Question

Assertion :Statement 1: $\sin x=a$ where $-1<a<0$ then for $xϵ[0,n\pi ]$ has $2(n-1)$ solutions $\forall nϵN$ Reason: Statement 2: $\sin x$ takes value $a$ exactly two times when we take one complete rotation covering all the quadrant starting from $x=0$

• A. Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1
• B. Both the statements are TRUE and STATEMENT 2 is NOT the correct explanation of STATEMENT 1
• C. STATEMENT 1 is TRUE and STATEMENT 2 is FALSE
• D. STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

Answer 1

The correct option is D STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

We do not get any solution if $n=1$
Considering the interval of $[0,2\pi ]$ that $n=2$ ..(one period of sin(x))
For $-1<a<0$
$\sin x=a$
Then 'a' belongs to the third and fourth quadrant.
$-1<sin\left(x\right)<0$ for all $ϵ(\pi ,2\pi )-\left\{\frac{3\pi }{2}\right\}$
Hence solution for the equation will be $xϵ(\pi ,2\pi )-\left\{\frac{3\pi }{2}\right\}$ .
However there are infinitely many points in the set $ϵ(\pi ,2\pi )-\left\{\frac{3\pi }{2}\right\}$.
Hence total number of solutions is obviously greater than $2(n-1)$.
In general
if n is odd, we get the solution as
$n>1$
$xϵ(\pi ,2\pi )\cup (3\pi ,4\pi )...\cup \left(\right(n-2)\pi ,(n-1\left)\pi \right)-\{\frac{3\pi }{2},\frac{7\pi }{2}...\}$
If n is even, we get the solutions as
$xϵ(\pi ,2\pi )\cup (3\pi ,4\pi )...\cup \left(\right(n-1)\pi ,(n\left)\pi \right)-\{\frac{3\pi }{2},\frac{7\pi }{2}...\}$
Hence Statement 1 is wrong.