Mathematics Test 218

Result

Mathematics Test 218

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Weekly Quiz Competition

Question 1
4 / -1

The slope of the tangent of the curve x=t²+3t-8, y=2t²-2t-5 at the point (2,-1) is

A
22/7

B
6/7

C
-6

D
none of these

Solution
Question 2
4 / -1

The equation of the sphere with centre (3, 6, -4) and touching the plane 2x - 2y - z - 10 = 0 is

A
x² + y² + z² - 6x + 12y - 8z + 45 = 0

B
x² + y² + z² + 6x - 9y + 8z + 35 = 0

C
x² + y² + z² - 6x + 9y + 8z + 35 = 0

D
x² + y² + z² - 6x - 12y + 8z + 45 = 0

Solution

As the required sphere touches the plane

2x - 2y - z - 10 = 0 .....(1)

∴ Radius of the required sphere = ⊥ from the center (3, 6, -4) of the required sphere to the plane (1)

∴ The required equation of the sphere is given by

(x - 3)² + (y - 6)² + (z + 4)² = 16 (Central form)

⇒ x² + 9 - 6x + y² + 36 - 12y + z² + 16 + 8z = 16

⇒ x² + y² + z² - 6x - 12y + 8z + 45 = 0
Question 3
4 / -1

If for real values of x, cos θ = x+1/x, then

A
θ is an acute angle

B
θ is a right angle

C
θ is an obtuse angle

D
No value of θ is possible

Solution
Question 4
4 / -1

A force F → = 2 î + ĵ − k̂ acts at a point A whose position vector is 2 î − ĵ. If point of application of F → moves from A to the point B with position vector 2 î + ĵ , then work done by F → is

A
4

B
20

C
2

D
none of these

Solution

W=F.ds ds=2i+j-(2i-j) ds =2j F=2i +j-k W=(2i+j-k).(2j)=2(0)+1(2)+(-1)(0)=2
Question 5
4 / -1

A
0

B
1

C
2

D
-1

Solution

As the given function is odd function. So by property six (see in ncert) the answer is zero
Question 6
4 / -1

Lim_x→-1 [1 + x + x² + ……….+ x¹⁰] is

A
1

B
0

C
-1

D
2

Solution

Given, Lim_x→-1 [1 + x + x² + ……….+ x¹⁰]

= 1 + (-1) + (-1)² + ……….+ (-1)¹⁰

= 1 – 1 + 1 – ……. + 1 = 1
Question 7
4 / -1

A square tank of capacity 250 cubic m has to be dug out. The cost of land is Rs 50 per sq.m. The cost of digging increases with the depth and for the whole tank is 400 (depth)² rupees. The dimensions of the tank for the least total cost are

A
depth=2.5m, side=10m

B
side=10m, depth=3.5m

C
side=10m, depth=10m

D
side=2.5m, depth=2.5m

Solution
Question 8
4 / -1

The g^-1({0}) for the function g(x)= ⌊x⌋ is ___________

A
{x | 0 ≤ x < 1}

B
{x | 0 < x ≤ 1}

C
{x | 0 < x < 1}

D
{x | 0 ≤ x ≤ 1}

Solution

g({0}) for the function g(x) is {x | 0 ≤ x ≤ 1}. Put g(x) = y and find the value of x in terms of y such that ⌊x⌋ = y.
Question 9
4 / -1

A
1,-1

B
-1,1

C
2,-2

D
-2,3
Question 10
4 / -1

Let f(x) = ax²+bx+c,a ≠ 0 and triangle = b²-4ac.If α+β, α²+β² and α³+β³ are in G.P., then :

A
∆≠0

B
b∆=0

C
c∆=0

D
bc∆≠0

Solution

α+β,α²+β²,α³+β³ are in G.P.

∴(α²+β²)²= (α+β)(α³+β³)

∴α⁴+β⁴+2α²β² = α4+β⁴+α³β+αβ³

∴α²β²−α³β−αβ³+α²β²= 0

∴α²β(β−α)−αβ²(β−α)=0

∴(α²β−αβ²)(β−α)=0

∴αβ(α−β)(β−α)=0

∴αβ(α−β)²=0

∴α=0 or β=0 or α−β=0

Case (1) α=0 or β=0,

⟹x=0 is a solution of given equation

∴a(0)²+b(0)+c=0

∴c=0

Case (2) α−β=0

⟹α=β

So, the equation has equal roots.

∴Δ=b²−4ac=0

∴c=0 or Δ=0

∴cΔ=0