1
IIT-JEE 2007 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

Let $$f(x)=2+\cos x$$ for all real $$x$$.

STATEMENT - 1 : For each real $$t$$, there exists a point $$c$$ in $$[t, t+\pi]$$ such that $$f^{\prime}(C)=0$$.

STATEMENT - 2 : $$f(t)=f(t+2 \pi)$$ for each real $$t$$.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
C
Statement-1 is True, Statement-2 is False
D
Statement-1 is False, Statement-2 is True
2
IIT-JEE 2007 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

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

A
no point
B
one point
C
two points
D
more than two points
3
IIT-JEE 2007 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

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

A
$$\frac{1}{e}$$
B
1
C
$$e$$
D
$$\log _{\mathrm{e}} 2$$
4
IIT-JEE 2007 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

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

A
$$\left(0, \frac{1}{e}\right)$$
B
$$\left(\frac{1}{e}, 1\right)$$
C
$$\left(\frac{1}{e}, \infty\right)$$
D
$$(0,1)$$

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