MPTET Varg 1 Mathematics Mock Test - 1 Free Online Test 2026

सही उत्तर: रेखांकित शब्द का विलोम है विमुख।

• सम्मुख= मुंह के सामने
• विमुख= मुंह के पीछे

स्पष्टीकरण

अन्यसंबंधित बिंदु

''Learning and knowledge are the greatest assets of a man'

• उक्त वाक्य का सही हिंदी रूपांतरण होगा -."सीख और ज्ञान मनुष्य की सबसे बड़ी सम्पत्ति है।"
• अन्य विकल्प अनुचित है।

स्पष्टीकरण
वाक्य के महत्वपूर्ण शब्दों का हिंदी अनुवाद -

• Learning - सीखना
• knowledge - ज्ञान
• greatest - महानतम।
• assets - सम्पति

अन्यसंबंधित बिंदु
assets शब्द से बने अन्य वाक्य -
The river is a real asset. - नदी एक वास्तविक संपत्ति है।

सही उत्तर 'संश्लिष्ट - विश्लिष्ट' है।

• दिए गए विकल्पों में से, 'संश्लिष्ट - विश्लिष्ट' परस्पर विलोम शब्द है।
• संश्लिष्ट का अर्थ: जुड़ा हुआ।
• विश्लिष्ट का अर्थ: अलग किया हुआ।

स्पष्टीकरण
अन्य विकल्पों का विश्लेषण:

अन्यसंबंधित बिंदु

• विलोमार्थक शब्द/ विलोम: 'विलोम’ शब्द का अर्थ है- उल्टा या विपरीत।
• अत: किसी शब्द का उल्टा अर्थ व्यक्त करने वाला शब्द विलोमार्थक शब्द कहलाता है।
• विलोमार्थक शब्दों को विपर्यायवाची, प्रतिलोमार्थक और विलोम शब्द भी कहते हैं।

कुछ महत्वपूर्ण विलोम शब्द:-

Explanation:

• Civil Disobedience movement is one of the most popular events in the history of the Independence movements of India.
• It was also called passive resistance.
• Civil Disobedience is defined as the refusal to obey the demands or commands of a government by the citizens.
• Civil Disobedience movement was started by Mahatama Gandhi in 1930.
• It was started when Gandhi left Sabarmati Ashram at Ahemadabad along with his 78 followers of Ashram for Dandi on 12 March 1930.
• After reaching Dandi, Gandhi broke the salt law.
• After this it hurts the monopoly of the British Government to make salt.
• The civil disobedience movement get Nationwide acceptance at that time.
• This movement becomes symbolic of people's defiance of the British government policies.

Thus, the Civil Disobedience movement took place in the year 1930.

Person's with Disabilities Act (PWD Act) was passed in 1995 but it came into force by 1996. It narrates about equal opportunities, protection fo rights and full participation for the Persons with Disabilities.

Explanation

• The children with special needs are provided Education, Vocational Training and Employment opportunities under the PWD act 1995.
• It provides 3% reservation for person with disabilities in the government services who are having the following disabilities:

1. Visual Impairment
2. Hearing Impairment
3. Mental Retardation
4. Mental Illness
5. Locompter Disabilities
6. Leprosy
7. Short Sightedness

• It also have the provision of Public building, rail compartments, buses, ships and aircrafts will be designed to give easy access to disabled people.
• Surveys, investigations and research shall be conducted to ascertain the cause of occurrence of disabilities.

Hence, it can be concluded that Disabilities covered under PWD Act, 1995 are 7.

Explanations:

Euclid's Third Postulate States that A circle can be drawn with any centre and any radius.

Hence, option 1, 2 and 3 are incorrect.

∴ option 4 is correct.

Given:
PQRS is trapezium having PQ || RS and RS = 6 cm and MN = 9 cm.
Formula used:
For a trapezium,
MN = 1/2 x (PQ/RS) 
Where, M and N are midpoints of diagonals of trapezium, i.e. PS and QR .
Calculation:
MN = 9 = 1/2 x (PQ/6) 
or 18 = PQ + 6
or PQ = 12 cm

∴ The answer is 12 cm .

Given:

A(1, 3), B(3, 9) and C(0.k) are collinear.

Concept:

If the points are collinear, then they will not form a triangle.

⇒ ΔABC = 0

Calculation:

We have,

x₁ = 1, y₁ = 3, x₂ = 3, y₂ = 9, x₃ = 0, y₃ = k

Putting the values

⇒

⇒ 1(9 - k) + 3(k - 3) + 0 = 0

⇒ 9 - k + 3k - 9 = 0

⇒ 2k + 0 = 0

⇒ 2k = 0

⇒ k = 0

∴ The value of k is 0

Given:
Greatest Integer Function [x]
Concept used:
Jump Discontinuity Of First Kind:
if Both Left-hand limit and Right-hand limit exist but are not equal to each other
Jump Discontinuity Of Second Kind:
if Both the Left-hand limit and Right-hand limit does not exist.
Calculations:
Let's Find the limits at x = 0 for [x].

⇒ Left-hand Limit and Right-hand limit Exists But not equal
∴ Jump Discontinuity of the First kind

Given:
To find mode from grouped frequency distribution.
Formula used:

Mode =

Where, I = the lower limit of the modal class.
h = the size of the class interval.
f1 = the frequency of the modal class.
f₀ denotes the frequency of the class preceding the modal class.
f2 denotes the frequency of the class succeeding the modal class.
Calculation:
l = 20, f₁ = 30, f₀ = 15, f₂ = 6, h = 30 - 20 = 10

Mode = = 23.85

∴ The answer is 23.85.

Given:

Concept used:

sin(a + b) = sin a cos b + sin b cos a

Calculation:

⇒

⇒

⇒

⇒ 1 + tanβcotθ

∴ The required answer is 1 + tanβcotθ.

Explanation:

If sin θ = x ⇒ θ = sin^-1x

sin(sin^-1 x) = x and sin^-1(sin y) = y, where -1 ≤ x ≤ 1 and -π/2 ≤ y ≤ π/2.

We have, 2 tan−1x + sin−1....(i)

Let tan θ = x ⇒ θ = tan^-1x

∴ (i) becomes:

2θ + sin^-1

= 2θ + sin^-1(sin 2θ) [∵ sin 2θ = ]

= 2θ + 2θ [∵ sin^-1(sin x) = x]

= 4θ

= 4tan^-1x

Concept Used:

Reflexive: For any element x in the set, x is related to itself (xRx).

Symmetric: If x is related to y (xRy), then y is related to x (yRx).

Transitive: If x is related to y (xRy) and y is related to z (yRz), then x is related to z (xRz).

Calculation:

As for 1∈A, (1, 1) ∉ R, ⇒ Relation is not reflexive.

As (1, 2) ∈ R ⇒ (2, 1) ∈ R and vice versa

Thus, Relation is Symmetric.

As (1, 2) ∈ R and (2, 1) ∈ R but (1, 1)∉R.

Thus, a relation is also not transitive.

Hence, a relation is symmetric but neither reflexive nor transitive.

Given:

0 ≤ y ≤ x² (this is represented by vertical strip)

And x varies from 0 to 1.

Now if we change the order of integration, we have to draw a horizon strip.

After changing the order of Integration

And, 0 ≤ y ≤ 1

∴

Explanation:

If z were function of y alone, the solution would have been z = A sin y + B cos y, where A and B are constants.
Since z is a function of x and y, A and B can be arbitrary functions of x.
Hence the solution of the given equations is,
z = f(x) sin y + ϕ(x) cos y

At, y = 0, z = e^x

⇒ e^x = ϕ(x)

At,

⇒ e^-x = f(x)
The solution is,
z = e^-x sin y + e^x cos y

Given:

Concept used:

Limit Comparision test:

if an and bn are two positive series such that

where c > 0 and finite then, either Both series converges or diverges together

P - Series test:

∑ is convergent for p > 1 and divergent for p ≤ 1

Calculations:

n^th term of the given series = u_n =

Let

∴ By comparison test, Σu_n and Σv_n both converge or diverge.

But Σv_n is convergent. [p series test - p = 2 > 1]

∴ Σu_n is convergent.

Given :

A monoid is a group if

Explanation:

A group is a set and a binary operation such that. For all x, y ∈ G, x * y ∈ G (closure).

There exists an identity element e ∈ G with x * e = e * x = x for all x ∈ G (identity).

For all x , y , z ∈ G we have (x * y) * z = x * (y * z) (associativity).

There exists an Inverse element c such that x * c = e = c * x for all x ∈ G (Inverse)

Now,

Monoid ⇒ Associative binary operation and identity

∴ A monoid(B,*) is called Group if to each element there exists an element c such that (a*c)=(c*a)=e. Here e is called an identity element and c is defined as the inverse of the corresponding element.

∴ option 1 is correct

Given:

Two points are (-4, 3) and (-5,6)

Concept used:

The equation of a line passing through the points (x_1, y₁) and (x₂, y₂) is given as

Calculation:

Here, x₁ = -4, y₁ = 3, x₂ = -5 and y₂ = 6

⇒ (y - 3) = {x - (-4)}

⇒ (y - 3) = (x + 4)

⇒ (y - 3) = -3(x + 4)

⇒ (y - 3) = -3x - 12

⇒ y + 3x + 9 = 0

∴ The required equation of the line is 3x + y + 9 = 0

Area of triangle ABC will be ‘0’;

⇒ 12λ – 12b + 4b + 12 – 24 – 8λ = 0

⇒ 4λ – 8b = 12

⇒ λ – 2b = 3

∴ Value of λ – 2b = 3

Calculation:
To calculate the Standard deviation, we first calculate the Variance

where x is a mean.

For variance, calculate , for each value of x_i
8 - 14 = -6
12 - 14 = -2
13 - 14 = -1
15 - 14 = 1
22 - 14 = 8

= 36 + 4 + 1 + 1 + 64/5
= 106/5
= 21.2

Standard deviation =

Given:
2cosec²38°cot²52° - sin²38° - sin²52° - cot²52°
Concept used:

cosecθcotθ = secθ
sin²θ + cos²θ = 1
tan²θ = sec²θ -1
Calculation:
2cosec²38°cot²52° - sin²38° - sin²52° - cot252°
⇒ 2cosec²38°cot²(90 - 38)° - sin²38° - sin²(90 - 38)° - cot²(90 - 38)°
⇒ 2cosec²38°tan²38° - sin²38° - cos²38° - tan²38°
⇒ 2sec²38° - 1(sin²38° + cos²38°) - 1(sec²38° -1)
⇒ 2sec²38° - 1 - sec²38° +1
⇒ sec²38°
∴ The required answer is sec238°.

Concept:

We know that,

Therefore,

Solution:

(1 + x)(1 - x)n = (1 + x) (ⁿC₀ - ⁿC₁x + ...........+ (-1)^{n- 1} × ⁿC_n-1 x^n-1 + (- 1)ⁿ × ⁿC_n xⁿ

Coefficient of xⁿ = (- 1)ⁿ nCn + (-1)n- 1 × nCn-1

= (- 1)ⁿ + (-1)^{n- 1}n

= (- 1)ⁿ - (- 1)ⁿ n

= (- 1)ⁿ ( 1 - n)

∴ The coefficient of xn in the expansion of (1 + x)(1 - x)n is (- 1)n ( 1 - n).

Concept:

Let f(x) =

There are three conditions that need to be met by a function f(x) in order to be continuous at x = a. These are:

1) f(a) is defined [you can’t have a hole in the function]

2) exists.

3)

Note:

If any of the three conditions of continuity is violated, the function is said to be discontinuous.

Calculation:

Given: f(x) = cot x

⇒ f(x) =

Check where the denominator becomes 0.

tan x = 0

x = nπ, n∈ Z

Hence the function is discontinuous at {x = nπ ∶ n ∈ Z}

The correct answer is option 1.

Given:

150 x ≡ 35 (mod 31)

Given that 150x and 35 is congruent modulo 31.

Concept Used:

Definition: Given an integer m, two integers a and b are congruent modulo m if m| (a - b).

Solution:

We know that for a, b integers

a ≡ b (mod m) if m| (a - b)

In this question given that

150 x ≡ 35 (mod 31)

⇒ 31|150x - 35

Putting x = 14, 22, 24, 12

⇒ 31| 150 x 24 - 35

⇒ 31| 3565

⇒ 1| 115

Only x = 24 will satisfy this modulo.

Option 3 is correct.

Concept:

If and are two non-zero vectors then scalar product will be-

…1)

Where,

θ is angle between

and, and …2)

Calculation:

Given,

Then,

⇒ cos θ = cos 60° ⇒ θ = 60°

Given:

G = Z37 × Z7

Concept used:

The order of the group Zm × Zn is mn

if m and n are relatively prime.

Calculations:

G = Z37 × Z7

here, 37 and 7 are relatively prime

⇒ o( Z37 × Z7) = 37 × 7 = 259

∴ option 4 is correct

Concept:

If the system of equations has non-zero solutions, then the determinant is zero.

Calculation:

The system will have a non-zero solution, if

Δ =

C₂ → C2 - C1

C3 → C₃ - C1

⇒ Δ =

[(a + 1)³ - a³ = a³ + 1 + 3(a + 1) - a³ = 3a² + 3a + 1 and

(a + 2)3 - a3 = a3 + 8 + 3(a + 2) - a3 = 6a2 + 12a + 8]

Expanding along R₃,

⇒ (6a² + 6a + 2) - (6a² + 12a + 8) = 0

⇒ -6a - 6 = 0

⇒ a = -1

∴ The value of a is -1.

Given:

The vertices of a triangle are (1, √3), (0, 0), and (2, 0)

Concept:

The centriod P(x , y) of a circle with vertices A(x₁ , y₁), B(x₂ , y₂), and C(x₃ , y₃) is given as,

Solution:

As per the question,

The vertices of a triangle are (1, √3), (0, 0), and (2, 0)

x1 = 1

y1 = √3

x2 = 0

y2 = 0

x3 = 2

y3 = 0

Hence, option 4 is correct.

Concept Used:

for vector and to be coplanar,

Calculation:

. . .(i)

cross product with from right side in (i)

cross product with from right side in (i)

Hence, given vector is neither coplanar with nor

Concept:

The general formula,

The relation between time and Laplace's transformation is

=

Calculation:

Given, the Laplace transform is

=

Other Related Points

Few relations between Laplace and time functions