Solved: First Mid Term Test - 2024 | Standard X Mathematics

FIRST MID TERM TEST - 2024

Standard X

MATHEMATICS

Time: 1.30 hrs

Reg.No.

Marks: 50

Part - I

I. Choose the correct answer: $7 \times 1 = 7$

If $A = \{a, b, p\}$, $B = \{2, 3\}$, $C = \{p, q, r, s\}$, then $n[(A \cup C) \times B]$ is
- a) 8
- b) 20
- c) 12
- d) 16
View Solution

Step 1: Find the union of sets A and C.
Given $A = \{a, b, p\}$ and $C = \{p, q, r, s\}$.
The union $A \cup C$ contains all unique elements from both sets.
$A \cup C = \{a, b, p, q, r, s\}$.

Step 2: Find the number of elements in $A \cup C$ and B.
Number of elements in $A \cup C$, denoted by $n(A \cup C)$, is 6.
Given $B = \{2, 3\}$. Number of elements in B, denoted by $n(B)$, is 2.

Step 3: Calculate $n[(A \cup C) \times B]$.
The number of elements in the Cartesian product of two sets is the product of the number of elements in each set.
$n[(A \cup C) \times B] = n(A \cup C) \times n(B)$
$n[(A \cup C) \times B] = 6 \times 2 = 12$.

c) 12 is the correct answer.
If $f: A \rightarrow B$ is a bijective function and if $n(B) = 7$, then $n(A)$ is equal to
- a) 7
- b) 49
- c) 1
- d) 14
View Solution

Step 1: Understand the definition of a bijective function.
A function is bijective if it is both injective (one-to-one) and surjective (onto).
- Injective means every element in the domain A maps to a unique element in the codomain B.
- Surjective means every element in the codomain B is mapped to by at least one element from the domain A.

Step 2: Relate bijectivity to the number of elements.
For a function between two finite sets to be bijective, there must be a perfect pairing between the elements of the two sets. This is only possible if the number of elements in the domain is equal to the number of elements in the codomain.
Therefore, if $f: A \rightarrow B$ is bijective, then $n(A) = n(B)$.

Step 3: Apply the given values.
We are given that $n(B) = 7$.
Since the function is bijective, $n(A) = n(B) = 7$.

a) 7 is the correct answer.
$f(x) = (x + 1)^3 - (x - 1)^3$ represents a function which is
- a) linear
- b) cubic
- c) reciprocal
- d) quadratic
View Solution

Step 1: Recall the binomial expansion formulas.
We need the formulas for $(a+b)^3$ and $(a-b)^3$.
$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

Step 2: Expand the terms in the function.
For $(x+1)^3$, let $a=x$ and $b=1$:
$(x+1)^3 = x^3 + 3(x^2)(1) + 3(x)(1^2) + 1^3 = x^3 + 3x^2 + 3x + 1$
For $(x-1)^3$, let $a=x$ and $b=1$:
$(x-1)^3 = x^3 - 3(x^2)(1) + 3(x)(1^2) - 1^3 = x^3 - 3x^2 + 3x - 1$

Step 3: Substitute the expansions back into $f(x)$ and simplify.
$f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)$
$f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1$
Combine like terms:
$f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1)$
$f(x) = 0 + 6x^2 + 0 + 2$
$f(x) = 6x^2 + 2$

Step 4: Identify the type of function.
The simplified function is a polynomial of degree 2 (the highest power of x is 2). A polynomial of degree 2 is called a quadratic function.

d) quadratic is the correct answer.
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
- a) 2025
- b) 5220
- c) 5025
- d) 2520
View Solution

Step 1: Understand the problem.
We need to find the Least Common Multiple (LCM) of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Step 2: Find the prime factorization of each number from 1 to 10.
1 = 1
2 = 2
3 = 3
4 = $2^2$
5 = 5
6 = $2 \times 3$
7 = 7
8 = $2^3$
9 = $3^2$
10 = $2 \times 5$

Step 3: Calculate the LCM.
To find the LCM, we take the highest power of each prime factor present in the numbers.
The prime factors are 2, 3, 5, and 7.
- Highest power of 2 is $2^3$ (from 8).
- Highest power of 3 is $3^2$ (from 9).
- Highest power of 5 is $5^1$ (from 5 and 10).
- Highest power of 7 is $7^1$ (from 7).

