Practicing Success
Let [a] denote the greatest integer less than or equal to a. The set of all values of x in $[-π/2,π/2)$ satisfying the inequality $|2x^2-4x-7|<\left[1+\frac{1}{2}\left(\frac{\cos θ}{\cos\frac{θ}{2}-\sin\frac{θ}{2}}\right)^2\right]$, is |
$(-1,3)$ $(1-\sqrt{5},1+\sqrt{5})$ $(1-\sqrt{5},-1)∪(3,1+\sqrt{5})$ none of these |
$(1-\sqrt{5},-1)∪(3,1+\sqrt{5})$ |
We have, $1+\frac{1}{2}\left(\frac{\cos θ}{\cos\frac{θ}{2}-\sin\frac{θ}{2}}\right)^2$ $=1+\frac{1}{2}\left(\frac{1-\sin^2θ}{1-\sin θ}\right)=1+\frac{1}{2}(1+\sin θ)=\frac{3}{2}+\frac{1}{2}\sin θ$ Now, $-\frac{π}{2}≤θ\frac{π}{2}$ $⇒-1≤\sin θ<1$ $⇒-\frac{1}{2}≤\frac{1}{2}\sin θ<\frac{1}{2}$ $⇒\frac{3}{2}-\frac{1}{2}≤\frac{3}{2}+\frac{1}{2}\sin θ<\frac{3}{2}+\frac{1}{2}$ $⇒1≤1+\frac{1}{2}\left(\frac{\cos θ}{\cos\frac{θ}{2}-\sin\frac{θ}{2}}\right)^2<2$ $⇒\left[1+\frac{1}{2}\left(\frac{\cos θ}{\cos\frac{θ}{2}-\sin\frac{θ}{2}}\right)^2\right]=1$ Thus, the given inequality reduces to $|2x^2-4x-7|<1$ $⇒-1<2x^2-4x-7<1$ $⇒-1 <2x^2-4x-7$ and $2x^2 -4x-7 <1$ $⇒ 2x^2-4x-6>0$ and $2x^2-4x-8<0$ $⇒x^2 -2x-3> 0$ and $x^2 - 2x - 4 <0$ $⇒(x < -1\, or,\, x> 3)$ and $1-\sqrt{5} <x<1+ \sqrt{5}$ $⇒x∈(1-\sqrt{5},-1)∪(3,1+\sqrt{5})$ |