Class 11 Commerce Maths - Relations And Functions

Find an explicit formula (in terms of n $\epsilon$ N)for $f(n)= (n - 1)+ 2n -1$

$f(n)=(n-1)+2n-1$

$f(n)=n+2n-1-1$

$f(n)=3n-2$

$f(n)=-2+3n$

where

$n\in N$

If f is a real function defined by $f(x) = \dfrac{x - 1}{x + 1}$ , then prove that
$f(2x) = \dfrac{3f(x) + 1}{f(x) + 3}$

$3f(x) + 1 = \dfrac{3(x-1)}{x+1} +1 = \dfrac{4x-2}{x+1}$

$f(x) +3 = \dfrac{x-1}{x+1}+3=\dfrac{4x+2}{x+1}$

Dividing we get

$\dfrac{4x-2}{4x+2}=\dfrac{2x-1}{2x+1}=f(2x)$

Find the least value of $x$ when $-5\le 2x-7$ .

Given,

$-5\le 2x-7$

or,

$2x\ge 7-5$

or,

$2x\ge 2$

or,

$x\ge 1$

From this we get the least value of

$x$ is

$1$ .

If $f(x)=x$ , then which of the following must be true?
(a) $f(a+b)=f(a)+f(b)$
(b) $f(2a)=2f(a)$
(c) $f(-a)=-f(a)$

$(i)$

$f(a+b)=f(a)+f(b)$ acn be possible because.

$f(a+b)=(a+b)$

$f(a)+f(b)$

$(ii)$

$f(2a)=2a[\because f(x)=x]$

$f(2a)=2f(a)$

$(iii)$

$f(-a)=-a$

$f(a)=-f(a)$

Three conditions are true.

If $f(x)=x+7$ and $g(x)=x-7, x\in R$ , find $fog(6)$ .

Given,

$f(x)=x+7$ and

$g(x)=x-7$

Now,

$fog(x)=f(g(x))=f(x-7)=x-7+7=x$

Therefore,

$fog(6)=6$

In eq $x^{2}+y=6x-14$
if $x=0 ,1,2$
then the values of $y$ are

Consider given the equation,

${{x}^{2}}+y=6x-14$

${{x}^{2}}-6x+y=-14$

Put, $x=0$

${{0}^{2}}-6\times 0+y=-14$

$y=-14$

Again put, x=1

${{1}^{2}}-6\times 1+y=-14$

$y=-14+6-1=-9$

Again put, x=2

${{2}^{2}}-6\times 2+y=-14$

$y=-2-4$

$y=-6$

Hence, this is the answer.

The length, l metres, of a garden is $78.5$ metres, correct to the nearest half metre. Complete this statement about the value of l.
__ $\leq$ l $<$ _.

The length is

$78.5m$ which is correct upto half meter

So it can be given as

$78.25m \leq l < 78.75m$

If $y=f(x)=\dfrac {ax-b}{bx-a}$ , show that $x=f(y)$

1417170_1664586_ans_1ed14b27475641b5899b72a626108480.jpg

$\displaystyle f' \, (1) \, + \, f \, (-1) \, = \, -\, f \, (0)$ , if $f \, (x) \, = \, x^5 \, +\, x^3 \, - \, 2x\, -\, 3$

$f(x) = x^{5} + x^{3} - 2x - 3$

$f'(x) = 5x^{4} + 3x^{2} - 2$

$\therefore f'(1) = 5(1)^{4} + 3(1)^{2} - 2$

$= 5 + 3 - 2$

$= 6$

$Also\ f(1) = (1)^{5} + (1)^{3} - 2(1) -3$

$= 1 + 1 - 2 - 3$

$= 2 - 2 - 3$

$= -3$

$\therefore f(0) = -3$

and

$f(-1) = (-1)^{5} + (-1)^{3} - 2(-1) - 3$

$= -1 - 1 + 2 - 3$

$\Rightarrow -3$

$f'(1) + f(-1) = 6 - 3 = 3=-f(0)$ .

Discuss the commutativity and associativity of binary operation $\ast$ . defined on $A=Q-\left\{ 1 \right\}$ by the rule $a\ast b=a-b+ab$ fo all $a,b\in A$ . Also find the identity element of $\ast$ in $A$ and hence find the invertible elements of $A$ .

