for x = 0<br />then<br />x-1<br />If f(x) = 3x2 - 7x +5<br />1<br />3<br />$1 (1) =<br />for x = 1<br />2<br />(1) - (2) - (3) - (4)<br />3<br />3
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Answer:
We have,
f(x)={
x
2
+3x+a,
bx+2,
x≤1
x>1
(LHD at x=1)=lim
x→ 1
−
x−1
f(x)−f(1)
=lim
h→ 0
1−h−1
f(1−h)−1
=lim
h→ 0
−h
[(1−h)
2
+3(1−h)+a]−(1+3+a)
=lim
h→ 0
−h
1+h
2
−2h+3−3h+a−4−a
=lim
h→ 0
−h
h
2
−5h
=lim
h→ 0
−1
h−5
=5
Now,
(RHD at x=1)=lim
x→ 1
+
x−1
f(x)−f(1)
=lim
h→ 0
1+h−1
f(1+h)−1
=lim
h→ 0
h
[b(1+h)+2]−(b+2)
=lim
h→ 0
h
b+bh+2−b−2
=lim
h→ 0
h
bh
=b
Since, f(x) is differentiable, so
(LHD at x=1)=(RHD at x=1)
b=5
And f(1)=1+3+a=4+a
Now,
LHL=lim
x→ 1
−
f(x)
LHL=lim
h→ 0
f(1−h)
LHL=lim
h→ 0
(1−h)
2
+3(1−h)+a
LHL=1+3+a=4+a
Now,
RHL=lim
x→ 1
+
f(x)
RHL=lim
h→ 0
f(1+h)
RHL=lim
h→ 0
b(1+h)+2
RHL=b+2
Since, f(x) is continuous, so
LHL=RHL
4+a=b+2
4+a=5+2
a=7−4=3
Hence, a=7 and b=5.
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