If equation sin -1 √ x + cos -1 √ x 2-1+ tan -1 tan y = k has atleast one solution, then k ∈ ((p π/2), (q π/2)), where p, q ∈ I, then value of (p+q) is greater than or equal to

Question

If equation sin -1 √ x + cos -1 √ x 2-1+ tan -1 tan y = k has atleast one solution, then k ∈ ((p π/2), (q π/2)), where p, q ∈ I, then value of (p+q) is greater than or equal to (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

sin -1 √ x + cos -1 √ x 2-1+ tan -1 tan y = k 0 ≤ x ≤ 1 and 0 ≤ x 2-1 ≤ 1 Hence, x=1 (π/2)+(π/2)+ tan -1 tan y = k (-π/2)< tan -1 tan y <(π/2) ∴ k ∈((π/2), (3 π/2)) Hence, (p+q)=4.

Q. If equation $sin^{- 1} x + cos^{- 1} x^{2} - 1 + tan^{- 1} tan y = k$ has atleast one solution, then $k \in$ $(\frac{p π}{2}, \frac{q π}{2})$ , where $p, q \in I$ , then value of $(p + q)$ is greater than or equal to

2

3

4

5

$sin^{- 1} x + cos^{- 1} x^{2} - 1 + tan^{- 1} tan y = k$
$0 \leq x \leq 1$ and $0 \leq x^{2} - 1 \leq 1$
Hence, $x = 1$
$\frac{π}{2} + \frac{π}{2} + tan^{- 1} tan y = k$
$\frac{- π}{2} < tan^{- 1} tan y < \frac{π}{2}$
$∴ k \in (\frac{π}{2}, \frac{3 π}{2})$
Hence, $(p + q) = 4$ .

Q. If equation sin−1x​+cos−1x2−1​+tan−1tany=k has atleast one solution, then k∈ (2pπ​,2qπ​), where p,q∈I, then value of (p+q) is greater than or equal to

2

3

4

5

sin−1x​+cos−1x2−1​+tan−1tany=k 0≤x≤1 and 0≤x2−1≤1 Hence, x=1 2π​+2π​+tan−1tany=k 2−π​<tan−1tany<2π​ ∴k∈(2π​,23π​) Hence, (p+q)=4.

Q. If equation $sin^{- 1} x + cos^{- 1} x^{2} - 1 + tan^{- 1} tan y = k$ has atleast one solution, then $k \in$ $(\frac{p π}{2}, \frac{q π}{2})$ , where $p, q \in I$ , then value of $(p + q)$ is greater than or equal to

$sin^{- 1} x + cos^{- 1} x^{2} - 1 + tan^{- 1} tan y = k$
$0 \leq x \leq 1$ and $0 \leq x^{2} - 1 \leq 1$
Hence, $x = 1$
$\frac{π}{2} + \frac{π}{2} + tan^{- 1} tan y = k$
$\frac{- π}{2} < tan^{- 1} tan y < \frac{π}{2}$
$∴ k \in (\frac{π}{2}, \frac{3 π}{2})$
Hence, $(p + q) = 4$ .