Mathematics - Online Test
Q1. 
then B equals
then B equals
Answer : Option D
Explaination / Solution:
Q2. General solution of is
Answer : Option B
Explaination / Solution:
Since
Q3. he locus of the equation is
Answer : Option A
Explaination / Solution:
We have this is a circle with centre (1/4,0) and radius =1/4.
Q4. 5th term from the end in the expansion of is
Answer : Option D
Explaination / Solution:
Q5. If A = { 1,2,3,4 } and B = { 4,5,6,7 } then A−B =
Answer : Option C
Explaination / Solution:
Given A={1,2,3,4} and B={4,5,6,7}
A-B set will contain those elements which are present in A but not in B
we can see
Now
so
Q6. The equation of the normal to the curve y = sinx at (0, 0) is
Answer : Option C
Explaination / Solution:
Since , therefore , slope of tangent at ( 0 , 0 ) = cos 0 = 1 and hence slope of normal at ( 0 , 0 ) is - 1 .
Q7. The area bounded by the parabola = x and the straight line 2y = x is
Answer : Option C
Explaination / Solution:
Q8. Solve : 30x < 200, when x is a natural number :
Answer : Option C
Explaination / Solution:
Q9. If k be an integer, then is equal to
Answer : Option A
Explaination / Solution:
Q10. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
Answer : Option A
Explaination / Solution:
Objective function is Z = 5x + 3 y ……………………(1).
The given constraints are : 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
The corner points obtained by drawing the lines 3x+5y=15 and 5x+2y=10 are (0,0),(0,3), (2,0) and (20/19,45/19)
Corner points | Z = 5x + 3 y |
O(0 , 0 ) | 0 |
B ( 2 , 0 ) | 10 |
C( 0 , 3 ) | 9 |
D ( 20/19 , 45/19 ) | 235/19 ……………….(Max.) |
Here , Z = 235/19 is maximum at ( 20/19 , 45/19 ) .
