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.
The line $$y=x$$ meets $$y=k e^{\mathrm{x}}$$ for $$k \leq 0$$ at
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.
The positive value of $$k$$ for which $$k e^{x}-x=0$$ has only one root is
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) | $$f(x) < 1$$ |
JEE Advanced Subjects
Browse all chapters by subject