If ∫x4+1x6+1dx=tan−1(f(x))−23tan−1(g(x))+c, where c is arbitrary constant, then the number of real root(s) of f(x)=g(x) is
A
0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A0 ∫x4+1x6+1dx=tan−1(f(x))−23tan−1(g(x))+c
Let I=∫(x2+1)2−2x2(x2+1)(x4−x2+1)dx⇒I=∫(x2+1)dx(x4−x2+1)−2∫x2dx(x6+1)⇒I=∫(1+1x2)dx(x2−1+1x2)−2∫x2dx(x3)2+1⇒I=∫(1+1x2)dx1+(x−1x)2−2∫x2dx(x3)2+1
Assuming I1=∫(1+1x2)dx1+(x−1x)2
Let (x−1x)=t ⇒I1=∫11+t2dt⇒I1=tan−1t=tan−1(x−1x) I2=∫x2dx(x3)2+1
Let x3=t ⇒I2=13tan−1(x3)
So, I=tan−1(x−1x)−23tan−1(x3)+cf(x)=x−1x,g(x)=x3 f(x) and g(x) are odd functions.
When f(x)=g(x), then x−1x=x3⇒x4−x2+1=0⇒(x2−12)2+34=0
So, no real roots.