Step 4: Multiply these highest powers together.
LCM = $2^3 \times 3^2 \times 5^1 \times 7^1$
LCM = $8 \times 9 \times 5 \times 7$
LCM = $72 \times 35$
LCM = 2520

d) 2520 is the correct answer.
Given $F_1 = 1$, $F_2 = 3$, $F_n = F_{n-1} + F_{n-2}$ then $F_5$ is
- a) 3
- b) 5
- c) 8
- d) 11
View Solution

Step 1: Understand the recurrence relation.
The formula $F_n = F_{n-1} + F_{n-2}$ means each term is the sum of the two preceding terms. We are given the first two terms to start the sequence.

Step 2: List the given terms.
$F_1 = 1$
$F_2 = 3$

Step 3: Calculate $F_3$ using the formula.
For n=3: $F_3 = F_{3-1} + F_{3-2} = F_2 + F_1$
$F_3 = 3 + 1 = 4$

Step 4: Calculate $F_4$.
For n=4: $F_4 = F_{4-1} + F_{4-2} = F_3 + F_2$
$F_4 = 4 + 3 = 7$

Step 5: Calculate $F_5$.
For n=5: $F_5 = F_{5-1} + F_{5-2} = F_4 + F_3$
$F_5 = 7 + 4 = 11$

d) 11 is the correct answer.
The solution of the system $x + y - 3z = -6$, $-7y + 7z = 7$, $3z = 9$ is
- a) $x = 1, y = 2, z = 3$
- b) $x = -1, y = 2, z = 3$
- c) $x = -1, y = -2, z = 3$
- d) $x = -1, y = -2, z = -3$
View Solution

Step 1: Solve the third equation for z.
The system of equations is:
(1) $x + y - 3z = -6$
(2) $-7y + 7z = 7$
(3) $3z = 9$
From equation (3):
$3z = 9 \implies z = \frac{9}{3} \implies z = 3$

Step 2: Substitute the value of z into the second equation to find y.
Substitute $z=3$ into equation (2):
$-7y + 7(3) = 7$
$-7y + 21 = 7$
$-7y = 7 - 21$
$-7y = -14$
$y = \frac{-14}{-7} \implies y = 2$

Step 3: Substitute the values of y and z into the first equation to find x.
Substitute $y=2$ and $z=3$ into equation (1):
$x + (2) - 3(3) = -6$
$x + 2 - 9 = -6$
$x - 7 = -6$
$x = -6 + 7 \implies x = 1$

The solution is $x = 1, y = 2, z = 3$.
a) $x = 1, y = 2, z = 3$ is the correct answer.
In $\Delta LMN$, $\angle L = 60^\circ$, $\angle M = 50^\circ$. If $\Delta LMN \sim \Delta PQR$, then the value of $\angle R$ is
- a) $40^\circ$
- b) $70^\circ$
- c) $30^\circ$
- d) $110^\circ$
View Solution

Step 1: Use the property of the sum of angles in a triangle.
The sum of the angles in any triangle is $180^\circ$. For $\Delta LMN$:
$\angle L + \angle M + \angle N = 180^\circ$
We are given $\angle L = 60^\circ$ and $\angle M = 50^\circ$.
$60^\circ + 50^\circ + \angle N = 180^\circ$
$110^\circ + \angle N = 180^\circ$
$\angle N = 180^\circ - 110^\circ = 70^\circ$

Step 2: Use the property of similar triangles.
When two triangles are similar, their corresponding angles are equal.
Given $\Delta LMN \sim \Delta PQR$, the correspondence is:
$\angle L = \angle P$
$\angle M = \angle Q$
$\angle N = \angle R$

Step 3: Find the value of $\angle R$.
From Step 1, we found that $\angle N = 70^\circ$.
Since $\angle N = \angle R$, we have $\angle R = 70^\circ$.

b) $70^\circ$ is the correct answer.

Part - II

II. Answer any 5 questions. (Q.No.14 is compulsory) $5 \times 2 = 10$

If $B \times A = \{(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)\}$, find A and B.

View Solution

Step 1: Understand the Cartesian Product $B \times A$.
The Cartesian product $B \times A$ is the set of all ordered pairs $(b, a)$ where $b \in B$ and $a \in A$.

Step 2: Identify Set B.
Set B is the set of all the first elements (coordinates) of the ordered pairs in $B \times A$.
First elements are: -2, -2, 0, 0, 3, 3.
Listing the unique elements, we get $B = \{-2, 0, 3\}$.

