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Statement -1: "cos"^(7) x + "sin"^(4) x ...

Statement -1: `"cos"^(7) x + "sin"^(4) x = 1` has only two nonzero solutions in the inteval `(-pi, pi)`
Statement-2: `"cos"^(5)x + "cos"^(2)x -2=0` is possible only when cos x = 1.

A

Statement -1 is true, Statement-2 is true, Statement -2 is a correct explanation for Statement-1.

B

Statement -1 is True, Statement-2 is True, Statement -2 is not a correct explanation for Statement -1.

C

Statement-1 is True, Statement-2 is False.

D

Statement -1 is False, Statement-2 is True.

Text Solution

Verified by Experts

The correct Answer is:
B
`"cos"^(5)x + "cos"^(2) x - 2 =0`
`hArr "cos"^(5) x + "cos"^(2)x = 2`
`hArr "cos"^(5) x = 1 " and cos"^(2)x = 1 hArr "cos" x = 1`
So, statement-2 is true.
Now, `"cos"^(7)x + "sin"^(4)x = 1`
`rArr "cos"^(7)x +(1-"cos"^(2)x)^(2) = 1`
`rArr "cos"^(7)x + "cos"^(4)x-2"cos"^(2)x = 0`
`rArr "cos"^(2)x("cos"^(5)x + "cos"^(2)x-2) =0`
`rArr "cos"^(2)x = 0"or, cos"x-1`
`rArr x = +-(pi)/(2) "or", x = 0 " " [because -pi lt x lt pi]`
So, `"cos"^(7)x + "sin"^(4)x = 1` has only two non-zero solutions in the interval `(-pi, pi)`
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