HPSC PGT Mathematics Mock Test - 7 - HPSC TGT/PGT MCQ

International Yoga Day is celebrated on 21^st June every year.

Similarly;

Kargil Victory day is commemorated on 26^th July every year.

Hence, "Kargil Victory Day : 26^th July" is the correct answer.

Important Points

Preparing the family tree using the following symbol:

According to the given information,

Hence, Q is the grandmother of S.

The pattern followed is, 

Logic: For all the other option sum of place value is 14 except “DJA”

Hence, “DJA” is the correct answer.

⁹⁵C₄ + ⁹⁵C₃ + ⁹⁶C₃ + ⁹⁷C₃ + ⁹⁸C₃ + ⁹⁸C₄ + ⁹⁹C₃ 
= ⁹⁶C₄ + ⁹⁶C₃ + ⁹⁷C₃ + ⁹⁸C₃ + ⁹⁹C₃
= ⁹⁷C₄ + ⁹⁷C₃ + ⁹⁸C₃ + ⁹⁹C₃ 
= ⁹⁸C₄ + ⁹⁸C₃ + ⁹⁹C₃ =  ⁹⁹C₄ + ⁹⁹C₃ = ¹⁰⁰C₄ 

When P = 1000 and t = 0 ., then ,
c = 1000, therefore, we have :

tan210°
= tan (180° + 30°)
= tan30°
= 1/√3

Diff. both sides w.r.t Z

It is distributive law.

S_p = Sq
⇒  p/2(2a+(p−1)d) = q/2(2a+(q−1)d)
⇒ p(2a+(p−1)d) = q(2a+(q−1)d)
⇒ 2ap + p²d − pd = 2aq + q²d − qd
⇒ 2a(p−q) + (p+q)(p−q)d − d(p−q) = 0
⇒ (p−q)[2a + (p+q)d − d] = 0
⇒ 2a + (p+q)d − d = 0
⇒ 2a + ((p+q) − 1)d = 0
⇒ S_p+q = 0

z lies inside or on the ellipse.

Clearly the minimum distance of z from the given point 4 is 1 and maximum distance is 9

The given system of equations does not have solution if :

⇒ (24 + 6λ - 14λ) = 0 ⇒ λ = 3

By definition , The angle θ between the planes A₁x + B₁y + C₁z + D₁ = 0 and A₂ x + B₂ y + C₂ z + D₂ = 0 is given by :

Total 3 digit numbers = 900
There are only 5 odd numbers (1,3,5,7,9)
So 3 digit numbers having containing only odd digits = 5*5*5 = 125
So, 3 digit numbers containing at least one even digit = 900–125 = 775

Required area :

∑a(b - c) = 0, then both roots of equation is 1. 
Product of roots 


Hence, roots are in H.P.

We know that the equation
ax² + by² + 2hxy + 2gx + 2fy + c = 0
represent a rectangular hyperbola if Δ ≠ 0, h² > ab and a + b = 0.
∴ The given equation represents a rectangular hyperbola if 4 + k = 0 i.e. k = -4 

sin3x/sin4x = sin3x/(1)⋅(1/sin4x)
= [(3x/1)sin3x/3x] ⋅ [1/4x(4x/sin4x)]
= 3x/4x[sin(3x)/3x(4x/sin(4x))]
= 3/4 [sin3x/3x/(4x/sin4x)]
Now, as x→0, (3x)→0 so sin3x/3x→1.(Using θ= 3x)
And, as x→0, (4x)→0 so 4x/sin4x→1. (Using θ = 4x)
Therefore the limit is 3/4
lim x→0 sin3x/sin4x
= lim x→0   3/4sin3x/3x(4x/sin4x)
= 3/4 lim x→0 sin3x/3x
lim x→0 4x/sinx
= 3/4(1)(1)
= 3/4

T_(r+1) = ¹¹C_r x^(11-2r) (m/r)^r (-1)^r
= ¹¹C_r x^(11-2r) m^r (-1)^r
Coefficient of x³ = 11 - 2r = 3
8 = 2r 
r = 4
T5 = ¹¹C₄ x³ m⁴ (-1)
Coefficient of x³ = ¹¹C₄ m⁴
= 330 m⁴

Distance between the foci of an ellipse = length of latus rectum
i.e.  (2b²)/a=2ae
e=b²/a²
But e=[1−b²/a²]^1/2
Then e=(1−e)^1/2
Squaring both sides, we get
e² +e−1=0
e=−1 ± (1 + 4)^1/2]/2
(∵ Eccentricity cannot be negative)
e=[(5)^1/2 − 1]/2

y = log(secx + tanx)
dy/dx = 1/(secx + tanx){(secxtanx) + sec^2x}
= secx(secx + tanx)/(secx + tanx)
= secx

We have 5x−3<3x+1
⇒ 5x − 3 + 3 < 3x + 1 + 3
⇒ 5x < 3x + 4
⇒ 5x − 3 × < 3x + 4 − 3x
⇒ 2x < 4 ⇒ x < 2
 When x is an integer the solutions of the given inequality are {.............,−4,−3,−2,−1,0,1}

Let α, α² be the roots of 3x² + px + 3 = 0. Then 

As α³ = 1 ⇒ α = 1, ω, ω²
We take α = ω so that α² = ω²
Thus - 1 = ω + ω² = α + α² 

Determinant = -2(-12 + 12) -2(-6 - 2) +1(-12 - 4)
= - 2(0) - 2(-8) + 1(16)
= 0 + 16 + 16
= 32

The function is clearly differentiable except possible at x = 2, 3