Test: Functions- 3 - 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.

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.

At the end of the 1st year, he will get Rs 50000, it will give him interest for 4 years compounded annually
Hence at the end of 5 years, this amount will become 50000(1.1)⁴
Similarly, the amount deposited in the 2nd year will give interest for 3 years. Hence it will become 50000(1.1)³
Similarly, we can calculate for the remaining years.
The total saving at the end of the 5th year would be a GP, given by
Net saving = 50000(1.1)⁴ +  50000(1.1)³  ..... 50000
Thus net saving = =  Rs 3,05,255

Option a = (a – b) (a + b) = a² – b²

3 – 4 × 2 + 4/8 – 2 = 3 – 8 + 0.5 – 2 = – 6.5
(using BODMAS rule)

The minimum would depend on the values of a and b. Thus, cannot be determined.

Again (a + b) or a/b can both be greater than each other depending on the values we take for a and b.
E.g. for a = 0.9 and b = 0.91, a + b > a/b.
For a = 0.1 and b = 0.11, a + b < a/b

We notice that the minimum value of the term in the LHS will be greater than or equal to 2 {at x=0; LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of x at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.