Algebra
- If (x – 1) and (x + 3) are the factors of x2 + k1x + k2 then
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f (x) = x² +k1x + k1
(x – 1) is a factor of f(x).
∴ f (1) = 0
⇒ 1 + k1 + k2 = 0
⇒ k1 + k2 = –1 ... (i)
Again, f (–3) = 0
⇒ (– 3)² + k1 (–3) + k2 = 0
⇒ 9 – 3k1 + k2 = 0
⇒ 3k1 – k2 = 9 ...(ii)
On adding both equations,
4k1 = 8
⇒ k1 = 2 From equation (i),
k1 + k2 = –1
⇒ 2 + k2 = –1
⇒ k2 = –1 – 2 = –3Correct Option: B
f (x) = x² +k1x + k1
(x – 1) is a factor of f(x).
∴ f (1) = 0
⇒ 1 + k1 + k2 = 0
⇒ k1 + k2 = –1 ... (i)
Again, f (–3) = 0
⇒ (– 3)² + k1 (–3) + k2 = 0
⇒ 9 – 3k1 + k2 = 0
⇒ 3k1 – k2 = 9 ...(ii)
On adding both equations,
4k1 = 8
⇒ k1 = 2 From equation (i),
k1 + k2 = –1
⇒ 2 + k2 = –1
⇒ k2 = –1 – 2 = –3
- The expression x4 – 2x2 + k will be a perfect square if the value of k is
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Expression = x4 – 2x2 + k
= (x2)2 – 2.x.2 .1 + (1)2 – (1)2 + k
For a perfect square,
–1 + k = 0 ⇒ k = 1Correct Option: A
Expression = x4 – 2x2 + k
= (x2)2 – 2.x.2 .1 + (1)2 – (1)2 + k
For a perfect square,
–1 + k = 0 ⇒ k = 1
- Find the value of x for which the expression 2 – 3x – 4x2 has the greatest value.
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Expression = 2 – 3x – 4x2
= – (4x2 + 3x – 2)= − (2x)2 + 2 × x × 3 + 3 2 − 3 2 − 2 4 4 4 = − 2x + 3 2 + 3 2 + 2 4 4
The value of expression will be maximum if,2x + 3 = 0 4 ⇒ 2x = − 3 4 ⇒ x = − 3 8 Correct Option: C
Expression = 2 – 3x – 4x2
= – (4x2 + 3x – 2)= − (2x)2 + 2 × x × 3 + 3 2 − 3 2 − 2 4 4 4 = − 2x + 3 2 + 3 2 + 2 4 4
The value of expression will be maximum if,2x + 3 = 0 4 ⇒ 2x = − 3 4 ⇒ x = − 3 8
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If x2 + 1 x + a2 is a perfect square, then a is 5
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x2 + 1 x + a2 5 = x2 + 2.x. 1 + 1 2 − 1 2 + a2 10 10 10 ∴ a2 − 1 2 = 0 ⇒ a2 = 1 2 10 10 ⇒ a = 1 10 Correct Option: C
x2 + 1 x + a2 5 = x2 + 2.x. 1 + 1 2 − 1 2 + a2 10 10 10 ∴ a2 − 1 2 = 0 ⇒ a2 = 1 2 10 10 ⇒ a = 1 10
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If x = 3t, y = 1 (t + 1), then the value of t for which x = 2y is 2
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x = 2y
⇒ 3t = 2 × 1 (t + 1) 2
⇒ 3t = t + 1 ⇒ 3t – t = 1⇒ 2t = 1 ⇒ t = 1 2 Correct Option: B
x = 2y
⇒ 3t = 2 × 1 (t + 1) 2
⇒ 3t = t + 1 ⇒ 3t – t = 1⇒ 2t = 1 ⇒ t = 1 2