Q64 / 125 · BITSAT 2016 · Mathematics
The set of all values of a for which the function $f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} x / 4-\sin ^{2} x / 4\right)+(a-1) x+$
$\sin 1$ does not possess critical points is
- A$[1, \infty)$
- Correct answerB(0,1)$\cup(1,4)$
- C(-2,4)
- D(1,3)$\cup(3,5)$
Solution
\(f(x)=\left(a^{2}=-3 a+2\right)\left(\cos ^{2} x / 4-\sin ^{2} x / 4\right)+(a-1) x+\sin 1\)
\(\Rightarrow \mathrm{f}(\mathrm{x})=(\mathrm{a}-1)(\mathrm{a}-2) \cos \mathrm{x} / 2+(\mathrm{a}-1) \mathrm{x}+\sin 1\)
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{2}(\mathrm{a}-1)(\mathrm{a}-2) \sin \frac{\mathrm{x}}{2}+(\mathrm{a}-1)\)
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{a}-1)\left[1-\frac{(\mathrm{a}-2)}{2} \sin \frac{\mathrm{x}}{2}\right]\)
If \(f(x)\) does not possess critical points, then \(f^{\prime}(x)\) = 0 for any \(x \in R\)
\(\Rightarrow(a-1)\left[1-\frac{(a-2)}{2} \sin \frac{x}{2}\right]\) = 0 for any \(x \in R\)
\(\Rightarrow \mathrm{a}=1\) and \(1-\left(\frac{\mathrm{a}-2}{2}\right) \sin \frac{\mathrm{x}}{2}=0\)
must not have any solution in \(\mathrm{R}\).
\(\Rightarrow \mathrm{a}=1\) and \(\sin \frac{\mathrm{x}}{2}=\frac{2}{\mathrm{a}-2}\) is not solvable in \(\mathrm{R}\).
\(\Rightarrow \mathrm{a} = 1\) and \(\left|\frac{2}{\mathrm{a}-2}\right|>1\left[\right.\) For \(\mathrm{a}=2, \mathrm{f}(\mathrm{x})=\mathrm{x}+\sin 1 \therefore \mathrm{f}^{\prime}(\mathrm{x})=1\)
\(\Rightarrow \mathrm{a}=1\) and \(0<\mathrm{a}<4 \Rightarrow \mathrm{a} \in(0,1) \cup(1,4)\).
Hence, the correct answer is option: 'B'.