If $$($$$$sin$$α$$)x2$$−$$2x+b$$≥$$2$$, for all real values of x≤$$1$$ and α∈(0$$, π$$/2)∪$$($$π$$/2$$, π$$)$$, then possible real values of b $$is/are$$

Question

If  $$($$$$sin$$α$$)x2$$−$$2x+b$$≥$$2$$,      for all real values  of x≤$$1$$      and α∈(0$$, π$$/2)∪$$($$π$$/2$$, π$$)$$,      then possible real  values of b      $$is/are$$

Infinity Learn · Accepted Answer

Given,  $$($$$$sin$$ α$$)x2$$−$$2x+b$$≥$$2$$.       Let  $$f(x)=($$$$sin$$α$$)x2$$−$$2x+b$$−$$2$$     Abscissa  corresponding to the vertex is given by             $$x=1sin$$α>$$1$$                 The  graph of $$f(x)=($$$$sin$$α$$)x2$$−$$2x+b$$−$$2$$,     ∀  x≤$$1$$,       is shown in the  figure. Therefore,  minimum of $$f(x)=($$$$sin$$ α$$)x2$$−$$2x+b$$−$$2$$      must be greater than  zero but minimum is at $$x=1$$.      That  is,  $$sin$$α−$$2+b$$−$$2$$≥$$0$$,     b≥$$4$$−$$sin$$α, α∈(0$$,π$$)

If $(\sin α) x^{2} - 2 x + b \geq 2,$ for all real values of $x \leq 1$ and $α \in (0, π / 2) \cup (π / 2, π),$ then possible real values of $b$ is/are

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If (sinα)x2−2x+b≥2, for all real values of x≤1 and α∈(0, π/2)∪(π/2, π), then possible real values of b is/are

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If $(\sin α) x^{2} - 2 x + b \geq 2,$ for all real values of $x \leq 1$ and $α \in (0, π / 2) \cup (π / 2, π),$ then possible real values of $b$ is/are