CAT Practice: Functions - 2 - Free MCQ Test with solutions for Quant Aptitude

f(g(x),h(x)) = 2

log_h(x) g (x) = 2

log_2x+1​ x³ − 2x² + 7x + 11 = 2

(2x+1)² = x³ − 2x² + 7x + 11

4x² + 4x + 1 = x³ − 2x² + 7x + 11

x³ − 6x² + 3x + 10 = 0

Also, x = -1 satisfies the above equation. Hence, (x+1) is a factor of the expression.

Using algebraic division and further factorization, we get

(x+1) (x-2) (x-5) = 0

Hence, the roots are -1, 2 and 5.

But if we consider -1, the base of the logarithm, 2x + 1 becomes -1, which is  not possible, hence the only possible values are 2 and 5.

Hence, sum = 7.

We know that f(g(x)) =x² + 12x +33
Thus f(g(f(x))) = f(x²) + 12f(x) + 33
Putting g(f(x)) = x² + 4x + 5
Thus f(x² + 4x + 5) = f(x)² + 12f(x) + 33
Given f(26) = 46
putting x² + 4x + 5 = 26
Or x² + 4x - 21 = 0
Or (x+7)(x-3) =0.
Putting x = 3 in f(x² + 4x + 5) = f(x)² + 12f(x) + 33 
f(26) = 46 = f(3)²+12f(3)+33
Or f(3)² + 12f(3) - 13 = 0
(f(3)+13)(f(3)-1) = 0 Thus f(3) = 1 or -13
1 > -13 so it is the answer

We see that the value of f(x) increases linearly w.r.t. the change in the value of x.
f(x) feels like 3x for values -2, 2, 3 and 4. But f(x) is actually a 4 degree polynomial.
So, f(x)= (x-a)(x-b)(x-c)(x-d) + g(x) can be the 4th degree polynomial. And for f(x) to be linear at given points, we see that -2, 2, 3 and 4 has to be the values of a,b,c and d such that the first term becomes 0 and g(x) drives the value of f(x).
g(x) has to be 3x as discussed before and hence f(x) is nothing but (x+2)(x-2)(x-3)(x-4)+ 3x.
f(x)=(x+2)(x-2)(x-3)(x-4)+ 3x satisfies all the given values of x and so we will find f(5) using this.
f(5)= (5+2)(5-2)(5-3)(5-4)+ 15= (7)(3)(2)(1)+15= 57.

number of bogies attached = b, proportionality constant = c, Resultant speed = s

We have R = cb² and s = k − R = k − cb²

100 = k − c(4)² or, 100 = k − 16c(i)

and 55 = k − c(5)² or 55 = k − 25c(ii)

Solving (i) and (ii) we get k = 180 and c = 5. Now we have S = 180 − 5b². If we put b = 6, S = 0

Therefore, at most we can attach 5 bogies to the engine.

We can see that f(x) is similar to the Log function.

i.e. Log(x/y) = Log(x) - Log(y)

So we need to break 4536 into its prime factors as follows,

4536 = 23.34.7

Log(4536) = 3Log 2 + 4 Log 3 + Log 7

Similarly f(4536) = 3f(2) + 4f(3) + f(7)

= 3 x 2.4 + 4 x 3.7 + 7

= 29

x>=a, so |x-a| = x-a

x<100, so |x-100| = 100-x

f(x) = (x-a) + (100-x) + |x-a-50| =100

or, |x-a-50| = a

 

From the graph we can can see that when x=a then

|x-a-50|=a

or, a= 50

Similarly when x=a+100

|x-a-50|=a

or, a= 50

So value of a is 50 when f(x) is 100.

In the question, it is given that the equation ∣x + a∣ + ∣x − 1∣ = 2 has an infinite number of solutions for any value of x. This is possible when x in |x+a| and x in |x-1| cancels out.

Case 1:

x + a < 0, x - 1 ≥ ≥  0

- a - x + x - 1 = 2

a = -3

Case 2:

x + a ≥ ≥  0 and x - 1 < 0

x + a - x + 1 = 2

a = 1

Largest value of a is 1.

Concept: we have to distribute 6 distinct balls into three different boxes:
Case 1 (1, 1, 4)

Hence, the total number of such functions = 90 + 360 + 90<br>
= 540

We have,
max{x, 2} – min{x, 2} = |x + 2| – |x – 2|
Case1–> For x ≥ 2
x – 2 = x + 2 – (x – 2)
x – 2 = 4
x = 6
Thus for x ≥ 2, x = 6 is a solution.
Case 2–> For x < 2
2 – x = (x + 2) – (2 – x)
2 – x = 2x
x = 2/3
Hence, 2 distinct values of x exist.

For n = 3,4,5 and x ∈ [n, n + 1) we have [x] = n, so the equation

Option B and C have √10 included, which is not part of the original set. And Option D has √18. So, it is not possible.
Option A is the answer.

We check where  or their compositions, become undefined.
First, f(x) is undefined at x = 1/2 and g(x) is undefined at x=1.
Next, for 
Since, the denominator can't be zero, x = -1 must also be excluded.

⇒ x = 1 is not possible, which is already excluded.
So the values at which h(x) = f(g(x)) +g(f(x)) is undefined are 

Let (x² + 3x) be equal to k. We have, 

 

Since x is real, the discriminant of the above quadratic has to be greater than or equal to zero.

We find that 3² + 4 ∗ 9700 ≥ 0 and therefore the quadratic has real roots.

The sum of the roots will be 
Option D is the correct answer.

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.