(Each question 2 marks) - MATHS STD 11 Science Questions - Vidyadip

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18 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

Let f = {(1, 1), (2, 3), (0, - 1), - 1, - 3)} be a function from Z to Z defined by f (x) = ax + b for some integers a, b. Determine a,b.

Answer

Here f(x) = ax + b
f = {(1, 1), (2, 3), (0, -1), (-1, -3)}
$ \Rightarrow $ f (1) = 1, f (2) = 3, f(0) = -1, f(-1) = -3
Now f(1) = 1 $ \Rightarrow a \times 1 + b = 1 \Rightarrow $ a+ b = 1 . . . (i)
$f(2) = 3 \Rightarrow a \times 2 + b = 3 \Rightarrow 2a + b = 3$ . . . (ii)
Subtracting (i) from (ii) we get
2a+b-(a+b)=3-1 $ \Rightarrow $ a = 2
Putting a = 2 in (i)
2 + b = 1 $ \Rightarrow $ b = -1

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Question 22 Marks

Let $f= \left\{ \left(x , \frac { x ^ { 2 } } { 1 + x ^ { 2 } } \right) : x \in R \right\}$ be a function from R into R. Determine the range of f.

Answer

Here $f ( x ) = \frac { x ^ { 2 } } { 1 + x ^ { 2 } }$
Put $y = \frac { x ^ { 2 } } { 1 + x ^ { 2 } }$ $\Rightarrow$ $y+y x^2=x^2 \Rightarrow x^2(1-y)=y$
$\Rightarrow x ^ { 2 } = \frac { y } { 1 - y } \Rightarrow x = \pm \sqrt { \frac { y } { 1 - y } }$
$\frac { y } { 1 - y } \geq 0$
$\Rightarrow \frac { y } { y - 1 } \leq 0$
$\Rightarrow 0 \leq \mathrm { y } < 1$
$\Rightarrow \mathrm { y } \in [ 0,1 )$
$\therefore$Range of f(x) = [0, 1)

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Question 32 Marks

Find the domain and the range of the real function f defined by f(x) = |x - 1|.

Answer

Here f (x) = |x - 1|
The function f (x) is defined for all values of x
$\therefore$Domain of f(x) = R
when x > 1
|x - 1| = x - 1 > 0
When x = 1
|x - 1| = 0
When x < 1
|x - 1| = - x + 1 > 0
Range of f(x) = all real numbers $ \geqslant 0$
$ = \left[ {0,\infty } \right)$

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Question 42 Marks

Find the domain and range of the real function f defined by $f(x) = \sqrt {x - 1} $

Answer

Here f (x) = $\sqrt {x - 1} $, f (x) assumes real values if $x - 1 \geqslant 0 \Rightarrow x \geqslant 1$
$ \Rightarrow x \in \left[ {1,\infty } \right)$
$\therefore$ Domain of f(x) $ = \left[ {1,\infty } \right)$
For $x \geqslant 1,f(x) \geqslant 0$
Range of f (x) = all real numbers $ \geqslant 0$
$ = \left[ {0,\infty } \right)$

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Question 52 Marks

Find the domain of the function $f(x) = \frac{{{x^2} + 2x + 1}}{{{x^2} - 8x + 12}}$.

Answer

Here $f(x) = \frac{{{x^2} + 2x + 1}}{{{x^2} - 8x + 12}}$
f (x) is a rational function of x.
f (x) assumes real values of all x except for those values of x for which
$x^2-8 x+12=0$
$\Rightarrow(x-6)(x-2)=0$
$\Rightarrow x=2,6$
$\therefore$ Domain of function $=R-\{2,6\}$.

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Question 62 Marks

If $f(x)=x^2$, find $\frac{{f(1.1) - f(1)}}{{(1.1 - 1)}}$

Answer

Here $f(x)=x^2$
At $x=1.1$
$\mathrm{f}(1.1)=(1.1)^2=1.21$
$\mathrm{f}(1)=(1)^2=1$
$\therefore \frac{f(1.1)-f(1)}{(1.1-1)}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1$

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Question 72 Marks

If the function t which maps temperature in degree Celcius into temperature in degree Fahrenheit is defined by t(C) = $\frac{9C}{5}$+ 32, then find the value of C, when t(C) = 212.