Step 3: Identify Set A.
Set A is the set of all the second elements (coordinates) of the ordered pairs in $B \times A$.
Second elements are: 3, 4, 3, 4, 3, 4.
Listing the unique elements, we get $A = \{3, 4\}$.

The sets are:
A = \{3, 4\}
B = \{-2, 0, 3\}
A relation 'f' is defined by $f(x) = x^2 - 2$ where $x \in \{-2, -1, 0, 3\}$.
i) List the elements of f.

ii) Is f a function?

View Solution

i) List the elements of f.
We need to evaluate $f(x) = x^2 - 2$ for each value in the domain $\{-2, -1, 0, 3\}$.
- For $x = -2$: $f(-2) = (-2)^2 - 2 = 4 - 2 = 2$. The ordered pair is $(-2, 2)$.
- For $x = -1$: $f(-1) = (-1)^2 - 2 = 1 - 2 = -1$. The ordered pair is $(-1, -1)$.
- For $x = 0$: $f(0) = (0)^2 - 2 = 0 - 2 = -2$. The ordered pair is $(0, -2)$.
- For $x = 3$: $f(3) = (3)^2 - 2 = 9 - 2 = 7$. The ordered pair is $(3, 7)$.
The elements of f are the set of these ordered pairs.

ii) Is f a function?
A relation is a function if every element in the domain is mapped to exactly one element in the range. In our case, each input value ($-2, -1, 0, 3$) has produced a single, unique output value ($2, -1, -2, 7$ respectively). Therefore, the relation 'f' is a function.

i) The elements of f are: $f = \{(-2, 2), (-1, -1), (0, -2), (3, 7)\}$.
ii) Yes, f is a function.
'a' and 'b' are two positive integers such that $a^b \times b^a = 800$. Find 'a' and 'b'.

View Solution

Step 1: Find the prime factorization of 800.
$800 = 8 \times 100$
$800 = 2^3 \times 10^2$
$800 = 2^3 \times (2 \times 5)^2$
$800 = 2^3 \times 2^2 \times 5^2$
$800 = 2^{3+2} \times 5^2$
$800 = 2^5 \times 5^2$

Step 2: Compare the prime factorization with the given equation.
We have the equation $a^b \times b^a = 800$.
We found that $800 = 2^5 \times 5^2$.
So, $a^b \times b^a = 2^5 \times 5^2$.

Step 3: Determine 'a' and 'b' by inspection.
By comparing the bases and exponents, we can see a direct match if we let:
$a = 2$ and $b = 5$.
Let's check this: $a^b \times b^a = 2^5 \times 5^2 = 32 \times 25 = 800$. This works.
Alternatively, let $a = 5$ and $b = 2$.
Let's check this: $a^b \times b^a = 5^2 \times 2^5 = 25 \times 32 = 800$. This also works.

Since the question asks for 'a' and 'b' without specifying order, the values are 2 and 5.
a = 2, b = 5 (or a = 5, b = 2).
Find the sum to infinity of $9 + 3 + 1 + \dots$.

View Solution

Step 1: Identify the type of series.
This is a Geometric Progression (GP) because the ratio between consecutive terms is constant.
First term, $a = 9$.

Step 2: Calculate the common ratio (r).
$r = \frac{\text{second term}}{\text{first term}} = \frac{3}{9} = \frac{1}{3}$.
(Check: $r = \frac{\text{third term}}{\text{second term}} = \frac{1}{3}$). The common ratio is constant.

Step 3: Check the condition for the sum to infinity.
The formula for the sum to infinity of a GP, $S_\infty = \frac{a}{1-r}$, is valid only if $|r| < 1$.
Here, $r = \frac{1}{3}$, and $|\frac{1}{3}| < 1$, so we can use the formula.

Step 4: Apply the formula.
$S_\infty = \frac{a}{1-r} = \frac{9}{1 - \frac{1}{3}}$
$S_\infty = \frac{9}{\frac{3}{3} - \frac{1}{3}} = \frac{9}{\frac{2}{3}}$
$S_\infty = 9 \times \frac{3}{2} = \frac{27}{2}$

The sum to infinity is $\frac{27}{2}$ or 13.5.
Find $a_6$ and $a_{13}$ of the sequence whose n^th term is given by $a_n = \frac{5n}{n+2}$.

