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Byju's Answer
Standard X
Mathematics
Solving Inequalities
Let a and b b...
Question
Let a and b be two integers such that 10a+b=5 and P(x)= x+ax+b. The integer n such that P(10).P(11)=P(n) is
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Solution
Let
a
and
b
be
two
integers
such
that
10
a
+
b
=
5
and
P
(
x
)
=
x
+
ax
+
b
.
The
integer
n
such
that
P
(
10
)
.
P
(
11
)
=
P
(
n
)
is
what
?
P
(
10
)
=
10
+
10
a
+
b
=
10
+
5
=
15
…
1
10
a
+
b
=
5
P
(
11
)
=
11
+
11
a
+
b
=
11
+
a
+
5
=
16
+
a
…
2
10
a
+
b
=
5
Now
,
P
(
10
)
.
P
(
11
)
=
P
(
n
)
⇒
15
×
16
+
a
=
n
+
na
+
b
=
n
+
n
-
10
a
+
5
10
a
+
b
=
5
⇒
240
+
15
a
=
n
+
na
-
10
a
+
5
⇒
n
+
na
-
25
a
-
235
=
0
⇒
n
=
235
+
25
a
a
+
1
…
3
We
see
here
that
'
n
'
is
just
a
special
case
of
'
x
'
such
that
P
(
n
)
=
P
(
10
)
.
P
(
11
)
so
,
we
can
put
eq
3
in
1
10
=
235
+
25
a
a
+
1
10
a
+
10
=
235
+
25
a
-
15
a
=
225
a
=
-
15
using
10
a
+
b
⇒
b
=
155
for
solving
for
a
&
b
we
assumed
n
=
10
⇒
P
(
n
)
=
P
(
10
)
.
P
(
11
)
⇒
P
(
10
)
=
P
(
10
)
.
P
(
11
)
⇒
P
(
11
)
=
1
Now
,
with
these
values
of
a
&
b
if
P
(
11
)
=
1
,
this
would
mean
n
=
10
P
(
11
)
=
16
+
a
Using
eq
2
P
(
11
)
=
16
+
-
15
putting
a
=
-
15
P
(
11
)
=
1
Therefore
,
n
=
10
Suggest Corrections
0
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