First slide

Introduction to ITF

Question

For $x, y, z, t \in R, \sin^{- 1} x + \cos^{- 1} y + \sec^{- 1} z \geq t^{2} - \sqrt{2 π} t + 3 π$

Moderate

Question

The value of x + y + z is equal to

A

1
B

0
C

2
D

-1

Solution

$\begin{aligned} \sin^{- 1} x & \in [- \frac{π}{2}, \frac{π}{2}] \\ \cos^{- 1} y & \in [0, π] \\ \sec^{- 1} z \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π] \\ \Rightarrow & \sin^{- 1} x + \cos^{- 1} y + \sec^{- 1} z \leq \frac{π}{2} + π + π = \frac{5 π}{2} \end{aligned}$
Also,
$\begin{aligned} t^{2} - \sqrt{2 π} t + 3 π & = t^{2} - 2 \sqrt{\frac{π}{2}} t + \frac{π}{2} - \frac{π}{2} + 3 π \\ = {(t - \sqrt{\frac{π}{2}})}^{2} + \frac{5 π}{2} \geq \frac{5 π}{2} \end{aligned}$
The given inequation exists if equality holds, i.e.,
$\begin{aligned} L.H.S. = R.H.S. = \frac{5 π}{2} \\ \Rightarrow & x = 1, y = - 1, z = - 1 and t = \sqrt{\frac{π}{2}} \end{aligned}$

Question

The principal value of $\cos^{- 1} (\cos 5 t^{2})$ is

A

$\frac{3 π}{2}$
B

$\frac{π}{2}$
C

$\frac{π}{3}$
D

$\frac{2 π}{3}$

Solution

$\begin{aligned} \sin^{- 1} x & \in [- \frac{π}{2}, \frac{π}{2}] \\ \cos^{- 1} y & \in [0, π] \\ \sec^{- 1} z \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π] \\ \Rightarrow & \sin^{- 1} x + \cos^{- 1} y + \sec^{- 1} z \leq \frac{π}{2} + π + π = \frac{5 π}{2} \end{aligned}$
Also,
$\begin{aligned} t^{2} - \sqrt{2 π} t + 3 π & = t^{2} - 2 \sqrt{\frac{π}{2}} t + \frac{π}{2} - \frac{π}{2} + 3 π \\ = {(t - \sqrt{\frac{π}{2}})}^{2} + \frac{5 π}{2} \geq \frac{5 π}{2} \end{aligned}$
The given inequation exists if equality holds, i.e.,
$\begin{aligned} L.H.S. = R.H.S. = \frac{5 π}{2} \\ \Rightarrow & x = 1, y = - 1, z = - 1 and t = \sqrt{\frac{π}{2}} \\ \Rightarrow & \cos^{- 1} (\cos 5 t^{2}) = \cos^{- 1} (\cos (\frac{5 π}{2})) = \frac{π}{2} \end{aligned}$

Question

The value of $\cos^{- 1} (min {x, y, z})$ is

A

0
B

$\frac{π}{2}$
C

$π$
D

$\frac{π}{3}$

Solution

$\begin{aligned} \sin^{- 1} x & \in [- \frac{π}{2}, \frac{π}{2}] \\ \cos^{- 1} y & \in [0, π] \\ \sec^{- 1} z \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π] \\ \Rightarrow & \sin^{- 1} x + \cos^{- 1} y + \sec^{- 1} z \leq \frac{π}{2} + π + π = \frac{5 π}{2} \end{aligned}$
Also,
$\begin{aligned} t^{2} - \sqrt{2 π} t + 3 π & = t^{2} - 2 \sqrt{\frac{π}{2}} t + \frac{π}{2} - \frac{π}{2} + 3 π \\ = {(t - \sqrt{\frac{π}{2}})}^{2} + \frac{5 π}{2} \geq \frac{5 π}{2} \end{aligned}$
The given inequation exists if equality holds, i.e.,
$\begin{aligned} L.H.S. = R.H.S. = \frac{5 π}{2} \\ \Rightarrow & x = 1, y = - 1, z = - 1 and t = \sqrt{\frac{π}{2}} \\ \Rightarrow & \cos^{- 1} (\cos 5 t^{2}) = \cos^{- 1} (\cos (\frac{5 π}{2})) = \frac{π}{2} \\ \cos^{- 1} (min {x, y, z}) = \cos^{- 1} (- 1) = π \end{aligned}$

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