View Solution

Step 1: Calculate $a_6$.
To find the 6th term, substitute $n=6$ into the formula $a_n = \frac{5n}{n+2}$.
$a_6 = \frac{5(6)}{6+2} = \frac{30}{8}$
Simplifying the fraction: $a_6 = \frac{15}{4}$.

Step 2: Calculate $a_{13}$.
To find the 13th term, substitute $n=13$ into the formula.
$a_{13} = \frac{5(13)}{13+2} = \frac{65}{15}$
Simplifying the fraction by dividing the numerator and denominator by 5: $a_{13} = \frac{13}{3}$.

$a_6 = \frac{15}{4}$ and $a_{13} = \frac{13}{3}$.
If $\Delta ABC \sim \Delta DEF$ such that BC = 3 cm, EF = 4 cm and area of $\Delta ABC = 54\text{ cm}^2$, find the area of $\Delta DEF$.

View Solution

Step 1: Recall the theorem on the areas of similar triangles.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. $$ \frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \left(\frac{BC}{EF}\right)^2 $$

Step 2: Substitute the given values into the formula.
Area($\Delta ABC$) = 54 cm²
BC = 3 cm
EF = 4 cm
$$ \frac{54}{\text{Area}(\Delta DEF)} = \left(\frac{3}{4}\right)^2 $$

Step 3: Solve for the area of $\Delta DEF$.
$$ \frac{54}{\text{Area}(\Delta DEF)} = \frac{9}{16} $$ Now, cross-multiply to solve for the unknown area:
$ \text{Area}(\Delta DEF) \times 9 = 54 \times 16 $
$ \text{Area}(\Delta DEF) = \frac{54 \times 16}{9} $
Since $54 \div 9 = 6$:
$ \text{Area}(\Delta DEF) = 6 \times 16 = 96 $

The area of $\Delta DEF$ is 96 cm².
Find the sum of $1 + 8 + 27 + \dots + 1000$.

View Solution

Step 1: Identify the pattern in the series.
The terms of the series can be written as cubes of natural numbers:
$1 = 1^3$
$8 = 2^3$
$27 = 3^3$
...
$1000 = 10^3$
So, the series is $1^3 + 2^3 + 3^3 + \dots + 10^3$.

Step 2: Recall the formula for the sum of the first n cubes.
The sum of the cubes of the first n natural numbers is given by the formula: $$ \sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 $$

Step 3: Apply the formula with the correct value of n.
In this series, the last term is $10^3$, so $n = 10$.
Sum = $ \left[\frac{10(10+1)}{2}\right]^2 $
Sum = $ \left[\frac{10 \times 11}{2}\right]^2 $
Sum = $ \left[\frac{110}{2}\right]^2 $
Sum = $ (55)^2 $

Step 4: Calculate the final value.
$ 55^2 = 55 \times 55 = 3025 $

The sum of the series is 3025.

Part - III

III. Answer any 5 questions. (Q.No.21 is compulsory) $5 \times 5 = 25$

Given $A = \{1,2,3\}$, $B = \{2,3,5\}$, $C = \{3,4\}$, $D = \{1,3,5\}$, check if $(A \cap C) \times (B \cap D) = (A \times B) \cap (C \times D)$ is true.

View Solution

Step 1: Calculate the Left Hand Side (LHS).
LHS = $(A \cap C) \times (B \cap D)$.
First, find the intersections:
- $A \cap C = \{1,2,3\} \cap \{3,4\} = \{3\}$
- $B \cap D = \{2,3,5\} \cap \{1,3,5\} = \{3,5\}$
Now, find the Cartesian product of these results:
$(A \cap C) \times (B \cap D) = \{3\} \times \{3,5\} = \{(3,3), (3,5)\}$

Step 2: Calculate the Right Hand Side (RHS).
RHS = $(A \times B) \cap (C \times D)$.
First, find the Cartesian products:
- $A \times B = \{1,2,3\} \times \{2,3,5\} = \{(1,2), (1,3), (1,5), (2,2), (2,3), (2,5), (3,2), (3,3), (3,5)\}$
- $C \times D = \{3,4\} \times \{1,3,5\} = \{(3,1), (3,3), (3,5), (4,1), (4,3), (4,5)\}$
Now, find the intersection of these two sets of ordered pairs:
$(A \times B) \cap (C \times D) = \{(3,3), (3,5)\}$ (These are the only pairs common to both sets).

