Algebra
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If a = c = e = 3,then 2a2 + 3c2 +4e2 = ? b d f 2b2 + 3d2 + 4f2
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a = c = e = 3 b d f
⇒ a = 3b ; c = 3d ; e = 3f∴ 2a2 + 3c2 +4e2 2b2 + 3d2 + 4f2 = 2 × 9b2 + 3 × 9d2 + 4 × 9f2 2b2 + 3d2 + 4f2 = 9(2b2 + 3d2 +4f2) = 9 2b2 + 3d2 + 4f2 Correct Option: D
a = c = e = 3 b d f
⇒ a = 3b ; c = 3d ; e = 3f∴ 2a2 + 3c2 +4e2 2b2 + 3d2 + 4f2 = 2 × 9b2 + 3 × 9d2 + 4 × 9f2 2b2 + 3d2 + 4f2 = 9(2b2 + 3d2 +4f2) = 9 2b2 + 3d2 + 4f2
- If a, b, c are real and a2 + b2 + c2 = 2 (a – b – c) – 3, then the value of 2a – 3b + 4c is
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a2 + b2 + c2 = 2a – 2b – 2c – 3
⇒ a2 – 2a + b2 + 2b + c2 + 2c + 1 + 1 + 1 = 0
⇒ (a2 – 2a + 1) + (b2 + 2b + 1) + (c2 + 2c + 1) = 0
⇒ (a – 1)2 + (b + 1)2 +(c + 1)2 = 0
⇒ a – 1 = 0 ⇒ a = 1
⇒ b + 1 = 0 ⇒ b = –1
and c + 1 = 0 ⇒ c = –1
∴ 2a – 3b + 4c = 2 + 3 – 4 = 1Correct Option: C
a2 + b2 + c2 = 2a – 2b – 2c – 3
⇒ a2 – 2a + b2 + 2b + c2 + 2c + 1 + 1 + 1 = 0
⇒ (a2 – 2a + 1) + (b2 + 2b + 1) + (c2 + 2c + 1) = 0
⇒ (a – 1)2 + (b + 1)2 +(c + 1)2 = 0
⇒ a – 1 = 0 ⇒ a = 1
⇒ b + 1 = 0 ⇒ b = –1
and c + 1 = 0 ⇒ c = –1
∴ 2a – 3b + 4c = 2 + 3 – 4 = 1
- If a, b, c are real and a2 + b2 + c2 = 2 (a – b – c) – 3, then the value of 2a – 3b + 4c is
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a2 + b2 + c2 = 2a – 2b – 2c – 3
⇒ a2 – 2a + b2 + 2b + c2 + 2c + 1 + 1 + 1 = 0
⇒ (a2 – 2a + 1) + (b2 + 2b + 1) + (c2 + 2c + 1) = 0
⇒ (a – 1)2 + (b + 1)2 +(c + 1)2 = 0
⇒ a – 1 = 0 ⇒ a = 1
⇒ b + 1 = 0 ⇒ b = –1
and c + 1 = 0 ⇒ c = –1
∴ 2a – 3b + 4c = 2 + 3 – 4 = 1Correct Option: C
a2 + b2 + c2 = 2a – 2b – 2c – 3
⇒ a2 – 2a + b2 + 2b + c2 + 2c + 1 + 1 + 1 = 0
⇒ (a2 – 2a + 1) + (b2 + 2b + 1) + (c2 + 2c + 1) = 0
⇒ (a – 1)2 + (b + 1)2 +(c + 1)2 = 0
⇒ a – 1 = 0 ⇒ a = 1
⇒ b + 1 = 0 ⇒ b = –1
and c + 1 = 0 ⇒ c = –1
∴ 2a – 3b + 4c = 2 + 3 – 4 = 1
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If for non-zero, x, x2 – 4x – 1 = 0, the value of x2 + 1 is x2
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x2 – 4x – 1 = 0
⇒ x2 –1 = 4x
Dividing by xx − 1 = 4 x
On squaring both sides,
x − 1 
2 = 16 x ⇒ x2 + 1 – 2 = 16 x2 ⇒ x2 + 1 = 16 + 2 = 18 x2
Second Method :
Using Rule 5,
Here, x2 – 4x – 1 = 0
⇒ x2 – 1 = 4 x⇒ x2 – 1 = 4 x
We know thatx2 + 1 = x − 1 2 + 2 x2 x
= 42 + 2 = 18Correct Option: D
x2 – 4x – 1 = 0
⇒ x2 –1 = 4x
Dividing by xx − 1 = 4 x
On squaring both sides,x − 1 2 = 16 x ⇒ x2 + 1 – 2 = 16 x2 ⇒ x2 + 1 = 16 + 2 = 18 x2
Second Method :
Using Rule 5,
Here, x2 – 4x – 1 = 0
⇒ x2 – 1 = 4 x⇒ x2 – 1 = 4 x
We know thatx2 + 1 = x − 1 2 + 2 x2 x
= 42 + 2 = 18
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The value of 4 + 3√3 is 7 + 4√3
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Expression = 4 + 3√3 7 + 4√3
Rationalising the denominator,= (4 + 3√3)(7 − 4√3) (7 + 4√3)(7 − 4√3) = 28 − 16√3 + 21 √3 − 12 × 3 49 − 48
= 28 + 5√3 – 36 = 5√3 – 8Correct Option: A
Expression = 4 + 3√3 7 + 4√3
Rationalising the denominator,= (4 + 3√3)(7 − 4√3) (7 + 4√3)(7 − 4√3) = 28 − 16√3 + 21 √3 − 12 × 3 49 − 48
= 28 + 5√3 – 36 = 5√3 – 8
