UP PGT Math Mock Test - 9 - UPTET MCQ

Slope of OP x slope of OQ = -1 
⇒ 4t₁.4t₂ = -1
Eq of tangent at 

Eq of tangent at 

Let (x₁ , y₁) is the point of intersection

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

• All three of them are positive in Quadrant I
• Sine only is positive in Quadrant II
• Tangent only is positive in Quadrant III
• Cosine only is positive in Quadrant IV
• 

If A ∪ B = A ∩ B then A = B.
Example: Let A = {1, 2, 3, 4} & B = {1, 2, 3, 4}
(AUB) = (A⋂B)
⇒{1,2,3,4} = {1,2,3,4}

a,b,c are in H.P
∴1/a , 1/b , 1/care in A.P
∴2/b= 1/a+1/c_______ (1)
∴ x/a+y/b+1/c=0 ________ (2)
∴ From (1)
1/a+1/c−2/b=0 ______ (3)
∴ From (2) & (3)
We infer that,
(x=1,y=−2)
∴ Point is (1,−2)

x² – 5xy + 6y² = 0 represent a pair of lines passing through origin
ax² + ay² = - c 

⇒ ax² + ay² + c = 0 represent a circle

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}

lim x → 2 [(2^x - 4) (2^x - 2) ((√2^x) + 2)]/[(√2^x - 2) (√2x - 2)]
= lim x → 2 [(2x - 4) (2 *4)]/(√2x - 2)
= 2 lim(x → 2) [(√2x)2 - (2)2]/(√2x - 2)
= 2 lim(x → 2) [(√2x)2 + 2) (√2x)2 - 2)]/(√2x - 2)
= 8

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).

We have, f(x) = |x−1|, which always gives non-negative values of f(x) for all x ∈ R.Therefore range of the given function is all non-negative real numbers i.e. [0,∞).

i) 6!/(6-3)!=6*5*4=120

.

ii) 2/6 * 120 = 40

.

iii) 2/6 * 120 = 40

.

iv) 1/6 * 120 = 20

R is symmetric and transitive but not reflexive 

We have , 
2x + y + 3z – 2 = 0 and x – 2y + 5 = 0. Let θ be the angle between the planes , then 

lim(x → 0) (tanx-x)/x² tanx
As we know that tan x = sinx/cosx
lim(x → 0) (sinx/cosx - x)/x²(sinx/cosx)
lim(x → 0) (sinx - xcosx)/(x² sinx)
lim(x → 0) cosx - (-xsinx + cosx)/(x²cosx + sinx (2x))
lim(x → 0) (cosx + xsinx - cosx)/x²cosx + 2xsinx)
lim(x → 0) sinx/(xcosx + 2sinx)
Hence it is 0/0 form, apply L hospital rule
lim(x → 0) cosx/(-xsinx + cosx + 2cosx)
⇒ 1/(0+1+2)
= 1/3

We have : 

When x= 0 and y = - 2 ,we have : c = 2,

Given, 4x - 3y = 5 and 2x² - 3y² = 12

 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.

The 2-digit number which when divided by 3 gives remainder 1 are: 10, 13, 16, ...97
Here a = 10, d = 13 - 10 = 3
t_n = 97
nth term of an AP is t_n = a + (n – 1)d
97 = 10 + (n – 1)³
⇒ 97 = 10 + 3n – 3
⇒ 97 = 7 + 3n
⇒ 3n = 97 – 7 = 90
∴ n = 90/3 = 30
Recall sum of n terms of AP, 
= 15[20 + 87] = 15 × 107 = 1605

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,