## Question 12:

### Part (a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $u = 3x^2$, $y = a^u$; $u' = kx$, $y' = a^u \ln a$ | M1 | Attempt use of chain rule, with $a^u$ correctly differentiated. No credit for just stating $\frac{d}{dx}(a^x) = a^x \ln a$ unless clearly used in a correct chain rule |
| $\frac{dy}{dx} = 6xa^u \ln a$ | A1 | Use chain rule to obtain correct derivative. Product of their $u'$ and $y'$. May still be in terms of $x$ and $u$ |
| $\frac{dy}{dx} = 6xa^{3x^2} \ln a$ **A.G.** | A1 | Obtain correct derivative. Now fully in terms of $x$ |

**OR:** M1 – attempt differentiation of $\ln y = 3x^2 \ln a$; A1 – obtain $\frac{1}{y}\frac{dy}{dx} = 6x \ln a$; A1 – obtain correct derivative **A.G.**

**OR:** M1 – attempt differentiation of $y = e^{(3\ln a)x^2}$; A1 – obtain $\frac{dy}{dx} = (6x \ln a)e^{(3\ln a)x^2}$; A1 – obtain correct derivative **A.G.**

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### Part (b):

| Answer | Mark | Guidance |
|--------|------|----------|
| When $x = 1$, $m = 6a^3 \ln a$ | B1 | Correct gradient of tangent |
| $y - a^3 = 6a^3 \ln a(x-1)$; $0 - a^3 = 6a^3 \ln a(\frac{1}{2} - 1)$ | M1* | Use $(\frac{1}{2}, 0)$ in attempt at equation of tangent through $(1, a^3)$, or vice versa. OR $\frac{0-a^3}{\frac{1}{2}-1} = 6a^3 \ln a$. M0 if gradient still in terms of $x$. Allow BOD if something other than $x=1$ was used to find the gradient |
| $a^3 = 3a^3 \ln a$; $a^3(3\ln a - 1) = 0$; $a = e^{\frac{1}{3}}$ | M1d* | Attempt to find $a$. Must go as far as attempting a value for $a$. Condone cancelling by $a^3$ rather than factorising |
| $a = e^{\frac{1}{3}}$ | A1 | Obtain correct value for $a$. Any equivalent exact form e.g. $a = \sqrt[3]{e}$ |

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### Part (c):

| Answer | Mark | Guidance |
|--------|------|----------|
| $u = 6x\ln a$, $v = a^{3x^2}$ | M1* | Attempt use of product rule. On given $\frac{dy}{dx}$, with both parts a function of $x$. If using parts as $u = 6x a^{3x^2}$ and $v = \ln a$, then must use product rule properly on $u$ (but condone $\ln a$ differentiating to $\frac{1}{a}$) |
| $\frac{d^2y}{dx^2} = (6\ln a)(a^{3x^2}) + (6x\ln a)(6xa^{3x^2}\ln a)$ | A1 | At least one term correct |
| $= a^{3x^2}(6\ln a)(1 + 6x^2 \ln a)$ | A1 | Fully correct second derivative |
| $\ln a > 0$ for $a > 1$; $a^{f(x)} > 0$ for all $a$ and $x$; $6x^2 \geq 0$, so $1 + 6x^2\ln a \geq 1$ | M1d* | Consider sign of each term – must consider each component of each term. Possibly factorised, or possibly considered term by term. Domains not needed for M1. Allow BOD if $> $ not $\geq$ |
| $\frac{d^2y}{dx^2} > 0$ for all $x$, hence curve is always convex | A1 | Correct working only. Second derivative must be correct. Domains must be seen. Inequality signs must be correct throughout |

**OR (substituting $a = e^{\frac{1}{3}}$):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 2xe^{x^2}$ | M1* | Attempt use of product rule. Substitute their $a$ into $\frac{dy}{dx}$ and attempt to differentiate |
| $\frac{d^2y}{dx^2} = 2e^{x^2} + 4x^2e^{x^2}$ | A1FT | At least one term correct, FT their $a$, any $a$ |
| (fully correct second derivative) | A1FT | Fully correct second derivative, FT their $a$ as long as of form $e^k$. Expect $6ke^{3kx^2} + 36k^2x^2e^{3kx^2}$ |
| $e^{x^2} > 0$ for all $x$, so $2e^{x^2} > 0$; $x^2 \geq 0$ for all $x$, so $4x^2e^{x^2} \geq 0$; hence $2e^{x^2} + 4x^2e^{x^2} > 0$ | M1d* | Consider sign of each term. Domains not needed for M1. Allow BOD if $>$ not $\geq$ |
| $\frac{d^2y}{dx^2} > 0$ for all $x$, hence curve is always convex | A1 | Correct working only. Second derivative must be correct. Domains must be seen. Inequality signs must be correct throughout |