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MATHEMATICS [4 marks sum]

Expansions · ICSE STD 9 (ICSE - English Medium). 24 questions.

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[4 marks sum]

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24 questions · timed · auto-graded

Question 14 Marks
If $a ^2+\frac{1}{ a ^2}=14$; find the value of $a +\frac{1}{ a }$
Answer
Using $(a+b)^2=a^2+2 a b+b^2$
$ \left(a+\frac{1}{a}\right)^2 $
$=a^2+2 a\left(\frac{1}{a}\right)+\left(\frac{1}{a}\right)^2$
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=a^2+2+\frac{1}{a^2} $
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2 $
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=14+2$
$\Rightarrow\left(a+\frac{1}{a}\right)^2=16$
$\Rightarrow a+\frac{1}{a}= \pm 4 .$
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Question 24 Marks
If $3 x-\frac{1}{3 x}=9$; find the value of $27 x^3-\frac{1}{27 x^3}$.
Answer
$3 x-\frac{1}{3 x}=9 $
Using $\left( a -\frac{1}{ a }\right)^3$
$= a ^3-\frac{1}{ a ^3}-3\left( a -\frac{1}{ a }\right) $, we get: 
$\left(3 x-\frac{1}{3 x}\right)^3 $
$ =(3 x)^3-\left(\frac{1}{3 x}\right)^3-3\left(3 x-\frac{1}{3 x}\right)$
$ \Rightarrow 729=27 x^3-\frac{1}{27 x^3}-3(9) $
$\Rightarrow 27 x^3-\frac{1}{27 x^3} $
$=729+27 $
$=756$
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Question 34 Marks
Evaluate the following :$1.81 \times 1.81 - 1.81 \times 2.19 + 2.19 \times 2.19$
Answer
$1.81 \times 1.81-1.81 \times 2.19+2.19 \times 2.19$
$=(1.81)^2-(1.81 \times 2.19)+(2.19)^2$
$=(1.81)^2-(1.81 \times 2.19)+(2.19)^2-(1.81 \times 2.19)+(1.81 \times 2.19)$
$=(1.81)^2-(1.81 \times 2.19)+(2.19)^2-(1.81 \times 2.19)$
$=(1.81-2.19)^2+(2.00-0.19)^2+(1.81 \times 2.19)$
$=(0.38)^2+\left(4-(0.19)^2\right)$
$= 0.1444+(4-0.0361)$
$= 0.1444+3.9639$
$= 4.1083$
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Question 44 Marks
Evaluate the following :$7.16 \times 7.16 + 2.16 \times 7.16 + 2.16 \times 2.16$
Answer
$7.16 \times7.16 + 2.16 \times7.16 + 2.16 \times 2.16$
$= (7.16)^2 + (2.16)(7.16) + (2.16)^2$
$= (7.16)^2 + (2.16)(7.16) + (2.16)^2 + (2.16)(7.16) - (2.16)(7.16)$
$= (7.16)^2 + (2.16)(7.16) + (2.16)^2 + (2.16)(7.16)$
$= (7.16 + 2.16)^2- (2.16)(7.16)$
$= (9.32)^2- 15.4656$
$= 86.8624 - 15.4656$
$= 71.3968.$
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Question 54 Marks
Evaluate the following :$(8.12)^3 - (3.12)^3$
Answer
$(8.12)^3 - (3.12)^3$
$= (8.12 - 3.12)^3 + 3(8.12)(3.12)(8.12 - 3.12)$
$= 5^3 + 3(8.12)(3.12) \times 5$
$= 125 + 15 \times (8.12)(3.12)$
$= 125 + 15 \times 25.3344$
$= 125 + 380.016$
$= 505.016.$
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Question 64 Marks
