UP PGT Math Mock Test - 5 - UPTET MCQ

Hence (C) is the correct answer.

A polynomial function has all exponents as integral whole numbers. 

Point should be on y axis so x=0
y²+8y+5 = 0
y²+8y+16 =5+16
(y+4)² = 21
y+4 = (21)^½
y = -4 +- (21)^½
A = (0, 4+(21)^½)       B=(0, 4-(21)^½)
AB= [(0-0)² + {(4+(21)^½)  - (4-(21)^½)}]
= (2(21)^½)
= 2(21)^½

• A subset is a set whose elements are all members of another set.
• Since all the elements of set A are the members of set E. So A is a subset of E.  

The given series is a geometric series in which a=2,r=3,l=4374.
Therefore,
Required sum = (lr−a)/(r−1)
​= (4374×3−2)/(3−1)
​= 6560

4x²−9y²−8x=32
⇒4(x²−2x)−9y2 = 32
⇒4(x²−2x+1)−9y² = 32 + 4 = 36
⇒(x−1)²]/9 − [y2]/4 = 1
⇒a²=9, b²=4
∴e=[1+b²/a²]^1/2 
= [(13)^1/2]/3

D. R’s of line are 2 : 1 : 2
D.R’s of Normal to plane is :3 : -2 : 6
Angle between line and plane

15cosec(x/3)= 2π/1/3=6 π

By verification f ' (2 -) ≠ f ' (2 +)
∴ f(x) is not differentiable at x = 2

• n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 12 + 9 − 4 = 17
• Now, n((A ∪ B)^c) = n(U) − n(A ∪ B) = 20 − 17 = 3

P(n) = 4 + 8 + 12 +……+ 4n = 2n (n + 1)
P(6) = 4 + 8 + 12 +……+ 4(6) 
= 2(6)[(6) + 1)]

6, a bare in H.P 
 are in A.P.

Let the digits at ones, tens and hundreds place be (a−d)a and (a+d) respectively. The, the number is
(a+d)×100+a×10+(a−d) = 111a+99d
The number obtained by reversing the digits is
(a−d)×+a×10+(a+d) = 111a−99d
It is given that the sum of the digits is 21.
(a−d)+a+(a+d) = 21                        ...(i)
Also it is given that the number obtained by reversing the digits is 594 less than the original number.
∴111a−99d = 111a+99d−396          ...(ii)
⟹ 3a = 21 and 198d = 396
⟹ a = 7 and d = -2
Original number = (a−d)×+a×10+(a+d)
= 100(9) + 10(7) + 5
= 975

In the case of strictly decreasing functions, the slope of the tangent line is always negative. This implies that the derivative of a strictly decreasing function is always negative. The derivative represents the rate of change of the function, and if the function is strictly decreasing, the rate of change is negative.

lim x→0 (1−cosx)/(xlog(1+x))
= lim x→0 {2sin² x/2}/[xlog(1+x)]
= lim x→0 {{2sin²x/2/(x/2)²}*/1/4}/[1/xlog(1+x)]
= lim x→0 [log(1+x)/x]1/2
= 2​

We have, α + β = -p,αβ = q
Now (ωα + ω²β)(ω²α + ωβ) = ω³α² +ω⁴αβ+ω²αβ + ω³β²
= α² +β² +(ω+ω²)αβ = α²+β² -αβ = (α + β)² -3αβ = p² -3q

Applying the above conditions, we get
nC2 + nC3+ nCn−2 + nCn−3 = 440
2(nC2 + nC3)=440
nC2 ​ + nC3=220
n(n−1)/2!+ [n(n−2)(n−1)]/3!=220
n(n−1)/2 ​(1+(n−2)/3=220
n(n−1)/2 (n+1)/3=220
n(n²−1)/6 ​= 220
n³−n=1320
n³−n−1320=0
The only real root of the above equation is 11.
Hence n=11

For x - axis, y = 0,

Therefore, 4x - x² = 0

Therefore, x = 0 or x = 4

Required area :

We have :

Using 

 y =−5x²+2x−7
Differentiating the above equation with respect to x, gives us
dy/dx=−10x+2
For maximum, dy/dx=0
2−10x=0
x=1/5​
Hence the above expression reaches its maximum value at x = 1/5

ANSWER :- c

Solution :- Number of such triangle = 6C3

= 6!/3!3!

=20

 is a constant function ⇒ bc ad 0 ⇒ ad bc.

The smallest number in the series is 1000, a 4-digit number.  

The largest number in the series is 4000, the only 4-digit number to start with 4.

The left-most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

So, The left-most digits have values 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

So, the next 3 digits have values 5.

Hence, there are numbers.

or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.