2+2ab+2bc+2ca\) \(=a^2+b^2+2ab+2bc+2ca+c^2\)

\(=(a+b)^2+2(a+b)c+c^2\)

= \((a+b+c)^2\)

Let the number of jambhul trees planted be x.

∴ Number of ashoka trees = x – 60 

According to the given condition, x + x – 60 = 200

∴ 2x = 200 + 60 

∴ 2x = 260

∴ x = 260/2 = 130

∴ 130 jambhul trees were planted.

(c) 4

In the given expression there are 4 terms.

(a) 4x² and –3xy

A term is the product of factors.

(x² + 2x) (2x + 3)

= x² (2x + 3) + 2x (2x + 3)

= 2x³ + 3x² + 4x² + 6x .

= 2x³ + 7x² + 6x

3x² + 4xy + 2y²

= 3 (5)² + 4(5) (2) + 2(2)²

= 75 + 40 + 8

= 123

(c) – 5 × x × x × y × y × z

Factors may be numerical as well as algebraic (literal).

(i) Given –3y

The coefficient of y is -3.

(ii) Given 2ab

The coefficient of a is 2b.

(iii) Given -7xyz

The coefficient of z is -7xy.

(iv) Given -3pqr

The coefficient of p is -3qr.

(v) Given 9xy²z

The coefficient of y² is 9xz.

(vi) Given x³ +1

The coefficient of x³ is 1.

(vii) Given − x²

The coefficient of x² is -1.

(i) Given -12x

The numerical coefficient of x is -12.

(ii) Given -7xy

The numerical coefficient of x is -7y.

(iii) Given xyz

The numerical coefficient of x is yz.

(iv) Given -7ax

The numerical coefficient of x is -7a.

(i) In (2x + 2) put x = 1 and y= 0
2x + 2y = 2 x 1 + 2 x 0
= 2+ 0 = 2

(ii) In (2x² + y² + 1) put x = 1 and y = 0
2x² + y² + 1 = 2(1)² + (0)² + 1
= 2 x 1 +0 + 1
= 2 + 1 = 3

(iii) In (2x²y+ 2x²y² +2) put x = 1 and y = 0
2x²y + 2x²y² + y² = 2(1)² x 0 + 2(1)²x (0)²+ (0)²
= 2 x 1 x 0 + 2 x 1 x 0 + 0
= 0 + 0 + 0 = 0

(iv) In (x² + xy + 5) put x = 1 and y = 0
x² + xy + 5 = (1)² + 1 x 0 +5
= 1 + 0 + 5 = 6

Correct option is A) x

Factors of 7y³ will be 7, y, y, y.

Coefficient of yz in 15/16 xyz is 15/16 x.

The sum of 4xy, 3.5xy, – 2xy, 2xz is 5.5xy + 2xz.

The number of terms in the expression 8x² + 2xy + 3x² + 2y² + 2x² is 3.

\(\frac{5}{6} x + \frac{7}{6} x + \frac{1}{6} x\)

= \(\frac{5x + 7x + x}{6}\)

= \(\frac{13x}{6}\)

Given 2x + (5x – 3y)

Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in the parentheses when we remove them.

= 2x + 5x – 3y

On simplifying, we get

= 7x – 3y

(2x² + 7x – 3) consists three terms, 2x², 7x and – 3.

In 6x²y + 9x² coefficient of y is 6x².

(i) In 6x² + 2x – 4 put x = 2

6x² + 2x – 4 = 6(2)² + 2(2) – 4
= 6 x 4 + 4 – 4
= 24 + 4 – 4
= 28 – 4 = 24

(ii) In x³ + 6x – 2 put x = 2

x³ + 6x – 2 = (2) + 6(2) – 2
= 8 + 12 – 2
= 20 – 2
= 18

(i) In x – 3 put x = 2

x – 3 = 2 – 3 = -1

(ii) In 2x – 5 put x = 2

2x – 5 = 2 x 2 – 5
= 4 – 5 = -1

(iii) In 9 – 6x put x = 2

9 – 6x = 9 – 6 x 2
= 9 – 12 = -3

(iv) In 3x² – 4x – 7 put x = 2

3x² – 4x – 7 = 3 x 2 x 2 – 4 x 2 – 7
= 12 – 8 – 7
= 12 – 15 = -3

(v) In 5x/2 – 4 put x = 2

5x/2 – 4 = (5 x 2)/2 – 4

= 5 – 4 = 1

(i) Coefficient of x in 4x is 4.

(ii) Coefficient of y² in 9x²y² is 9x²

Coefficient of x² in 9x²y² is 9y²

(iii) Coefficient of x³ in \(\frac{-8}{5}\)y³y³ is \(\frac{-8}{5}\)x³ 

Coefficient of x³y³ in \(\frac{-8}{5}\)x³y³ in \(\frac{-8}{5}\)y³

(iv) Coefficient of a² in \(\frac{9a^2b^2}{13}\) is \(\frac{9}{13}\)b²

Coefficient of b² in \(\frac{9a^2b^2}{13}\) is \(\frac{9}{13}\)a²

Number of terms in 5xy + 5y + 2 will be 3.

