TS Polycet (Polytechnic) 2024 Previous Question Paper with Answers And Model Papers With Complete Analysis

51) The roots of the quadratic equation x²-4x+4=0 are x² – 4x + 4 = 0 అను వర్గ సమీకరణ మూలాలు

A) 4, 1
B) 2, 2
C) -2, -2
D) 4, 2

View Answer

B) 2, 2

Explanation:To find the roots of the quadratic equation $x^2 - 4x + 4 = 0$ , we can either factor the equation or use the quadratic formula.
Step 1: Try factoring the quadratic equation
The equation is:
$x^2 - 4x + 4 = 0$
We look for two numbers that multiply to give $4$ (the constant term) and add up to $-4$ (the coefficient of $x$ ).
The numbers are $-2$ and $-2$ , since:
$-2 \times -2 = 4 \quad \text{and} \quad -2 + (-2) = -4$
So, we can factor the quadratic equation as:
$(x - 2)(x - 2) = 0$
Step 2: Solve for the roots
Now, solving for $x$ :
$x - 2 = 0 \quad \Rightarrow \quad x = 2$
Thus, the equation has a repeated root: $x = 2$ .
Final Answer:
The roots of the quadratic equation $x^2 - 4x + 4 = 0$ are $2, 2$ .

52) The sum of the roots of the quadratic equation 3x² – 5x + 2 = 0 is 3x² – 5x + 2 = 0 అనే వర్గ సమీకరణ మూలాల మొత్తము

A) $\frac53$
B) $\frac{-5}3$
C) $\frac{-3}5$
D) $\frac35$

View Answer

A) $\frac53$

Explanation:To find the sum of the roots of the quadratic equation $3x^2 - 5x + 2 = 0$ , we can use the formula for the sum of the roots of a quadratic equation.
The general form of a quadratic equation is:
$ax^2 + bx + c = 0$
For this equation, we have:
– $a = 3$
– $b = -5$
– $c = 2$
Sum of the roots:
The sum of the roots of a quadratic equation is given by the formula:
$\text{Sum of the roots} = -\frac{b}{a}$
Substitute the values of $b$ and $a$ :
$\text{Sum of the roots} = -\frac{-5}{3} = \frac{5}{3}$
Final Answer:
The sum of the roots of the equation $3x^2 - 5x + 2 = 0$ is $\frac{5}{3}$ .
The correct answer is:1. $\frac{5}{3}$ .

53) Sum of the areas of two squares is 625 m2. If the difference of their perimeters is 20 m, find the sides of the two squares. రెండు చతురస్రాల వైశాల్యాల మొత్తము 625 చ.మీ. వాని చుట్టు కొలతల బేధము 20 మీ. అయిన, ఆ రెండు చతురస్రాల భుజాలను కనుగొనుము.