Step 3: Compare LHS and RHS.
LHS = $\{(3,3), (3,5)\}$
RHS = $\{(3,3), (3,5)\}$
Since LHS = RHS, the statement is true.

Yes, the statement $(A \cap C) \times (B \cap D) = (A \times B) \cap (C \times D)$ is true.
If $f(x) = 2x + 3$, $g(x) = 1 - 2x$ and $h(x) = 3x$, prove that $fo(goh) = (fog)oh$.

View Solution

Step 1: Calculate the Left Hand Side (LHS): $fo(goh)$
First, find the composition $goh$.
$goh(x) = g(h(x)) = g(3x)$
Substitute $3x$ into the function $g(x) = 1 - 2x$:
$goh(x) = 1 - 2(3x) = 1 - 6x$.
Next, find $fo(goh)$.
$fo(goh)(x) = f(goh(x)) = f(1 - 6x)$
Substitute $(1 - 6x)$ into the function $f(x) = 2x + 3$:
$f(1 - 6x) = 2(1 - 6x) + 3 = 2 - 12x + 3 = 5 - 12x$.
So, LHS = $5 - 12x$.

Step 2: Calculate the Right Hand Side (RHS): $(fog)oh$
First, find the composition $fog$.
$fog(x) = f(g(x)) = f(1 - 2x)$
Substitute $(1 - 2x)$ into the function $f(x) = 2x + 3$:
$f(1 - 2x) = 2(1 - 2x) + 3 = 2 - 4x + 3 = 5 - 4x$.
Next, find $(fog)oh$.
$(fog)oh(x) = (fog)(h(x)) = (fog)(3x)$
Substitute $3x$ into the function $(fog)(x) = 5 - 4x$:
$(fog)(3x) = 5 - 4(3x) = 5 - 12x$.
So, RHS = $5 - 12x$.

Step 3: Compare LHS and RHS.
LHS = $5 - 12x$
RHS = $5 - 12x$
Since LHS = RHS, the associative property of function composition is proven for these functions.

We have shown that $fo(goh)(x) = (fog)oh(x) = 5 - 12x$. Hence, proved.
Use Euclid's division algorithm to find the HCF of 84, 90 and 120.

View Solution

Step 1: Find the HCF of the first two numbers (84 and 90).
We use Euclid's Division Lemma: $a = bq + r$, where $0 \le r < b$.
Let $a = 90$ and $b = 84$.
$90 = 84 \times 1 + 6$. The remainder is 6.
Now, let $a = 84$ and $b = 6$.
$84 = 6 \times 14 + 0$. The remainder is 0.
The HCF is the last non-zero remainder, which is 6. So, HCF(84, 90) = 6.

Step 2: Find the HCF of the result from Step 1 and the third number (120).
We need to find HCF(6, 120).
Let $a = 120$ and $b = 6$.
$120 = 6 \times 20 + 0$. The remainder is 0.
The HCF is the divisor when the remainder is 0, which is 6. So, HCF(6, 120) = 6.

The HCF of 84, 90, and 120 is 6.
Find the sum to n terms of the series: $5 + 55 + 555 + \dots$

View Solution

Step 1: Let $S_n$ be the sum and take the common factor out.
$S_n = 5 + 55 + 555 + \dots$ (n terms)
$S_n = 5(1 + 11 + 111 + \dots \text{ (n terms)})$

Step 2: Multiply and divide by 9.
This is a standard trick to convert the terms into a more manageable form.
$S_n = \frac{5}{9} (9 + 99 + 999 + \dots \text{ (n terms)})$

Step 3: Rewrite each term in the parenthesis.
$S_n = \frac{5}{9} ((10 - 1) + (100 - 1) + (1000 - 1) + \dots \text{ (n terms)})$
$S_n = \frac{5}{9} ((10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1))$

Step 4: Separate the terms into two series.
$S_n = \frac{5}{9} [(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{ (n times)})]$

Step 5: Sum each series.
The first series is a Geometric Progression (GP) with $a=10, r=10, n=n$.
The sum of this GP is $S_{GP} = \frac{a(r^n - 1)}{r-1} = \frac{10(10^n - 1)}{10-1} = \frac{10}{9}(10^n - 1)$.
The second series is simply the sum of 'n' ones, which is $n$.

