KVS PGT Mathematics Mock Test - 4 - KVS PGT/TGT/PRT MCQ

 The interest provides access to fresh encounters, knowledge, and abilities.

Key Points 

• Interest opens up new opportunities and teaches the youngster the value and excitement of exploring new things and learning new things.
• Following a passion may inspire innovation, experimentation, exploration, and prudent risk-taking.
• The measurement of interest method is raising interest in-class teaching as through this method teachers can find out the increase and decrease of the level of interest among students regarding various things like studying, playing, talking, sharing, etc.

Hence, it is concluded that the measurement of interest is the correct option.

Diagnostic evaluation has a placement function and so takes place before instruction. It is used to determine underlying causes of learning difficulties. Diagnostic evaluation may include administration of standardised achievement tests, standardised diagnostic tests, teacher-made tests, observation and checklists. Scoring and interpretation can be normative (based on norms) or criterion-referenced. Results are usually shown as an individual profile of sub-skills.

The parameters which must be considered while choosing the teaching methodologies include:
Capacity of the students: Choosing lecture method to teach 6th grade student will not help the teacher to attain his goal of teaching because the child cannot understand the concept just by mere lecturing. Hence to such grade, demonstration method of teaching must be chosen.
Content: Based upon the content, the teacher should decide the methodology.
Nature of the subject: Before deciding the method of teaching, the instructor must keep in mind the nature of the subject because the methods of teaching of science-based subjects is different from those of humanities.

In Cartesian coordinate system Shortest distance between the lines

x = |z| cos θ and y = |z| sin θ satisfies infinite values of θ and for any infinite values of θ is the value of Arg z. Thus, for any unique value of θ that lies in the interval - π < θ ≤ π and satisfies the above equations x = |z| cos θ and y = |z| sin θ is known as the principal value of Arg z or Amp z and it is denoted as arg z or amp z.
 
We know that, cos (2nπ + θ) = cos θ and sin (2nπ + θ) = sin θ (where n = 0, ±1, ±2, ±3, .............), then we get,
 
Amp z = 2nπ + amp z where - π < amp z ≤ π

ρ = (b(xy) * b(yx))
But sign of ρρ is same as sign of b(xy), b(yx)
Therefore, ρ = 2/7

 Let Tr and Tr+1 denote the rthand(r+1)th terms in
the expansion of (1+x)¹⁹. 
 Tr = ¹⁹C_r-1 x^r-1 and T_r+1 = ¹⁹C_r x^r .
∴T_r+1/T_r = ¹⁹C_r xr/(19C_r-1 x^r-1)
⇒T_r+1/T_r = ¹⁹C_r  ¹⁹C_r-1 x
⇒T_r+1/T_r = 19!/(19−r)!r! × x[(19−r+1)!(r−1)]/10!
⇒T_r+1/Tr = x(20−r)/r
⇒T_r+1/T_r = (20−r)/r × 1/2   [∵x not equal to 1/2]
Now
T_r+1/T_r > 1
⇒ (20−r)/r × 1/2 > 1
⇒ 20 > 3r
r > 20/3
∴ (6+1)^th i.e. 7th term is the greatest term.

The given matrix is a skew – symmetric matrix.,therefore , A = - A’.

S₁ = 0 and S₁ = S₁₁

Asymptotes are equally inclined to the axes of hyperbola Find the bisector of the asymptotes which bisects the angle containing the origin.

R is reflexive
∴ (3, 3), (6,6), (9, 9), (12, 12) ∈ R 
again ∴ (6, 12) ∈ R but (12, 6) ∉ R ⇒ R is not symmetric
R is transitive
[∴ (3, 6) ∈ R, (6, 12) ∈ R, (3, 12) ∈ R others are clear 

3A= A+ 2A
⇒ tan 3A = tan (A + 2A)
⇒ tan 3A = (tan A + tan 2A) / (1 – tan A . tan 2A)
⇒ tan A + tan 2A = tan 3A – tan 3A x tan 2A . tan A
⇒ tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A

Calculation:

Given: A is a matrix of order 3 × 5 and B is a matrix of order 5 × 3

Number of rows in A = 3

Number of column in A = 5

Number of rows in B = 5

Number of column in B = 3

The order of AB = number of row is A × number of columns in B
= 3 × 3
And, 
The order of BA = number of row is B × number of columns in A
= 5 × 5
Hence, option (3) is correct.