A) 20 m, 5 m
B) 15 m, 10 m
C) 20 m, 15 m
D) 25 m, 5 m

View Answer

C) 20 m, 15 m

Explanation:We are given the following information:
– The sum of the areas of two squares is 625 m².
– The difference in their perimeters is 20 m.
Let the sides of the two squares be $x$ and $y$ .
Step 1: Use the sum of areas
The area of a square is given by $\text{side}^2$ .
So, the sum of the areas of the two squares is:
$x^2 + y^2 = 625$
Step 2: Use the difference of perimeters
The perimeter of a square is given by $4 \times \text{side}$ .
So, the difference in the perimeters of the two squares is:
$4x - 4y = 20$
Simplifying this equation:
$x - y = 5$
Step 3: Solve the system of equations
We now have the following system of equations:
– 1. $x^2 + y^2 = 625$
– 2. $x - y = 5$
We can solve this system by substituting $x = y + 5$ (from the second equation) into the first equation.
Substitute $x = y + 5$ into $x^2 + y^2 = 625$ :
$(y + 5)^2 + y^2 = 625$
Expanding the equation:
$(y^2 + 10y + 25) + y^2 = 625$
Simplifying:
$2y^2 + 10y + 25 = 625$
Subtract 625 from both sides:
$2y^2 + 10y - 600 = 0$
Divide the entire equation by 2:
$y^2 + 5y - 300 = 0$
Now, solve this quadratic equation using the quadratic formula:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
For the equation $y^2 + 5y - 300 = 0$ , the values of $a$ , $b$ , and $c$ are:
– $a = 1$
– $b = 5$
– $c = -300$
Substitute these values into the quadratic formula:
$y = \frac{-5 \pm \sqrt{5^2 - 4(1)(-300)}}{2(1)}$
$y = \frac{-5 \pm \sqrt{25 + 1200}}{2}$
$y = \frac{-5 \pm \sqrt{1225}}{2}$
$y = \frac{-5 \pm 35}{2}$
So, we have two possible solutions for $y$ :
– $y = \frac{-5 + 35}{2} = \frac{30}{2} = 15$
– $y = \frac{-5 - 35}{2} = \frac{-40}{2} = -20$ (which is not possible because side lengths cannot be negative)
Therefore, $y = 15$ .
Step 4: Find $x$
Since $x - y = 5$ , we substitute $y = 15$ :
$x - 15 = 5$
$x = 20$
Final Answer:
The sides of the two squares are 20 m and 15 m.

54) How many two digit numbers are divisible by 3? 3 చే భాగించబడే రెండంకెల సంఖ్యలు ఎన్ని?

A) 25
B) 28
C) 30
D) 36

View Answer

C) 30

Explanation:To find how many two-digit numbers are divisible by 3, we follow these steps:
Step 1: Find the smallest and largest two-digit numbers divisible by 3.
– The smallest two-digit number is 10, and the largest is 99.
– We need to find the smallest and largest two-digit numbers divisible by 3.
Smallest two-digit number divisible by 3:
– Divide 10 by 3: $10 \div 3 = 3.33$ , so the next multiple of 3 is 12.
– Hence, the smallest two-digit number divisible by 3 is 12.
Largest two-digit number divisible by 3:
– Divide 99 by 3: $99 \div 3 = 33$ , so 99 is divisible by 3.
– Hence, the largest two-digit number divisible by 3 is 99.
Step 2: Count the numbers divisible by 3.
The numbers divisible by 3 in the range from 12 to 99 form an arithmetic sequence where:
– First term $a_1 = 12$
– Common difference $d = 3$
– Last term $a_n = 99$
We can use the formula for the nth term of an arithmetic sequence:
$a_n = a_1 + (n - 1) \cdot d$
Substitute the known values:
$99 = 12 + (n - 1) \cdot 3$
Solve for $n$ :
$99 - 12 = (n - 1) \cdot 3$
$87 = (n - 1) \cdot 3$
$n - 1 = \frac{87}{3} = 29$
$n = 30$
Final Answer:
There are 30 two-digit numbers divisible by 3.
So, the correct answer is:3. 30.

55) If P(x) = 11x⁸ − 5x⁶ + 4x⁴ – 7x² +6, then the degree of P(x) is P(x) = 11x⁸ − 5x⁶ + 4x⁴ – 7x² +6, అయిన, P(x) యొక్క పరిమాణము

A) 8
B) 6
C) 4
D) 2

View Answer

A) 8

Explanation:The degree of a polynomial is the highest power of $x$ in the polynomial.
Given the polynomial:
$P(x) = 11x^8 - 5x^6 + 4x^4 - 7x^2 + 6$
Let’s look at the powers of $x$ :
– The highest power of $x$ is $x^8$ (from the term $11x^8$ ).
Therefore, the degree of $P(x)$ is 8.
Final Answer:
The degree of $P(x)$ is 8.
So, the correct answer is:1. 8.

Spread the love

Leave a Reply Cancel reply