If a continuous functions $$f$$ 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 $$f(x)=k e^{x}-x$$ for all real $$x$$ where $$k$$ is a real constant.
For $$k > 0$$, the set of all values of $$k$$ for which $$k e^{x}-x=0$$ has two distinct roots is
Let $$f(x) = {{{x^2} - 6x + 5} \over {{x^2} - 5x + 6}}$$.
Match the conditions/expressions in Column I with statements in Column II.
| Column I | Column II | ||
|---|---|---|---|
| (A) | If $$ - 1 < x < 1$$, then $$f(x)$$ satisfies | (P) | $$0 < f(x) < 1$$ |
| (B) | If $$1 < x < 2$$, then $$f(x)$$ satisfies | (Q) | $$f(x) < 0$$ |
| (C) | If $$3 < x < 5$$, then $$f(x)$$ satisfies | (R) | $$f(x) > 0$$ |
| (D) | If $$x > 5$$, then $$f(x)$$ satisfies | (S) | $$f(x) < 1$$ |
In the following [x] denotes the greatest integer less than or equal to x.
Match the functions in Column I with the properties Column II.
| Column I | Column II | ||
|---|---|---|---|
| (A) | $$x|x|$$ | (P) | continuous in ($$-1,1$$). |
| (B) | $$\sqrt{|x|}$$ | (Q) | differentiable in ($$-1,1$$) |
| (C) | $$x+[x]$$ | (R) | strictly increasing in ($$-1,1$$) |
| (D) | $$|x-1|+|x+1|$$ | (S) | not differentiable at least at one point in ($$-1,1$$) |
For $x>0, \mathop {\lim }\limits_{x \to 0}\left((\sin x)^{1 / x}+(1 / x)^{\sin x}\right)$ is :
JEE Advanced Subjects
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