Step 6: Combine the results and simplify.
$S_n = \frac{5}{9} \left[ \frac{10}{9}(10^n - 1) - n \right]$
Distribute the $\frac{5}{9}$:
$S_n = \frac{50}{81}(10^n - 1) - \frac{5n}{9}$

The sum to n terms is $S_n = \frac{50}{81}(10^n - 1) - \frac{5n}{9}$.
Solve the following system of linear equations in three variables: $3x - 2y + z = 2$, $2x + 3y - z = 5$, $x + y + z = 6$.

View Solution

Step 1: Label the equations.
(1) $3x - 2y + z = 2$
(2) $2x + 3y - z = 5$
(3) $x + y + z = 6$

Step 2: Eliminate one variable (e.g., z) using two pairs of equations.
Add equation (1) and (2) to eliminate z:
$(3x - 2y + z) + (2x + 3y - z) = 2 + 5$
$5x + y = 7$ --- (4)
Add equation (2) and (3) to eliminate z:
$(2x + 3y - z) + (x + y + z) = 5 + 6$
$3x + 4y = 11$ --- (5)

Step 3: Solve the new system of two equations.
We have a system with equations (4) and (5):
(4) $5x + y = 7$
(5) $3x + 4y = 11$
From equation (4), express y in terms of x: $y = 7 - 5x$.
Substitute this into equation (5):
$3x + 4(7 - 5x) = 11$
$3x + 28 - 20x = 11$
$-17x = 11 - 28$
$-17x = -17 \implies x = 1$

Step 4: Back-substitute to find the other variables.
Substitute $x=1$ into the expression for y:
$y = 7 - 5(1) = 7 - 5 \implies y = 2$.
Substitute $x=1$ and $y=2$ into one of the original equations (e.g., (3)):
$x + y + z = 6 \implies 1 + 2 + z = 6 \implies 3 + z = 6 \implies z = 3$.

The solution is x = 1, y = 2, z = 3.
The sum of three consecutive terms that are in an A.P is 27 and their product is 288. Find the three terms.

View Solution

Step 1: Represent the three consecutive terms in an A.P.
Let the three terms be $a-d$, $a$, and $a+d$, where 'a' is the middle term and 'd' is the common difference.

Step 2: Use the sum information to find 'a'.
Given that the sum is 27:
$(a-d) + a + (a+d) = 27$
$3a = 27 \implies a = \frac{27}{3} \implies a = 9$.
So, the middle term is 9. The terms are $9-d$, 9, $9+d$.

Step 3: Use the product information to find 'd'.
Given that the product is 288:
$(9-d) \times 9 \times (9+d) = 288$
Divide both sides by 9:
$(9-d)(9+d) = \frac{288}{9} = 32$
Using the difference of squares formula $(x-y)(x+y) = x^2 - y^2$:
$9^2 - d^2 = 32$
$81 - d^2 = 32$
$d^2 = 81 - 32 = 49$
$d = \pm\sqrt{49} \implies d = \pm 7$.

Step 4: Find the three terms for both values of 'd'.
- If $d = 7$: The terms are $a-d = 9-7=2$, $a=9$, $a+d = 9+7=16$. The terms are 2, 9, 16.
- If $d = -7$: The terms are $a-d = 9-(-7)=16$, $a=9$, $a+d = 9+(-7)=2$. The terms are 16, 9, 2.

In both cases, the set of three terms is the same. The three terms are 2, 9, and 16.
A function $f: [-5, 9] \rightarrow R$ is defined as follows: $$ f(x) = \begin{cases} 6x + 1 & \text{if } -5 \le x < 2 \\ 5x^2 - 1 & \text{if } 2 \le x < 6 \\ 3x - 4 & \text{if } 6 \le x \le 9 \end{cases} $$ Find:
i) $f(7) - f(1)$

ii) $\frac{2f(-2) - f(6)}{f(4) + f(-2)}$

View Solution

Step 1: Calculate the individual function values needed.
We need to find $f(7), f(1), f(-2), f(6), f(4)$ by determining which interval each x-value falls into.
- To find $f(7)$: $x=7$ is in the interval $6 \le x \le 9$. Use $f(x) = 3x - 4$.
$f(7) = 3(7) - 4 = 21 - 4 = 17$.
- To find $f(1)$: $x=1$ is in the interval $-5 \le x < 2$. Use $f(x) = 6x + 1$.
$f(1) = 6(1) + 1 = 7$.
- To find $f(-2)$: $x=-2$ is in the interval $-5 \le x < 2$. Use $f(x) = 6x + 1$.
$f(-2) = 6(-2) + 1 = -12 + 1 = -11$.
- To find $f(6)$: $x=6$ is in the interval $6 \le x \le 9$. Use $f(x) = 3x - 4$.
$f(6) = 3(6) - 4 = 18 - 4 = 14$.
- To find $f(4)$: $x=4$ is in the interval $2 \le x < 6$. Use $f(x) = 5x^2 - 1$.
$f(4) = 5(4^2) - 1 = 5(16) - 1 = 80 - 1 = 79$.

