Question:
If a function f(x) defined by
\(f(x) = \begin{cases} ae^{x} + be^{-x}, & -1 \le x \lt 1 \\ cx^2, & 1 \le x \le 3 \\ ax^2 + 2cx, & 3 \lt x \le 4 \end{cases}\)
be continuous for some \(a, b, c \in R\) and \(f'(0) + f'(2) = e\), then the value of \(a\) is :
If a function f(x) defined by
\(f(x) = \begin{cases} ae^{x} + be^{-x}, & -1 \le x \lt 1 \\ cx^2, & 1 \le x \le 3 \\ ax^2 + 2cx, & 3 \lt x \le 4 \end{cases}\)
be continuous for some \(a, b, c \in R\) and \(f'(0) + f'(2) = e\), then the value of \(a\) is :
\(f(x) = \begin{cases} ae^{x} + be^{-x}, & -1 \le x \lt 1 \\ cx^2, & 1 \le x \le 3 \\ ax^2 + 2cx, & 3 \lt x \le 4 \end{cases}\)
be continuous for some \(a, b, c \in R\) and \(f'(0) + f'(2) = e\), then the value of \(a\) is :
Show Hint
Impose continuity at x = 1 and x = 3, then use the derivative condition to solve for a.
Updated On: Nov 5, 2026
- \(\frac{e}{e^2 - 3e - 13}\)
- \(\frac{1}{e^2 - 3e + 13}\)
- \(\frac{e}{e^2 - 3e + 13}\)
- \(\frac{e}{e^2 + 3e + 13}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Continuity at \(x=1\): \(ae + \frac{b}{e} = c\). (1)
Continuity at \(x=3\): \(9c = 9a + 6c \Rightarrow c = 3a\). (2)
From (1) and (2): \(b = e(c - ae) = e(3a - ae) = ae(3-e)\).
\(f'(x) = ae^x - be^{-x}\) on \((-1,1)\), so \(f'(0) = a - b\). On \((1,3)\), \(f'(x)=2cx\), so \(f'(2) = 4c = 12a\).
\(f'(0)+f'(2) = a - b + 12a = 13a - ae(3-e) = a(e^2 - 3e + 13) = e\).
Hence \(a = \frac{e}{e^2 - 3e + 13}\).
Answer: (3) \(\frac{e}{e^2 - 3e + 13}\)
Continuity at \(x=3\): \(9c = 9a + 6c \Rightarrow c = 3a\). (2)
From (1) and (2): \(b = e(c - ae) = e(3a - ae) = ae(3-e)\).
\(f'(x) = ae^x - be^{-x}\) on \((-1,1)\), so \(f'(0) = a - b\). On \((1,3)\), \(f'(x)=2cx\), so \(f'(2) = 4c = 12a\).
\(f'(0)+f'(2) = a - b + 12a = 13a - ae(3-e) = a(e^2 - 3e + 13) = e\).
Hence \(a = \frac{e}{e^2 - 3e + 13}\).
Answer: (3) \(\frac{e}{e^2 - 3e + 13}\)
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