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Answers
(2) Simplify : √225 – 5 ⁴√81 + √9 + ¹⁰√1024
⇒ √225 – 5 (⁴√81) + √9 + (¹⁰√1024)
⇒ 15 – 5 ( 3 ) + 3 + 2
⇒ 15 – 15 + 5
⇒ 5
(3) If x + y + z = 0, then show that x³ + y³ + z³ = 3xyz.
We know the algebraic identity :
⇒ x³ + y³ + z³ + 3xyz = (x + y + z) (x² + y² + z² – xy – yz – xz)
Let's substitute the value :
⇒ x³ + y³ + z³ + 3xyz = 0 (x² + y² + z² – xy – yz – xz)
⇒ x³ + y³ + z³ + 3xyz = 0
⇒ x³ + y³ + z³ = 0 – 3xyz
⇒ x³ + y³ + z³ = – 3xyz
Hence, proved!
(4) (i) If a die is rolled once, find the probability of getting :
(A) A composite number
Composite number is a number that has more than two factors.
⇒ P (E) = Favourable outcomes/Total outcomes
⇒ P (getting a composite number) = 2/6
⇒ P (getting a composite number) = 1/3
(B) An even number
⇒ P (getting an even number) = 3/6
⇒ P (getting an even number) = 1/6
(ii) Card is drawn at random from a well shuffled pack of 52 cards. Find the probability of getting :
(A) King or queen
⇒ P (E) = Favourable outcomes/Total outcomes
⇒ P (king or queen) = 8/52
⇒ P (king or queen) = 2/13
(B) A face card
⇒ P (face card) = 12/52
⇒ P (face card) = 3/13
(5) Write abscissa and ordinate from the following points :
A (0, 2)
Abscissa = 0 ; Ordinate = 2
B (1, – 5)
Abscissa = 1 ; Ordinate = – 5
C (3, 4)
Abscissa = 3 ; Ordinate = 4
D (5, 0)
Abscissa = 5 ; Ordinate = 0
(6) Write on which axis (or) on which quadrant the following points lies in :
P (0, 0) : Origin
Q (2, – 1) : IV quadrant
R (3, 0) : X axis
S (0, – 5) : Y axis
T (2, 2) : I quadrant
U (– 3, 5) : II quadrant
(8) Expand : (2x – 3y – 3z)²
We know the identity :
⇒ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Here :
⇒ a = 2x
⇒ b = – 3y
⇒ c = – 3z
Expanding :
⇒ (2x – 3y – 3z)² = (2x)² + (– 3y)² + (– 3z)² + 2(2x)(– 3y) + 2(– 3y)(– 3z) + 2(– 3z)(2x)
⇒ (2x – 3y – 3z)² = 4x² + 9y² + 9z² – 12xy + 18yz – 12zx
(9) Factorise :
(A) x² – 36
⇒ a² – b² = (a + b) (a – b)
⇒ (x + 4) (x – 4)
(B) z² – 12z – 45
Splitting the middle term :
⇒ z² – 12z – 45
⇒ z² – 15z + 3z – 45
⇒ z (z – 15) + 3 (z – 15)
⇒ (z – 15) (z + 3)
(10) Find the remainder when x³ – 2x² + 4x – 18 is divided by x – 2.
Substituting x = 2 in the polynomial :
⇒ x³ – 2x² + 4x – 18
⇒ (2)³ – 2 (2)² + 4 (2) – 18
⇒ 8 – 2 (4) + 8 – 18
⇒ 8 – 8 + 8 – 18
⇒ – 10
∴ – 10 is the remainder when x³ – 2x² + 4x – 18 is divided by x – 2.
(11) Find two irrational numbers between 2 and 3.
We know that :
⇒ 2² < 6 < 3² — 4 < 6 < 9
⇒ 2² < 7 < 3² — 4 < 7 < 9
Also :
⇒ √4 < √6 < √9 — 2 < √6 < 3
⇒ √4 < √7 < √9 — 2 < √7 < 3
So :
⇒ 2 < √6 < 3
⇒ 2 < √7 < 3
∴ The irrational numbers between 2 and 3 are √6 and √7.
(14) Find the probability of selecting a vowel from the word MATHEMATICS.
⇒ P (E) = Favourable outcomes/Total outcomes
⇒ P (a vowel) = 4/11
(15) Simplify each of the following :
(A) (3 + √3) (2 + √2)
⇒ 3(2 + √2) + √3(2 + √2)
⇒ 6 + 3√2 + 2√3 + √6
(B) (√5 + √2)²
⇒ (a + b)² = a² + b² + 2ab
⇒ (√5)² + (√2)² + 2 (√5) (√2)
⇒ 5 + 2 + 2√10
⇒ 7 + 2√10
(C) (5 – √2) (5 + √2)
⇒ (a – b) (a + b) = a² – b²
⇒ (5)² – (√2)²
⇒ 25 – 2
⇒ 23
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N0TE : (1), (7), (12) and (13) are in the attachments.
