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CAT 2025 Lesson : Functions - Piecewise & Non-Standard Input

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6.6 Piecewise Functions

In these type of questions, function provides different outputs for different ranges of inputs. A good way to solve these is to form a table and/or assume values across different ranges.

Example 18 & 19

f1(x)=xf_1(x) = xf1​(x)=x, if 0≤x≤10 \le x \le 10≤x≤1 f2(x)=f1(−x)f_2(x) = f_1( - x)f2​(x)=f1​(−x) for all xxx
              =1= 1=1, if x≥1x \ge 1x≥1 f3(x)=−f2(x)f_3(x) = - f_2(x)f3​(x)=−f2​(x) for all xxx
               =0= 0=0 otherwise f4=f3(−x)f_4 = f_3( - x)f4​=f3​(−x) for all xxx
[CAT 2004]

How many of the following products are necessarily zero for every
x−f4(x)f2(x),f2(x)f3(x),f2(x)f4(x)x - f_4(x)f_2(x), f_2(x)f_3(x), f_2(x)f_4(x)x−f4​(x)f2​(x),f2​(x)f3​(x),f2​(x)f4​(x)?

(
111) 0               (222) 1               (333) 2               (444) 3

Solution

As f1(x)f_1(x)f1​(x) is a piecewise function with 333 different outputs across 333 different ranges, we shall form a table with the 333 ranges in the columns and take 333 reference values for each range.

Note: If a positive and negative values are to be taken, avoid taking values such as
+2+2+2 and −2-2−2 wherein each values' modulus give the same result. The of -222, 0.50.50.5 and 333 that have been used are quite distinct.

x<0\bm{x \lt 0}x<0 0≤x≤1 \bm{ 0 \leq x \leq 1}0≤x≤1 x>1\bm{x \gt 1}x>1
Reference Value x=−2x = -2x=−2 x=0.5x = 0.5x=0.5 x=3x = 3x=3
f1(x)f_1(x)f1​(x) 0 00 1 11 3 33
f2(x)=f1(−x)f_2(x) = f_1(-x)f2​(x)=f1​(−x) 2 22 0 00 0 00
f3(x)=−f2(x)f_3(x) = -f_2(x)f3​(x)=−f2​(x) −2 -2−2 0 00 0 00
f4(x)=f3(−x)=−f2(−x)=−f1(x)f_4(x) = f_3(-x) = -f_2(-x) = -f_1(x)f4​(x)=f3​(−x)=−f2​(−x)=−f1​(x) 0 00 −1 -1−1 −3 -3−3


f2(x)f3(x)f_2(x)f_3(x)f2​(x)f3​(x) at x=−2x = -2x=−2 , provides a value of 2×−2=−4 2 \times - 2 = -42×−2=−4

f1(x)f2(x)f_1(x)f_2(x)f1​(x)f2​(x) and f2(x)f4(x)f_2(x)f_4(x)f2​(x)f4​(x) across the 333 ranges is always 000. Therefore, 2\bm{2}2 of the products are always 000.

Answer: (
333) 222

191919) Which of the following is necessarily true?

(
111) f4(x)=f1(x)f_4(x) = f_1(x)f4​(x)=f1​(x) for all xxx         (222) f1(x)=−f3(−x)f_1(x) = -f_3(-x)f1​(x)=−f3​(−x) for all xxx
(
333) f2(−x)=f4(x)f_2(-x) = f_4(x)f2​(−x)=f4​(x) for all xxx         (444) f1(x)+f3(x)=0f_1(x) + f_3(x) = 0f1​(x)+f3​(x)=0 for all xxx

Option (1) is not true at
x=0.5x = 0.5x=0.5 and x=3x = 3x=3.

Option (2):
−f3(−x)=−(−f2(−x))=f2(−x)=f1(−(−x))=f1(x)-f_3(-x) = -(-f_2(-x)) = f_2(-x) = f_1(-(-x)) = f_1(x)−f3​(−x)=−(−f2​(−x))=f2​(−x)=f1​(−(−x))=f1​(x) . This is always true.

Option (3): We note from the table that
f4(x)=f3(−x)=−f2(−x)f_4(x) = f_3(-x) = -f_2(-x)f4​(x)=f3​(−x)=−f2​(−x) . Therefore, this is also incorrect.

Option (4) is not true for any of the
xxx-values.

Answer: (
222) f1(x)=−f3(−x)f_1(x) = -f_3(-x)f1​(x)=−f3​(−x) for all xxx

6.7 Non-standard input for the function

In these questions equate the required input to the function's input, solve for
xxx and the substitute the xxx-value in the output.

Example 20

If f(2x+3x−1)=x+4x−2f (\dfrac{2x + 3}{x - 1}) = \dfrac{x +4 }{x - 2 }f(x−12x+3​)=x−2x+4​ , then f(5)f(5)f(5) = ?

Solution

In these question, we first equate the input term to 555, solve for xxx and then substitute xxx in the output.

