1. **Problem statement:** (i) Find the derivative $\frac{dy}{dx}$ if $y = \ln(2x^2 + 5)$. (ii) Given $f(x) = \frac{21x}{3x^2 + k}$ with $k > 1$ a positive integer, find the greatest $k$ such that $\int_1^k f(x) \, dx < 7 \ln 8$. 2. **Formula and rules:** - Derivative of $\ln u$ is $\frac{1}{u} \cdot \frac{du}{dx}$. - For integrals of rational functions, substitution is often useful. 3. **Part (i) derivative:** - Let $u = 2x^2 + 5$. - Then $\frac{du}{dx} = 4x$. - So $\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} = \frac{4x}{2x^2 + 5}$. 4. **Part (ii) integral:** - Given $f(x) = \frac{21x}{3x^2 + k}$. - Substitute $t = 3x^2 + k$, then $\frac{dt}{dx} = 6x$ or $dx = \frac{dt}{6x}$. - Rewrite integral: $$\int_1^k \frac{21x}{3x^2 + k} dx = \int_1^k \frac{21x}{t} \cdot \frac{dt}{6x} = \int_1^k \frac{21}{6t} dt = \frac{7}{2} \int_1^k \frac{1}{t} dt$$ - But limits in $t$ change: when $x=1$, $t=3(1)^2 + k = 3 + k$; when $x=k$, $t=3k^2 + k$. - So integral becomes: $$\frac{7}{2} \int_{3+k}^{3k^2 + k} \frac{1}{t} dt = \frac{7}{2} [\ln(3k^2 + k) - \ln(3 + k)] = \frac{7}{2} \ln \left( \frac{3k^2 + k}{3 + k} \right)$$ 5. **Inequality:** - Given $\int_1^k f(x) dx < 7 \ln 8$. - Substitute integral result: $$\frac{7}{2} \ln \left( \frac{3k^2 + k}{3 + k} \right) < 7 \ln 8$$ - Divide both sides by 7: $$\frac{1}{2} \ln \left( \frac{3k^2 + k}{3 + k} \right) < \ln 8$$ - Multiply both sides by 2: $$\ln \left( \frac{3k^2 + k}{3 + k} \right) < 2 \ln 8 = \ln 8^2 = \ln 64$$ - Exponentiate both sides: $$\frac{3k^2 + k}{3 + k} < 64$$ 6. **Solve inequality:** - Multiply both sides by $(3 + k)$ (positive since $k > 1$): $$3k^2 + k < 64(3 + k) = 192 + 64k$$ - Rearrange: $$3k^2 + k - 64k - 192 < 0$$ $$3k^2 - 63k - 192 < 0$$ 7. **Find roots of quadratic:** - Solve $3k^2 - 63k - 192 = 0$. - Divide by 3: $$k^2 - 21k - 64 = 0$$ - Use quadratic formula: $$k = \frac{21 \pm \sqrt{21^2 + 4 \times 64}}{2} = \frac{21 \pm \sqrt{441 + 256}}{2} = \frac{21 \pm \sqrt{697}}{2}$$ - Approximate $\sqrt{697} \approx 26.4$. - Roots: $$k_1 = \frac{21 - 26.4}{2} = -2.7$$ $$k_2 = \frac{21 + 26.4}{2} = 23.7$$ 8. **Inequality solution:** - Since leading coefficient positive, inequality $<0$ holds between roots. - So $-2.7 < k < 23.7$. - Given $k$ positive integer $>1$, greatest integer $k$ is $23$. **Final answers:** (i) $\frac{dy}{dx} = \frac{4x}{2x^2 + 5}$. (ii) Greatest possible $k$ is $23$.