(i) trinomial

(ii) 8,

(iii) 12x³ + 14y²,

(iv) variable

In (x² – y²) put x = 4 and y = 2

x² – y² = (4)² – (2)²

= 16 – 4 

= 12

(i) In (a² – b²) put a = 2 and b = -2
a² – b² = (2)² – (-2)
= 4 – 4 = 0

(ii) In (a² – ab + b²) put a = 2 and b = -2
a² – ab + b² = (2)² – 2x(-2) + (-2)²
= 4 + 4 + 4 = 12

(iii) In a² + b² put a = 2 and b = -2
a² + b² = (2)² + (-2)²
= 4 + 4 = 8

Subtracting 2a + 3b from 5a + 2b, we get 3a – b.

(i) In 4p + 5 put p =-1

4p + 5 

= 4 x (-1) + 5

= -4 + 5 

= 1

(ii) In -3p² + 4p +8 put p = -1

-3p² + 4p + 8 

= -3(-1)²+ 4(-1) + 8

= -3 x 1 – 4 + 8

= -3 – 4 + 8 

= 1

(iii) In 3(p – 2) + 6 put p = -1

3(p – 2) + 6 

= 3 (-1 – 2) + 6

= 3(-3) + 6 

= -9 + 6 

= -3

Let should be added to (2p + 6) to get 3p – q + 6.

(2p + 6) + R = (3p – 9 + 6)
R = (3p – q + 6) – (2p + 6)
= 3p – q + 6 – 2p – 6
= 3p – 2p – q + 6 – 6
= p – q + 0

Required expression = p – q

Let A should be subtracted from 7x – 8y to get( x+y + z)

(7x – 8y) – A = x + y + z
A = (7x – 8y) – (x + y + z)
= 7x – 8y – x – y – 2
= 6x – 9y – 2

(i) Given x = 1, y = -1, z = 2, a = -2, b = 1, c = -2

Consider ax + by + cz

On substituting the given values

= (-2) (1) + (1) (-1) + (-2) (2)

= –2 – 1 – 4

= –7

(ii) Given x = 1, y = -1, z = 2, a = -2, b = 1, c = -2

Consider ax² + by² – cz

On substituting the given values

= (-2) × 1² + 1 × (-1)² – (-2) × 2

= 4 + 1 – (-4)

= 5 + 4

= 9

(iii) Given x = 1, y = -1, z = 2, a = -2, b = 1, c = -2

Consider axy + byz + cxy

= (-2) × 1 × -1 + 1 × -1 × 2 + (-2) × 1 × (-1)

= 2 + (-2) + 2

= 4 – 2

= 2

If x = 4 then value of 2x + 8 will be 16.

Correct option is B) 6

Degree of the monomial 9xy²z³ = 1 + 2 + 3 = 6

(i) False

(ii) True

(iii) True

(iv) False

(i) 3x – 5 – x + 9 = 2x +4
Put x = 3
2x + 4 = 2(3) + 4 = 6 +4 = 10

(ii) 2 – 8x + 4x + 4 = 6 – 4x
Put x = 3
6 – 4x = 6 – 4(3) = 6 – 12 = -6

(iii) 3a +5 – 8a +1 = -5a + 6
Put a = -1
-5a + 6 = -5( -1) +6
= 5 + 6 = 11

(iv) 10 – 3b – 4 – 5b = 6 – 8b
Put b = -2
6 – 8b = 6 – 8(-2)
= 6 + 16 = 22

(v) 2a – 2b – 4 – 5 + a= 3a – 2b – 9
Put a = – 1 and b = -2
3a – 2b – 9 = 3(-1) – 2(- 2) – 9
= – 3 + 4 – 9 = -8

(i) x² – (-5x²) = x² + 5x²

= (1 + 5) x²
= (6)x² = 6x²

(ii) (a + b) (a – b)

= a + b – a + b

= a – a + b + b
= (1 – 1)a + (1 + 1)b
= 0a + 2b = 2b

(iii) (4x² – 3xy + 8) – (x² + 5x + 4)

= 4x² – 3xy + 8 – x² – 5x – 4
= 4x² – x² – 3xy – 5x + 8 – 4
= (4 – 1)x² – 3xy – 5x + 4
= 3x² – 3xy – 5x + 4

(iv) (3xy – 2x² – 2y²) – (5x² – 7xy + 5y²)

= 3xy – 2x² – 2y² – 5x² + 7xy – 5y²
= 3xy + 7xy – 2x² – 5x² – 2y² – 5y²
= (3 + 7)xy + (-2 – 5)x² + (-2 – 5)y²
= 10xy – 7x² – 7y²
= -7x² – 7y² + 10xy

(v) (5p² + 2q² – pq²) – (4pq – 5q² – 3p²)

= 5p² + 2q² – pq² – 4pq +5q² + 3p²
= 5p² + 3p² + 2q² +5q² – pq² – 4pq
= (5 + 3)p² + (2 + 5)q² – pq² – 4pq
= 8p² + 7q² – pq² – 4pq

(vi) (5x – 10) – (x² + 10x – 5)

= 5x -10 – x² – 10x +5
= -x² + 5x – 10x – 10 + 5
= -x² + (5-10)x – 5
= -x² – 5x – 5