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Let f (x) = (1 – x)2 + sin2 x + x2 for all x ∈ IR, and g (x) = integral from 1 to x [2 (t – 1)] / [t + 1] – ln t] f (t) dt for all x ∈ (1, ∞]. Consider the statements. P: There exists some x ∈ IR such that f (x) + 2x = 2 (1 + x2). Q: There exists some x ∈ IR such that 2 f (x) + 1 = 2x (1 + x). Then,

Consider the statements P: There exists some x ∈ IR such that f (x) + 2x = 2 (1 + x2) Q: There exists some x ∈ IR such that 2 f (x) + 1 = 2x (1... View Article

It is a continuous function f defined on the real line R, assume positive and negative values in R then the equation f(x) = 0 has root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R. Consider f(x) = ke^x – x for all real x where k is a real constant. For k > 0, the set of all values of k for which ke^x – x = 0 has two distinct roots is

1) (0, 1 / e) 2) (1 / e, 1) 3) (1 / e, ∞) 4) (0, 1) Solution: (1) (0, 1 / e) If k > 0, then for 2 distinct roots local minima... View Article

It is a continuous function f defined on the real line R, assume positive and negative values in R then the equation f(x) = 0 has root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R. Consider f(x) = kex – x for all real x where k is a real constant. The positive value of k for which kex – x = 0 has only one root is

The positive value of k for which kex – x = 0 has only one root is 1) 1 / e 2) 1 3) e 4) loge 2 Solution: (1) 1 / e If k > 0,... View Article

It is a continuous function f defined on the real line R, assume positive and negative values in R then the equation f(x) = 0 has root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R. Consider f (x) = kex – x for all real x where k is a real constant.The line y = x meets y = kex for k ≤ 0 at

1) No point 2) One point 3) Two points 4) More than two points Solution: (2) One point f (x) = kex – x f’ (x) = kex - 1 < 0 if... View Article

Why are gases not ideal?

No real gas can be an ideal gas. There are certain assumptions made for an ideal gas. In the ideal gas, intermolecular forces and volume occupied... View Article