Test: Functions - CAT MCQ

Let the value of f(2) and f(3) be ‘x’ and ‘y’ respectively,

⇒ f(24) = f(8).f(3) = f(2).f(2).f(2).f(3) = 54

⇒ x³y = 54

⇒ x³y = 27 × 2

⇒ x³y = 3³ × 2

On comparing, x = 3 and y = 2

⇒ f(2) = 3

⇒ f(3) = 2

⇒ f(18) = f(9).f(2) = f(3).f(3).f(2)

⇒ f(18) = 2 × 2 × 3

⇒ f(18) = 12

∴ The value of f(18) equals 12

The maximum value of f(x) will occur when
2x² = 52 – 5x i.e. when 2x² + 5x − 52 = 0
⇒ 2x² + 13x − 8x − 52 = 0
⇒ (2x + 13) (x – 4) = 0 ⇒ x = – 13/2 or 4. But x is any positive real number. So, x = 4.
Hence, maximum value of f (x) = 2(4²) = 32

Calculation:

⇒ f(x + 2) = f(x) + f(x + 1)

⇒ f(15) = f(13) + f(14)

⇒ f(13) + f(14) = 617  - (1)

⇒ f(12) + f(13) = f(14) - (2)

⇒ f(11) + f(12) = f(13) - (3)

By solving,

⇒ f(12) + f(13) = 617 - f(13)   [ using (1) and  (2) ]

⇒ 2f(11) + 3f(12) = 617  [ usining (3) ] 

⇒ f(12) = 145

⇒ f(12) = f(11) + f(10)

∴ f(10) = 54

The lines represented by A where a > 0 and when a < 0 are given in the following figures

The area enclosed by A and D would be zero if d < |a|. In choice (b), d = 1 and a = – 2 i.e., d < |a|.
If a > 0, then the only case when the area enclosed by A and D will be zero, is when d = 0.

[log₁₀ x] = 0, for any value of x ∈ {1, 2, ......9), ...(1)
Similarly [log₁₀ x] = 1, for x ∈ {10, 11, 12 ... 99}...(2)
and [log₁₀x] = 2, for
x ∈ {100, 101, 102, ... 999} ...(3)
Now consider, 1 ≤ n ≤ 9, then
[log₁₀1] + [log₁₀2] + [log₁₀3] ... [log₁₀ n] = 0 (i.e., ≠ n)
Hence the expression given in the question cannot be satisfied.
Now consider, 10 ≤ n ≤ 99, then [log₁₀1] + [log₁₀2] ... [log₁₀n]
From (1) and (2), the above expression becomes (0 + 0 ... 9 times) + (1 + 1 + ... (n – 9) times) = n – 9
Using the same approach, for
100 ≤  n ≤ 999, [log ₁₀1] + [log₁₀2] ... [log ₁₀n] = 90 + 2(n – 99)
If can be seen that, only for the third case i.e., 100 ≤  n ≤  999, can the expression given in the question be satisfied.
Hence 90 + 2(n – 99) =  n
⇒ n = 198 – 90 = 108
∴ 107 ≤ n < 111.

Other two vertices will make two right angled triangles with AB as the common hypotenuse. So they must lie on the circle with AB as the diameter and O as the centre. Radius of that circle will be 5 units.
There will be 5 such pairs in which both the coordinates are integers.
[(5, 0), (–5, 0), [(4, 3), (4, – 3)],
[(–3, 4), (3, –4)] [(–3, –4), (3, 4)] and [(0, 5), (0, –5)]

Let us assume f(0) = K, where 'K' is a constant.
Then, f(0 + y) = f(0.y) = f(0) = K
and  f(x + 0) = f(x.0) = f(0) = K.
This proves that the function is a constant function.
Thus, the value of
f(–49) = f (49) = 7
Hence, f(–49) + f(49) = 14.

The sum of the roots of ax² + bx + c = 0 is -b/a
By differentiating we get 2ax + b = 0
ax² + bx + c attains its maximum value at x = -b/2a
∴ -b/2a = 2 ⇒ -b/a = 4
Hence, the sum of the roots = 4.

Convert into sine form this is a increasing function

Hence, f(x) lies in the range of (0, ∞)

The graphs of the two functions are shown below: If a > 0 in any parabolic function then parabola open up side

From the above figure, it is obvious that the graphs of the two functions intersect at three points.

Given that 


Similarly, we can find the values of F(3), F (4), F (5) as  respectively..

From this we can say that every term except [F(1)], of the series [F(1)] + [F (2)] + .... + [F (50)] is equal to 1 as for 'n' > 0, F (n) lies between 1 and 2.
Therefore, [F (1)] + [F (2)] + .... +  [F (50)] = 51.
Hence, option (a) is the correct choice.

Base of the logarithmic function 5 – |x| > 0 and 5 – |x| ≠ 1
So, x ∈ (–5, –4) ∪ (–4, 4) ∪ (4, 5) ...(i)
Also, (x–1) (x–2) (x–4) must be greater than zero as well
So, x ∈ (1, 2) ∪ (4, ∞) ...(ii)
Combining (i) and (ii) : x ∈ (1, 2) ∪ (4, 5)

There are following three cases arise :
Case (I) :
When f (x ) = 1, is correct
then f (y ) = 1 and f (z ) = 2

clearly the mapping f is not injective (i.e., not one-one) Hence this case is not possible.
Case (II) :
When f (y) ≠ 1, is correct
then f (x) ≠ 1 and f (z) = 2
Hence z mapped to 2 but x and y or mapped to 2 or 3 or one of them mapped to 2 and the other mapped to 3.

Clearly in all the above 4 sub-cases, we see that the mapping f is not injective (i.e. not one-one). Hence, this case is not possible.
Case (III) :
If f (z) ≠ 2, is correct
then f(x) ≠  1 and f (y) =1
Hence y mapped 1 but x mapped 2 or 3
Whereas y mapped 1 or 3.
The possible four mapping are as follows :

Clearly in sub case (c), the mapping is injective (i.e., one-one). Hence this case is possible and f ^-1 (1) = y

Each of the answer choices in the form of the product of two factors on the left and a “≥ 0” or “ ≤ 0” on the right.
The product will be negative when the two factors have opposite signs, and it will be positive when the factors have the same sign. Choice (a), for example, has a “≥0”, so you’ll be looking for other factors to have the same sign.


The graph of x ≤ 0 and y ≤ 2x looks like this.

Together, they make the graph in the figure.

Let the roots of the equation x³ – Ax² + Bx – C = 0 be α, β, γ respectively.
So the roots of x³ + Px² + Qx – 18 = 0 will be α + 2, β + 2, γ + 2.
(α +2)(β +2)(γ +2) = 18
⇒ 4(α + β + γ) + 2 (αβ + βγ +  γα) + αβγ + 8 =18
⇒ 4A + 2B + C = 10

Since a, b, c are in H.P.

Now log (a + c) + log (a – 2b + c)
= log [(a + c){(a + c) – 2b}]
= log [(a + c)² – 2b (a + c)]
= log [(a + c)² – 2 × 2ac]
= log (a – c)² or log (c – a)²
= 2 log (a – c) or 2 log (c – a)
∴ log (a + c) + log (a – 2b + c) = 2 log (c – a)

The number of triangles with vertices on different lines
= ^PC₁ × ^PC₁ x ^PC₁ = p³
The number of triangles with 2 vertices on one line and the third vertex on any one of the other two lines 

∴ the required number of triangles = P³ + 3P²(P - 1)
= 4p³ – 3p² = p² (4p – 3)
(The work “maximum” shows that no selection of points from each of the three lines are collinear).

mx^m = nxⁿ