i) Calculate $f(7) - f(1)$.
Using the values from Step 1:
$f(7) - f(1) = 17 - 7 = 10$.

ii) Calculate $\frac{2f(-2) - f(6)}{f(4) + f(-2)}$.
Substitute the values from Step 1 into the expression.
- Numerator: $2f(-2) - f(6) = 2(-11) - 14 = -22 - 14 = -36$.
- Denominator: $f(4) + f(-2) = 79 + (-11) = 68$.
- The fraction is $\frac{-36}{68}$.
Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.
$\frac{-36 \div 4}{68 \div 4} = \frac{-9}{17}$.

i) $f(7) - f(1) = 10$
ii) $\frac{2f(-2) - f(6)}{f(4) + f(-2)} = -\frac{9}{17}$

Part - IV

IV. Answer the following question. $1 \times 8 = 8$

a) Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{3}{5}$ of the corresponding sides of the triangle PQR (Scale factor $ < 1 $).
(OR)
b) Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{7}{3}$ of the corresponding sides of the triangle PQR (Scale factor $ > 1 $).

View Solution

Since this is a construction question, the solution provides the step-by-step method for drawing the required triangle.

Solution for a) Scale Factor $\frac{3}{5}$ (< 1)

Step 1: Construct the given triangle $\Delta PQR$ with any measurements.

Step 2: Draw a ray QX starting from Q, making an acute angle with the side QR, on the side opposite to vertex P.

Step 3: On the ray QX, mark 5 (the larger number in the scale factor) points $Q_1, Q_2, Q_3, Q_4, Q_5$ at equal intervals. (i.e., $QQ_1 = Q_1Q_2 = \dots = Q_4Q_5$).

Step 4: Join the last point $Q_5$ to R. This forms the line segment $Q_5R$.

Step 5: From point $Q_3$ (the smaller number in the scale factor), draw a line parallel to $Q_5R$. This parallel line will intersect the side QR at a point. Name this point R'.

Step 6: From point R', draw a line parallel to the side PR. This parallel line will intersect the side PQ at a point. Name this point P'.

Step 7: The triangle $\Delta P'QR'$ is the required similar triangle, whose sides are $\frac{3}{5}$ of the corresponding sides of $\Delta PQR$.

Solution for b) Scale Factor $\frac{7}{3}$ (> 1)

Step 1: Construct the given triangle $\Delta PQR$ with any measurements.

Step 2: Draw a ray QX starting from Q, making an acute angle with the side QR, on the side opposite to vertex P.

Step 3: On the ray QX, mark 7 (the larger number in the scale factor) points $Q_1, Q_2, \dots, Q_7$ at equal intervals.

Step 4: Join the point $Q_3$ (the smaller number in the scale factor) to R. This forms the line segment $Q_3R$.

Step 5: Extend the sides QR and QP of the original triangle.

Step 6: From point $Q_7$, draw a line parallel to $Q_3R$. This parallel line will intersect the extended side QR at a point. Name this point R'.

Step 7: From point R', draw a line parallel to the side PR. This parallel line will intersect the extended side QP at a point. Name this point P'.

Step 8: The triangle $\Delta P'QR'$ is the required similar triangle, whose sides are $\frac{7}{3}$ of the corresponding sides of $\Delta PQR$.

🧮 Maths Question Papers (EM & TM) 10th Standard - 1st Mid Term Exam 2024 - Original Question Papers & Answer Keys 1st Mid Term Test 2024 - Original Question Paper | Theni District | Mr. M. Senthikumar (English Medium)

FIRST MID TERM TEST - 2024

Standard X

MATHEMATICS

Solution for a) Scale Factor \(\frac{3}{5}\) (< 1)

Solution for b) Scale Factor \(\frac{7}{3}\) (> 1)