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Notely Maths 12
Class 12 Maths
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1.1 Q-9

Question Statement

Determine whether the given functions are even or odd:

i. f(x)=x3+xf(x) = x^{3} + x
ii. f(x)=(x+2)2f(x) = (x+2)^{2}
iii. f(x)=xx2+5f(x) = x \sqrt{x^{2} + 5}
iv. f(x)=xβˆ’1x+1,,xβ‰ 1f(x) = \frac{x - 1}{x + 1}, , x \neq 1
v. f(x)=x23+6f(x) = x^{\frac{2}{3}} + 6
vi. f(x)=x3βˆ’xx2+1f(x) = \frac{x^{3} - x}{x^{2} + 1}

Background and Explanation

A function f(x)f(x) is:

  • Even if f(βˆ’x)=f(x)f(-x) = f(x) for all xx in its domain.
  • Odd if f(βˆ’x)=βˆ’f(x)f(-x) = -f(x) for all xx in its domain. To determine whether a function is even or odd, substitute βˆ’x-x for xx in the function and compare the result with the original function.

Solution

i. f(x)=x3+xf(x) = x^{3} + x

  • Step 1: Substitute βˆ’x-x for xx:
f(βˆ’x)=(βˆ’x)3+(βˆ’x)=βˆ’x3βˆ’x f(-x) = (-x)^{3} + (-x) = -x^{3} - x
  • Step 2: Compare with f(x)f(x):
f(βˆ’x)=βˆ’(x3+x)=βˆ’f(x) f(-x) = - (x^{3} + x) = -f(x)
  • Conclusion: Since f(βˆ’x)=βˆ’f(x)f(-x) = -f(x), the function is odd.

ii. f(x)=(x+2)2f(x) = (x+2)^{2}

  • Step 1: Substitute βˆ’x-x for xx:
f(βˆ’x)=(βˆ’x+2)2=(xβˆ’2)2 f(-x) = (-x+2)^{2} = (x-2)^{2}
  • Step 2: Compare with f(x)f(x):
f(x)=(x+2)2 f(x) = (x+2)^{2}

These are not equal, so the function is neither even nor odd.

  • Conclusion: The function is neither even nor odd.

iii. f(x)=xx2+5f(x) = x \sqrt{x^{2} + 5}

  • Step 1: Substitute βˆ’x-x for xx:
f(βˆ’x)=(βˆ’x)(βˆ’x)2+5=βˆ’xx2+5 f(-x) = (-x) \sqrt{(-x)^{2} + 5} = -x \sqrt{x^{2} + 5}
  • Step 2: Compare with f(x)f(x):
f(βˆ’x)=βˆ’f(x) f(-x) = -f(x)
  • Conclusion: Since f(βˆ’x)=βˆ’f(x)f(-x) = -f(x), the function is odd.

iv. f(x)=xβˆ’1x+1,,xβ‰ 1f(x) = \frac{x - 1}{x + 1}, , x \neq 1

  • Step 1: Substitute βˆ’x-x for xx:
f(βˆ’x)=βˆ’xβˆ’1βˆ’x+1=βˆ’(x+1)βˆ’(xβˆ’1)=x+1xβˆ’1 f(-x) = \frac{-x - 1}{-x + 1} = \frac{-(x + 1)}{-(x - 1)} = \frac{x + 1}{x - 1}
  • Step 2: Compare with f(x)f(x):
f(x)=xβˆ’1x+1 f(x) = \frac{x - 1}{x + 1}

These are not equal, so the function is neither even nor odd.

  • Conclusion: The function is neither even nor odd.

v. f(x)=x23+6f(x) = x^{\frac{2}{3}} + 6

  • Step 1: Substitute βˆ’x-x for xx:
f(βˆ’x)=(βˆ’x)23+6=x23+6 f(-x) = (-x)^{\frac{2}{3}} + 6 = x^{\frac{2}{3}} + 6
  • Step 2: Compare with f(x)f(x):
f(βˆ’x)=f(x) f(-x) = f(x)
  • Conclusion: Since f(βˆ’x)=f(x)f(-x) = f(x), the function is even.

vi. f(x)=x3βˆ’xx2+1f(x) = \frac{x^{3} - x}{x^{2} + 1}

  • Step 1: Substitute βˆ’x-x for xx:
f(βˆ’x)=(βˆ’x)3βˆ’(βˆ’x)(βˆ’x)2+1=βˆ’x3+xx2+1 f(-x) = \frac{(-x)^{3} - (-x)}{(-x)^{2} + 1} = \frac{-x^{3} + x}{x^{2} + 1}
  • Step 2: Compare with f(x)f(x):
f(βˆ’x)=βˆ’(x3βˆ’xx2+1)=βˆ’f(x) f(-x) = -\left( \frac{x^{3} - x}{x^{2} + 1} \right) = -f(x)
  • Conclusion: Since f(βˆ’x)=βˆ’f(x)f(-x) = -f(x), the function is odd.

Key Formulas or Methods Used

  • Even Function: f(βˆ’x)=f(x)f(-x) = f(x)
  • Odd Function: f(βˆ’x)=βˆ’f(x)f(-x) = -f(x)

Summary of Steps

  1. For i: Prove f(βˆ’x)=βˆ’f(x)f(-x) = -f(x), hence odd.
  2. For ii: Prove f(βˆ’x)β‰ f(x)f(-x) \neq f(x) and f(βˆ’x)β‰ βˆ’f(x)f(-x) \neq -f(x), hence neither even nor odd.
  3. For iii: Prove f(βˆ’x)=βˆ’f(x)f(-x) = -f(x), hence odd.
  4. For iv: Prove f(βˆ’x)β‰ f(x)f(-x) \neq f(x) and f(βˆ’x)β‰ βˆ’f(x)f(-x) \neq -f(x), hence neither even nor odd.
  5. For v: Prove f(βˆ’x)=f(x)f(-x) = f(x), hence even.
  6. For vi: Prove f(βˆ’x)=βˆ’f(x)f(-x) = -f(x), hence odd.