Question:

Step 1: Analyze option (a). Let $g(x) = \sin x - x$. Then $g'(x) = \cos x - 1 \leq 0$ for all $x$, so $g$ is non-increasing. Since $g(0) = 0$ and $g$ is decreasing for $x > 0$, $g(x) < 0$ for $x > 0$. Thus $\sin x - x = 0$ has no real root in $\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$. Statement (a) is <b>incorrect</b>. Step 2: Analyze option (b). Let $h(x) = \tan x - x$. Then $h'(x) = \sec^2 x - 1 = \tan^2 x \geq 0$, so $h$ is non-decreasing. $h(0) = 0$ and $h$ is increasing for $x > 0$, so $h(x) > 0$ for $x \in \left(\dfrac{\pi}{6}, \dfrac{\pi}{3}\right)$. Thus $\tan x - x = 0$ has no real root in $\left(\dfrac{\pi}{6}, \dfrac{\pi}{3}\right)$. Statement (b) is <b>incorrect</b>. Step 3: Analyze option (c). By the Intermediate Value Theorem, if $f$ is continuous on $[a, b]$, then $f$ takes all values between $f(a)$ and $f(b)$. The value $\dfrac{2f(a) + 3f(b)}{5}$ is a weighted average of $f(a)$ and $f(b)$ (with weights $2/5$ and $3/5$ summing to 1), so it lies between $f(a)$ and $f(b)$. By IVT, there exists $c \in [a, b]$ such that $f(c) = \dfrac{2f(a) + 3f(b)}{5}$. Statement (c) is <b>correct</b>. Step 4: Analyze option (d). The statement requires $f$ to be continuous on $(a, b)$ for IVT to guarantee a root. If $f$ is not continuous, $f(a)$ and $f(b)$ having opposite signs does not necessarily imply a root exists in $(a, b)$. Statement (d) is <b>incorrect</b> (missing continuity condition). Step 5: The incorrect statements are (a), (b), and (d), corresponding to options 1, 2, and 4.