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Question
Let f(x)=⎧⎪⎨⎪⎩limn→∞ex2−1+[(a+b)x−(a−b)]x2nx2n+1+cosx−1,x∈R−{0}k,x=0
If f(x) is continuous for all x∈R, then
Let f(x)=⎧⎪⎨⎪⎩limn→∞ex2−1+[(a+b)x−(a−b)]x2nx2n+1+cosx−1,x∈R−{0}k,x=0
If f(x) is continuous for all x∈R, then
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Solution
The correct option is C k=−2
Given : f(x)=⎧⎪⎨⎪⎩limn→∞ex2−1+[(a+b)x−(a−b)]x2nx2n+1+cosx−1,x∈R−{0}k,x=0
⇒f(x)=⎧⎪
⎪
⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪
⎪
⎪⎩ex2−1cosx−1,0<x2<1k,x=0(a+b)x−(a−b)x,x2>1
Checking the continuity at x=1, we get
limx→1−ex2−1cosx−1=e−1cos1−1limx→1+(a+b)x−(a−b)x=2b⇒b=e−12(cos1−1)
Checking the continuity at x=−1, we get
limx→−1+ex2−1cosx−1=e−1cos1−1limx→−1−(a+b)x−(a−b)x=−(a+b)−a+b−1⇒2a=e−1(cos1−1)∴a=b
Now, as f(x) is continuous for all x∈R, so
f(0)=limx→0ex2−1cosx−1
Using L'Hospital's Rule, we get
⇒k=limx→0ex2×2x−sinx=−2
f′(0+)=limh→0f(h)−f(0)h⇒f′(0+)=limh→0eh2−1cosh−1+2h⇒f′(0+)=limh→03−eh2−2cosh2hsin2h2⇒f′(0+)=limh→02(3−eh2−2cosh)h3
Using L'Hospital's Rule, we get
⇒f′(0+)=limh→02(−eh22h+2sinh)3h2⇒f′(0+)=limh→02(−eh24h2−2eh2+2cosh)6h⇒f′(0+)=limh→04(−eh2+cosh)6h⇒f′(0+)=limh→04(−eh22h−sinh)6⇒f′(0+)=0
Similarly,
f′(0−)=0∴f′(0)=0
Given : f(x)=⎧⎪⎨⎪⎩limn→∞ex2−1+[(a+b)x−(a−b)]x2nx2n+1+cosx−1,x∈R−{0}k,x=0
⇒f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ex2−1cosx−1,0<x2<1k,x=0(a+b)x−(a−b)x,x2>1
Checking the continuity at x=1, we get
limx→1−ex2−1cosx−1=e−1cos1−1limx→1+(a+b)x−(a−b)x=2b⇒b=e−12(cos1−1)
Checking the continuity at x=−1, we get
limx→−1+ex2−1cosx−1=e−1cos1−1limx→−1−(a+b)x−(a−b)x=−(a+b)−a+b−1⇒2a=e−1(cos1−1)∴a=b
Now, as f(x) is continuous for all x∈R, so
f(0)=limx→0ex2−1cosx−1
Using L'Hospital's Rule, we get
⇒k=limx→0ex2×2x−sinx=−2
f′(0+)=limh→0f(h)−f(0)h⇒f′(0+)=limh→0eh2−1cosh−1+2h⇒f′(0+)=limh→03−eh2−2cosh2hsin2h2⇒f′(0+)=limh→02(3−eh2−2cosh)h3
Using L'Hospital's Rule, we get
⇒f′(0+)=limh→02(−eh22h+2sinh)3h2⇒f′(0+)=limh→02(−eh24h2−2eh2+2cosh)6h⇒f′(0+)=limh→04(−eh2+cosh)6h⇒f′(0+)=limh→04(−eh22h−sinh)6⇒f′(0+)=0
Similarly,
f′(0−)=0∴f′(0)=0
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