Application of Derivatives 2 Question 32
32. Given $A=x: \frac{\pi}{6} \leq x \leq \frac{\pi}{3}$ and $f(x)=\cos x-x(1+x)$. Find $f(A)$.
$(1980,2 M)$
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Answer:
Correct Answer: 32. (a, c)
Solution:
- Given, $A=x: \frac{\pi}{6} \leq x \leq \frac{\pi}{3}$
and $f(x)=\cos x-x-x^{2}$
$\Rightarrow \quad f^{\prime}(x)=-\sin x-1-2 x=-(\sin x+1+2 x)$
which is negative for $x \in \frac{\pi}{6}, \frac{\pi}{3}$
$\therefore \quad f^{\prime}(x)<0$
or $f(x)$ is decreasing.
Hence, $f(A)=f \frac{\pi}{3}, f \frac{\pi}{6}$
$$ =\frac{1}{2}-\frac{\pi}{3} \quad 1+\frac{\pi}{3}, \frac{\sqrt{3}}{2}-\frac{\pi}{6} \quad 1+\frac{\pi}{6} $$
