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Page No 80:
Question 1:
Evaluate each of the following using identities:
(i)
(ii) (2x + y) (2x − y)
(iii) (a2b - b2a)2
(iv) (a - 0.1) (a + 0.1)
(v) (1.5x2 − 0.3y2) (1.5x2 + 0.3y2)
Answer:
In the given problem, we have to evaluate expressions by using identities.
(i) Given
We shall use the identity
By applying identity we get
Hence the value of is
(ii) We have been given
We shall use the identity
Here,,
By applying identity we get
Hence the value ofis
(iii) The given expression is
We shall use the identity
Here
By applying identity we get
Hence the value of is
(iv) The given expression is
We shall use the identity
Here
By applying identity we get
Hence the value ofis
(v) The given expression is
We shall use the identity
Here
By applying identity we get
Hence the value ofis
Page No 80:
Question 2:
Evaluate each of the following using identities:
(i) (399)2
(ii) (0.98)2
(iii) 991 ☓ 1009
(iv) 117 ☓ 83
Answer:
In the given problem, we have to evaluate expressions by using identities.
(i) Given
We can write as
We shall use the Identity
Where ,
By applying in identity we get
Hence the value of is
(ii) We have been given
We can write as
We shall use the identity
Where,
By applying in identity we get
Hence the value of is
(iii) The given expression is
We have
So we can express and in the terms of as
We shall use the identity
Here
By applying in identity we get
Hence the value of is
(iv) The given expression is
We have
So we can express and in the terms of 100 as
We shall use the identity
Here
By applying in identity we get
Hence the value of is
Page No 80:
Question 3:
Simplify each of the following:
(i) 175 × 175 + 2 × 175 × 25 + 25 × 25
(ii) 322 × 322 − 2 × 322 × 22 + 22 × 22
(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24
(iv)
Answer:
In the given problem, we have to simplify expressions
(i) Given
Put
Hence the equation becomes,
That is
Hence the value of is
(ii) We have been given
Put
Hence the equation becomes
That is
Hence the value of is
(iii) Given
Put
Hence the equation becomes
That is
Hence the value of is
(iv) We have been given
Put
Hence the equation becomes
Hence the value of is
Page No 80:
Question 4:
If find the value of .
Answer:
In the given problem, we have to find
Given
On squaring both sides we get,
Hence the value of is .
Page No 80:
Question 5:
If , find the value of
Answer:
In the given problem, we have to find
Given
On squaring both sides we get,
We shall use the identity
Hence the value of is.
Page No 80:
Question 6:
If , find the value of and .
Answer:
In the given problem, we have to find and
We have
On squaring both sides we get,
We shall use the identity
Again squaring on both sides we get,
We shall use the identity
Hence the value ofis and is .
Page No 80:
Question 7:
If 9x2 + 25y2 = 181 and xy = −6, find the value of 3x + 5y.
Answer:
In the given problem, we have to find
We have been given and
Let us take
We shall use the identity
By substituting and we get,
Hence the value of is.
Page No 80:
Question 8:
If 2x + 3y = 8 and xy = 2, find the value of 4x2 + 9y2
Answer:
In the given problem, we have to find
We have been given and
Let us take
On squaring both sides we get,
We shall use the identity
By substituting we get,
Hence the value of is
Page No 80:
Question 9:
If 3x − 7y = 10 and xy = -1, find the value of 9x2 + 49y2
Answer:
In the given problem, we have to find
We have been given and
Let us take
On squaring both sides we get,
We shall use the identity
By substituting we get,
Hence the value of is.
Page No 80:
Question 10:
Simplify each of the following products:
(i)
(ii)
Answer:
(i) In the given problem, we have to find product of
We have been given
On rearranging we get,
We shall use the identity
By substituting,we get,
We shall use the identity
Hence the value of is
(ii) In the given problem, we have to find product of
We have been given
On rearranging we get
We shall use the identity
By substituting,, we get ,
Hence the value of is .
Page No 80:
Question 11:
If , find the value of
Answer:
In the given problem, we have to find
Given
Adding and subtracting 2 on left hand side
Hence the value of is
Page No 80:
Question 12:
If , find the value of
Answer:
In the given problem, we have to find
Given
Adding and subtracting 2 on left hand side,
Hence the value of is
Page No 80:
Question 13:
Simplify each of the following products:
(i)
(ii)
(iii)
(iv)
Answer:
(i) In the given problem, we have to find product of
On rearranging we get
We shall use the identity
By substituting ,
Hence the value of is
(ii) In the given problem, we have to find product of
On rearranging we get
We shall use the identity
Hence the value of is
(iii) In the given problem, we have to find product of
Taking as common factor
We shall use the identity
Hence the value of is
(iv) In the given problem, we have to find product of
On rearranging we get
We shall use the identity
Hence the value of is.
Page No 80:
Question 14:
Prove that a2 + b2 + c2 −ab−bc−ca is always non-negative for all values of a, b, and c.
Answer:
In the given problem, we have to prove is always non negative for all that is we have to prove that
Consider,
Hence is always non negative for all
Note: Square of all negative numbers is always positive or non negative.
Page No 84:
Question 1:
Write the following in the expanded form:
(i) (a +2b + c)2
(ii) (2a − 3b − c)2
(iii) (−3x + y + z)2
(iv) (m + 2n − 5p)2
(v) (2 + x − 2y)2
(vi) (a2 + b2 + c2)2
(vii) (ab + bc + ca)2
(viii)
(ix)
(x)
(xi) (2x − y + z)2
(xii) (−2x + 3y + 2z)2
Answer:
In the given problem, we have to find expended form
(i) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is
(ii) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is
(iii) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is.
(iv) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is
(v) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is.
(vi) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is.
(vii) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form ofis .
(viii) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is
(ix) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is
(x) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is
(xi) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is
(xii) Given
We shall use the identity
Here
By applying in identity we get
Hence the expended form of is.
Page No 85:
Question 2:
If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + bc + ca.
Answer:
In the given problem, we have to find value of
Given and
Squaring the equation, we get
Now putting the value of in above equation we get,
Taking 2 as common factor we get
Hence the value of is .
Page No 85:
Question 3:
If a2 + b2 + c2 = 16 and ab + bc + ca = 10, find the value of a + b + c.
Answer:
In the given problem, we have to find value of
Given
Multiply equation with 2 on both sides we get,
Now adding both equation and we get
We shall use the identity
Hence the value of is .
Page No 85:
Question 4:
If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2.
Answer:
In the given problem, we have to find value of
Given
Squaring both sides of we get,
Substituting in above equation we get,
Hence the value of is.
Page No 85:
Question 5:
Find the value of 4x2 + y2 + 25z2 + 4xy − 10yz − 20zx when x= 4, y = 3 and z = 2.
Answer:
In the given problem, we have to find value of
Given
We have
This equation can also be written as
Using the identity
Hence the value of is .
Page No 85:
Question 6:
Simplify:
(i) (a +b + c)2 + (a − b + c)2
(ii) (a +b + c)2 − (a − b + c)2
(iii) (a +b + c)2 + (a − b + c)2 + (a +b − c)2
(iv) (2x + p − c)2 − (2x − p + c)2
(v) (x2 + y2 − z2) − (x2 − y2 + z2)2
Answer:
In the given problem, we have to simplify the expressions
(i) Given
By using identity
Hence the equation becomes
Taking 2 as common factor we get
Hence the simplified value of is
(ii) Given
By using identity
Hence the equation becomes
Taking 4 as common factor we get
Hence the simplified value of is.
(iii) Given
By using identity , we have
Taking 3 as a common factor we get
Hence the value ofis
.
(iv) Given
By using identity , we get
By cancelling the opposite terms, we get
Taking as common a factor we get,
Hence the value of is
(v) We have (x2 + y2 − z2) − (x2 − y2 + z2)2
Using formula, we get
(x2 + y2 − z2) − (x2 − y2 + z2)2
By canceling the opposite terms, we get
Taking as common factor we get
Hence the value of is.
Page No 85:
Question 7:
Simplify each of the following expressions:
(i)
(ii) (x + y− 2z)2 − x2 − y2 − 3z2 + 4xy
(iii) (x2− x + 1)2 − (x2 + x + 1)2
Answer:
In the given problem, we have to simplify the value of each expression
(i) Given
We shall use the identity for each bracket
By arranging the like terms we get
Now adding or subtracting like terms,
Hence the value of is
(ii) Given
We shall use the identity for expanding the brackets
Now arranging liked terms we get,
Hence the value of is
(iii) Given
We shall use the identity for each brackets
Canceling the opposite term and simplifies
Hence the value of is .
Page No 91:
Question 1:
Find the cube of each of the following binomials expressions:
(i)
(ii)
(iii)
(iv)
Answer:
In the given problem, we have to find cube of the binomial expressions
(i) Given
We shall use the identity
Here
By applying the identity we get
Hence cube of the binomial expression is
(ii) Given
We shall use the identity
Here
By applying the identity we get
Hence cube of the binomial expression of is
(iii) Given
We shall use the identity .
Here,
By applying identity we get
Hence cube of the binomial expression of is
(iv) Given
We shall use the identity
Here
By applying in identity we get
Hence cube of the binomial expression of is .
Page No 91:
Question 2:
If a + b = 10 and ab = 21, find the value of a3 + b3
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,
Hence the value of is .
Page No 91:
Question 3:
If a − b = 4 and ab = 21, find the value of a3 −b3
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,
Hence the value of is .
Page No 91:
Question 4:
If , find the value of
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,
Hence the value of is
Page No 91:
Question 5:
If , find the value of
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,
Hence the value of is
Page No 91:
Question 6:
If , find the value of
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,
Hence the value of is .
Page No 91:
Question 7:
If = 51, find the value of
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,
In order to find we are using identity
Here and
Hence the value of is .
Page No 91:
Question 8:
If , find the value of
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,
In order to find we are using identity
Here and
Hence the value of is .
Page No 91:
Question 9:
If , find the value of
Answer:
In the given problem, we have to find the value of
Given
Cubing on both sides of we get
We shall use identity
Hence the value of is .
Page No 91:
Question 10:
If 3x − 2y = 11 and xy = 12, find the value of 27x3 − 8y3
Answer:
In the given problem, we have to find the value of
Given,
In order to find we are using identity
Here putting,,
Hence the value of is.
Page No 91:
Question 11:
Evaluate each of the following:
(i) (103)3
(ii) (98)3
(iii) (9.9)3
(iv) (10.4)3
(v) (598)3
(vi) (99)3
Answer:
In the given problem, we have to find the value of numbers
(i) Given
In order to find we are using identity
We can write as
Hence where
The value of is
(ii) Given
In order to find we are using identity
We can write as
Hence where
The value of is
(iii) Given
In order to find we are using identity
We can write as
Hence where
The value of is
(iv) Given
In order to find we are using identity
We can write as
Hence where
The value of is
(v) Given
In order to find we are using identity
We can write as
Hence where
The value of is
(vi) Given
In order to find we are using identity
We can write as
Hence where
The value of is .
Page No 91:
Question 12:
Evaluate each of the following:
(i) 1113 − 893
(ii) 463+343
(iii) 1043 + 963
(iv) 933 − 1073
Answer:
In the given problem, we have to find the value of numbers
(i) Given
We can write as
We shall use the identity
Here
Hence the value of is
(ii) Given
We can write as
We shall use the identity
Here
Hence the value of is
(iii) Given
We can write as
We shall use the identity
Here
Hence the value of is
(iv) Given
We can write as
We shall use the identity
Here
Hence the value of is .
Page No 92:
Question 13:
If , calculate and
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,
Again squaring on both sides we get,
We shall use the identity
Again cubing on both sides we get,
We shall use identity
Hence the value of is respectively.
Page No 92:
Question 14:
Find the value of 27x3 + 8y3, if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy =
Answer:
In the given problem, we have to find the value of
(i) Given
On cubing both sides we get,
We shall use identity
Hence the value of is
(ii) Given
On cubing both sides we get,
We shall use identity
Hence the value of is .
Page No 92:
Question 15:
Find the value of 64x3 − 125z3, if 4x − 5z = 16 and xz = 12.
Answer:
From given problem we have to find the value of
Given
On cubing both sides of we get
We shall use identity
Hence the value of is .
Page No 92:
Question 16:
Simplify each of the following:
(i) (x+3)3 + (x−3)3
(ii)
(iii)
(iv) (2x − 5y)3 − (2x + 5y)3
Answer:
In the given problem, we have to simplify equation
(i) Given
We shall use the identity
Here
By applying identity we get
Hence simplified form of expression is .
(ii) Given
We shall use the identity
Here
By applying identity we get
By rearranging the variable we get
Hence the simplified value of is
(iii) Given
We shall use the identity
Here
By applying identity we get
By rearranging the variable we get,
Hence the simplified value of is
(iv) Given
We shall use the identity
Here
By applying the identity we get
By rearranging the variable we get,
Hence the simplified value of is .
Page No 92:
Question 17:
If find and
Answer:
In the given problem, we have to find the value of
Given
By adding and subtracting in left hand side of we get,
Again by adding and subtracting in left hand side of we get,
Now cubing on both sides of we get
we shall use identity
Hence the value of is respectively.
Page No 92:
Question 18:
If , find the value of
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Here putting,
In order to find we are using identity
In order to find we are using identity
Here and
Hence the value of is .
Page No 95:
Question 1:
Find the following products:
(i) (3x + 2y) (9x2 − 6xy + 4y2)
(ii) (4x − 5y) (16x2 + 20xy + 25y2)
(iii) (7p4 + q) (49p8 − 7p4q + q2)
(iv)
(v)
(vi)
(vii)
(viii)
(ix) (1 − x) (1+ x + x2)
(x) (1 + x) (1 − x + x2)
(xi) (x2 − 1) (x4 + x2 + 1)
(xii) (x3 + 1) (x6 − x3 + 1)
Answer:
(i) In the given problem, we have to find the value of
Given
We shall use the identity
We can rearrange the as
Hence the Product value of is
(ii) Given
We shall use the identity
We can rearrange the as
Hence the Product value of is
(iii) Given
We shall use the identity
We can rearrange the as
Hence the Product value of is
(iv) Given
We shall use the identity
We can rearrange the as
Hence the Product value of is
(v) Given
We shall use the identity
We can rearrange the as
Hence the Product value of is
(vi) Given
We shall use the identity ,
we can rearrange the as
Hence the Product value of is
(vii) Given
We shall use the identity,
We can rearrange the as
Hence the Product value of is
(viii) Given
We shall use the identity
We can rearrange the as
Hence the Product value of is
(ix) Given
We shall use the identity
We can rearrange the as
Hence the Product value of is
(x) Given
We shall use the identity
We can rearrange the as
Hence the Product value of is
(xi) Given
We shall use the identity
We can rearrange the as
Hence the Product value of is
(xii) Given
We shall use the identity,
We can rearrange the as
Hence the Product value of is .
Page No 95:
Question 2:
If x = 3 and y = − 1, find the values of each of the following using in identify:
(i) (9y2 − 4x2) (81y4 +36x2y2 + 16x4)
(ii)
(iii)
(iv)
(v)
Answer:
In the given problem, we have to find the value of equation using identity
(i) Given
We shall use the identity
We can rearrange the as
Now substituting the value in we get,
Hence the Product value of is
(ii) Given
We shall use the identity
We can rearrange the as
Now substituting the value in we get,
Hence the Product value of is
(iii) Given
We shall use the identity,
We can rearrange the as
Now substituting the value in
Taking Least common multiple, we get
Hence the Product value of is
(iv) Given
We shall use the identity
We can rearrange the as
Now substituting the value in we get,
Taking Least common multiple, we get
Hence the Product value of is
(v) Given
We shall use the identity,
We can rearrange the as
Now substituting the value in
Taking Least common multiple, we get
Hence the Product value of is .
Page No 95:
Question 3:
If a + b = 10 and ab = 16, find the value of a2 − ab + b2 and a2 + ab + b2
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
We can rearrange the identity as
Now substituting values in as,
We can write as
Now rearrange as
Thus
Now substituting values
Hence the value of is respectively.
Page No 95:
Question 4:
If a + b = 8 and ab = 6, find the value of a3 + b3
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Hence the value of is .
Page No 95:
Question 5:
If a – b = 6 and ab = 20, find the value of a3 − b3
Answer:
In the given problem, we have to find the value of
Given
We shall use the identity
Hence the value of is .
Page No 95:
Question 6:
If x = −2 and y = 1, by using an identity find the value of the following
(i) (4y2 − 9x2) (16y4 + 36x2y2 + 81x4)
(ii)
(iii)
Answer:
(i) In the given problem, we have to find the value of using identity
Given
We shall use the identity
We can rearrange the as
Now substituting the value in we get,
Taking 64 as common factor in above equation we get,
Hence the Product value of is
(ii) In the given problem, we have to find the value of using identity
Given
We shall use the identity
We can rearrange the as
Now substituting the value in we get,
Hence the Product value of is = 0.
(iii) Given
We shall use the identity,
We can rearrange the as
Now substituting the value in
Hence the Product value of is
Page No 98:
Question 1:
Find the following products:
(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 − 6xy − 4yz − 6zx)
(ii) (4x − 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy + 6yz − 8zx)
(iii) (2a − 3b − 2c) (4a2 + 9b2 +4c2 + 6ab − 6bc + 4ca)
(iv) (3x − 4y + 5z) (9x2 +16y2 + 25z2 + 12xy −15zx + 20yz)
Answer:
In the given problem, we have to find Product of equations
(i)Given
We shall use the identity
Hence the product of is
(ii) Given
We shall use the identity
Hence the product of is
(iii) Given
We shall use the identity
Hence the product of is
(iv) Given
We shall use the identity
Hence the product of is
Page No 98:
Question 2:
Evaluate:
(i) 253 − 753 + 503
(ii) 483 − 303 − 183
(iii)
(iv) (0.2)3 − (0.3)3 + (0.1)3
Answer:
In the given problem we have to evaluate the following
(i) Given
We shall use the identity
Let Take
Hence the value of is
(ii) Given
We shall use the identity
Let Take
Hence the value of is
(iii) Given
We shall use the identity
Let Take
Applying least common multiple we get,
Hence the value of is
(iv) Given
We shall use the identity
Let Take
Hence the value of is
Page No 99:
Question 3:
If x + y + z = 8 and xy +yz +zx = 20, find the value of x3 + y3 + z3 −3xyz
Answer:
In the given problem, we have to find value of
Given
We shall use the identity
We know that
Here substituting we get
Hence the value of is .
Page No 99:
Question 4:
If a + b + c = 9 and ab +bc + ca = 26, find the value of a3 + b3+ c3 − 3abc
Answer:
In the given problem, we have to find value of
Given
We shall use the identity
We know that
Here substituting we get,
Hence the value of is .
Page No 99:
Question 5:
If a + b + c = 9 and a2+ b2 + c2 =35, find the value of a3 + b3 + c3 −3abc
Answer:
In the given problem, we have to find value of
Given
We shall use the identity
We know that
Here substituting we get
Hence the value of is .
Page No 99:
Question 1:
If x + = 3, then find the value of .
Answer:
We have to find the value of
Given
Using identity
Here
By substituting the value of we get,
By transposing + 2 to left hand side, we get
Hence the value of is .
Page No 99:
Question 2:
If a − b = 5 and ab = 12, find the value of a2 + b2.
Answer:
We have to find the value
Given
Using identity
By substituting the value of we get ,
By transposing – 24 to left hand side we get
Hence the value of is .
Page No 99:
Question 3:
If a + b = 7 and ab = 12, find the value of a2 + b2
Answer:
We have to find the value of
Given
Using identity
By substituting the value of we get
By transposing +24 to left hand side we get ,
Hence the value of is .
Page No 99:
Question 4:
If , then find the value of .
Answer:
We have to find the value of
Given
Using identity
Here
By substituting the value of We get,
By transposing + 2 to left hand side, we get
Cubing on both sides we get,
Using identity
Here
Put we get
By transposing 21 to left hand side we get ,
Hence the value of is .
Page No 99:
Question 5:
If , then write the value of
Answer:
We have to find the value of
Given
Using identity
Here
By substituting the value of we get
By transposing – 2 to left hand side we get
By taking least common multiply we get
By multiplying 4 on both sides we get
Hence the value of is
Page No 99:
Question 6:
If , find the value of .
Answer:
We have to find the value of
Given
Using identity
Here
By substituting we get
Hence the value of is
Page No 99:
Question 7:
If a + b + c = 0, then write the value of
Answer:
We have to find the value of
Given
Using identity
Put
Hence the value of is
Page No 99:
Question 1:
If (a – b)2 + (b – c)2 + (c – a)2 = 0, then __________.
Answer:
Hence, if (a – b)2 + (b – c)2 + (c – a)2 = 0, then a = b = c.
Page No 99:
Question 2:
If
Answer:
Hence, if
Page No 100:
Question 3:
If
Answer:
Hence, if
Page No 100:
Question 4:
If a2 – 2a – 1 = 0, then = __________.
Answer:
Hence, if a2 – 2a – 1 = 0, then = 6.
Page No 100:
Question 5:
If a2 + b2 + c2 = 24 and ab + bc + ca = – 4, then a + b + c = ________.
Answer:
Hence, if a2 + b2 + c2 = 24 and ab + bc + ca = – 4, then a + b + c = ±4.
Page No 100:
Question 6:
If x + y + z = 0, then (x + y)3 + (y + z)3 + (z + x)3 = ________.
Answer:
Hence, if x + y + z = 0, then (x + y)3 + (y + z)3 + (z + x)3 = −3xyz.
Page No 100:
Question 7:
If x + y + z = 5 and xy + yz + zx = 7, then x3 + y3 + z3 – 3xyz = _________.
Answer:
Hence, if x + y + z = 5 and xy + yz + zx = 7, then x3 + y3 + z3 – 3xyz = 20.
Page No 100:
Question 8:
If (a + b + c) {(a – b)2 + (b – c)2 + (c – a)2} = k(a3 + b3 + c3 – 3abc), then k = ________.
Answer:
Hence, if (a + b + c) {(a – b)2 + (b – c)2 + (c – a)2} = k(a3 + b3 + c3 – 3abc), then k = 2.
Page No 100:
Question 9:
If
Answer:
Hence, if
Page No 100:
Question 10:
If
Answer:
Hence, if
Page No 100:
Question 11:
If
Answer:
Hence, if
Page No 100:
Question 12:
If
Answer:
Hence, if
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