\(f(x) = x^{100} + sin\,x - 1\)
\(f'(x) = 100\,x^{99} + cos\,x\)
For strictly increasing, we have
\(100\,x^{99} + cos\,x > 0\)
⇒ \(100\,x^{99} > - cos\,x \)
⇒ \(100\,x^{99} > -1\) \((\because -1\le cos\,x\le 1)\)
⇒ \(x^{99} > \frac{-1}{100}\)
⇒ \(x> \left(\frac{-1}{100}\right)^{\frac1{99}}\)
⇒ \(x > -0.96 \) (approx)
Hence, for values greater than -0.96, the given function f(x) is strictly increasing.