Explanation
y=2−sin3x1.........................(i)
Let f(x)=2−sin3x1
Domain of f(x) is (−α,α)
From (1):
2−sin3x=y1
sin3x=2−y1
sin3x=y2y−1
3x=sin−1(y2y−1)
⇒x=31sin−1(y2y−1)
\therefore [y > 0 \text{ as } -1 \le \sin 3x \le 1]
For x to be real:
−1≤y2y−1≤1
-y \le 2y - 1 \le y \quad [y > 0]
2y−1≥−y & 2y−1≤y
y≥31, y≤1
Range of f(x)=[31,1]
(b) y=x2+x+1x2+x+2 [i.e. y > 1] ............(1)
⇒yx2+yx+y=x2+x+2
⇒x2(y−1)+x(y−1)+(y−2)=0
x∈R⇒D≥0
(y−1)2−4(y−1)(y−2)≥0
⇒(y−1)(y−1−4(y−2))≥0
⇒(y−1)(−3y+7)≥0
1≤y≤37 ...................(2)
From (1) and (2):
1≤y≤37
(c) f(x)=sinx−cosx
=2×21(sinx−cosx)
=2×{(21sinx−21cosx)}
=2sin(x−4π)
Range of sin .......... [−1,1]
Range of f(x)=[−2,2]
(d) f(x)=cot−1(−x)−tan−1x+sec−1x
∴(cot−1x=π−cot−1x)
⇒f(x)=π−cot−1x−tan−1x+sec−1x
π−2π=cot−1x+tan−1x+sec−1x
Range of sec−1x:
(0 < \sec^{-1}x < \pi)
\cup \left( \frac{\pi}{2} < \sec^{-1}x < \pi \right)
\cup \left( \pi < \sec^{-1}x < \frac{3\pi}{2} \right)
2π≤f(x)≤π∪π≤f(x)≤23π
⇒[2π,π)∪(π,23π]
