Tag: introduction to sets

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Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

If $ P ( X ) = x ^ { 3 } - 3 x ^ { 2 } + 2 x + 5 $ and P ( a ) = P ( b ) = P ( c ) = 0 then the value of ( 2 - a ) ( 2 - b ) ( 2 - c ) is

  1. 3

  2. 5

  3. 7

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since P(a) = P(b) = P(c) = 0, a, b, and c are the roots of the cubic polynomial P(x) = x^3 - 3x^2 + 2x + 5. This means P(x) can be factored as (x - a)(x - b)(x - c). We need the value of (2 - a)(2 - b)(2 - c), which is precisely P(2). Substituting x = 2 into P(x) gives 2^3 - 3(2^2) + 2(2) + 5 = 8 - 12 + 4 + 5 = 5.

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

If f : R $\rightarrow$ R, g : R $\rightarrow$ R and h : R $\rightarrow$ R is such that $f(x) = x^2, g(x) = tan  x$ and $h(x) = log  x$, then the value of [ho(gof)], if $x = \displaystyle \dfrac{\pi}{2}$ will be

  1. 0

  2. 1

  3. -1

  4. 10
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ho(gof) =(hof)(f(x))$
$=(hog)(x^2)=(hof) (\dfrac{\pi}{4}) = h(g(\dfrac{\pi}{4}))$
$= h(tan \dfrac{ \pi}{4}) = h(1) = log 1 =0$

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

The number of elements of an identity function defined on a set containing four elements is______

  1. $\displaystyle 2^{2}$
  2. $\displaystyle 2^{4}$
  3. $\displaystyle 2^{8}$
  4. $\displaystyle 2^{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If an element is related to itself, it is called an identity function. That is $ f(x) = x $

So, if  the set has $ 4 $ elements, then the function will also have $ 4 = 2^2 $ elements.

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

Let $f(-2, 2)\rightarrow(-2, 2)$ be a continuous function given $f(x)=f{(x}^{2})$. Given $f(0)=\dfrac{1}{2}$ then the $4f(\dfrac{1}{2})$

  1. $4$
  2. $2$
  3. $-2$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given f(x) = f(x^2), by repeated substitution we have f(x) = f(x^4) = f(x^8) and so on. For any x in (-1, 1), x^(2^n) approaches 0 as n goes to infinity. Since f is continuous and f(0) = 1/2, f(x) must be equal to 1/2 for all x in (-1, 1). Thus, f(1/2) = 1/2, and 4 * f(1/2) = 4 * (1/2) = 2.

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

Let $f\left( x \right) = p{x^2} + qx - \left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right),\,\left( {p,q,a,b,c \in R} \right)(a,b,c$ are distinct). If both roots of $f(x)=0$ are non-real, then 

  1. $2\left( {p + q} \right) - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] > 0$
  2. $2\left( {p + q} \right) - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] < 0$
  3. $p - 2q - 2 - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] < 0$
  4. $p - 2q - 2 - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] > 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression (a-b)^2 + (b-c)^2 + (c-a)^2 is always positive for distinct a, b, c. If the roots of px^2 + qx - K = 0 are non-real, the discriminant q^2 + 4pK < 0. The options involve complex algebraic manipulations of these coefficients.