Chhattisgarh board of secondary education - CLASS 10 - Mathematics - september assignment (Question with answer) - full details

 Chhattisgarh Board Of Secondary Education, Raipur  Academic Session 2020-21

Month- September 

Assignment - 1  

Class - X  

Subject-Mathematics  

             Total marks - 20 

Instruction:- Please attempt the questions as per given instruction.

Q.1 If divisor is 𝟑𝐱^𝟐 − 𝟐𝐱 + 𝟐 quotient is 𝐱 + 𝟏 and remainder is

3 then find the divident.

Ans-

Dividend = Divisor x Quotient + Remainder.

Here,

Divisor x Quotient = ( x +1) ( 𝟑𝐱^𝟐 − 𝟐𝐱 + 𝟐)

= 3x³ - 2x² + 2x + 3x² – 2x + 2

= 3x³ + x² + 2

Adding the remainder,

 = 3x³ + x²  + 2 + 3

= 3x³ + x² + 5   Ans

 

= Dividend

Q.2 Find the value of K, in quadratic equation 9𝐱^𝟐 − 𝐊𝐱 + 𝟏𝟔 = 𝟎 If roots are equal.

Ans. 

Given Quadratic Equation is

9x² –Kx + 16 = 0

Comparing equations with ax² + bx  + c  = 0

a  = 9, b = -K, c = 16

Given that the Equation have real roots

b² – 4ac  = 0

(-K)² – 4 (9) (16) = 0

K² – 576 = 0

K² = 576

K = 24 (or) -24    Ans

Q.3 Find the quadratic equation whose roots are (𝟔 + √𝟓 ) and (𝟔 − √𝟓 )

Ans.

 

We know that if m and n are the roots of a quadratic equations ax² + bx + c = 0

Then the sum of the roots is (m + n)

Product of the roots is (mn)

 then the quadratic equation becomes x² – (m + n)x + mn = 0

Here it is given that the roots of the quadratic equation are

 m = (6 + √5) and

n = (6 - √5) 

 

therefore,

 

The sum of the roots is :

m + n = (6 + √5)  + (6 - √5  = 6 + √5 + 6 - √5 = 6 + 6 = 12

 

And the product of the roots is:

 mn  = (6 + √5) (6 - √5)   

= (6)² – (√5)²             (a² – b² = (a – b ) (a + b))

= 36 – 5 = 31     

 

Therefore, the required quadratic equation is

 

x² – (m + n)x + mn = 0

x² – (12)x + 31 = 0

x² – 12x + 31 = 0   Ans

 

Hence, x ² – 12x + 31 = 0 is the quadratic equation whose roots are (6 + √5 and (6 - √5)

Q.4 For what value of K, system of equation 𝟓𝐱 − 𝟕𝐲 − 𝟓 = 𝟎 and 𝟐𝐱 + 𝐤𝐲 − 𝟏 = 𝟎, have unique solution.

Ans.

𝟓𝐱 − 𝟕𝐲 − 𝟓 = 𝟎  ...(i)

𝟐𝐱 + 𝐤𝐲 − 𝟏 = 𝟎 ...(ii)

Here,

By comparing equation (i) and equation (ii) with

a₁ x + b₁ y + c₁ = 0

a₂x + b₂y + c₂ = 0

We get the value of 

a₁  = 5 , b₁ = -7, c₁ = -5

a₂ = 2 , b₂ = K , c₂ = -1

System has a unique solution.

    

Q.5 Solve the equation- (NOT- using graphical method)

𝟒𝐱 − 𝟓𝐲 + 𝟏𝟔 = 𝟎

𝟐𝐱 + 𝐲 − 𝟔 = 𝟎

Ans.

4x - 5y + 16 = 0 ...(i)

2x + y - 6 = 0 ...(ii)

in Equation (ii) multiply by 2, we get

4x + 2y - 12 = 0 ...(iii)

then equation (i) - equation (ii), we get

-7y + 28 = 0

y = 28/7 = 4

y = 4

Putiing the value of y =4 in equation (i), we get the value of x

4x - 5(4) + 16 = 0

x  = 1

so, by solving these equation we get

x =1 and y = 4.

Answer (x,y) = (1,4)

Q.6 Solve the equation-

Ans.

 

24x = 5x² - 5 
5x² - 24x - 5 = 0
5x² -25x + x - 5 = 0
5x( x - 5) + 1 (x -5 ) = 0
(x - 5) = 0, (5x +1) = 0
Ans      x = 5, -1/5         

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