JEE MATHEMATICS MCQ: FUNCTIONS

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Analyzing +4 | -1

(\sin \alpha) x^2-2 x+b-2 \geq 0 \forall x \in(-\infty, 1]

and

\alpha \in\left(0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right)

, then the possible real value of

b

View answer and explanation

Correct option: D

Option D (4) is correct because the minimum value of the quadratic must be greater than or equal to zero, leading to b >= 4.

More Questions from FUNCTIONS

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f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}