Mathematics Test 135

Result

Mathematics Test 135

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

Let f(x) be a continuous function such that f(a – x) + f(x) = 0 for all x[0,a]. Then, the value of the integral is equal to

A
a

B
a/2

C
fA

D
1/2 fA

Solution
Question 2
4 / -1

The circles which can be drawn to pass through (1,0) & (3,0) and touching the y-axis, intersect at an angle θ. The value of cos θ is equal to

A
1/2

B
1/√2

C
1/4

D
√3/2

Solution

Equation of line joining A (1,0) and B (3,0) is y = 0. Equation of family of circles passing through A and B is:
(x – 1)(x – 3) + (y – 0) (y – 0) + λy = 0
x² + y² – 4x + λy + 3 = 0
If above circle touches y-axis then x = 0 is a tangent. Substituting x = 0:
y² + λy + 3 = 0
Discriminant of above quadratic must be zero.
λ² – 12 = 0

Hence the circles are:

Thus, the coordinates of C₁ and C₂ arerespectively. Also the radii of each of the circles is r₁ = r₂ = 2
We know that the angle of intersection of two circles of radii r₁ and r₂ is given by
where d is the distance between their centers.

Since between two lines whenever there is angle θ there is always the angle π–θ, therefore,
Question 3
4 / -1

a, b, c are positive numbers and abc² has the greatest value 1/ 64. Then

A

B

C

D
none of these

Solution

Also for the greatest value of abc² the numbers have to be equal, i.e a = b = c/2
Also given that maximum value = 1/64
so, a + b + c = 1
i.e. a = b = 1/4, c = 1/2
Question 4
4 / -1

If A and B are two square matrices such that B = –A^–1 BA, then (A+B)² is equal to

A
O

B
A² + B²

C
A² + 2AB + B²

D
A+B

Solution

B = –A^–1 BA
AB = A (–A^–1BA)

AB = –BA
AB + BA = 0
Now, (A + B)² = A² + B² + AB + BA
(A + B)² = A² + B², (∵AB + BA = 0)
Question 5
4 / -1

The set of all values of the parameter a for which the points of minimum of the function y = 1 + a² x – x³
Satisfy the inequality

A
an empty set

B
(-3√3, - 2√3)

C
(2√3,3√3)

D

Solution
Question 6
4 / -1

The value of is

A

B

C

D

Solution
Question 7
4 / -1

The function

A
an even function

B
an odd function

C
a periodic function

D
none of these

Solution

f (x) is an even function.
NOTE: Integration of every even function is odd function and integration of every odd function is even function, provided the integrands are integrable and the limit is 0 to x. Similarly differentiation of every odd function is even function and that of every even function is odd function, provided the they are differentiable. Here which is an odd function, therefore is even function.
Question 8
4 / -1

If for a variable line the condition a^–2 + b^–2 = c^–2(c is a constant), is satisfied, then the locus of foot of the perpendicular drawn from origin to this is:

A

B
x² + y² = 2c²

C
x² + y² = c²

D
x² – y² = c²

Solution

The foot of perpendicular to the above line from O(0,0) is given by:
Question 9
4 / -1

The eccentricity of the hyperbola whose latus rectum is half of its transverse axis, is

A

B

C

D

Solution

Length of latus rectum of hyperbola = Length of the transverse axis
Question 10
4 / -1

A tangent having slope of to the ellipse intersects the major and minor axes in points A and B respectively. If C is the center of the ellipse then the area of the triangle ABC is:

A
24 sq. units

B
12 sq. units

C
36 sq. units

D
48 sq. units

Solution

From symmetry of ellipse it is obvious that area of the triangle ABC will be same with respect either tangent.