Composition of Inverse Trigonometric Functions and Trigonometric Functions
For any posit...
Question
For any positive integer n, define fn:(0,∞)→R as fn(x)=n∑j=1tan−1(11+(x+j)(x+j−1)) for all x∈(0,∞) (Here, the inverse trigonometric function tan−1x assumes values in (−π2,π2))
Then, which of the following statement(s) is (are) TRUE?
A
5∑j=1tan2(fj(0))=55
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
10∑j=1(1+f′j(0))sec2(fj(0))=10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
For any fixed positive integer n,limx→∞tan(fn(x))=1n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
For any fixed positive integer n,limx→∞sec2(fn(x))=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D For any fixed positive integer n,limx→∞sec2(fn(x))=1 fn(x)=n∑j=1tan−1((x+j)−(x+j−1)1+(x+j)(x+j−1))=n∑j=1[tan−1(x+j)−tan−1(x+j−1)]=tan−1(x+n)−tan−1xfn(x)=tan−1[n1+(x+n)x]⇒tan[fn(x)]=[n1+(x+n)x]limx→∞tan(fn(x))=0∴(C) is wrong.limx→∞sec2(fn(x))=limx→∞(1+tan2(fn(x))=1+limx→∞[n1+(x+n)x]2=1∴(D) is correct.5∑j=1tan2(fj(0))=[51+0]2=25∴(A) is wrong.f′j(x)=11+(n1+(x+n)x)2×(2x+n)f′j(0)=n1+n210∑j=1(1+f′j(0))sec2(fj(0))=10∑j=1(1+n1+n2)(1+[n1+0]2)=10∑j=1(1+n+n21+n2)(1+n2)=1+10+100=111∴(B) is wrong.