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CAT 2025 Lesson : Functions - Quadratic & Log

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5. Algebraic Functions

5.1 Solving Quadratic Functions

These questions typically provide the f(x)f(x) values for certain xx-values and would require you to find the quadratic expression. For this, we substitute the xx-values and find the coefficients/constants.

Example 6

Where f(x)f(x) is a quadratic expression, f(5)=f(−5),f(0)=15f(5) = f(-5), f(0) = 15 and f(3)=33f(3) = 33, then what is f(10)=?f(10) = ?

Solution

Let f(x)=ax2+bx+cf(x) = ax^2 + bx + c

f(5)=f(−5)f(5) = f(-5)
⇒ 25a+5b+c=25a−5b+c 25a + 5b + c = 25a - 5b + c
⇒ 10b=0 10b = 0
⇒ b=0 \bm{b = 0}

f(0)f(0) = c=15\bm{c = 15}

f(3)=(a×9)+0+15=33f(3) = (a \times 9) + 0 + 15 = 33
⇒ a=2\bm{a = 2}

∴f(x)=2x2+15\therefore f(x) = 2x^2 + 15

f(10)=200+15=215f(10) = 200 + 15 = 215

Answer: 215

5.2 Solving Logarithmic Functions

We have to remember the basic rule of logarithm listed below.

Rule: logb a \mathrm{log}_b \space a is defined when a>0,b>0a > 0, b > 0 and b≠1b \ne 1

Example 7

log∣x∣−4x2−3x−10 \mathrm{log}_{|x| - 4}x^2 - 3x - 10. What is the domain of xx?

Solution

We need to find the acceptable values of xx wherein x2−3x−10>0, ∣x∣−4>0x^2 - 3x - 10 > 0, \space |x| - 4 > 0 and ∣x∣−4≠1|x| - 4 \ne 1

If x2−3x−10>0x^2 - 3x - 10 > 0
⇒ (x−5)(x+2)>0 (x - 5)(x + 2) > 0


Therefore, x>5x > 5 or x<−2x < -2.

If ∣x∣−4>0|x| - 4 > 0, then x>4x > 4 or x<−4x < -4.

If ∣x∣−4≠1|x| - 4 \ne 1 ⇒ ∣x∣≠5|x| \ne 5 ⇒ x≠5,−5x \ne 5, -5

Merging the two conditions we get the domain to be (−∞,−4) ⋃ (5,∞)(- \infty, - 4)\space \bigcup \space(5, \infty), where x≠−5x \ne -5

Answer: (−∞,−4) ⋃ (5,∞),(-\infty, -4)\space \bigcup \space(5, \infty), where x≠−5x \ne -5

Example 8

What is the domain of f(x)=log(x4−x)f(x) = \mathrm{log}(x^4 - x)

Solution

For a valid domain, x4−x>0x^4 - x > 0
⇒ x(x3−1)>0 x(x^3 - 1) > 0
⇒ x(x−1)(x2+x+1)>0 [x2+x+1is rejected as it has imaginary roots] x(x - 1)(x^2 + x + 1) > 0 \space [x^2 + x + 1 \text{is rejected as it has imaginary roots}]
⇒ 0 0 and 11 are the deflection points.


x4−xx^4 - x is positive in the range of (−∞,0) ⋃ (1,∞)(- \infty, 0)\space \bigcup \space(1, \infty)

Answer: (−∞,0) ⋃ (1,∞)(- \infty, 0)\space \bigcup \space(1, \infty)

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