21.
The domain of $$f\left( x \right) = {\sin ^{ - 1}}\left( {\frac{{1 + {x^2}}}{{2x}}} \right) + \sqrt {1 - {x^2}} \,$$ is :
A
$$\left\{ 1 \right\}$$
B
$$\left( { - 1,\,1} \right)$$
C
$$\left\{ {1,\, - 1} \right\}$$
D
none of these
Answer :
$$\left\{ {1,\, - 1} \right\}$$
View Solution
Discuss Question
Here $$\left| {\frac{{1 + {x^2}}}{{2x}}} \right| \leqslant 1$$ and $$1 - {x^2} \geqslant 0.$$ The first inequality implies $$x=1,\,-1,$$ which satisfy the second inequation.
22.
Let $$f$$ be a function from $$R$$ to $$R$$ given by $$f\left( x \right) = \frac{{{x^2} - 4}}{{{x^2} + 1}}.$$ Then $$f\left( x \right)$$ is :
A
one-one and into
B
one-one and onto
C
many-one and into
D
many-one and onto
Answer :
many-one and into
View Solution
Discuss Question
$$f\left( x \right) = f\left( { - x} \right).$$ So, $$f$$ is many-one. Also, $$f\left( x \right) = 1 - \frac{5}{{{x^2} + 1}} > 1 - 5 = - 4.$$ So, $$f$$ is into.
23.
Let
\[\begin{array}{l}
{f_1}\left( x \right) = \left\{ \begin{array}{l}
x,\,\,\,0 \le x \le 1\\
1,\,\,\,x > 1\\
0,\,\,\,{\rm{otherwise}}
\end{array} \right.\\
{f_2}\left( x \right) = {f_1}\left( { - x} \right){\rm{ for\,\,all\,\, }}x\\
{f_3}\left( x \right) = - {f_2}\left( x \right){\rm{ for\,\,all\,\, }}x\\
{f_4}\left( x \right) = {f_3}\left( { - x} \right){\rm{ for\,\, all\,\, }}x
\end{array}\]
Which of the following is necessarily true ?
A
$${f_4}\left( x \right) = {f_1}\left( x \right){\text{ for all }}x$$
B
$${f_1}\left( x \right) = - {f_3}\left( { - x} \right){\text{ for all }}x$$
C
$${f_2}\left( { - x} \right) = {f_4}\left( x \right){\text{ for all }}x$$
D
$${f_1}\left( x \right) + {f_3}\left( x \right) = 0{\text{ for all }}x$$
Answer :
$${f_1}\left( x \right) = - {f_3}\left( { - x} \right){\text{ for all }}x$$
24.
The entire graphs of the equation $$y = {x^2} + kx - x + 9$$ is strictly above the $$x$$-axis if and only if
A
$$k < 7$$
B
$$ - 5 < k < 7$$
C
$$k > - 5$$
D
None of these.
Answer :
$$ - 5 < k < 7$$
View Solution
Discuss Question
$$y = {x^2} + \left( {k - 1} \right)x + 9 = {\left( {x + \frac{{k + 1}}{2}} \right)^2} + 9 - {\left( {\frac{{k - 1}}{2}} \right)^2}$$
For entire graph to be above $$x$$-axis, we should have
$$9 - {\left( {\frac{{k - 1}}{2}} \right)^2} > 0$$
$$\eqalign{
& \Rightarrow {k^2} - 2k - 35 < 0 \Rightarrow \left( {k - 7} \right)\left( {k + 5} \right) < 0 \cr
& {\text{i}}{\text{.e,}} - 5 < k < 7 \cr} $$
25.
If $$F\left( x \right) = {\left( {f\left( {\frac{x}{2}} \right)} \right)^2} + {\left( {g\left( {\frac{x}{2}} \right)} \right)^2}$$ where $$f''\left( x \right) = - f\left( x \right)$$ and $$g\left( x \right) = f'\left( x \right)$$ and given that $$F\left( 5 \right) = 5,$$ then $$F\left( {10} \right)$$ is equal to
A
5
B
10
C
0
D
15
Answer :
5
View Solution
Discuss Question
$$\eqalign{
& F\left( x \right) = {\left( {f\left( {\frac{x}{2}} \right)} \right)^2} + {\left( {g\left( {\frac{x}{2}} \right)} \right)^2} \cr
& \Rightarrow F'\left( x \right) = 2f\left( {\frac{x}{2}} \right).f'\left( {\frac{x}{2}} \right).\frac{1}{2} + 2g\left( {\frac{x}{2}} \right).g'\left( {\frac{x}{2}} \right).\frac{1}{2} \cr
& = f\left( {\frac{x}{2}} \right).f'\left( {\frac{x}{2}} \right) + f'\left( {\frac{x}{2}} \right).f''\left( {\frac{x}{2}} \right)\,\left[ {\because g\left( x \right) = f'\left( x \right) \Rightarrow g'\left( x \right) = f''\left( x \right)} \right] \cr
& = f\left( {\frac{x}{2}} \right).f'\left( {\frac{x}{2}} \right) - f'\left( {\frac{x}{2}} \right)f\left( {\frac{x}{2}} \right) \cr
& = 0\,\,\left[ {\because f''\left( x \right) = - f\left( x \right)} \right] \cr
& \Rightarrow F\left( x \right)\,{\text{is}}\,{\text{a}}\,{\text{constant function}}{\text{.}} \cr
& \therefore F\left( x \right) = F\left( 5 \right) = 5\forall x \in R \Rightarrow F\left( {10} \right) = 5 \cr} $$
26.
