OAVS TGT Math Mock Test - 8 - OTET MCQ

Alpha(a) and beta(b) are roots of x^2 + px + 1

This implies that sum of roots= a+b = -p/1=-p

And the product of roots = ab = 1/1=1

Similarly ,

Gamma(c) and delta(d) are roots of x^2 + qx + 1

So c+d=-q and cd =1.

The above results can be obtained once we know that any quadratic equation has two roots and hence can be written as (x-p)(x-q)=0 where a and b are the roots .

So x^2 -(p+q)x + pq =0

Comparing this with the general form of quadratic equation :ax^2 + bx + c= 0 we get

Sum of the roots =p+q= -b/a

And product of the roots = pq = c/a}
 

RHS=(a-c)(b-c)(a+d)(b+d)

=(c^2-ac-bc +ab)(d^2 +bd +ad + ab)

We know ab=1

So RHS= (c^2-ac-bc +1)(d^2 +bd +ad + 1)

= (c^2)(d^2) +(a+b)c^2(d) + c^2 -(d^2)c(a+b) -(a+b)^2(cd) -(a+b)c + d^2 + bd + ad + 1

= 1 + ac+bc + c^2 - da-db - (a^2 + b^2 + 2(1)) -ac -bc + bd + ad +1

Cancelling off all the common terms,

We get c^2 +d^2-a^2-b^2

= c^2+d^2-a^2-b^2
=c^2 -a^2 + d^2 -b^2 + 2(1) -2(1){ ab=cd =1}
=c^2 + d^2 + 2cd - a^2 - b^2 - 2ab
LHS=(c+d)^2-(a+b)^2
Therefore,
But we know -p= a+b and -q = c+d
LHS= q^2-p^2

The graph of the polynomial
f (x) = 2x - 5

Method.

Let us consider, y = f (x) = 2x - 5

Now, when x = 0, y = - 5

and when y = 0, x = 5/2

So, we get two points (0, - 5) and (5/2, 0).

We plot the two points on a graph paper considering two rectangular axes XOX' and YOY'. Then, we find a straight line as shown in the graph.

We see that the line intersects x-axis at the point (5/2, 0).

Given, four points such that no three of them are collunear.
we kow that line can be formed using two points. therfore, 

Sum of the zeroes of the polynomial  And Product of the zeroes of the polynomial

P(x) = 2x² + 5x + 1
Sum of roots = -5/2
Product of roots = 1/2
Therefore substituting these values, 
α + β +αβ 
=(α + β) + αβ
= -5/2 + 1/2
= -4/2 
= -2

x² + 15x – 8x – 120 = 0

x (x + 15) – 8 (x + 15) = 0

x = 8 and x = – 15 (rejected)

R and r be the radii of the two cones.

H and h be the heights of the two cones.

⇒

⇒

⇒

⇒ which proves that assertion is wrong.

Therefore, A is false but R is true.

Let the radius of the circle be rr and side of the square be a. Then, according to question,

Now, ratio of their areas, 

Let, PQ be a chord of a circle with centre O and radius 10cm. Draw OR ⊥PQ = .

Now, OP = 10cm and OR = 5cm

In right triangle ORP, we get

OP² = PR² + OR²

PR2 = OP² - OR²

PR² = 10² - 5² = 75

Since, the perpendicular from the centre to a chord bisects the chord.

Therefore, PQ = 2 x PR

= 2 x 8.66 = 17.32 cm

class mark= (upper limit+lower limit)/2
                 = (15.5+20.5)/2
                 =36/2
                 =18
 

A linear polynomial has 1 zero.

A quadratic polynomial has 2 zeroes.

A cubic polynomial has 3 zeroes.

In general, any polynomial has as many zeroes as its degree.

Explanation:-

The area covered by the minute hand in 60 minutes is equal to area of the circle of radius 7 cm 

 The area covered by minute hand in 30 minutes will be half of the area of the circle. 

We know, area of the circle = π r^2

Half of the area of the circle = ½ π r^2 

Here, π = 22/7  and r = 7

=>  ½ *22/7 * (7) *(7) 

=>  11 * 7 

=> 77 cm square

4^th potato = 2 × (5 + 3 + 3 + 3)

= 28 m.

A fair dice has number 1,2,3,4,5,6 only . So there are no number greater than 6
No. of favourable outcomes=0
Total no. of outcomes=6
Probability of getting no. higher than 6=No. of favourable outcomes/Total no. of outcomes=0/6=0
 

Here, in the denominator of the given fraction the highest power of prime factor 5 is 3, therefore, the decimal expansion of the rational number  will terminate after 3 decimal places.

Least positive integer divisible by 20 and 24 is LCM (20, 24). 20
= 2²×5 24=2³×3
∴ LCM (20, 24) = 2³x 3 x 5 = 120

We have radius of cylinder and cone = r
Height of cone and cylinder = h
Volume of cone = 
Volume of cylinder = πr²h

Let AB be the tower and AC be its shadow 

And AB = AC= x m
Let the angle of elevation of the sun be θ.
Then ∠ACB = θ
In right angled triangle ABC 

Therefore, the angle of elevation of the Sun is 45°.

Mode= 
Where L=lower limit of the modal class =20
f1= modal class frequency.=27
f2 = just after the modal class frequency.=21 
f0 = just previous the modal class frequency=23
C = Class size=10
Mode= 
 =20+4=24

Given: a₂₀ - a₁₃ and d = 5
⇒ a₂₀ - a₁₃ = a + (20 - 1) d - [a + (13 - 1) d] = a + (20 - 1) x 5 - [a + (13 - 1) x 5]
⇒ a₂₀ - a₁₃ - a = a + 95 - [a + 60]
= a+95 - a - 60 = 35 

∠DBA = ∠DCA = 58° …(1)

[Angles in same segment]  ABCD is a cyclic quadrilateral : 

Sum of opposite angles = 180 degrees 

∠A +∠C = 180° 

75° + ∠C = 180° 

∠C = 105° 

Again, 

∠ACB + ∠ACD = 105° 

∠ACB + 58° = 105° 

or ∠ACB = 47° …(2) 

Now,  ∠ACB = ∠ADB = 47° [Angles in same segment] 

Also, ∠D = 77° (Given) 

Again From figure, ∠BDC + ∠ADB = 77° 

∠BDC + 47° = 77° 

∠BDC = 30° 

In triangle DPC 

∠PDC + ∠DCP + ∠DPC = 180° 

30° + 58° + ∠DPC = 180° 

or ∠DPC = 92° 

Zero of the polynomial is x -2 =0 =>x = 2 put the value of x = 2 (2²)² + a(2)³ + 2(2)² - 3(2) = 16 + 8a + 8 - 6 = 18 + 8a = 0 = 8a = -18 = a = -18/8 = a = -9/4

The distance around the circle is called its circumference

Taking n = 1,

a₁ = 9 – 5(1) = 9 – 5 = 4

Taking n = 2,

a₂ = 9 – 5(2) = 9 – 10 = -1

Taking n = 3,

a₃ = 9 – 5(3) = 9 – 15 = -6

Therefore the series is 4, -1, -6, …

So, a = 4, d = a₂ – a₁ = -1 – 4 = -5

Now, we have to find the sum of the first 15 terms of the AP

⇒ S₁₅ = 15 × (-31)

⇒ S₁₅ = -465

Hence, the sum of 15 terms is -465.