Given # is a binary operation on

$Q-\left\{ 1 \right\}$

defined by

$a*b=a-b+ab$

Commutativity:

for any

$a,b \in A$

We have

$a*b=a-b+ab$ and

$b*a=b-a+ba$

Since,

$a-b+ab \neq b-a+ab$

$\therefore a*b \neq b*a$

So,

$*$ is not commutative on

$A$

Associativity:

Let

$a,b,c \in A(a*b)*c=(a-b+ab)*c$

$\Rightarrow (a*b)*c=(a-b+ab)-c+(a-b+ab)c$

$\Rightarrow (a*b)*c=a-b+ab-c+ac-bc+abc$

$a*(b*c)=a*(b-c+bc)$

$\Rightarrow a*(b*c)=a-(b-c+bc)+a(b-c+bc)$

$\Rightarrow a*(b*c)=a-b+c-bc+ab-ac+abc$

$\Rightarrow (a*b)*c\neq a*(b*c)$

$*$ is not associative on

$A$

Identity Element:

Let

$e$ be the identity elements in

$A$

the

$a*e=a=e*a a\in Q-\left\{ 1 \right\}$

$a-e+ae=a$

$(a-1)e=0$

$e=0$

$($ as

$a\neq 1)$ [so

$'0'$ si identity element in

$A]$

Inverse of an Element:

Let

$a$ be an arbitary element of

$A$ as

$b$ be the inverse of

$a$ .

$\because$

$a*b=e=b*a$

$\Rightarrow a*b=e$

$\Rightarrow a-b+ab=0[\because e=0]$

$a=b(1-a)$

$b=a/1-a$

since

$b\in Q-1$

So, every elements of

$A$ is invertible.

If $f (x)=a+bx+cx^2$ , show that
$\displaystyle \underset{0}{\overset{1}{\int}} f(x)dx=\dfrac{1}{6} \left[f(0)+4f \left(\dfrac{1}{2}\right)+f(1)\right]$

$f(x) = a + bx + cx^2$

$\therefore f(0) = a$

$\therefore f(1/2) = a + \dfrac{b}{2} + \dfrac{c}{4}$

$\therefore f(1) = a + b + c$

$\therefore \displaystyle \int_0^1 f(x) dx = \int_0^1 (a + bx + cx^2) dx$

$= ax + \dfrac{bx^2}{2} + \dfrac{cx^3}{3} |_0^1$

$= a(1-0) + \dfrac{b}{2} (1 - 0) + \dfrac{c}{3} (1 - 0)$

$= a + \dfrac{b}{2} + \dfrac{c}{3} \,\,\, ......(1) \, \rightarrow LHS$

$\therefore RHS = \dfrac{1}{6} [f(0) + 4f (1/2) + f(1)]$

$= \dfrac{1}{6} \left[a + 4 \left(a + \dfrac{b}{2} + \dfrac{c}{4} \right) + a + b + c \right]$

$= \dfrac{1}{6} (a + 4a + a + 2b + c + b + c)$

$= a + \dfrac{3b}{6} + \dfrac{2c}{6}$

$= a + \dfrac{b}{2} + \dfrac{c}{3} \rightarrow RHS$

$\therefore LHS = RHS$

Hence proved.

Divide polynomial $P(x) = x^{4} + 4x^{3} + 5x^{2} - 7x - 3, by \ S(x) = x^{2} - 1$ .

1372804_1127027_ans_a9f7d945fced4412840eaaa2a1971c19.jpg

Let $f(x)=x^2$ and $g(x)=2x+1$ be two real functions. Find $(f+g)(x),(f-g)(x),(fg)(x), \left(\dfrac{f}{g}\right)(x).$

