NIMCET 2014 — Mathematics PYQ
NIMCET | Mathematics | 2014If x is so small that x2 and higher powers of x can be neglected, then (1−x)1/5(9+2x)1/2(3+4x) is approximately equal to:
Choose the correct answer:
- A.
9+1574x
- B.
9+574x
(Correct Answer) - C.
3+1574x
- D.
3+574x
9+574x
Explanation
Concept:
Expansion of (1+x)n when |x| < 1:
-
(1+x)n=1+nx+2!n(n−1)x2+…
Calculation:
Let E=(1−x)1/5(9+2x)1/2(3+4x)
⇒E=(9+2x)1/2(3+4x)(1−x)−1/5
⇒E=[91/2(1+92x)1/2][3(1+34x)](1−x)−1/5
⇒E=[3(1+92x)1/2][3(1+34x)](1−x)−1/5
Expanding and neglecting higher powers of x:
⇒E=9(1+92×21x)(1+34x)(1−(−51)x)
⇒E=9(1+91x)(1+34x)(1+51x)
Neglecting the higher powers during multiplication:
⇒E=9(1+91x+34x)(1+51x)
⇒E=9(1+913x+51x)
⇒E=9(1+4574x)
⇒E=9+574x
Explanation
Concept:
Expansion of (1+x)n when |x| < 1:
-
(1+x)n=1+nx+2!n(n−1)x2+…
Calculation:
Let E=(1−x)1/5(9+2x)1/2(3+4x)
⇒E=(9+2x)1/2(3+4x)(1−x)−1/5
⇒E=[91/2(1+92x)1/2][3(1+34x)](1−x)−1/5
⇒E=[3(1+92x)1/2][3(1+34x)](1−x)−1/5
Expanding and neglecting higher powers of x:
⇒E=9(1+92×21x)(1+34x)(1−(−51)x)
⇒E=9(1+91x)(1+34x)(1+51x)
Neglecting the higher powers during multiplication:
⇒E=9(1+91x+34x)(1+51x)
⇒E=9(1+913x+51x)
⇒E=9(1+4574x)
⇒E=9+574x