Let $$f\left( x \right)$$ be a function whose domain is $$\left[ { - 5,\,7} \right].$$ Let $$g\left( x \right) = \left| {2x + 5} \right|.$$ Then the domain of $$\left( {f\,o\,g} \right)\left( x \right)$$ is :
A
$$\left[ { - 5,\,1} \right]$$
B
$$\left[ { - 4,\,0} \right]$$
C
$$\left[ { - 6,\,1} \right]$$
D
none of these
Answer :
$$\left[ { - 6,\,1} \right]$$
View Solution
Discuss Question
$$\eqalign{
& f\left\{ {g\left( x \right)} \right\} = f\left( {\left| {2x + 5} \right|} \right)\,\,\,\,\,\,\,\,\,\therefore \left| {2x + 5} \right| \in \left[ { - 5,\,7} \right] \cr
& \therefore - 5 \leqslant \left| {2x + 5} \right| \leqslant 7 \cr
& \Rightarrow \left| {2x + 5} \right| \leqslant 7 \cr
& \Rightarrow - 7 \leqslant 2x + 5 \leqslant 7 \cr
& \Rightarrow - 6 \leqslant x \leqslant 1 \cr} $$
27.
If $$f\left( {x + y,\,x - y} \right) = xy$$ then the arithmetic mean of $$f\left( {x,\,y} \right)$$ and $$f\left( {y,\,x} \right)$$ is :
A
$$x$$
B
$$y$$
C
0
D
none of these
Answer :
0
View Solution
Discuss Question
$$\eqalign{
& {\text{Let }}x + y = p,\,\,x - y = q \cr
& {\text{Then }}f\left( {p,\,q} \right) = \frac{{p + q}}{2}.\frac{{p - q}}{2} = \frac{{{p^2} - {q^2}}}{4} \cr
& \therefore \,\,f\left( {x,\,y} \right) = \frac{{{x^2} - {y^2}}}{4}{\text{ and }}f\left( {y,\,x} \right) = \frac{{{y^2} - {x^2}}}{4} \cr} $$
28.
If the real-valued function $$f\left( x \right) = px + \sin \,x$$ is a bijective function then the set of possible values of $$p\, \in \,R$$ is :
A
$$R - \left\{ 0 \right\}$$
B
$$R$$
C
$$\left( {0,\, + \infty } \right)$$
D
none of these
Answer :
$$R - \left\{ 0 \right\}$$
View Solution
Discuss Question
Clearly, $$p \ne 0.$$ If $$p=0$$ then $$f\left( x \right) = \sin \,x$$ which is many-one.
29.
A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
Interval
Function
(a)
$$\left( { - \infty ,\,\infty } \right)$$
$${x^3} - 3{x^2} + 3x + 3$$
(b)
$$\left[ {2,\,\infty } \right)$$
$$2{x^3} - 3{x^2} - 12x + 6$$
(c)
$$\left( { - \infty ,\,\frac{1}{3}} \right]$$
$$3{x^2} - 2x + 1$$
(d)
$$\left( { - \infty ,\, - 4} \right)$$
$${x^3} + 6{x^2} + 6$$
A
a
B
b
C
c
D
d
Answer :
c
View Solution
Discuss Question
Clearly function $$f\left( x \right) = 3{x^2} - 2x + 1$$ is increasing when $$f'\left( x \right) = 6x - 2 \geqslant 0 \Rightarrow x \in \left[ {\frac{1}{3},\infty } \right)$$
$$\therefore f\left( x \right)$$ is incorrectly matched with $$\left( { - \infty ,\frac{1}{3}} \right)$$
30.
Let $$f\left( x \right) = {\log _{{x^2}}}25$$ and $$g\left( x \right) = {\log _x}5$$ then $$f\left( x \right) = g\left( x \right)$$ holds for $$x$$ belonging to :
A
$$R$$
B
$$\left( {0,\,1} \right) \cup \left( {1,\, + \infty } \right)$$
C
$$\phi $$
D
none of these
Answer :
$$\left( {0,\,1} \right) \cup \left( {1,\, + \infty } \right)$$
View Solution
Discuss Question
$$\eqalign{
& {\text{Domain of }}f = {D_1} = \left( { - \infty ,\, - 1} \right) \cup \left( { - 1,\,0} \right) \cup \left( {0,\,1} \right) \cup \left( {1,\, + \infty } \right) \cr
& {\text{Domain of }}g = {D_2} = \left( {0,\,1} \right) \cup \left( {1,\, + \infty } \right) \cr
& {\text{So, }}{D_1} \cap {D_2} = \left( {0,\,1} \right) \cup \left( {1,\, + \infty } \right) \cr
& {\text{In this set, }}{\log _{{x^2}}}25 = \frac{1}{2} \times 2\,{\log _x}5 = {\log _x}5 \cr} $$