Consider the system of equations $sin x cos 2 y = (a^{2} - 1)^{2} + 1, cos x sin 2 y = a + 1$ The number of values of $a$ for which the system has a solution is

- 1

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

2

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

3

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

infinite

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

Open in App

Solution

The correct option is D $- 1$
$sin x cos 2 y = (a^{2} - 1)^{2} + 1$ -----------(1)

$cos x sin 2 y = a + 1$ -----------(2)

Adding Equation (1) & (2), we get

$sin (2 y + x) = (a^{2} - 1)^{2} + a + 2$

From the range of $s i n θ$

$- 1 \leq (a^{2} - 1)^{2} + a + 2 \leq 1$

This is only possible for $a = - 1$

Suggest Corrections

Similar questions

sin x cos 2 y = (a^{2} - 1)^{2} + 1, cos x sin 2 y = a + 1

y \in [0, 2 π]

a

sin x cos x 2 y = (a^{2} - 1)^{2} + 1

cos x sin 2 y = a + 1

sin x . cos 2 y = (a^{2} - 1)^{2} + 1, cos x . sin 2 y = a + 1

x ϵ [0, 2 π]

Consider the system of equations
ax + y + z = 1
x + ay + z = 1
x + y + az = 1, then which of the following statement(s) is/are correct?

⎡ ⎢ ⎣ \begin{matrix} 2 & 1 & - 4 4 & 3 & - 12 1 & 2 & - 8 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} a 57 \end{matrix} ⎤ ⎥ ⎦

a

Join BYJU'S Learning Program