1
IIT-JEE 2007
+4
-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 $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

The line $$y=x$$ meets $$y = k{e^x}$$ for $$k \le 0$$ at

A
no point
B
one point
C
two points
D
more than two points
2
IIT-JEE 2007
+4
-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 $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is 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
$$\left( {0,{1 \over e}} \right)$$
B
$$\left( {{1 \over e},1} \right)$$
C
$$\left( {{1 \over e},\infty } \right)$$
D
$$\left( {0,1} \right)$$
3
IIT-JEE 2007
+4
-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 $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

The positive value of $$k$$ for which $$k{e^x} - x = 0$$ has only one root is

A
$${1 \over e}$$
B
$$1$$
C
$$e$$
D
$${\log _e}2$$
4
IIT-JEE 2007
+3
-0.75
Let $$F(x)$$ be an indefinite integral of $$si{n^2}x.$$

STATEMENT-1: The function $$F(x)$$ satisfies $$F\left( {x + \pi } \right) = F\left( x \right)$$
for all real $$x$$. because

STATEMENT-2: $${\sin ^2}\left( {x + \pi } \right) = {\sin ^2}x$$ for all real $$x$$.

A
Statement-1 is True, Statement-2 is True; Statement-2 is 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.
EXAM MAP
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