If sin- x2-2x-b - 2 for all the real values of x-1 and -0-2-2- then the possible real values of b lies in

The correct option is A (3,4) The minimum value of the quadratic takes place at nbsp; 1sinα nbsp;which is less than 1.Thus, we check at 1 and the ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Principal Solution of Trigonometric Equation

If sinα x2-...

Question

If $(s i n α) x^{2} - 2 x + b \geq 2$ for all the real values of $x \geq 1$ and $α ϵ (0, π / 2) \cup (π / 2, π)$ , then the possible real values of b lies in

(2, 4)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(3, 4)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(4, 5)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(5, 8)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

t \in [- \frac{π}{2}, \frac{π}{2}]

2 sin t = \frac{1 - 2 x + 5 x^{2}}{3 x^{2} - 2 x - 1}

\forall x \in R - {- \frac{1}{3}, 1}

f : [0, \frac{π}{2}] \to [0, 1]

f (0) = 0, f (\frac{π}{2}) = 1

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x i f x \leq - \frac{π}{2} A sin x + B i f - \frac{π}{2} < x < \frac{π}{2}; cos x i f x \geq \frac{π}{2} \end{matrix}

A

B

f (x)

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, if x \neq \frac{π}{2} 3 if x = \frac{π}{2} \end{matrix}

x ?

(c o s p - 1) x^{2} + c o s p x + s i n p = 0

Join BYJU'S Learning Program