Q. 1 C4.0( 4 Votes )

Check wheth

–x – (x + 1) + (x + 2) + (x + 3) = 4

We will calulate LHS in every case and compare it with RHS

Taking x = 1,

= -1-(1 + 1) + (1 + 2) + (1 + 3)

= -1-2 + 3 + 4

= 4

Taking x = 2,

= -2-(2 + 1) + (2 + 2) + (2 + 3)

= -2-3 + 4 + 5

= 4

Taking x = 3,

= -3-(3 + 1) + (3 + 2) + (3 + 3)

= -3-4 + 5 + 6

= 4

Taking x = 4,

= -4-(4 + 1) + (4 + 2) + (4 + 3)

= -4-5 + 6 + 7

= 4

Taking x = 5,

= -5-(5 + 1) + (5 + 2) + (5 + 3)

= -5-6 + 7 + 8

= 4

Taking x = -1,

= 1-(-1 + 1) + (-1 + 2) + (-1 + 3)

= 1 + 0 + 1 + 2

= 4

Taking x = -2,

= 2-(-2 + 1) + (-2 + 2) + (-2 + 3)

= 2 + 1 + 0 + 1

= 4

Taking x = -3,

= 3-(-3 + 1) + (-3 + 2) + (-3 + 3)

= 3 + 2-1 + 0

= 4

Taking x = -4,

= 4-(-4 + 1) + (-4 + 2) + (-4 + 3)

= 4 + 3-2-1

= 4

Taking x = -5,

= 5-(-5 + 1) + (-5 + 2) + (-5 + 3)

= 5 + 4-3-2

= 4

As in each case in LHS = 4 so, LHS = RHS

Hence, above equation is an identity.

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