Standard +0.3 This question involves routine application of the quotient rule to find g'(x), then showing it's negative (which is straightforward since the numerator is constant and denominator is always positive for x in domain). Finding where g(a) > 0 requires basic inequality solving. The multi-part structure adds some length, but each part uses standard techniques without requiring novel insight or complex reasoning.
13. The function \(g\) is defined by
$$\mathrm { g } ( x ) = \frac { 2 e ^ { x } - 5 } { e ^ { x } - 4 } \quad x \neq k , x > 0$$
where \(k\) is a constant.
a. Deduce the value of \(k\).
b. Prove that
$$\mathrm { g } ^ { \prime } ( x ) < 0$$
For all values of \(x\) in the domain of g .
c. Find the range of values of \(a\) for which
$$\mathrm { g } ( a ) > 0$$
Deduce value of \(k\) for \(g(x) = \frac{2e^x - 5}{e^x - 4}\), \(x \neq k\), \(x > 0\)
(1)
B1 Deduces \(k = \ln 4\) or \(x \neq \ln 4\)
Part (b):
Answer
Marks
Guidance
Prove that \(g'(x) < 0\) for all values of \(x\) in domain of \(g\)
(3)
M1 Attempts to differentiate via quotient rule/product rule/chain rule; A1 \(\frac{-3e^x}{(e^x - 4)^2}\) or equivalent; A1 Correct solution only. States that as \(-3e^x < 0\) and \((e^x - 4)^2 > 0\) so \(g'(x) < 0\)
Part (c):
Answer
Marks
Guidance
Find range of values of \(a\) for which \(g(a) > 0\)
(2)
M1 Attempts to solve either \(2e^x - 5 = 0\) or \(e^x - 4 = 0\) or using inequalities; A1 \(a > \ln 4\), \(0 < a < \ln\frac{5}{2}\)
**Part (a):**
Deduce value of $k$ for $g(x) = \frac{2e^x - 5}{e^x - 4}$, $x \neq k$, $x > 0$ | (1) | B1 Deduces $k = \ln 4$ or $x \neq \ln 4$
**Part (b):**
Prove that $g'(x) < 0$ for all values of $x$ in domain of $g$ | (3) | M1 Attempts to differentiate via quotient rule/product rule/chain rule; A1 $\frac{-3e^x}{(e^x - 4)^2}$ or equivalent; A1 Correct solution only. States that as $-3e^x < 0$ and $(e^x - 4)^2 > 0$ so $g'(x) < 0$
**Part (c):**
Find range of values of $a$ for which $g(a) > 0$ | (2) | M1 Attempts to solve either $2e^x - 5 = 0$ or $e^x - 4 = 0$ or using inequalities; A1 $a > \ln 4$, $0 < a < \ln\frac{5}{2}$
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13. The function $g$ is defined by
$$\mathrm { g } ( x ) = \frac { 2 e ^ { x } - 5 } { e ^ { x } - 4 } \quad x \neq k , x > 0$$
where $k$ is a constant.\\
a. Deduce the value of $k$.\\
b. Prove that
$$\mathrm { g } ^ { \prime } ( x ) < 0$$
For all values of $x$ in the domain of g .\\
c. Find the range of values of $a$ for which
$$\mathrm { g } ( a ) > 0$$
\hfill \mbox{\textit{Edexcel PMT Mocks Q13 [6]}}