Evaluate the following :$(5.45)^3 + (3.55)^3$
Answer
$(5.45)^3 + (3.55)^3$
$= (5.45 + 3.55)^3 - 3(5.45)(3.55)(5.45 + 355)$
$= (9)^3 - 3(4 + 1.45)(4 - 1.45)(9)$
$= 81 - 3(16 - (1.45)^2)(9)$
$= 81 - 27(16 - 2.1025)$
$= 81 - 27 \times 13.8975$
$= 81 - 522.3825$
$= 206.6175.$
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Question 74 Marks
Evaluate the following :$(3.29)^3 + (6.71)^3$
Answer
$(3.29)^3 + (6.71)^3$
$= (3.29 + 6.71)^3 - 3(3.29) (6.71) (3.029 + 6.71)$
$= (10)^3- 3(3.29) (6.71) (10)$
$= 1000 - 30 (5 - 1.71) (5 + 1.71)$
$= 1000 - 30 (5)^2 - (1.71)^2$
$= 1000 - 30 (25 - 2.9241)$
$= 1000 - 30 \times 22.0759$
$= 1000 - 662.277$
$= 337.723.$
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Question 84 Marks
Simplify:$(3a + 2b - c)(9a^2 + 4b^2 + c^2 - 6ab + 2bc +3ca)$
Answer
$(3a + 2b - c)(9a^2 + 4b^2 + c^2 - 6ab + 2bc +3ca)$
$= 3a(9a^2 + 4b^2 + c^2 - 6ab + 2bc + 3a) + 2b (9a^2+ 4b^2 + c^2 - 6ab + 2bc + 3ca) - $$c(9a^2 + 4b^2 + c^2 - 6ab + 2bc + 3ca)$
$= 27a^3 + 12ab^2 + 3ac^2 - 18a^2b + 6abc + 9a^2c + 18a^2b + 8b^3 + 2bc^2 - 12ab^2 + $$4b^2c + 6abc - 9a^2c - 4b^2c - c^3 + 6abc - 2bc^2 - 3ac^2$
$= 27a^3 + 8b^3 - c^3 + 18abc.$
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Question 94 Marks
Simplify:$(1 + x)(1 - x)(1 - x + x^2)(1 + x + x^2)$
Answer
$(1 + x)(1 - x)(1 - x + x^2)(1 + x + x^2)$
$= (1 + x)(1 - x)(x^2 + 1 - x)(x^2 + 1 + x)$
$= (1^2 - x^2)[(x^2 + 1 - x)^2 - x^2] \dots.....($Using $a^2- b^2 = (a + b)(a - b))$
$= (1 - x^2)[x^4 + 2x^2 + 1 - x^2]$
$= (1 - x^2)(x^4 + x^2 + 1)$
$= 1(x^4 + x^2 + 1) -x^2(x^4 + x^2 + 1)$
$= x^4+ x^2+ 1 - x^6 - x^4 - x^2$
$= 1 - x^6.$
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Question 104 Marks
Simplify:$\left(a+\frac{1}{a}\right)^3-\left(a-\frac{1}{a}\right)^3$
Answer
$\left(a+\frac{1}{a}\right)^3-\left(a-\frac{1}{a}\right)^3 $
$ =(a) 3+\left(\frac{1}{a}\right)^3+3(a)\left(\frac{1}{a}\right)\left(a+\frac{1}{a}\right)-\left[(a)^3-\left(\frac{1}{a}\right)^3=-3(a)\left(\frac{1}{a}\right)\left(a-\frac{1}{a}\right)\right] $
$ =a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right)-\left[a^3-\frac{1}{a^3}-3\left(a-\frac{1}{a}\right)\right]$
$ =a^3+\frac{1}{a^3}+3 a+\frac{3}{a}-a^3+\frac{1}{a^3}+3 a-\frac{3}{a} $
$ =\frac{2}{a^3}+6 a$
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Question 114 Marks
If $m - n = -2$ and $m^3 - n^3 = -26$, find $mn$.
Answer
$m - n = -2, m^3 - n^3 = -26$
$(m - n)^3 = m^3 - n^3 - 3mn (m - n)$
$\Rightarrow (-2)^3 = -26 - 3mn (-2)$
$\Rightarrow 6mn$
$= - 8 + 26$
$= 18$
$\Rightarrow mn = 3.$
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Question 124 Marks
If $5 x+\frac{1}{5 x}=7 ;$ find the value of $125 x^3+\frac{1}{125 x^3}$.
