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Consider f(x)=|1-x|,1 le xle2 and g(x)=f...

Consider `f(x)=|1-x|,1 le xle2 and g(x)=f(x)+b sin.(pi)/(2)x, 1 le xle 2` then which of the following is correct?

A

Rolle's theorem is applicable to both f and g with `b=(3)/(2)`.

B

LMVT is not applicable to f and Rolle's theorem is applicable to g with `b=(1)/(2)`

C

LMVT is applicable to f and Rolle's theorem is applicable to g with b = 1.

D

Rolle's theorem is not applicable to both f and g for any real b.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will analyze the functions \( f(x) \) and \( g(x) \) step by step. ### Step 1: Define the function \( f(x) \) The function is given as: \[ f(x) = |1 - x| \] For \( 1 \leq x \leq 2 \), we can simplify \( f(x) \): - When \( x = 1 \), \( f(1) = |1 - 1| = 0 \) - When \( x = 2 \), \( f(2) = |1 - 2| = 1 \) Since \( 1 \leq x \leq 2 \), we can express \( f(x) \) as: \[ f(x) = x - 1 \quad \text{for } 1 \leq x \leq 2 \] ### Step 2: Define the function \( g(x) \) The function \( g(x) \) is defined as: \[ g(x) = f(x) + b \sin\left(\frac{\pi}{2} x\right) \] Substituting \( f(x) \): \[ g(x) = (x - 1) + b \sin\left(\frac{\pi}{2} x\right) \] ### Step 3: Evaluate \( g(x) \) at the endpoints Now we will evaluate \( g(x) \) at the endpoints \( x = 1 \) and \( x = 2 \): - For \( x = 1 \): \[ g(1) = (1 - 1) + b \sin\left(\frac{\pi}{2} \cdot 1\right) = 0 + b \cdot 1 = b \] - For \( x = 2 \): \[ g(2) = (2 - 1) + b \sin\left(\frac{\pi}{2} \cdot 2\right) = 1 + b \cdot 0 = 1 \] ### Step 4: Apply Rolle's Theorem For Rolle's Theorem to be applicable, we need \( g(1) = g(2) \): \[ b = 1 \] ### Step 5: Check the conditions for Mean Value Theorem (MVT) - \( f(x) \) is continuous on the closed interval \([1, 2]\) and differentiable on the open interval \( (1, 2) \). - \( g(x) \) is continuous on \([1, 2]\) and differentiable on \( (1, 2) \) when \( b = 1 \). ### Conclusion - Since \( f(1) \neq f(2) \), Rolle's Theorem is not applicable to \( f(x) \). - However, Mean Value Theorem is applicable to \( f(x) \) and Rolle's Theorem is applicable to \( g(x) \) when \( b = 1 \). Thus, the correct option is: **Mean Value Theorem is applicable to \( f \) and Rolle's Theorem is applicable to \( g \).**

To solve the problem, we will analyze the functions \( f(x) \) and \( g(x) \) step by step. ### Step 1: Define the function \( f(x) \) The function is given as: \[ f(x) = |1 - x| \] For \( 1 \leq x \leq 2 \), we can simplify \( f(x) \): ...
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