Mathematics Test

Result

Mathematics Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

What is \(\int_{0}^{1} {xe}^{{x}} {dx}\) equal to?

A
1

B
-1

C
0

D
e

Solution

Let \(I=\int_{0}^{1} {x}{e}^{{x}} {dx}\) ----(1)
This will be solved by using integration's by parts rule.
We know that,
\(\int uv dx = u\int v dx - ∫ [u'\int v\ dx] dx \)
By applying above rule in equation (1), we get,
\(\Rightarrow {I}={x} \int_{0}^{1}{e}^{{x}} {dx}-\int_{0}^{1}\left(\frac{{dx}}{{dx}} \int {e}^{{x}} {dx}\right) {dx}\)
\(\Rightarrow I=\left[{xe}^{{x}}\right]_{0}^{1}-\int_{0}^{1} 1 \times {e}^{{x}} {dx}\quad (\because \int e^x=e^x)\)
\(\Rightarrow I=\left[{xe}^{{x}}\right]_{0}^{1}-\left[{e}^{{x}}\right]_{0}^{1}\)
\(\Rightarrow I=(e-0)-(e-1)=1\)
Question 2
1 / -0

The greatest value of λ ≥ 0 for which both the equations 2x² + ( λ − 1)x + 8 = 0 and x² − 8x + λ + 4 = 0 have real roots is

A
9

B
15

C
12

D
16

Solution
Question 3
1 / -0

A
10!

B
¹⁰C₁₀

C
¹⁰C₁

D
10¹⁰

Solution
Question 4
1 / -0

A
2[√x sin √x + cos √x] + C

B
[√ x sin √x - cos √x] + C

C
2[√x sin √x - cos √x] + C

D
None of these

Solution
Question 5
1 / -0

A
Invertible

B
Injective but not surjective.

C
Surjective but not injective

D
Neither injective nor surjective.

Solution
Question 6
1 / -0

If P(1, 0), Q(– 1, 0) and R(2, 0) are three given points, then the locus of point S, which satisfies the relation (SQ)² + (SR)² = 2(SP)², is (where SQ, SR and SP are line segments)

A
a straight line || to x-axis

B
a circle through the origin

C
a circle with centre at the origin

D
a straight line || to y-axis

Solution

Let the point be (h, k)
So, SQ² = (h + 1)² + k²
SR² = (h - 2)² + k²
SP² = (h - 1)² + k²
Substituting in the given equation and replacing the values of h and k by x and y, we get
x = - 3/2 (A straight line parallel to y-axis).
Question 7
1 / -0

The orthocenter of the triangle with vertices

A

B

C

D

Solution
Question 8
1 / -0

The orthocenter of the triangle formed by the lines xy = 0 and x + y = 1 is

A
(1/2,1/2)

B
(1/3,1/3)

C
(0,0)

D
(1/4,1/4)

Solution

The triangle formed by the given lines is a right angle triangle right angled at (0, 0). Hence, it is the orthocenter.
Question 9
1 / -0

The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x - 2y = 7. Then PQRS must be a

A
rectangle

B
square

C
cyclic quadrilateral

D
rhombus

Solution

The slope of the first line: - 1/3
Slope of second line is: 3
Hence the diagonals are perpendicular, which is the property of a Rhombus.
Question 10
1 / -0

If the vertices P, Q, R of a triangle PQR are rational points, then which of the following points of the triangle PQR is not always a rational point?

A
Centroid

B
Incentre

C
Circumcentre

D
Orthocentre

Solution

Incentre of is not ΔPQR always a rational point.
Question 11
1 / -0

If f(x) = x^a + a^x and g(x) = x^a – a^x, find the value of f(x) + g(x).

A
2x^a

B
5x^a

C
(x^a)²

D
x^a

Solution

f(x) + g(x) = 2x^a
Question 12
1 / -0

Find the period of the given function:
f(x) = tan(x/3) + sin(2x)

A
π/2

B
π

C
3π

D
3π/2

Solution

Given, f(x) = tan(x/3) + sin(2x).
Here, tan(x/3) is periodic with period 3π and sin2x is periodic with period π.
Hence, f(x) will be periodic with period 3π.
Question 13
1 / -0

A
1.404

B
1.336

C
1.487

D
1.527

Solution

We know that amount = P(1+ r/100)ⁿ. Substituting r = 12 and n = 3 in it, we get A = P[1.12]³ or A/P = 1.404.
Question 14
1 / -0

If

L(X, Y) = Square of the larger number,

S(X, Y) = Square of the smaller number,

MODD (X, Y) = Absolute value of the difference (X - Y) and

X * Y = Product of X and Y,

then the value of MODD [{L (1, 2) * S(1, - 3)}, {S(0, 2) * MODD (3, - 4)}] *MODD[{S(3, -1) * L(1, - 2)}, { MODD (1, 2) * L(1, 2)}] will be

A
30

B
108

C
241

D
50

Solution
Question 15
1 / -0

A
3/7

B
1/5

C
1/3

D
3/5

Solution