Some functions and their graphs -i - class-XI

Given $p = q$
multiply both sides by $x$, we get
$px = qx$, which is required value.

An identity function is a function that always returns the same value that was used as its argument.

Hence $f:R\rightarrow R,f(x)=x$ is an identity function

Identity funtion returns the same value as of input.

$f(x)=7$ is constant function as its values do not depend on the variable $x$.

An identity function is of the form $f(x) = cx$

An identity function $f(x)$ is of the form $f(x) = cx$

A constant function is a function that has the same output value no matter what your input value is. Because of this, a constant function has the form y=k where k is a constant(a single value that doesn't change).

Identify function ,$f(x)=x$

Let $f(x)=ax+b$ [Since $f$ is given to be linear function] where $a$ and $b$ are constants.

$f(x)=1,\quad f(y)\neq 1,\quad f(z)\neq 1$

f($x$)=$\frac{ax+b}{cx+d}$ is a constant function,

$f(n+1)=f(n)$
$\Rightarrow f\left ( n \right )=f\left ( n-1 \right )=f\left ( n-2 \right )=...........=f\left ( 1 \right )$
$\therefore f\left ( 7 \right )=5\Rightarrow f\left ( 35 \right )=5$

The given function is f(x) = 2 * tan^-1(x) + sin^-1(2x / (1 + x^2)). For x >= 1, the second term simplifies to pi - 2 * tan^-1(x), making the total function equal to pi, which is a constant function. For x <= -1, the second term simplifies to -pi - 2 * tan^-1(x), which also gives a constant value. The question specifies a positive output, and since pi is positive, the interval where this holds is [1, infinity).

Since f(x) is a polynomial and f(3x) = f'(x) * f''(x), comparing degrees shows that if f(x) has degree n, the left side has degree n while the right side has degree 2n - 3, meaning n = 3. Thus f(x) is a cubic polynomial of the form ax^3 + bx^2 + cx + d. Solving the functional equation coefficients leads to specific relations, satisfying f''(2) - f'(2) = 0.

let $f(x)=ax+b$

Given f(x) is a quartic polynomial with extreme values at x=2, f'(2)=0. The limit condition lim(x->0) (f(x)/x^2 + 1) = 3 implies f(0)=0 and f'(0)=0, and f''(0)/2 = 2, so f''(0)=4. Using these conditions, one can determine the coefficients of f(x).

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

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).

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.

$f:Z \rightarrow Z$ 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.

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.

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.

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

A constant function $ f(x) $ will have the result as a constant for any value of $ x $

So, if $ f(100) = 100 $
then $ f(2007 ) $ is also $ 100 $

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.

Derivative of an Identity function gives a Constant function

$f(x)=f(x+a)$
A constant function is a period with any value of $'a'$ as a period

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.

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.

The polynomial which satisfies $f(x)f(1/x)=f(x)+f(1/x)$ is $ \pm x^n+1$ (standard result)
Given that $f(2) = 9 \ \Rightarrow \pm 2^n + 1 = 9 \ \Rightarrow 2^n = 8 $
(-ve sign not possible here)
$ \Rightarrow n=3$
Hence the function is $ f(x)=x^3+1$
$ \Rightarrow f(1) = 2, ; f(3)=28 , ; f(5)=126$
$ f(4) = 65, ; f(6) = 217$
Using these, we see only option B and C are correct. 

$(x-y)f(x+y)-(x+y)f(x-y)$ $=2y((x-y)(x+y))$

$g(x) g(y) = g (x) + g (y) + g (xy) - 2$

Since the rational function $f$ has the vertical asymptote at $x=4$, then the denominator of $f$ contains the term $(x-4)$.

Let $t$ be the number of minutes since the tap opened. 

Since the water increases at $20$ gallons per minute, and the sugar increases at $2$ pound per minute, these are constant rates of change. 

This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. 

We can write an equation independently for each:

$\text{Water}: \: W(t)=200+20t$ in gallons

$\text{Sugar}: \: S(t)=10+2t$ in pounds​​

The concentration $C$ will be the ratio of pounds of sugar to gallons of water.

$C(t)=\dfrac{10+2t}{200+20t}$​​

The concentration after $14$ minutes is given by evaluating $C(t)$ at $t=14$

$\therefore C(14)=\dfrac{10+28}{200+280}=\dfrac{19}{240}$

This means the concentration is $19$ pounds of sugar to $240$ gallons of water.

At the beginning, the concentration is $C(0)=\dfrac{10+0}{200+0}=\dfrac{1}{20}$.

Since $\dfrac{19}{240}\approx 0.08 > \dfrac{1}{20}=0.05$, the concentration is greater after $14$ minutes than at the beginning.