$(f+g)(x)=f(x)+g(x)$

$=x^{2}+2x+1$

$(f-g)(x)=f(x)-g(x)$

$=x^{2}-(2x+1)$

$=x^{2}-2x-1$

$(fg)(x)=f(x).g(x)$

$=x^{2}\left ( 2x+1 \right )$

$=2x^{3}+x^{2}$

$\left ( \cfrac{f}{g} \right )(x)=\cfrac{f(x)}{g(x)}$

$\left ( \; g(x)\neq 0 \right )$

$=\cfrac{x^{2}}{2x+1}$

$\left ( \; x\neq -\cfrac{1}{2} \right )$

$f=\{ (3,4),(4,3)\}\ \ g=\{(3,7),(4,8)\}$ Find $f-g$

$f=\{(3,4),(4,3)\}$

$g=\{(3,7),(4,8)\}$

$f-g=\{ (3,-3),(4,-5)\}$

If $f(x)=\dfrac {x-1}{x+1}$ then show that

$f\left (\dfrac {1}{x}\right)=-f(x)$

1417189_1664621_ans_360578f4ddd9436eadd8ab624824c1ee.jpg

If $f(x)=(x-a)^2 (x-b)^2$ , find $f(a+b)$

1417171_1664585_ans_9a3dc08ad84d4591919499f3b2b3252c.jpg

If $f(x)$ be defined on $[-2, 2]$ and is given by $f\left( x \right)=\begin{cases} -1,\quad -2\le x\le 0 \\ x-1,\quad 0<x\le 2 \end{cases}$ and $g(x)=f(|x|)+|f(x)|$ . Find $g(x)$

1418889_1664716_ans_b537ebee5e3b4870a5dbc91a4a13054b.jpg

If $f(x)=\dfrac {x-1}{x+1}$ then show that $f\left (\dfrac {1}{x}\right) =-{f(x)}$

1428677_1664624_ans_dc94712124dc42fbb798f8396f0ae2c6.jpg

Let $f, g$ be two real functions defined by $f(x)=\sqrt {x+1}$ and $g(x)=\sqrt {9-x^2}$ . Then, described each of the following functions:
$f+g$

1419018_1664722_ans_104de14c6b8846d89424248053e8ac41.jpg

Find $f+g, f-g, cf (c \in R, c\neq 0), \ fg, \dfrac {1}{f}$ and $\dfrac {1}{g}$ in each of the following:
$f(x)=x^3+1$ and $g(x)=x+1$

1418661_1664705_ans_bc72bb250f5e4133b6c3b335ecf22315.jpg

If $f:R\rightarrow R,$ defined as
$f(x)=x^3+5$
Find $f^{-1}(x)$

According to question,

$f:R\rightarrow R,f(x)=x^3+5$

One-one/onto: Let

$a,b\epsilon R$

$f(a)=f(b)$

$\Rightarrow a^3+5=b^3+5$

$\Rightarrow a^3=b^3$

$a=b$

So,

$f(a)=f(b)$

$\Rightarrow a=b,\forall a,b \epsilon R$

$\therefore f$ is one-one function.

Onto/into:

Let

$y\epsilon R$ (co-domain)

$f(x)=y$

$\\Rightarrow x^3+5=y$

$\Rightarrow x^3=y-5$

$\Rightarrow x=(y-5)^{1/3}\epsilon R, \forall x\epsilon R$

So pre-image for each value of

$y$ exists in domain

$R$ .

Thus, range of

$f=$ co-domain of

$f$ .

So, function

$'f'$ is onto function.

Hence, we can say that

$f$ is one-one onto function.

So,

$f^{-1}:R\rightarrow R$ defined as

$f^{-1}(y)=x\Leftrightarrow f(x)=y$ ..........(i)

$\Rightarrow f(x)=y$

$\Rightarrow x^3+5=y$

$x^3=y-5$

$\Rightarrow x=(y-5)^{1/3}$

$\Rightarrow f^{-1}(y)=(y-5)^{1/3}$ [from(i)]

$\Rightarrow f^{-1}(x)=(x-5)^{1/3}$

If $f:R\rightarrow R,$ defined as
$f(x)=2x-3$
Find $f^{-1}(x)$

Given function,

$f:R\rightarrow R, f(x)=2x-3$

One-one/many-one:

Let

$a,b\epsilon R$

$f(a)=f(b)$

$\Rightarrow 2a-3=2b-3$

$\Rightarrow 2a=2b$

$\Rightarrow a=b$

So,

$f(a)=f(b)\Rightarrow a=b,\forall a,b \epsilon R$

$\therefore f$ is one-one function.

onto/into:

Let

$y\epsilon R$ (co-domain)

$f(x)=y$

$\Rightarrow 2x-3=y$

$\Rightarrow x=\dfrac{y+3}{2}\epsilon R,\forall y\epsilon R$

So, pre-image for each value of

$y$ exists in domain

$R$ . Thus, function

$'f'$ is onto function.

