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The number of value/values of x for whic...

The number of value/values of x for which sin `y=x^(2)-2x` si possible is

A

0

B

1

C

infinite

D

none of these

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To solve the problem of finding the number of values of \( x \) for which \( \sin y = x^2 - 2x \) is possible, we will follow these steps: ### Step 1: Understand the Equation We start with the equation: \[ \sin y = x^2 - 2x \] We need to determine for which values of \( x \) this equation holds true. ### Step 2: Rewrite the Quadratic Expression We can rewrite the quadratic expression \( x^2 - 2x \) as: \[ x^2 - 2x = (x - 1)^2 - 1 \] This transformation helps us analyze the range of the expression more easily. ### Step 3: Determine the Range of the Quadratic Expression From the rewritten form, we have: \[ \sin y = (x - 1)^2 - 1 \] The term \( (x - 1)^2 \) is always non-negative (i.e., \( \geq 0 \)). Therefore, the minimum value of \( (x - 1)^2 \) is 0, which occurs when \( x = 1 \). Thus, the minimum value of \( (x - 1)^2 - 1 \) is: \[ 0 - 1 = -1 \] As \( (x - 1)^2 \) can take any non-negative value, \( (x - 1)^2 - 1 \) can take any value greater than or equal to \(-1\). ### Step 4: Compare with the Range of the Sine Function The sine function, \( \sin y \), has a range of: \[ [-1, 1] \] This means \( \sin y \) can take values from \(-1\) to \(1\). ### Step 5: Determine Valid Values of \( x \) For \( \sin y = (x - 1)^2 - 1 \) to hold, we need: \[ -1 \leq (x - 1)^2 - 1 \leq 1 \] This simplifies to: 1. \( (x - 1)^2 - 1 \geq -1 \) which is always true since it simplifies to \( (x - 1)^2 \geq 0 \). 2. \( (x - 1)^2 - 1 \leq 1 \) simplifies to: \[ (x - 1)^2 \leq 2 \] Taking the square root, we find: \[ -\sqrt{2} \leq x - 1 \leq \sqrt{2} \] Therefore: \[ 1 - \sqrt{2} \leq x \leq 1 + \sqrt{2} \] ### Step 6: Conclusion The interval \( [1 - \sqrt{2}, 1 + \sqrt{2}] \) contains infinitely many values of \( x \). Therefore, the number of values of \( x \) for which \( \sin y = x^2 - 2x \) is possible is infinite. ### Final Answer The number of values of \( x \) for which \( \sin y = x^2 - 2x \) is possible is **infinite**. ---

To solve the problem of finding the number of values of \( x \) for which \( \sin y = x^2 - 2x \) is possible, we will follow these steps: ### Step 1: Understand the Equation We start with the equation: \[ \sin y = x^2 - 2x \] We need to determine for which values of \( x \) this equation holds true. ...
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