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Solve the following.
Consider the functions and , where . Find . The graphs of and have a common tangent at . Show that…
Consider the functions $f(x) = -(x-h)^2 + 2k$ and $g(x) = e^{x-2} + k$, where $h, k \in \mathbb{R}$.
(a)[1.0]
Find $f'(x)$.
(b)[1.0]
The graphs of $f$ and $g$ have a common tangent at $x = 3$. Show that: $2h = e + 6$.
Answer
Answer (a)
Official answer key$f'(x) = -2(x - h)$
Final answer: $-2(x-h)$
Answer (b)
Official answer key$\because f(x)$ and $g(x)$ have common tangent at $x = 3$.
$\Rightarrow$ slopes of the tangents to the two curves at $x = 3$ are equal.
$\Rightarrow f'(x) = g'(3)$ [$\because g(x) = e^{x-2} + k$, $g'(x) = e^{x-2} \Rightarrow g'(3) = e$]
$\Rightarrow -2(3 - h) = e$
$\therefore 2h = e + 6$
Hence, proved.
From ISC 2025 Practice Mathematics, question 91.