Answer

Here it is given that, $t(C) =$ $\frac{9 C}{5}$$+ 32$
Put $t(C) = 212$, we get
$212 =$ $\frac{9 C}{5}$ + 32
$\Rightarrow$ $\frac{9 C}{5}$$= 212 - 32$
$\Rightarrow$ $\frac{9 C}{5}$ = 180
$\therefore$ C = $\frac{5 \times 180}{9}$
$C = 5$ $\times$ 20 = 100

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Question 82 Marks

If the function t which maps temperature in degree Celcius into temperature in degree Fahrenheit is defined by t(C) = $\frac{9C}{5}$+ 32, then find t(-10).

Answer

Here it is given that, $t(C) =$ $\frac{9 C}{5}$$+ 32$
Put C = -10, we get
$t(-10) =$ $\frac{9 \times(-10)}{5}$$+ 32$
= $\frac{-9 \times 10}{5}$ $+ 32 = - 9$ $\times$ $2 + 32$
= -18 + 32 = 14

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Question 92 Marks

If the function t which maps temperature in degree Celcius into temperature in degree Fahrenheit is defined by t(C) = $\frac{9C}{5}$+ 32, then find t(28).

Answer

We have, $t(C) =$ $\frac{9 C}{5}$$+ 32$
On putting C = 28, we get
$t(28) = $$\frac{9 \times 28}{5}$ $+ 32 =$ $\frac{252}{5}$ + $\frac{32}{1}$
= $\frac{252+160}{5}$ = $\frac{412}{5}$ = 82.4

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Question 102 Marks

If A $\times$ B = {(p, q), (p, r), (m, q), (m, r)}, find A and B.

Answer

A = first entry elements = {p, m}
B = 2nd entry elements = {q, r}.

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Question 112 Marks

Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find : (A $\times$ B) $\cup$ (A $\times$ C)

Answer

We have, (A $\times$ B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} and (A $\times$ C)
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Now, (A $\times$ B) $\cup$ (A $\times$ C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

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Question 122 Marks

Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find : A $\times$ (B $\cup$ C)

Answer

We have, (B $\cup$ C) = {3, 4, 5, 6},
Therefore, A $\times$ (B $\cup$ C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

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Question 132 Marks

Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find : (A $\times$ B) $\cap$ (A $\times$ C)

Answer

Here we have, A = {1,2,3}, B = {3,4} and C = {4,5,6}
Now (A $\times$ B) = {(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)} and (A $\times$ C) = {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)}
Therefore, (A $\times$ B) $\cap$ (A $\times$ C) = {(1, 4), (2, 4), (3, 4)}

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Question 142 Marks

Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find : A $\times$ (B $\cap$ C)

Answer

Here we have,
A = {1,2,3}, B = {3,4} and C = {4,5,6}
By the definition of the intersection of two sets, (B $\cap$ C) = {4}
Therefore, A $\times$ (B $\cap$ C) = {(1, 4), (2, 4), (3, 4)}

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Question 152 Marks

Find the domain of the function f(x) = $\begin{equation} \frac{x^{2}+3 x+5}{x^{2}-5 x+4} \end{equation}$

Answer

Since $x^2-5 x+4=(x-4)(x-1)$, therefore the function $f$ is defined for all real numbers except at $x=4$ and $x=1$.
Therefore, the domain of $f$ is $R -\{1,4\}$

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Question 162 Marks

Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a linear function from Z into Z. Find f(x).

Answer

Since f is a linear function, f(x) = mx + c.Since (1, 1), (0, – 1) $\in$ R, f (1) = m + c = 1 and f (0) = c = –1.
This gives m = 2 and hence f(x) = 2x – 1

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Question 172 Marks

Let N be the set of natural numbers. Define a real valued function f : $N \rightarrow N$ by f (x) = 2x + 1. Using this definition, complete the table given below

x	1	2	3	4	5	6	7
y	f (1) = ...	f (2) = ...	f (3) = ...	f (4) = ...	f (5) = ...	f (6) = ...	f (7) = ...

Answer

Given function is, f (x) = 2x + 1
The completed table is given by

x	1	2	3	4	5	6	7
y	f (1) = 3	f (2) = 5	f (3) = 7	f (4) = 9	f (5) = 11	f (6) = 13	f (7) =15

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Question 182 Marks

Let N be the set of natural numbers and the relation R be defined on N such that R = {(x, y) : y = 2x, x, y $\in$ N}. What is the domain, codomain and range of R? Is this relation a function?

Answer

Since the relation R is defined on N.Therefore,
The domain of R is the set of natural numbers N. The co-domain is also N. The range is the set of even natural numbers.
Since every natural number n has one and only one image, this relation is a function.

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