It is clear that

$'f'$ is one-one into function.

Hence

$f^{-1}:R\rightarrow R$ exists.

Let

$x\epsilon R$ (domain of

$f$ )

and

$y\epsilon R$ (co-domain of

$f$ )

Let

$f(x)=y,$ then

$f^{-1}(y)=x$

$\Rightarrow f(x)=y$

$\Rightarrow 2x-3=y$

$\Rightarrow x=\dfrac{y+3}{2}\epsilon R$

$\Rightarrow f^{-1}(y)=\dfrac{y+3}{2}$

$\Rightarrow f^{-1}(x)=\dfrac{x+3}{2},\forall x\epsilon R$

If statement

$P(n) : 1^2 + 3^2 + 5^2 + ...+ (2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}$ ,

then test the authenticity of $P(4)$ .

Let the given statement is

$P(n)$ , where

$n \epsilon N$
So,

$P(n) = 1^2 + 3^2 + 5^2 + ...+ (2n-1)^2$

$= \dfrac{n(2n-1)(2n+1)}{3}$

Putting

$n=4$

$L.H.S. = P(4) : = 1^2 + 3^2 + 5^2 + ...+ (2 \times 4 -1)^2$

$= 1^2 + 3^2 + 5^2 + 7^2$

$= 1+ 9 + 25 + 49$

$=84$

$R.H.S= \dfrac{4(2 \times 4 -1)(2 \times 4 + 1)}{3} = \dfrac{(4(8-1)(8+1)}{3}$

$= \dfrac{4 \times 7 \times 9}{3} = 84$

$L.H.S. = R.H.S.$
Hence,

$P(4)$ is true.

Find a, b, c when $f(x) = ax^2 + bx + c$ and $f(0) = 6, f(2) = 11, f(-3) = 6$ Determine the value of $f(1)$ .

Given

$f(x) = ax^2 + bx + c$

$\therefore f(0) = a.0^2 + b.0 + c = 6$
or

$0.a + 0.b + 1.c = 6$ (i)
and

$f(2) = a(2)^2 + b(2) + c = 11$

$\Rightarrow 4.a + 2.b + 1.c = 11$ (ii)
and

$f(-3) = a(-3)^2 + b(-3) + c = 6$

$\Rightarrow 9.a - 3.b + c = 6$ (iii)
Since (i), (ii) and (iii) are three equations in a, b, c solving these by Cramer's rule

$\therefore D = \begin{vmatrix} 0& 0 & 1\\ 4 & 2 & 1\\ 9 & -3 & 1\end{vmatrix} = 1. (-12 - 18) = - 30$

$D_1 = \begin{vmatrix}6 & 0 & 1\\ 11 & 2 & 1\\ 6 & -3 & 1\end{vmatrix} = 6(2 + 3) - 0 + 1 (-33 - 12)$

$= - 15$

$D_2 = \begin{vmatrix}0 & 6 & 1\\ 4 & 11 & 1\\ 9 & 6 & 1\end{vmatrix}$

$= 0 - 6 (4 - 9) + 1 (24 - 99)$

$= 20 - 75$

$= - 45$

$D_3 = \begin{vmatrix}0 & 0 & 6\\ 4 & 2 & 11\\ 9 & -3 & 6\end{vmatrix}$

$= 6 (-12 - 18)$

$= - 180$
Hence by Cramer's rule

$\displaystyle a = \frac{D_1}{D} = \frac{- 15}{-30} = \frac{1}{2}$

$\displaystyle b = \frac{D_2}{D} = \frac{-45}{-30} = \frac{3}{2}$
and

$\displaystyle c = \frac{D_3}{D } = \frac{- 180}{-30} = 6$
Hence

$a = \dfrac{1}{2} , b = \dfrac{3}{2} , c = 6$

$\therefore f(x) = \dfrac{1}{2} x^2 + \dfrac{3}{2} x + 6$
and

$f(1) = \dfrac{1}{2} (1)^2 +\dfrac{3}{2} (1) + 6$

$= 8$

If $f(x)$ be a function such that $f(x-1)+f(x+1)=\sqrt{3} f(x)$ and $f(5) =10,$ then the sum of digits of the value of $\displaystyle \sum_{r=0}^{19} f(5+12r)$ is ......