Answer
$5 x+\frac{1}{5 x}=7 $
$ \text { Using }\left(a+\frac{1}{a}\right)^3$
$=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right) \text {, we get: }$
$ \left(5 x+\frac{1}{5 x}\right)^3$
$=(5 x)^3+\left(\frac{1}{5 x}\right)^3+3\left(5 x+\frac{1}{5 x}\right) $
$ \Rightarrow 343=125 x^2+\frac{1}{125 x^3}+3(7)$
$ \Rightarrow 125 x^3+\frac{1}{125 x^3} $
$=343-21 $
$=322 .$
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Question 134 Marks
If $a + b = 5$ and $ab = 2$, find $a^3+ b^3$.
Answer
Using $(a + b)^2 = a^2 + 2ab + b^2$
$a^2+ b^2 = (a + b)^2- 2ab$
$\Rightarrow a^2 + b^2 = (5)^2 - 2(2)$
$\Rightarrow a^2 + b^2 = 25 - 4$
$\Rightarrow a^2 + b^2= 21$
$a^3 + b^3$
$= (a + b) (a^2 + b^2 - ab)$
$= (5) (21 - 2)$
$= (5) (19)$
$= 95.$
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Question 144 Marks
If $a + 2b + c = 0$; then show that $a^3 + 8b^3 + c^3 = 6abc$
Answer
$a + 2b + c = 0 \dots...(i)$
$\Rightarrow (a + 2b) + c = 0$
$\Rightarrow (a + 2b)^3 + c^3+ 3(a + 2b) c(a + 2b + c) = 0$
$\Rightarrow a^3+ 8b^2+ 6ab (a + 2b) + c^3+ 0 = 0$
$\Rightarrow a^3 + 8b^3 + c^3+ 6ab (a + 2b) = 0 \dots....(2)$
Using $(1),$ we get $a + 2b = -c$
From $(2),$
$a^3 + 8b^3+ 6ab (-c) = 0$
$\Rightarrow a^3 + 8b^3+ c^3 = 6abc.$
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Question 154 Marks
If $a + b + c = 0$; then show that $a^3 + b^3 + c^3 = 3abc$.
Answer
$a + b + c = 0 \dots...(i)$
$\Rightarrow (a + b) + c = 0$
Cubing both sides
$\Rightarrow (a + b)^3 + c^3 + 3(a + b) (c) (a+ b + c) = 0$
$\Rightarrow a^3 + b^3+ 3ab (a + b) + c^3 + 0 = 0$
$\Rightarrow a^3 + b^3+ c^3 + 3ab (a + b) = 0 \dots...(2)$
Using $(i),$ we get,
$a + b = -c$ From $(2),$
$a^3 + b^3 + c^3 + 3ab (-c) = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$.
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Question 164 Marks
Find the cube of: $a-\frac{1}{a}+b$
Answer
Using $(a+b+c)^3 $
$ =a^3+b^3+c^3+3 a^2 b+3 a^2 c+3 b^2 a+3 c^2 a+6 a b c$
$\left(a-\frac{1}{a}+b\right) $
$ = a^3+\left(-\frac{1}{a}\right)^3+b^3+3 a^2\left(-\frac{1}{a}\right)+3 a^2+3\left(-\frac{1}{a}\right)^2 b+3\left(-\frac{1}{a}\right)^2 a+3 b^2a+3 b^2\left(-\frac{1}{a}\right)+6 a\left(-\frac{1}{a}\right) b$
$=a^3-\frac{1}{a^3}+b^3-3 a+3 a^2 b+\frac{3 b}{a^2}+\frac{3}{a}+3 b^2 a-\frac{3 b^2}{a}-6 b .$
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Question 174 Marks
If $\left(a+\frac{1}{a}\right)^2=3$; then show that $a^3+\frac{1}{a^3}=0$
Answer
$\left(a+\frac{1}{a}\right)^2=3 $
$\Rightarrow a+\frac{1}{a}=\sqrt{3}$
Now, $\left(a+\frac{1}{a}\right)^3$
$=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right) $
$\Rightarrow(\sqrt{3})^3 $
$=a^3+\frac{1}{a^3}+3(\sqrt{3}) $
$\Rightarrow a^3+\frac{1}{a^3} $
$=3 \sqrt{3}-3 \sqrt{3} $
$=0$
 
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Question 184 Marks
If $a +\frac{1}{ a }= p$; then show that $a ^3+\frac{1}{ a ^3}= p \left( p ^2-3\right)$
Answer
$a+\frac{1}{a}=p $
$\left(a+\frac{1}{a}\right)^3$
$=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right) $
$ \Rightarrow p^3=a^3+\frac{1}{a^3}+3(p) $
$=a^3+\frac{1}{a^3}$
$ =p^3-3 p $
$=p\left(p^2-3\right) .$
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Question 194 Marks
If $x + y + z = p$ and $xy + yz + zx = q$; find $x^2 + y^2 + z^2$.
