Subject

Mathematics

Class

JEE Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

1.

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is

  • At least 750 but less than 1000

  • At least 1000

  • Less then 500

  • At least 500 but less than 750


B.

At least 1000

Number of ways of selecting 4 novels from 6 novels
= 6C4
Number of ways of selecting 1 dictionary from
3 dictionaries = 3C1
Required arrangements = 6C4 × 3C1 × 4! = 1080
=> At least 1000


2.

The Boolean expression ~ (p v q )v (-p ∧q ) is equivalent to

  • ~q

  • ~ p

  • p

  • q


B.

~ p

~ (p v q )v (p ∧q )

By property (-~ p ∧- q )v (-p ∧ q )

~p


3.

Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the
parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and,
CPB = θ then a value of tan θ is

  • 4/3

  • 1/2

  • 2

  • 3


C.

2

y2 = 16x

Tangent at P (16,6) is 2y = x + 16..... (1)
Normal at P (16,16) is y = -2x + 48... (2)
i.e., A is (-16,0); B is (24,0)
Now Centre of circle is (4,0)

Now, mPC = 4/3
mPB = -2
i.e., tan θ = 43+21-83 = 2


4.

Tangents are drawn to the hyperbola 4x2 – y2 = 36 at the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of △PTQ is

  • 365

  • 455

  • 543

  • 603


B.

455

Clearly, PQ is a chord of contact,
i.e., the equation of PQ is T = 0
=> y = –12

Solving with curve, 4x2 - y2 = 36

 x = ± 35, y = - 12i.e., P(35, - 12); Q (-35, - 12); T (0.3)Area of PQT is = 12 x 65 x 15455


5.

For each t ∈R, let [t] be the greatest integer less than or equal to t. Then

limx0+ x1x+2x+......+15x

  • does not exist (in R)

  • is equal to 0

  • is equal to 15

  • is equal to 120


D.

is equal to 120

limx0+ x1x+2x +.......+15x Now 1x - 1 < 1x1x2x-1<2x2x15x-1 <15x15xlimx0+ x   15x -15 < limx0+ x1x + 2x + ..... + 15xlimx0+ x   15 xlimx0+ x  15-5x < L limx0+ 15L 120


6.

Two sets A and B are as under:

A = {(a-b)∈ RxR:|a-5|<1 and |b-5|<1}

B = {(a,b)∈ Rx R: 4(a-6)2 + 9 (b-5)2 ≤ 36},then

  • Neither A ⊂ B nor B ⊂ A

  • B ⊂ A

  • A ⊂ B

  • A  B = ϕ (an empty set)


C.

A ⊂ B

As, |a-5| < 1 and |b-5| < 1

 4 <a, b<6 and (a-6)29 + (b-5)24  1

Taking axes as a-axis and b-axis

The set A represents square PQRS inside set B representing ellipse and hence A ⊂ B


7.

The sum of the coefficients of all odd degree terms in the expansion of x + x3 -15 + x - x3-15, (x>1) is

  • 2

  • -1

  • 0

  • 1


A.

2

x  + x3 -15 + x  -x3 -15 =2 [C02x5 + C25 x3 (x3-1) + C45 x (x3-1)2] = 2 [x5 + 10 (x6 - x3) + 5x(x6-2x3 +1)] = 2[ x5 + 10x6 - 10x3 + 5x7 -10x4 + 5x]= 2[5x7 + 10x6 + x5 - 10x4-10x3 + 5x]

Sum of odd degree terms coefficients
= 2(5 + 1 – 10 + 5)
= 2


8.

If the tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 then the value of c is

  • 95

  • 195

  • 185

  • 85


A.

95

Equation of tangent at (1,7) to curve x2 = y -6 is

x - 1 =12(y +7) -6

2x - y +5 = 0 ... (i)

Centre of circle  = (-8,-6)

Radius of circle = 64 + 36-c = 100-c

 Line (i) touches the circle 2(-8) -(-6) +54 + 1 = 100-C5 = 100-c c = 95


9.

PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45o, 30o and 30o, then the height of the tower (in m) is

  • 502

  • 100

  • 50

  • 1003


B.

100

Let ST = h (height of tower)

PT = ST = h

STQT = tan 30°QT = h3Now PT2 + QT2 = 20024h2 = 2002h = 100


10.

If α, β ∈ C are the distinct roots, of the equation x2 -x + 1 = 0, then α101 + β107 is equal to

  • 2

  • -1

  • 0

  • 1


D.

1

x2-x + 1 = 0

Roots are -ω, -ω2

Let α = -ω, β = -ω2

α101 + β107 = (-ω)101 + (-ω2)107

= -( ω101214)
= - (ω2 + ω)
= 1