$f(x-1)+f(x+1)=\sqrt{3}f(x)$

$\Rightarrow f(x)+f(x+2)=\sqrt{3}f(x+1)$
Putting,

$x=x+2$ ,

$f(x+1)+f(x+3)=\sqrt{3}f(x+2)$

$f(x-1)+2f(x+2)+f(x+3)=\sqrt{3}\left [\sqrt{3}f(x+1) \right ]$

$f(x-1)+f(x+3)=f(x+1)$
Putting,

$x=x+2$ , again

$f(x+1)+f(x+5)=f(x+3)$

$f(x-1)+f(x+5)=0$

$f(x+5)=-f(x-1)$

$f(x)=-f(x+6)$

$f(x+12)=f(x)$

$\sum_{r-0}^{19}f(5+12r)=20f(5)=20\times 10=200$
Hence, sum of digits

$=2+0+0=2$

If $f(x) \varepsilon [1, 2]$ when $x\varepsilon R$ and for a fixed positive real number p, $f(x+p)=1+\sqrt{2f(x)-{f(x)}^2}$ for all $x \varepsilon R$ then prove that f(x) is a periodic function .

$f(x+p)=1+\sqrt{2f(x)-f(x)^2}$

put

$x=x+p$

$f(x+2p)=1+\sqrt{2f(x+p)-f(x+p)^2}$

$f(x+2p)=1+\sqrt{(1+\sqrt{2f(x)-f(x)^2}(1-\sqrt{2f(x)-f(x)^2}})$

$f(x+2p)=1+\sqrt{1-2f(x)+f(x)^2}=1+f(x)-1$

$f(x+2p)=f(x)$

hence

$f(x)$ is a periodic function with period

$=2p$

Relations And Functions - Class 11 Commerce Maths - Extra Questions

Find an explicit formula (in terms of n $\epsilon$ N)for $f(n)= (n - 1)+ 2n -1$

If f is a real function defined by $f(x) = \dfrac{x - 1}{x + 1}$ , then prove that
$f(2x) = \dfrac{3f(x) + 1}{f(x) + 3}$

Find the least value of $x$ when $-5\le 2x-7$ .

If $f(x)=x$ , then which of the following must be true?
(a) $f(a+b)=f(a)+f(b)$
(b) $f(2a)=2f(a)$
(c) $f(-a)=-f(a)$

If $f(x)=x+7$ and $g(x)=x-7, x\in R$ , find $fog(6)$ .

In eq $x^{2}+y=6x-14$
if $x=0 ,1,2$
then the values of $y$ are

The length, l metres, of a garden is $78.5$ metres, correct to the nearest half metre. Complete this statement about the value of l.
__ $\leq$ l $<$ _.

If $y=f(x)=\dfrac {ax-b}{bx-a}$ , show that $x=f(y)$

$\displaystyle f' \, (1) \, + \, f \, (-1) \, = \, -\, f \, (0)$ , if $f \, (x) \, = \, x^5 \, +\, x^3 \, - \, 2x\, -\, 3$

Discuss the commutativity and associativity of binary operation $\ast$ . defined on $A=Q-\left\{ 1 \right\}$ by the rule $a\ast b=a-b+ab$ fo all $a,b\in A$ . Also find the identity element of $\ast$ in $A$ and hence find the invertible elements of $A$ .

If $f (x)=a+bx+cx^2$ , show that
$\displaystyle \underset{0}{\overset{1}{\int}} f(x)dx=\dfrac{1}{6} \left[f(0)+4f \left(\dfrac{1}{2}\right)+f(1)\right]$

Divide polynomial $P(x) = x^{4} + 4x^{3} + 5x^{2} - 7x - 3, by \ S(x) = x^{2} - 1$ .

Let $f(x)=x^2$ and $g(x)=2x+1$ be two real functions. Find $(f+g)(x),(f-g)(x),(fg)(x), \left(\dfrac{f}{g}\right)(x).$

$f=\{ (3,4),(4,3)\}\ \ g=\{(3,7),(4,8)\}$ Find $f-g$

If $f(x)=\dfrac {x-1}{x+1}$ then show that

$f\left (\dfrac {1}{x}\right)=-f(x)$

If $f(x)=(x-a)^2 (x-b)^2$ , find $f(a+b)$

If $f(x)$ be defined on $[-2, 2]$ and is given by $f\left( x \right)=\begin{cases} -1,\quad -2\le x\le 0 \\ x-1,\quad 0<x\le 2 \end{cases}$ and $g(x)=f(|x|)+|f(x)|$ . Find $g(x)$

If $f(x)=\dfrac {x-1}{x+1}$ then show that $f\left (\dfrac {1}{x}\right) =-{f(x)}$

Let $f, g$ be two real functions defined by $f(x)=\sqrt {x+1}$ and $g(x)=\sqrt {9-x^2}$ . Then, described each of the following functions:
$f+g$

Find $f+g, f-g, cf (c \in R, c\neq 0), \ fg, \dfrac {1}{f}$ and $\dfrac {1}{g}$ in each of the following:
$f(x)=x^3+1$ and $g(x)=x+1$

If $f:R\rightarrow R,$ defined as
$f(x)=x^3+5$
Find $f^{-1}(x)$

If $f:R\rightarrow R,$ defined as
$f(x)=2x-3$
Find $f^{-1}(x)$

If statement

$P(n) : 1^2 + 3^2 + 5^2 + ...+ (2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}$ ,

then test the authenticity of $P(4)$ .

Find a, b, c when $f(x) = ax^2 + bx + c$ and $f(0) = 6, f(2) = 11, f(-3) = 6$ Determine the value of $f(1)$ .

If $f(x)$ be a function such that $f(x-1)+f(x+1)=\sqrt{3} f(x)$ and $f(5) =10,$ then the sum of digits of the value of $\displaystyle \sum_{r=0}^{19} f(5+12r)$ is ......

If $f(x) \varepsilon [1, 2]$ when $x\varepsilon R$ and for a fixed positive real number p, $f(x+p)=1+\sqrt{2f(x)-{f(x)}^2}$ for all $x \varepsilon R$ then prove that f(x) is a periodic function .

Class 11 Commerce Maths Extra Questions

Relations And Functions - Class 11 Commerce Maths - Extra Questions

Find an explicit formula (in terms of nϵ\epsilon N)for f(n)=(n−1)+2n−1f(n)= (n - 1)+ 2n -1

If f is a real function defined by f(x)=x−1x+1f(x) = \dfrac{x - 1}{x + 1}, then prove thatf(2x)=3f(x)+1f(x)+3f(2x) = \dfrac{3f(x) + 1}{f(x) + 3}

Find the least value of xx when −5≤2x−7-5\le 2x-7.

If f(x)=xf(x)=x, then which of the following must be true?(a) f(a+b)=f(a)+f(b)f(a+b)=f(a)+f(b)(b) f(2a)=2f(a)f(2a)=2f(a)(c) f(−a)=−f(a)f(-a)=-f(a)

If f(x)=x+7f(x)=x+7 and g(x)=x−7,x∈Rg(x)=x-7, x\in R, find fog(6)fog(6).

In eq x2+y=6x−14x^{2}+y=6x-14if x=0,1,2x=0 ,1,2 then the values of yy are

The length, l metres, of a garden is 78.578.5 metres, correct to the nearest half metre. Complete this statement about the value of l.________≤\leq l << _______.

If y=f(x)=ax−bbx−ay=f(x)=\dfrac {ax-b}{bx-a}, show that x=f(y)x=f(y)

Prove that f′(1)+f(−1)=−f(0)\displaystyle f' \, (1) \, + \, f \, (-1) \, = \, -\, f \, (0) , if f(x)=x5+x3−2x−3 f \, (x) \, = \, x^5 \, +\, x^3 \, - \, 2x\, -\, 3

Discuss the commutativity and associativity of binary operation ∗\ast. defined on A=Q−{1}A=Q-\left\{ 1 \right\} by the rule a∗b=a−b+aba\ast b=a-b+ab fo all a,b∈Aa,b\in A. Also find the identity element of ∗\ast in AA and hence find the invertible elements of AA.

If f(x)=a+bx+cx2f (x)=a+bx+cx^2, show that1∫0f(x)dx=16[f(0)+4f(12)+f(1)]\displaystyle \underset{0}{\overset{1}{\int}} f(x)dx=\dfrac{1}{6} \left[f(0)+4f \left(\dfrac{1}{2}\right)+f(1)\right]

Divide polynomial P(x)=x4+4x3+5x2−7x−3,by S(x)=x2−1P(x) = x^{4} + 4x^{3} + 5x^{2} - 7x - 3, by \ S(x) = x^{2} - 1.

Let f(x)=x2f(x)=x^2 and g(x)=2x+1g(x)=2x+1 be two real functions. Find (f+g)(x),(f−g)(x),(fg)(x),(fg)(x).(f+g)(x),(f-g)(x),(fg)(x), \left(\dfrac{f}{g}\right)(x).

f={(3,4),(4,3)} g={(3,7),(4,8)}f=\{ (3,4),(4,3)\}\ \ g=\{(3,7),(4,8)\} Find f−gf-g

If f(x)=x−1x+1f(x)=\dfrac {x-1}{x+1} then show thatf(1x)=−f(x)f\left (\dfrac {1}{x}\right)=-f(x)

If f(x)=(x−a)2(x−b)2f(x)=(x-a)^2 (x-b)^2, find f(a+b)f(a+b)

If f(x)f(x) be defined on [−2,2][-2, 2] and is given by f(x)={−1,−2≤x≤0x−1,0<x≤2f\left( x \right)=\begin{cases} -1,\quad -2\le x\le 0 \\ x-1,\quad 0<x\le 2 \end{cases} and g(x)=f(|x|)+|f(x)|g(x)=f(|x|)+|f(x)|. Find g(x)g(x)

If f(x)=x−1x+1f(x)=\dfrac {x-1}{x+1} then show that f(1x)=−f(x)f\left (\dfrac {1}{x}\right) =-{f(x)}

Let f,gf, g be two real functions defined by f(x)=√x+1f(x)=\sqrt {x+1} and g(x)=√9−x2g(x)=\sqrt {9-x^2}. Then, described each of the following functions:f+gf+g

Find f+g,f−g,cf(c∈R,c≠0), fg,1ff+g, f-g, cf (c \in R, c\neq 0), \ fg, \dfrac {1}{f} and 1g\dfrac {1}{g} in each of the following:f(x)=x3+1f(x)=x^3+1 and g(x)=x+1g(x)=x+1

If f:R→R,f:R\rightarrow R, defined asf(x)=x3+5f(x)=x^3+5Find f−1(x)f^{-1}(x)

If f:R→R,f:R\rightarrow R, defined asf(x)=2x−3f(x)=2x-3Find f−1(x)f^{-1}(x)

If statementP(n):12+32+52+...+(2n−1)2=n(2n−1)(2n+1)3P(n) : 1^2 + 3^2 + 5^2 + ...+ (2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3},then test the authenticity of P(4)P(4).

Find a, b, c when f(x)=ax2+bx+cf(x) = ax^2 + bx + c and f(0)=6,f(2)=11,f(−3)=6f(0) = 6, f(2) = 11, f(-3) = 6 Determine the value of f(1)f(1).

If f(x)f(x) be a function such that f(x−1)+f(x+1)=√3f(x) f(x-1)+f(x+1)=\sqrt{3} f(x) and f(5)=10,f(5) =10, then the sum of digits of the value of 19∑r=0f(5+12r) \displaystyle \sum_{r=0}^{19} f(5+12r) is ......

If f(x)ε[1,2] f(x) \varepsilon [1, 2] when xεRx\varepsilon R and for a fixed positive real number p, f(x+p)=1+√2f(x)−f(x)2f(x+p)=1+\sqrt{2f(x)-{f(x)}^2} for all xεRx \varepsilon R then prove that f(x) is a periodic function .

Class 11 Commerce Maths Extra Questions

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Find an explicit formula (in terms of n $\epsilon$ N)for $f(n)= (n - 1)+ 2n -1$

If f is a real function defined by $f(x) = \dfrac{x - 1}{x + 1}$ , then prove that
$f(2x) = \dfrac{3f(x) + 1}{f(x) + 3}$

Find the least value of $x$ when $-5\le 2x-7$ .

If $f(x)=x$ , then which of the following must be true?
(a) $f(a+b)=f(a)+f(b)$
(b) $f(2a)=2f(a)$
(c) $f(-a)=-f(a)$

If $f(x)=x+7$ and $g(x)=x-7, x\in R$ , find $fog(6)$ .

In eq $x^{2}+y=6x-14$
if $x=0 ,1,2$
then the values of $y$ are

The length, l metres, of a garden is $78.5$ metres, correct to the nearest half metre. Complete this statement about the value of l.
__ $\leq$ l $<$ _.

If $y=f(x)=\dfrac {ax-b}{bx-a}$ , show that $x=f(y)$

$\displaystyle f' \, (1) \, + \, f \, (-1) \, = \, -\, f \, (0)$ , if $f \, (x) \, = \, x^5 \, +\, x^3 \, - \, 2x\, -\, 3$

Discuss the commutativity and associativity of binary operation $\ast$ . defined on $A=Q-\left\{ 1 \right\}$ by the rule $a\ast b=a-b+ab$ fo all $a,b\in A$ . Also find the identity element of $\ast$ in $A$ and hence find the invertible elements of $A$ .

If $f (x)=a+bx+cx^2$ , show that
$\displaystyle \underset{0}{\overset{1}{\int}} f(x)dx=\dfrac{1}{6} \left[f(0)+4f \left(\dfrac{1}{2}\right)+f(1)\right]$

Divide polynomial $P(x) = x^{4} + 4x^{3} + 5x^{2} - 7x - 3, by \ S(x) = x^{2} - 1$ .

Let $f(x)=x^2$ and $g(x)=2x+1$ be two real functions. Find $(f+g)(x),(f-g)(x),(fg)(x), \left(\dfrac{f}{g}\right)(x).$

$f=\{ (3,4),(4,3)\}\ \ g=\{(3,7),(4,8)\}$ Find $f-g$

If $f(x)=\dfrac {x-1}{x+1}$ then show that

$f\left (\dfrac {1}{x}\right)=-f(x)$

If $f(x)=(x-a)^2 (x-b)^2$ , find $f(a+b)$

If $f(x)$ be defined on $[-2, 2]$ and is given by $f\left( x \right)=\begin{cases} -1,\quad -2\le x\le 0 \\ x-1,\quad 0<x\le 2 \end{cases}$ and $g(x)=f(|x|)+|f(x)|$ . Find $g(x)$

If $f(x)=\dfrac {x-1}{x+1}$ then show that $f\left (\dfrac {1}{x}\right) =-{f(x)}$

Let $f, g$ be two real functions defined by $f(x)=\sqrt {x+1}$ and $g(x)=\sqrt {9-x^2}$ . Then, described each of the following functions:
$f+g$

Find $f+g, f-g, cf (c \in R, c\neq 0), \ fg, \dfrac {1}{f}$ and $\dfrac {1}{g}$ in each of the following:
$f(x)=x^3+1$ and $g(x)=x+1$

If $f:R\rightarrow R,$ defined as
$f(x)=x^3+5$
Find $f^{-1}(x)$

If $f:R\rightarrow R,$ defined as
$f(x)=2x-3$
Find $f^{-1}(x)$

If statement

$P(n) : 1^2 + 3^2 + 5^2 + ...+ (2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}$ ,

then test the authenticity of $P(4)$ .

Find a, b, c when $f(x) = ax^2 + bx + c$ and $f(0) = 6, f(2) = 11, f(-3) = 6$ Determine the value of $f(1)$ .

If $f(x)$ be a function such that $f(x-1)+f(x+1)=\sqrt{3} f(x)$ and $f(5) =10,$ then the sum of digits of the value of $\displaystyle \sum_{r=0}^{19} f(5+12r)$ is ......

If $f(x) \varepsilon [1, 2]$ when $x\varepsilon R$ and for a fixed positive real number p, $f(x+p)=1+\sqrt{2f(x)-{f(x)}^2}$ for all $x \varepsilon R$ then prove that f(x) is a periodic function .