OAVS TGT Math Mock Test - 7 - OTET MCQ

A cubic polynomial is a polynomial of the form . A cubic polynomial is a polynomial of degree 3. A univariate cubic polynomial has the form . An equation involving a cubic polynomial is called a cubic equation. A closed-form solution known as the cubic formula exists for the solutions of an arbitrary cubic equation.

We have as the roots which means x + and x- are the factors of the quadratic equation. Multiplying x+ and x- and applying a²-b² we get the equation x²-5.

17/16 cannot be the probability of an event because probability of an event cannot be greater than 1.

In equation x²+x−5 = 0
a = 1,b = 1,c = −5
∴ b²−4ac = (1)²−4×1×(−5) = 1 + 20 = 21
Since b²−4ac > 0 therefore, x²+x−5 = 0 has two distict roots.

Given that the perimeter of one face of a cube is 40 cm. Let each side of the cube be a cm. Each face of a cube is a square. Therefore, the perimeter of one face of the cube is 4a. We can write:4a = 40 cm
a = 10 cmTherefore, the length of each side of the cube is 10 cm. The volume of the cube is given by the formula V = a^3, where a is the length of each side of the cube. Substituting the value of a, we get:V = 10^3 = 1000 cm^3Therefore, the volume of the cube is 1000 cm^3, which corresponds to option B

The mode of a set of data values is the value that appears most often .So mode is the most occurrence of observation. 
Converting the distribution in continuous grouped and applying the formula
Mode= 
Modal class is 400-500
Where L=lower limit of the modal class =400
f1= modal class frequency=32
f2 = just after the modal class frequency=20
f0 = just previous the modal class frequency=20
C = Class size=100
Mode= 
=400+50=450

sin3θ = cos(θ - 26°)

=> cos(90° - 3θ) = cos(θ - 26°)

=> 90° - 3θ = θ - 26°

=> 3θ + θ = 90° + 26°

=> 4θ = 116°

=> θ = 116°/4

=> θ = 29°

As AB = AC so angle ACB = angle(ABC) as angles opposite to equal sides r equal.

In triangle PBC and Triangle QCB we see that:
i) PB = QC (given)
ii) angle(PBC) = (angle)QCB (proved earliar)
iii) BC = BC (common)

So, triangle PBC is congruent to triangle QCB by SAS axiom of congruency.

Here Side of square ABCD = AB = 3 x diameter of circular design
⇒ AB = 3 x (2 x 7) = 42 cm
∴ Area of square = 42 x 42 = 1764 cm²
And Area of one circular design   
And Area of one circular design
∴ Area of 9 circular designs = 154 x 9 = 1386 sq. cm
∴ Area of remaining portion of handkerchief = 1764 - 1386 = 378 sq. cm

The given polynomial is

Also, α and β are the zeros of p(x).

and  α + β = α × β

The probability that the letter is a consonant = (26 - 5)/26 = 21/26.

Let the first prize be a.
The seven prizes form an AP with first term a and common difference d = −20
Now the sum of all seven prizes = Rs. 700

Let required radius be R. Then according to the question,

class mark of 1st interval=212

class mark of 2nd interval=236.5

Step-by-step explanation:

First Class interval=200-224

class mark=upper limit +lower limit divided by 2..

class mark of first interval is =200+224 divided by 2=212

Second class interval=224-249

class mark of second interval =224 +249 divided by 2

answer= 236.5

As the number 2m will always be even, so if we add 1 to it then, the number will always be odd.

The middle most value of the data set is called median.

Given ,g(x)=x+2=0
x=-2

P(x)=x³-2ax²+16
p(-2)=(-2³)-2×a×(-2²)+16
= -8 - 8a +16
= 8 - 8a
⇒ -8a = -8
⇒ a=1
Hence, value of a=1

So option B is correct answer. 

Fif. 9.11
i.e., ∠ACB = 30°
In right triangle ABC, we have

Hence, the height of the pole is 10 m.

One circle measures 360 degrees. The half of the circle is called a semi-circle.
Thus, the measure of semi-circle
= 360/2 = 180 degree.

x = 1

However at x = 1 the expression is not defined.

Hence the above equation has no real roots.

Using the result, HCF × LCM = Product of two natural numbers
 ⇒ LCM (a, b) = 1800/12 = 150

In right triangle ABC,

Hence, the the height of the pole is 10 m.