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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Limits
NDA 2015 · Mathematics · Limits
Given that _ x arrow ∞ ((2+x^2)/(1+x)-A x-B)=3 . What is the value...
Given that
lim
x
→
∞
(
2
+
x
2
1
+
x
−
A
x
−
B
)
=
3
\lim _{x \rightarrow \infty}\left(\frac{2+x^2}{1+x}-A x-B\right)=3
lim
x
→
∞
(
1
+
x
2
+
x
2
−
A
x
−
B
)
=
3
. What is the value of B?
Options
A
–1
B
3
C
–4
D
–3
Correct answer
C. –4
Step-by-step solution
lim
x
→
∞
(
2
+
x
2
1
+
x
−
A
x
−
B
)
=
3
⇒
lim
x
→
∞
(
2
+
x
2
−
A
x
−
B
−
A
x
2
−
B
x
1
+
x
)
=
3
⇒
lim
x
→
∞
[
(
1
−
A
)
x
2
−
(
A
+
B
)
x
+
2
−
B
1
+
x
]
=
3
\begin{aligned} & \lim _{\mathrm{x} \rightarrow \infty}\left(\frac{2+\mathrm{x}^2}{1+\mathrm{x}}-\mathrm{Ax}-\mathrm{B}\right)=3 \\ \Rightarrow & \lim _{\mathrm{x} \rightarrow \infty}\left(\frac{2+\mathrm{x}^2-\mathrm{Ax}-\mathrm{B}-\mathrm{Ax}^2-\mathrm{Bx}}{1+\mathrm{x}}\right)=3 \\ \Rightarrow & \lim _{\mathrm{x} \rightarrow \infty}\left[\frac{(1-\mathrm{A}) \mathrm{x}^2-(\mathrm{A}+\mathrm{B}) \mathrm{x}+2-\mathrm{B}}{1+\mathrm{x}}\right]=3 \end{aligned}
⇒
⇒
x
→
∞
lim
(
1
+
x
2
+
x
2
−
Ax
−
B
)
=
3
x
→
∞
lim
(
1
+
x
2
+
x
2
−
Ax
−
B
−
Ax
2
−
Bx
)
=
3
x
→
∞
lim
[
1
+
x
(
1
−
A
)
x
2
−
(
A
+
B
)
x
+
2
−
B
]
=
3
Applying L'Hospital rule,
2
x
(
1
−
A
)
−
(
A
+
B
)
=
3
2 x(1-A)-(A+B)=3
2
x
(
1
−
A
)
−
(
A
+
B
)
=
3
Comparing coefficients
2
(
1
−
A
)
=
0
∴
A
=
1
and
−
(
A
+
B
)
=
3
∴
B
=
−
3
−
1
=
−
4
A
+
B
=
−
3
A
=
1.
B
=
−
4
\begin{array}{ll} & 2(1-A)=0 \\ \therefore \quad & A=1 \text{ and } \\ & -(A+B)=3 \\ \therefore \quad & B=-3-1=-4 \\ & A+B=-3 \\ & A=1 . B=-4 \end{array}
∴
∴
2
(
1
−
A
)
=
0
A
=
1
and
−
(
A
+
B
)
=
3
B
=
−
3
−
1
=
−
4
A
+
B
=
−
3
A
=
1.
B
=
−
4
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Given that _ x arrow ∞ ((2+x^2)/(1+x)-A x-B)=3 . What is the value of A?
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If G(x)= 25-x^2 , then what is _ x arrow 1 (G(x)-G(1))/(x-1) equal to?
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