Mathematics - Online Test
Q1. If the direction cosines of a straight line are < k , k , k > , then what will be the value of k
Answer : Option B
Explaination / Solution:
We know that if l , m , n are direction cosines of a lines then l2+m2 +n2= 1
so, k2 + k2 +k2 = 1
3 k2 = 1
k2 = 1/3
taking squareroot on both sides , we have
Q2. A square matrix is called a diagonal matrix if for
Answer : Option B
Explaination / Solution:
In a diagonal matrix all elements except diagonal elements are zero.i.e.
Q3.
The function f (x) = , for all real x, is
Answer : Option A
Explaination / Solution:
f(x) = x2
f'(x) = 2x for all x in R.
Since f ‘(x) = 2x > 0 for x >0, and f ‘ (x) = 2x< 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing . Infact , f is strict increasing on [ 0 , ) and strict decreasing on (- ].
f'(x) = 2x for all x in R.
Since f ‘(x) = 2x > 0 for x >0, and f ‘ (x) = 2x< 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing . Infact , f is strict increasing on [ 0 , ) and strict decreasing on (- ].
Q4. Let A = { a,b,c } , B = { b,c,d } , C = { a,b,d,e } , then A∩(B∪C) is
Answer : Option B
Explaination / Solution:
Q5. Let x,y∈R, then x + iy is a purely imaginary number if
Answer : Option B
Explaination / Solution:
Purely imaginary number is a complex number which has only imaginary part ( iy)
But if y=0 the complex number iy will become 0 which is real.
Hence the condition for a number to be purely imaginary is x=0 and
Q6. One root of the equation 
Answer : Option D
Explaination / Solution:
Q7. The term independent of x in the expansion of is
Answer : Option A
Explaination / Solution:
Q8. Identify solution set for | 4 − x | + 1 < 3?
Answer : Option A
Explaination / Solution:
So solution set is ( 2 , 6 )
Q9. is
Answer : Option C
Explaination / Solution:
Q10. Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then
Answer : Option A
Explaination / Solution:
Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and each of these occur at the corner point (vertex) of R.