If
(2x+3x−1)=5 \left(\dfrac{2x + 3}{ x - 1}\right) = 5(x−12x+3​)=5 ⇒ 2x+3=5x−52x + 3 = 5x - 52x+3=5x−5 ⇒ x=83x = \dfrac{8}{3}x=38​

The question states
(2x+3x−1)=x+4x−2 \left(\dfrac{2x +3}{x - 1}\right) = \dfrac{x + 4}{x - 2}(x−12x+3​)=x−2x+4​

When
x=83x = \dfrac{8}{3}x=38​, f(2x+3x−1)=f(5)=83+483−2=20323f \left(\dfrac{2x + 3}{x - 1}\right) = f(5) = \dfrac{\dfrac{8}{3} + 4}{\dfrac{8}{3} - 2} = \dfrac{\dfrac{20}{3}}{\dfrac{2}{3}}f(x−12x+3​)=f(5)=38​−238​+4​=32​320​​

⇒
f(5)=10f(5) = 10f(5)=10

Answer:
101010

Example 21

If 3f(x+2)+4f(1x+2)=4x,x≠−23 f(x + 2) + 4f( \dfrac{1}{x + 2}) = 4x, x \neq - 23f(x+2)+4f(x+21​)=4x,x=−2 , the f(4)f(4)f(4) is
[XAT 2008]

(
111) 777
(
222) 527 \dfrac{52}{7}752​
(
333) 888
(
444) 567\dfrac{56}{7}756​
(
555) None of the above

Solution

The 222 functions have non-standard inputs with 111 being the reciprocal of the other. The equation looks like a linear equation with 222 variables. We need to form 222 equations to solve it.

If the first input equals
444, i.e. x+2=4x + 2 = 4x+2=4 ⇒ x=2x = 2x=2

Substituting
x=2x = 2x=2, we get 3f(4)+4f(14)=83 f(4) + 4f(\dfrac{1}{4}) = 83f(4)+4f(41​)=8 -----(a)

If the second input equals
444, i.e. 1x+2=4\dfrac{1}{x + 2} = 4x+21​=4 ⇒ x=−74x = - \dfrac{7}{4}x=−47​

Substituting
xxx === −74 - \dfrac{7}{4}−47​, we get 3f(14)+4f(4)=−73 f\left(\dfrac{1}{4}\right) + 4f(4) = -73f(41​)+4f(4)=−7 -----(b)

4×(b)−3×(a) 4 \times (b) - 3 \times (a)4×(b)−3×(a) ⇒ 4f(4)=−524f(4) = - 524f(4)=−52

⇒
f(4)f(4)f(4) === −-− 527\dfrac{52}{7}752​

Answer: (
555) None of the above

6.8 Substitute and get the answer

In these questions, substitute small integral values for
xxx, starting from 111 and find the option that is true.

Example 22

If f(x)=3x+22x−3f(x) = \dfrac{3x + 2}{2x - 3}f(x)=2x−33x+2​, then which of the following is true?

(
111)f(x)=f(−x)f(x) = f(-x)f(x)=f(−x)
(
222) f(x)=−1f(−x)f(x) = \dfrac{-1}{f(-x)}f(x)=f(−x)−1​
(
333) x=f(f(x))x = f(f(x))x=f(f(x))
(
444) f(x)=−f(−x)f(x) = -f(-x)f(x)=−f(−x)

Solution

In these questions, substitute values for xxx and proceed by eliminating the answer options.

f(1)=3+22−3=−5f(1) = \dfrac{3 + 2}{2 - 3} = -5f(1)=2−33+2​=−5 and f(−1)=−3+2−2−3=15f(-1) = \dfrac{-3 + 2}{-2 - 3} =\dfrac{1}{5}f(−1)=−2−3−3+2​=51​

As
f(1)≠f(−1)f(1) \neq f(-1)f(1)=f(−1) , option (111) is rejected
As
f(1)≠−f(−1)f(1) \neq -f(-1)f(1)=−f(−1) , option (444) is rejected

As
f(1)=−1f(−1)f(1) = \dfrac{-1}{f(-1)}f(1)=f(−1)−1​ , we keep option (222) on hold and check if option (333) can be rejected.

f(f(1))=f(−5)=−15+2−10−3=1f(f(1)) = f(-5) = \dfrac{-15 + 2}{-10 - 3} = 1f(f(1))=f(−5)=−10−3−15+2​=1,

∴ Option (
333) cannot be rejected. We need to substitute a different value, say x=2x = 2x=2 and check for options 222 and 333 alone.

f(2)=8f(2) = 8f(2)=8 and f(−2)=47f(-2) = \dfrac{4}{7}f(−2)=74​

Clearly,
f(2)≠−1f(−2)f(2) \neq \dfrac{-1}{f(-2)}f(2)=f(−2)−1​ and option (222) is rejected. And, the correct answer is option (3\bm{3}3).

Note that f(f(2))=f(8)=24+216−3=2f(f(2)) = f(8) = \dfrac{24 + 2}{16 - 3} = 2f(f(2))=f(8)=16−324+2​=2 . So, option (333) is true.

Answer: (
333) x=f(f(x))x = f(f(x))x=f(f(x))

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