Question:
Let \( f(x) \) and \( g(x) \) be two differentiable functions for \( 0 \leq x \leq 1 \) such that \( f(0)=2, g(0)=0, f(1)=6 \). If there exists a real number \( c \in (0,1) \) such that \( f'(c)=2g'(c) \), then \( g(1) \) is equal to
Let \( f(x) \) and \( g(x) \) be two differentiable functions for \( 0 \leq x \leq 1 \) such that \( f(0)=2, g(0)=0, f(1)=6 \). If there exists a real number \( c \in (0,1) \) such that \( f'(c)=2g'(c) \), then \( g(1) \) is equal to
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Convert condition into a function and apply Mean Value Theorem.
Updated On: Apr 30, 2026
- \(0\)
- \(-1\)
- \(4\)
- \(-2\)
- \(2\)
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The Correct Option is
Solution and Explanation
Concept:
Use Mean Value Theorem.
Step 1: Define function. Let: \[ h(x)=f(x)-2g(x) \]
Step 2: Apply MVT. Given: \[ h'(c)=0 \] So: \[ h(1)=h(0) \]
Step 3: Compute values. \[ f(1)-2g(1)=f(0)-2g(0) \] \[ 6-2g(1)=2-0 \] \[ 6-2g(1)=2 \Rightarrow 2g(1)=4 \Rightarrow g(1)=2 \]
Step 1: Define function. Let: \[ h(x)=f(x)-2g(x) \]
Step 2: Apply MVT. Given: \[ h'(c)=0 \] So: \[ h(1)=h(0) \]
Step 3: Compute values. \[ f(1)-2g(1)=f(0)-2g(0) \] \[ 6-2g(1)=2-0 \] \[ 6-2g(1)=2 \Rightarrow 2g(1)=4 \Rightarrow g(1)=2 \]
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