HPSC PGT Mathematics Mock Test - 9 - HPSC TGT/PGT MCQ

Given:

The ratio of the present age of Annu and Raj = 4 : 5

Then after 8 years ratio of their age = 5 : 6

Calculation:

Let be assume the present age of Annu and Raj is 4x and 5x respectively

⇒ (4x + 8)/(5x + 8) = 5/6

⇒ 24x + 48 = 25x + 40

⇒ x = 8

⇒ The age of Annu = 4x = 4 × 8 = 32

⇒ The age of Raj = 5x = 5 × 8 = 40

∴ The required result will be 32 years, and 40 years respectively.

By using the symbols in the table given below, we can draw the following family tree:

We do not know the gender of Shanti. 
Clearly, Shaurya is the cousin of Karan.
Hence, ‘Cousin’ is the correct answer.

NEP 2020 aims to provide a holistic and multidisciplinary education that focuses on the overall development of students.

8-6 = 4x-2x
⇒ x = 1

U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {1, 3, 5, 7, 9}
A ∩ B = {1, 3, 5, 7, 9}
(A ∩ B)’ = U - (A ∩ B)
(A ∩ B)’ = {0, 2, 4, 6, 8, 10, 11}

Given expansion is 
∴ General term


Since, we have to find coefficient of x^-10 ∴ -12 + 2r = -10  ⇒ r = 1
Now, then  coefficient  of x ^-10 is 12 C₁(a)¹¹ (b)¹ = 12a¹¹b

f(x) = sinx
f(x) = sinx is a one-one function.
codomain = range
therefore, f(x) = sinx is an onto function.

Constant Function is defined as the real valued function.
f : R→R, y = f(x) = c for each x∈R and c is a constant.

So, this function basically associate each real number to a constant value.

It is a linear function where f(x₁) = f(x₂) for all x₁,x₂ ∈ R

For f : R→R, y = f(x) = c for each x ∈ R
Domain = R
Range = {c}
The value of c can be any real number.

Sum (S_n) = a x (rⁿ -1)/(r-1)
5460 = 4 x (4ⁿ -1)/3
16380 = 4ⁿ⁺¹ - 4

16384 = 4ⁿ⁺¹

4⁷ = 4ⁿ⁺¹

7 = n + 1

n = 6

If a line makes angles 90^∘, 135^∘, 45^∘ with the x, y and z – axes respectively, then the direction cosines of this line is given by :

touches the another circle

Now, Central first circle will be
And its radius will be 3 units.
Also,centre of second circle
And radius,

As both touches each other
So,

(cot α₁) (cot α₂)........... cot α_n =1
(cos α₁) (cos α₂) .......... cos α_n = (sin α₁) (sin α₂) .......... sin α_n
Let y = (cos α₁) (cos α₂) ................ (cos α_n) (to be maximum)
Squaring y² = (cos² α₁) (cos² α₂) ................ (cos² α_n)
= (cos α₁ sin α₁) (cos α₂ sin α₂) ................ (cos α_n sin α_n)    using equation 1

[sin 2α₁. sin2α₂ .........sin 2α_n]
0 < sin2α₁. sin2α₂ .........sin 2α_n < 1

As the quadrants lies in first and second quadrant
y = -a    Focus(0,a)
x^{2 = 4ay}
 

The Required point is the radical centre of the three given circles. The radical axes of the three circles taken in pairs are 3x - 24 = 0,16y + 120 = 0 and - 3x + 16y + 80 = 0. On solving, the required point is (8, -15/2).

D is not a trigonometric identity.

If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two lines; and θ is the acute angle between the two lines; then the cosine of the angle between these two lines is given by : 

Since, ω³ = 1, hence

 is called the unit vector of a given vector 

a₁ + a_2n = a₂ + a_2n-1 = a_n + a_n-1 = K(say)

By demorgan's law
n( A' ∩ B' )=n(A ∪ B)' = n(U) - n(A ∪ B)
=100 - [ n(A) + n(B) - n(A ∩ B) ]
=100 - 60 = 40

 lt x1^+   f(x) = log(4x -3) (x²- 2x + 5)
= ln(x² - 2x +5)/ln(4x-3)
lt(h → 0)   ln(1+h)² - 2(1+h) + 5)/ln(4(1+h) - 3)
lt(h → 0)   ln(1+h² + 2h - 2 -2h + 5)/ln(4 + 4h - 3)
ln(h → 0)    ln(4 + h²)/(1+4h)
Divide and multiply the denominator by 4h
ln(h → 0)    ln(4 + h²)/[((1+4h)/4h) * 4h]
As we know that (1+4h)/4h = 1
ln 4/(4*0)   = + ∞ (does not exist)
 lt x→ 1^-   f(x) = log(4x -3) (x²- 2x + 5)
lt(h → 0)   ln(1-h)² - 2(1-h) + 5)/ln(4(1-h) - 3)
lt(h → 0)   ln(1+h² - 2h - 2 + 2h + 5)/ln(4 - 4h - 3)
ln(h → 0)    ln(4 + h²)/(1 - 4h)
Divide and multi[ly the denominator by (-4h)
ln(h → 0)    ln(4 + h²)/[((1+4h)/-4h) * (-4h)
As we know that (1-4h)/(-4h) = 1
ln 1/(-4*0)   = - ∞ (does not exist)

f (x) is onto ∴ S = range of f (x)

Expand (x + ai)ⁿ and (x – ai)ⁿ then multiply.

 N!/(n-5)! = 60×(n-1)!/(n-1-3)!
n!/(n-5)! = 60×(n-1)!/(n-4)!
n(n-1)!/(n-5)!=60×(n-1)!/(n-4)×(n-5)!
n=60/(n-4)
n(n-4)=60
n^2-4n-60=0
(n-10)(n+6)=0
n=10 and n is not equal to -6.