Answer
Given $x + y + x = p$ and $xy ++ yz + zx = q$
$(x + y + x)^2$
$= x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
$\Rightarrow x^2 + y^2 + z^2$
$= (x + y + z)^2 - 2xy + 2yz + 2zx$
$\Rightarrow x^2 + y^2 + z^2$
$= (x + y + z)^2 - 2(xy + yz + zx)$
$\Rightarrow x^2+ y^2 + z^2$
$= (p)^2 - 2(q)$
$\Rightarrow x^2 + y^2+ z^2$
$= p^2 - 2q.$
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Question 204 Marks
If $x + y + z = 12$ and $xy + yz + zx = 27$; find $x^2 + y^2 + z^2$.
Answer
$(x + y + z)^2= (12)^2$
$\Rightarrow x^2+ y^2 + z^2 + 2xy + 2yz + 2zx = 144$
$\Rightarrow x^2 + y^2 + z^2 + 2(xy + yz + zx) = 144$
$\Rightarrow x^2 + y^2 + z^2 + 2(27) = 144$
$\Rightarrow x^2 + y^2 + z^2$
$= 144 - 54$
$= 90.$
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Question 214 Marks
If $2x + 3y = 10$ and $xy = 5$; find the value of $4x^2 + 9y^2$
Answer
Given $2x + 3y = 10$ and $xy = 5$
$4x^2 +9y^2$
$= (2x^2) + (3y)^2$
$= (2x + 3y)^2 - 2(2x)(3y)\dots......[\because (a + b)^2 + a^2+ b^2 + 2ab$, so, $a^2 + b^2 = (a + b)^2 - 2ab]$
$= (10)^2 - 12(5)$
$= 100 - 60$
$= 40.$
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Question 224 Marks
If $x + y = 1$ and $xy = -12$; find:$x - y$
Answer
$(x + y)^2 = (1)^2$
$\Rightarrow x^2 + y^2 + 2xy$
$= 1$
$\Rightarrow x^2 + y^2$
$= 1 - 2(-12)$
$= 1 + 24$
$= 25$
Now, $(x - y)^2$
$= x^2 + y^2 - 2xy$
$= 25 - 2(-12)$
$= 25 + 24$
$= 49$
$\Rightarrow x - y$
$= ±7.$
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Question 234 Marks
If $m - n = 0.9$ and $mn = 0.36$, find:$m + n$
Answer
Given $m - n = 0.9$ and $mn = 0.36$
$(m - n)^2 = m^2 - 2mn + n^2$
$\Rightarrow (0.9)^2 = m^2 - 2mn + n^2$
$\Rightarrow 0.81 = m^2 + n^2 - 2(0.36)$
$\Rightarrow 0.81 = m^2 + n^2 - 0.72$
$\Rightarrow m^2 + n^2 = 1.53$
So, $(m + n)^2 = m^2 + 2mn + n^2$
$\Rightarrow (m + n)^2= m^2 + n^2 + 2mn$
$\Rightarrow (m + n)^2 = 1.53 + 2(0.36)$
$\Rightarrow (m + n)^2 = 2.25$
$\Rightarrow m + n = ± 1.5.$
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Question 244 Marks
If $p + q = 8$ and $p - q = 4$, find:$pq$
Answer
$(p + q)^2 = (8)^2$
$p^2 + q^2 + 2pq = 64 \dots...(i)$
$(p - q)^2 = (4)^2$
$p^2 + q^2 - 2pq = 16$
$p^2 + q^2 = 16 + 2pq \dots...(ii)$
Using $(ii)$ in $(i)$, we get :
$16 + 2pq + 2pq = 64$
$\Rightarrow 4pq$
$= 64 - 16$
$= 48$
$\Rightarrow pq = 12.$
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[4 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip