Q. 1 A5.0( 3 Votes )
Check whether the equations are identities. Write the patterns got form each,on taking x = 1, 2, 3, 4, 5 and x = –1, –2, –3, –4, –5.
–x + (x + 1) = 1
Answer :
Let’s take as x = 1,2,3,4,5 (positive numbers)
And as x = -1,-2,-3,-4,-5 (negative numbers)
(i) –x + (x + 1) = 1
We will calculate LHS in every case and compare it with RHS taking x = 1, = -1 + (1 + 1) = -1 + 2
= 1
Taking x = 2, = -2 + (2 + 1) = -2 + 3
= 1
taking x = 3,
= -3 + (3 + 1)
= -3 + 4 = 1
taking x = 4,
= -4 + (4 + 1)
= -4 + 5 = 1
taking x = 5,
= -5 + (5 + 1)
= -5 + 6 = 1
now taking x as negative numbers Taking x = -1, = -(-1) + (-1 + 1)
= 1 – 1 + 1 = 1
Taking x = -2, = -(-2) + (-2 + 1)
= 2 + 1 -2 = 1
Taking x = -3, = -(-3) + (-3 + 1)
= 3 -3 + 1 = 1
Taking x = -4, = -(-4) + (-4 + 1)
= 4 -4 + 1 = 1
Taking x = -5, = -(-5) + (-5 + 1)
= 5 – 5 + 1 = 1
As in each case LHS = 1 so, LHS = RHS
Hence, above equation is an identity.
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Check whether the equations are identities. Write the patterns got form each,on taking x = 1, 2, 3, 4, 5 and x = –1, –2, –3, –4, –5.
–x + (x + 1) = 1
Kerala Board Mathematics Part IITaking different numbers, positive negative and zero, as x, y, z and compute x + (y + z) and (x + y) + z. Check whether the equation, x + (y + z) = (x + y) + z holds for all these numbers.
Kerala Board Mathematics Part II