1.

Number of solutions of the equation Sin x + cos x=x^2-2x+√35

Answer»

Here ,

The given equation is ;

sinx + cosx = x² - 2x + √35 .

Here ,

=> LHS = sinx + cosx

=> LHS = √2[ sinx•(1/√2) + cosx•(1/√2) ]

=> LHS = √2[ sinx•sin45° + cosx•sin45° ]

=> LHS = √2sin(x + 45°)

Also ,

We know that , -1 ≤ sin∅ ≤ 1 .

Thus ,

=> -1 ≤ sin(x + 45°) ≤ 1

=> -1•√2 ≤ √2•sin(x + 45°) ≤ 1•√2

=> -√2 ≤ √2sin(x + 45°) ≤ 2

=> -√2 ≤ LHS ≤ √2

Now ,

=> RHS = x² - 2x + √35

=> RHS = x² - 2x + 1² - 1² + √35

=> RHS = (x - 1)² + 1 + √35

Also ,

We know that , x² ≥ 0 .

Thus ,

=> (x - 1)² ≥ 0

=> (x - 1)² + 1 + √35 ≥ 1 + √35

=> RHS ≥ 1 + √35

Observing LHS and RHS , we can conclude that there exist no real number for which LHS and RHS would be equal . Thus , there is no real solution of the given equation .


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