Number of integral solutions of the equation operatornamesgn( sin -1[(π x/6)])=1, where [ x ] denotes the greatest integer less than or equal to x and operatornamesgn x denotes signum function of x.

Question

Number of integral solutions of the equation operatornamesgn( sin -1[(π x/6)])=1, where [ x ] denotes the greatest integer less than or equal to x and operatornamesgn x denotes signum function of x. (A) 2 (B) 3 (C) 5 (D) 7

Tardigrade Admin · Accepted Answer

We have sin -1[(π x/6)]>0 ⇒[(π x/6)]=1 ⇒ 1 ≤ (π x/6)<2 ⇒ (6/π) ≤ x<(12/π) ∴ x =2,3 only. Hence two integral solution will satisfy above equation

Q. Number of integral solutions of the equation $sgn (sin^{- 1} [\frac{π x}{6}]) = 1$ , where $[x]$ denotes the greatest integer less than or equal to $x$ and $sgn x$ denotes signum function of $x$ .

2

3

5

7

We have $sin^{- 1} [\frac{π x}{6}] > 0 \Rightarrow [\frac{π x}{6}] = 1$
$\Rightarrow 1 \leq \frac{π x}{6} < 2 \Rightarrow \frac{6}{π} \leq x < \frac{12}{π}$
$∴ x = 2, 3$ only.
Hence two integral solution will satisfy above equation

Q. Number of integral solutions of the equation sgn(sin−1[6πx​])=1, where [x] denotes the greatest integer less than or equal to x and sgnx denotes signum function of x.

2

3

5

7

We have sin−1[6πx​]>0⇒[6πx​]=1 ⇒1≤6πx​<2⇒π6​≤x<π12​ ∴x=2,3 only. Hence two integral solution will satisfy above equation

Q. Number of integral solutions of the equation $sgn (sin^{- 1} [\frac{π x}{6}]) = 1$ , where $[x]$ denotes the greatest integer less than or equal to $x$ and $sgn x$ denotes signum function of $x$ .

We have $sin^{- 1} [\frac{π x}{6}] > 0 \Rightarrow [\frac{π x}{6}] = 1$
$\Rightarrow 1 \leq \frac{π x}{6} < 2 \Rightarrow \frac{6}{π} \leq x < \frac{12}{π}$
$∴ x = 2, 3$ only.
Hence two integral solution will satisfy above equation