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If f(x)=(acos x-cos b x)/(x^2),x!=0a n d...

If `f(x)=(acos x-cos b x)/(x^2),x!=0a n df(0)=4` is continous at `x=0,` then the ordered pair `(a ,b)` is `(+-1,3)` b. `(1,+-3)` c. `(-1,-3)` d. `(-1,3)`

A

`+-1,3)`

B

`(1,+-3)`

C

`(-1,-3)`

D

`(1,3)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the function \( f(x) = \frac{a \cos x - \cos(bx)}{x^2} \) for \( x \neq 0 \) and ensure that it is continuous at \( x = 0 \) with \( f(0) = 4 \). ### Step-by-Step Solution: 1. **Understanding Continuity**: For \( f(x) \) to be continuous at \( x = 0 \), we need: \[ \lim_{x \to 0} f(x) = f(0) = 4 \] 2. **Finding the Limit**: We need to evaluate: \[ \lim_{x \to 0} \frac{a \cos x - \cos(bx)}{x^2} \] 3. **Applying L'Hôpital's Rule**: Since both the numerator and denominator approach 0 as \( x \to 0 \), we can apply L'Hôpital's Rule. We differentiate the numerator and denominator: - The derivative of the numerator \( a \cos x - \cos(bx) \) is: \[ -a \sin x + b \sin(bx) \] - The derivative of the denominator \( x^2 \) is: \[ 2x \] Thus, we have: \[ \lim_{x \to 0} \frac{-a \sin x + b \sin(bx)}{2x} \] 4. **Evaluating the New Limit**: Again, as \( x \to 0 \), both the numerator and denominator approach 0. We apply L'Hôpital's Rule again: - The derivative of the numerator \( -a \sin x + b \sin(bx) \) is: \[ -a \cos x + b^2 \cos(bx) \] - The derivative of the denominator \( 2x \) is: \[ 2 \] Therefore, we have: \[ \lim_{x \to 0} \frac{-a \cos x + b^2 \cos(bx)}{2} \] 5. **Substituting \( x = 0 \)**: Now we substitute \( x = 0 \): \[ \frac{-a \cdot 1 + b^2 \cdot 1}{2} = \frac{-a + b^2}{2} \] 6. **Setting the Limit Equal to 4**: We set the limit equal to 4: \[ \frac{-a + b^2}{2} = 4 \] Multiplying both sides by 2 gives: \[ -a + b^2 = 8 \quad \text{(Equation 1)} \] 7. **Finding Another Equation**: We also know that for the limit to exist, the numerator must approach 0 as \( x \to 0 \). Thus: \[ a \cos(0) - \cos(0) = a - 1 = 0 \quad \Rightarrow \quad a = 1 \quad \text{(Equation 2)} \] 8. **Substituting \( a \) in Equation 1**: Substituting \( a = 1 \) into Equation 1: \[ -1 + b^2 = 8 \quad \Rightarrow \quad b^2 = 9 \quad \Rightarrow \quad b = \pm 3 \] 9. **Final Ordered Pair**: Thus, the ordered pair \( (a, b) \) is: \[ (1, 3) \quad \text{or} \quad (1, -3) \] ### Conclusion: The correct answer is: \[ \text{The ordered pair } (a, b) = (1, \pm 3) \]
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