, which does not exist at x = 2 . However , we find that  , at x = 2 . Hence , there is a vertical tangent to the given curve at x = 2 .The point on the curve corresponding to x = 2 is (2 , 0). Hence , the equation of the tangent at x = 2 is x = 2

Let  E₁ : Head appears
E₂ : Tail appears
A : A reports that head appears

∴ Rwquired probability = 
By Bayes’ Theorem

If we substitute the value of x=0 in numerator and denominator, we get the indeterminate form 0/0 . We should use L'hopital's rule.
lim x→ 0 (√1+2x-√1-2x)/sinx
Differentiate it we get
lim x→ 0 (1/√1+2x) + (1/√1-2x)/cosx
= (1/√1+2(0)) + (1/√1-2(0))/cos 0
=2

LHD = limh→0 ((−h)² sin(−1/h) − 0)/−h
limh→0 −h² sin(1/h)/−h
= limh→0 h × sin(1/h)
= 0
RHD = limh→0 (h² sin(1/h) − 0)/h
= limh→0 h × sin(1/h)
= 0
RHD=LHD
∴ f(x) is differentiable at x= 0

Let (k,k+3) be the point on the line x−y+3=0
Equation of chord of contact is S₁=0
⇒yy1=4(x+x1)
⇒y(k+3)=4(x+k)
⇒4x−3y−k(y−4)=0
Therefore, straight line passes through fixed point (3,4)

n(A) = 4
B = {5, 6}
n(B) = 2
n(A x B) = n(A)*n(B) = 4*2 = 8

No. of ways to select 4 senior and 3 U-19 players = ⁶C₄ * ⁵C3 = 150
No. of ways to select 5 senior and 2 U-19 players = ⁶C₅ * ⁵C2 = 60
No. of ways to select 6 senior and 1 U-19 players = ⁶C₆ * ⁵C1 = 5 
Total no. of ways to select the team = 150 + 60 + 5 = 215

 Let α,β be roots of ax³+bx+c=0
γ,δ be roots of px²+2qx+r=0
α/β = γ/δ …………(1) and  
β/α = δ/γ ………….(2)
(1) + (2)
⇒ α/β + β/α = γ/δ + δ/γ
= [(2²+β²)/αβ + 2]= [γ²+δ²]/γδ + 2
⇒ [(α)²+(β)²+2αβ]/αβ ​= [γ²+δ²+2γδ]/γδ
​= [(α+β)²]/αβ = [(γ+δ)²]/γδ
⇒ (4b²/a²)/(c/a) = (4q²/p²)/(r/p)
⇒ b²/ac = q²/pr.

Even if f(x) is differentiable everywhere, |f(x)| need not be differentiable everywhere. An example being where the function changes its sign from negative to positive. f(x)=x at x=0B. 
∣f∣= {f if f>0                −f if f<0}
​So,∣f∣² = {f² if f>0        −f² if f<0}
​Differentiating ∣f∣²,
{2ff′ if f>0     −2ff′ if f<0}
​At f=0, LHD=RHD=∣f(0)∣²=0, so function is differentiable.

Correct Answer :- d

Explanation:- 3(6-6) -2(6-9) +3(4-6)

= 3(0) + 6 - 6

= 0

CosA + Cos(120o-A) + Cos(120°+A)
 cosA + 2cos(120° - a + 120° + a)/(2cos(120° - a - 120° - a)
we know that formula
(cos C+ cosD = 2cos (C+D)/2.cos (C-D) /2)
⇒ cosA + 2cos120° cos(-A)
⇒ cosA+ 2cos (180° - 60°) cos(-A)
⇒ cosA + 2(-cos60°) cosA
⇒ cos A - 2 * 1/2cos A
⇒ cosA-cosA
⇒ 0

(1 + x)ⁿ  = 1 + ⁿC₁x¹  + ⁿC₂x²+..............................+ⁿCnxⁿ
Three consecutive terms coefficients
ⁿCₐ  : ⁿCₐ₊₁  :  ⁿCₐ₊₂  :::  6 : 33 :  110
⇒  ⁿCₐ  = 6K  =>  n!/(a!)(n-a)!  = 6K  => n! = 6K (a!)(n-a)!
ⁿCₐ₊₁   = 33K   => n!/(a+1)!(n-a-1)! = 33K  => n!  = 33K (a+1)!(n-a-1)!
ⁿCₐ₊₂ = 110K  => n!/(a + 2)!(n-a-2)! = 110K  => n! = 110K (a + 2)!(n-a-2)!
6K (a!)(n-a)!  = 33K (a+1)!(n-a-1)!
⇒ 2  (a!)(n-a)(n-a - 1)! = 11 (a + 1)a! (n-a-1)!
⇒ 2(n-a) = 11(a + 1)
⇒ 2n - 2a = 11a + 11
⇒ 2n = 13a + 11
⇒ 13a = 2n - 11
33K (a+1)!(n-a-1)!  = 110K (a + 2)!(n-a-2)!
⇒ 3 (a+1)!(n-a-1)(n-a-2)!  = 10 (a + 2)(a + 1)!(n-a-2)!
⇒ 3 (n - a - 1) = 10(a + 2)
⇒ 3n - 3a - 3 = 10a + 20
⇒  3n = 13a + 23
⇒ 13a = 3n - 23
2n - 11 = 3n - 23
⇒ n = 12

On comparing the given equations with: 
, we get: