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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Functions
NDA 2024 · Mathematics · Functions
Let f(x) and g(x) be two functions such that g(x)=x-(1)/(x) and f o...
Let
f
(
x
)
f(x)
f
(
x
)
and
g
(
x
)
g(x)
g
(
x
)
be two functions such that
g
(
x
)
=
x
−
1
x
g(x)=x-\frac{1}{x}
g
(
x
)
=
x
−
x
1
and
f
f
f
o
g
(
x
)
=
x
3
−
1
x
3
g(x)=x^3-\frac{1}{x^3}
g
(
x
)
=
x
3
−
x
3
1
. What is
g
[
f
(
x
)
−
3
x
]
g[f(x)-3 x]
g
[
f
(
x
)
−
3
x
]
equal to?
Options
A
x
3
−
1
x
3
x^3-\frac{1}{x^3}
x
3
−
x
3
1
B
x
3
+
1
x
3
x^3+\frac{1}{x^3}
x
3
+
x
3
1
C
x
2
−
1
x
2
x^2-\frac{1}{x^2}
x
2
−
x
2
1
D
x
2
+
1
x
2
x^2+\frac{1}{x^2}
x
2
+
x
2
1
Correct answer
A.
x
3
−
1
x
3
x^3-\frac{1}{x^3}
x
3
−
x
3
1
Step-by-step solution
∵
f
(
x
)
\because f(x)
∵
f
(
x
)
and g(x) are two functions
g
(
x
)
=
x
−
1
x
and
f
o
g
(
x
)
=
x
3
−
1
x
3
f
(
x
−
1
x
)
=
x
3
−
1
x
3
⇒
f
(
x
−
1
x
)
=
(
x
−
1
x
)
3
+
3
(
x
−
1
x
)
\begin{aligned} g(x) & =x-\frac{1}{x} \text{ and } f o g(x)=x^3-\frac{1}{x^3} \\ f\left(x-\frac{1}{x}\right) & =x^3-\frac{1}{x^3} \\ \Rightarrow f\left(x-\frac{1}{x}\right) & =\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right) \end{aligned}
g
(
x
)
f
(
x
−
x
1
)
⇒
f
(
x
−
x
1
)
=
x
−
x
1
and
f
o
g
(
x
)
=
x
3
−
x
3
1
=
x
3
−
x
3
1
=
(
x
−
x
1
)
3
+
3
(
x
−
x
1
)
f(x)=x^3+3 x \quad ....(i)
g
[
f
(
x
)
−
3
x
]
=
f
(
x
)
−
3
x
−
1
f
(
x
)
−
3
x
g
(
f
(
x
)
−
3
x
]
=
x
3
+
3
x
−
3
x
−
1
x
3
+
3
x
−
3
x
=
x
3
−
1
x
3
g
(
f
(
x
)
−
3
x
]
=
x
3
−
1
x
3
\begin{aligned} g[f(x)-3 x] & =f(x)-3 x-\frac{1}{f(x)-3 x} \\ g(f(x)-3 x] & =x^3+3 x-3 x-\frac{1}{x^3+3 x-3 x} \\ & =x^3-\frac{1}{x^3} \\ g(f(x)-3 x] & =x^3-\frac{1}{x^3}\end{aligned}
g
[
f
(
x
)
−
3
x
]
g
(
f
(
x
)
−
3
x
]
g
(
f
(
x
)
−
3
x
]
=
f
(
x
)
−
3
x
−
f
(
x
)
−
3
x
1
=
x
3
+
3
x
−
3
x
−
x
3
+
3
x
−
3
x
1
=
x
3
−
x
3
1
=
x
3
−
x
3
1
$
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