Elementary Algebra
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If a = √3 - √2 , b = √3 + √2 , then the value of √3 + √2 √3 - √2 a2 + b2 is : b a
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As we know from given question,
a = √3 - √2 √3 + √2
Multiply and divide by √3 - √2, We will geta = √3 - √2 x √3 - √2 √3 + √2 √3 - √2 a = ( √3 - √2)2 3 - 2
a = 3 + 2 - 2√6
a = 5 - 2√6
Again We know from the given question,b = √3 + √2 = 5 + 2√6 √3 - √2
We will get the value by adding a and b,
⇒ a + b = 5 - 2√6 + 5 + 2√6
⇒ a + b = 10;
We will get the value by multiplying a and b,
a x b = ( 5 - 2√6 ) x ( 5 + 2√6 ) = 5 x 5 - 4 x 6 = 25 - 24 = 1Correct Option: B
As we know from given question,
a = √3 - √2 √3 + √2
Multiply and divide by √3 - √2, We will geta = √3 - √2 x √3 - √2 √3 + √2 √3 - √2 a = ( √3 - √2)2 3 - 2
a = 3 + 2 - 2√6
a = 5 - 2√6
Again We know from the given question,b = √3 + √2 = 5 + 2√6 √3 - √2
We will get the value by adding a and b,
⇒ a + b = 5 - 2√6 + 5 + 2√6
⇒ a + b = 10;
We will get the value by multiplying a and b,
a x b = ( 5 - 2√6 ) x ( 5 + 2√6 ) = 5 x 5 - 4 x 6 = 25 - 24 = 1
Now we will calculate the value ofa2 + b2 b a = a3 + b3 ab = ( a + b )3 - 3ab( a + b ) ab
= 103 - 3 x 10 = 1000 - 30 = 970
- In the xy–coordinate system, if (a, b) and (a + 3, b + k) are two points on the line defined by the equation x = 3y – 7, then k =?
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Given in the question,
Points (a, b) and [(a + 3), (b + k)] will satisfy the equation x – 3y + 7 = 0.
Points (a, b) in the equation x – 3y + 7 = 0, We will get
∴ a – 3b + 7 = 0 ...................... (i)
Points (a + 3), (b + k) in the equation x – 3y + 7 = 0, We will get
a + 3 – 3 (b + k) + 7 = 0
⇒ a + 3 – 3b – 3 k + 7 = 0 (Rearrange the equation)Correct Option: B
Given in the question,
Points (a, b) and [(a + 3), (b + k)] will satisfy the equation x – 3y + 7 = 0.
Points (a, b) in the equation x – 3y + 7 = 0, We will get
∴ a – 3b + 7 = 0 ...................... (i)
Points (a + 3), (b + k) in the equation x – 3y + 7 = 0, We will get
a + 3 – 3 (b + k) + 7 = 0
⇒ a + 3 – 3b – 3 k + 7 = 0 (Rearrange the equation)
⇒ a – 3b + 7 + 3 – 3 k = 0 (a – 3b + 7 = 0 , Put the value from Equation (1) ), We will get
⇒ 0 + 3 – 3k = 0
⇒ 3k = 3
⇒ k = 1
- In xy–plane, P and Q are two points having co– ordinates (2, 0) and (5, 4) respectively. Then the numerical value of the area of the circle with radius PQ is
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According to question ,
PQ = √( 5 - 2 )2 + ( 4 - 0 )2
PQ = √9 + 16 = 5Correct Option: D
According to question ,
PQ = √( 5 - 2 )2 + ( 4 - 0 )2
PQ = √9 + 16 = 5
∴ Area of circle = πr2
Area of circle = 25 π sq. units
- If ax + by = 6, bx – ay = 2 and x2 + y2 = 4, then the value of (a2 + b2) would be :
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As we know that given in question,
ax + by = 6........................ (i)
bx – ay = 2........................ (ii)
On squaring the both equation and adding,
a2 x2 + b2 y2 + 2abxy + b2 x2 + a2 y2 – 2abxy = 36 + 4
⇒ x2 (a2 + b2) + y2 (a2 + b2) = 40
⇒ (a2 + b2) (x2 + y2) = 40
As per given question, x2 + y2 = 40 , Put this value in above equation.Correct Option: A
As we know that given in question,
ax + by = 6 ........................ (i)
bx – ay = 2 ........................ (ii)
On squaring the both equation and adding,
a2 x2 + b2 y2 + 2abxy + b2 x2 + a2 y2 – 2abxy = 36 + 4
⇒ x2 (a2 + b2) + y2 (a2 + b2) = 40
⇒ (a2 + b2) (x2 + y2) = 40
As per given question, x2 + y2 = 40 , Put this value in above equation, we will get
⇒ (a2 + b2) × 4 = 40
⇒ a2 + b2 = 10
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If a + 1 = 1, then the value of a3 is : a
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from the given question,
a + 1 = 1 a
a2 + 1 = a
a2 + 1 - a = 0
Multiplying both side by (a + 1) and solve the equation.
Correct Option: C
from the given question,
a + 1 = 1 a
a2 + 1 = a
a2 + 1 - a = 0
Multiplying both side by (a + 1) and we will get,
(a + 1) (a2 – a + 1) = 0 x ( a + 1 )
a x (a2 – a + 1) + 1 x (a2 – a + 1) = 0 x ( a + 1 )
a3 - a2 + a + a2 - a + 1 = 0
a3 + 1 = 0
a3 = –1
