- A$8$
- ✓$4$
- C$-8$
- D$11$
For $x = 7$, $3f(7) + 2f(11) = 70 + 30 = 100$
For $x = 11$, $3f(11) + 2f(7) = 140$
$\frac{{f(7)}}{{ - 20}} = \frac{{f(11)}}{{ - 220}} = \frac{{ - 1}}{{9 - 4}}$ ==> $f(7) = 4$.
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$\left\{(\mathrm{x}, \mathrm{y}): \frac{\mathrm{a}}{\mathrm{x}^2} \leq \mathrm{y} \leq \frac{1}{\mathrm{x}}, 1 \leq \mathrm{x} \leq 2,0<\mathrm{a}<1\right\}$ is
$\left(\log _e 2\right)-\frac{1}{7}$ then the value of $7 a-3$ is equal to:
$(i)$ $f (x)$ is bounded on $a \le x \le b.$
$(ii)$ The equation $f (x) = 0$ has at least one solution in $a < x < b.$
$(iii)$ The maximum and minimum values of $f (x)$ on $a \le x \le b$ occur at points where $f ' (c) = 0$.
$(iv)$ There is at least one point $c$ with $a < c < b$ where $f ' (c) > 0$.
$(v)$ There is at least one point $d$ with $a < d < b$ where $f ' (c) < 0.$