Test: Functions (April 9) - CAT MCQ

y — axis by definition.

Given: f(x) = x² + 4x + 4

Replace x by -x,

⇒ f(-x) = (-x)² + 4(-x) + 4

= x² - 4x + 4                       (∵ (-x)² = x²)

⇒ f(-x) ≠ ± f(x)

Hence function is neither odd nor even.

The minimum value of the function would occur at the minimum value of (x² - 2x + 5) as this quadratic function has imaginary roots.

Thus, minimum value of the argument of the log is 4. So minimum value of the function is log₂ 4 = 2.

y – axis by definition.

x^–8 is even since f(x) = f(–x) in this case.

dy/dx = 2x + 10 = 0 fi x = –5.

Since the denominator x² – 3x + 2 has real roots, the maximum value would be infinity.

f(1) = 0, f(2) = 1,
f(3) = f(1) – f(2) = –1
f(4) = f(2) – f(3) = 2
f(5) = f(3) – f(4) = –3
f(6) = f(4) – f(5) = 5
f(7) = f(5) – f(6) = –8
f(8) = f(6) – f(7) = 13
f(9) = f(7) – f(8) = –21

13

f(1) = 0, f(2) = 1,
f(3) = f(1) – f(2) = –1
f(4) = f(2) – f(3) = 2
f(5) = f(3) – f(4) = –3
f(6) = f(4) – f(5) = 5
f(7) = f(5) – f(6) = –8
f(8) = f(6) – f(7) = 13
f(9) = f(7) – f(8) = –21

For any ⁿC_r, n should be positive and r ≥ 0.
Thus, for positive x, 5 – x ≥ 0
fi x = 1, 2, 3, 4, 5.