Tag: graphs of the form y=ax^2+bx+c

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Questions Related to graphs of the form y=ax^2+bx+c

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 $y^2 = ax^2 +bx+c$, then $y^2 \dfrac{d^2y}{dx^2}$ is

  1. a constant function

  2. a function of x only

  3. a function of y only

  4. a function of both x and y
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Differentiating y^2 = ax^2 + bx + c gives 2y*y' = 2ax + b. Differentiating again gives 2(y')^2 + 2y*y'' = 2a. Substituting y' = (2ax+b)/(2y) into the equation allows one to solve for y^2*y''. The result is a constant.

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 $fxln\left(1+\dfrac{1}{x}\right)dx=p(x)ln\left(1+\dfrac{1}{x}\right)+\dfrac{1}{2}x-\dfrac{1}{2}ln(1+x)+c$, being arbitary costant, then

  1. $p(X)=\dfrac{1}{2}x^{2}$
  2. $p(x)=0$
  3. $p(x)=1$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is an integration problem involving parts. Integrating ln(1 + 1/x) = ln((x+1)/x) = ln(x+1) - ln(x) leads to the form provided. Comparing the result with the given expression identifies p(x).

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 $f(x)$ is cubic polynomial with real coefficient such that $f''(3) = 0, f'(5) = 0$. If $f(3) = 1$ and $f(5) = -3$, then $f(1)$ is equal to

  1. $2$
  2. $3$
  3. $5$
  4. $6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since f(x) is a cubic polynomial, its second derivative f''(x) is a linear function. Given f''(3) = 0, we can write f''(x) = c(x - 3). Integrating twice and using the conditions f'(5) = 0, f(3) = 1, and f(5) = -3 allows us to find the exact cubic polynomial and evaluate f(1) to be 5.

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

Let $\displaystyle f(x)=ax^{2}+bx+c,$ where $a,b,c$ are rational, and $f: Z\rightarrow Z,$ where $Z$ is the set of integers. Then $a+b$ is

  1. a negative integer

  2. an integer

  3. nonintegral rational number

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$f:Z \rightarrow Z$ is defined as $f(x)=ax^2+bx+c$

which implies for integer inputs, the function gives integer outputs.

$ \Rightarrow f(0)=c=Z _1$ ...(1) (where $Z _1$ is some integer)

Similarly, $f(1)=a+b+c=Z _2$ ...(2) (where $Z _2$ is some integer)

(2) - (1) gives $a+b =Z _2-Z _1$, which is also an integer.

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 $f\quad \left( x \right) ={ x }^{ 2 }+2bx+{ 2c }^{ 2 }\quad and\quad g\quad (x)\quad ={ -x }^{ 2 }\quad -2cx+{ b }^{ 2 }\quad are\quad such\quad that\quad min\quad f\quad (x)\quad >\quad max\quad g\quad (x),\quad then$ relation between b and c, is

  1. none relation

  2. 0 < c < b/2
  3. $\left| c \right| <\frac { \left| b \right| }{ \sqrt { 2 } } $
  4. $\left| c \right| >\sqrt { 2 } \left| b \right| $
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $f\left(x\right)$ is a polynomial such that $ f\left(a\right) f\left(b\right)<0$, then number of zeros lieing between $a$ and $b$ is 

  1. $one$
  2. $at least one$
  3. $two$
  4. $at most 2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Intermediate Value Theorem, if f(x) is a continuous polynomial function such that f(a) and f(b) have opposite signs (their product is less than zero), then there must exist at least one root between a and b. Therefore, the correct statement is that there is at least one zero lying between a and b.

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.