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PASSAGE  1
If a continuous function f defined on the real line R, assumes positive and negative values in R then the equation f(x) = 0 has a 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.
Q. The line y = x meets y = ke^{x} for k < 0 at
For k = 0, line y = x meets y = 0, i.e., xaxis only at one point.
For k < 0, y = kex meets y = x only once as shown in the graph.
PASSAGE  1
If a continuous function f defined on the real line R, assumes positive and negative values in R then the equation f(x) = 0 has a 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.
Q. The positive value of k for which ke^{x} – x = 0 has only one root is
Let f (x) = ke^{x} – x
Now for f (x) = 0 to have only one root means the line y = x must be tangential to the curve y = ke^{x}.
Let it be so at (x_{1}, y_{1}) then
PASSAGE  1
If a continuous function f defined on the real line R, assumes positive and negative values in R then the equation f(x) = 0 has a 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.
Q. For k > 0, the set of all values of k for which ke^{x} – x = 0 has two distinct roots is
∵ For y = x to be tangent to the curve y = kex, k = 1/e
∴ For y = kex to meet y = x at two points we should
PASSAGE  2
Let f (x) = (1 – x)^{2} sin^{2}x + x^{2} for all x ∈ I R and let
Q.
Consider the statements:
P : Th er e exists some x ∈ R such th at f (x) + 2 x = 2(1 + x^{2})
Q : There exists some x ∈ R such that 2 f (x) + 1 = 2x (1 + x) Then
For the statement P
f(x) + 2x = 2 (1 + x^{2})
⇒ (1 – x)^{2} sin^{2}x + x^{2} + 2x = 2(1+ x^{2})
⇒ (1 – x)^{2} sin^{2}x = x^{2} – 2x + 1 + 1
⇒ (1 – x)^{2} sin^{2}x = (1 – x)^{2} + 1
⇒ (1 – x)^{2} cos^{2}x = –1
Which is not possible for any real value of x.
∴ P is not true.
Also let H(x) = 2f(x) + 1 – 2x (1 + x)
H(0) = 2f(0) + 1 – 0 = 1
and H(1) = 2f(1) + 1 – 4 = – 3
⇒ H(x) has a solution in (0, 1)
∴ Q is true.
PASSAGE  2
Let f (x) = (1 – x)^{2} sin^{2}x + x^{2} for all x ∈ I R and let
Q. Which of the following is true?
∴ h(x) is decreasing function.
PASSAGE  3
(the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) = f(1) = 0 and satisfies
Q. Which of the following is true for 0 < x < 1?
We have f ''(x) – 2f '(x) +f(x) > e^{x}
Let g(x) = e^{–x}f(x)
Then we have g ''(x) > 1> 0
So g is concave upward.
Also g(0) = g(1) = 0
PASSAGE  3
(the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) = f(1) = 0 and satisfies
Q. If the function e^{–x} f(x) assumes its minimum in the interval which of the following is true?
g(x) = e^{–x}f(x)
⇒ g'(x) = e^{–x}f '(x) – e^{–x}f(x) = e^{–x} (f '(x) – f(x))
As = 1/4 is point of local minima in [0, 1]
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