Determinants Practice Questions - DPP for JEE

 Page 2


= 
2. (a) We must have f (x) = k (x – a)
2
 where k is a constant. Now in order
that the given determinant [Say, D(x)] be divisible by f(x) we must
show that both D(x) and D'(x) vanish at x = a. Now
Also
 = 
which clearly gives D' (a) = 0, since first and third row become
identical.
3. (a)
 atleast two of A, B, C are equal.
Hence the triangle is isosceles or equilateral.
4. (a) Since, –1 < x < 0
? [x] = –1
0 < y < 1     ?  [y] = 0
1 < z < 2      ?  [z] = 1
Page 3


= 
2. (a) We must have f (x) = k (x – a)
2
 where k is a constant. Now in order
that the given determinant [Say, D(x)] be divisible by f(x) we must
show that both D(x) and D'(x) vanish at x = a. Now
Also
 = 
which clearly gives D' (a) = 0, since first and third row become
identical.
3. (a)
 atleast two of A, B, C are equal.
Hence the triangle is isosceles or equilateral.
4. (a) Since, –1 < x < 0
? [x] = –1
0 < y < 1     ?  [y] = 0
1 < z < 2      ?  [z] = 1
? Given determinant =  = 1 = [z]
5. (b) We have
 = 
[Interchanging rows and columns]
=   [Taking –1 common from each row]
= –?
 is purely imaginary.
6. (d) Given system of equations can be written in matrix form as AX =
B where
Since, system is consistent and has infinitely many solutions
?  (adj. A) B = 0
?  
?   – 6 – 9 + b = 0 ? b = 15
and  6(10 – a) + 9(a – 6) – 2(b) = 0
?  60 – 6a + 9a – 54 – 30 = 0
?  3a = 24 ? a = 8
Hence, a = 8, b = 15.
7. (d) The given determinant  vanishes, i.e.,
Page 4


= 
2. (a) We must have f (x) = k (x – a)
2
 where k is a constant. Now in order
that the given determinant [Say, D(x)] be divisible by f(x) we must
show that both D(x) and D'(x) vanish at x = a. Now
Also
 = 
which clearly gives D' (a) = 0, since first and third row become
identical.
3. (a)
 atleast two of A, B, C are equal.
Hence the triangle is isosceles or equilateral.
4. (a) Since, –1 < x < 0
? [x] = –1
0 < y < 1     ?  [y] = 0
1 < z < 2      ?  [z] = 1
? Given determinant =  = 1 = [z]
5. (b) We have
 = 
[Interchanging rows and columns]
=   [Taking –1 common from each row]
= –?
 is purely imaginary.
6. (d) Given system of equations can be written in matrix form as AX =
B where
Since, system is consistent and has infinitely many solutions
?  (adj. A) B = 0
?  
?   – 6 – 9 + b = 0 ? b = 15
and  6(10 – a) + 9(a – 6) – 2(b) = 0
?  60 – 6a + 9a – 54 – 30 = 0
?  3a = 24 ? a = 8
Hence, a = 8, b = 15.
7. (d) The given determinant  vanishes, i.e.,
Expanding along C
1
, we get
(x – 4)(x – 5)
2
 – (x – 5)(x – 4)
2
 – {(x – 3)(x – 5)
2
– (x – 5)(x – 3)
2
} + (x –
3)(x – 4)
2 
–
 
(x – 4)(x – 3)
2
 = 0
? (x – 4)(x – 5)(x – 5 – x + 4)– (x – 3)(x – 5)(x – 5 – x + 3)+(x –
3)(x – 4) (x – 4 –x + 3) = 0
? – (x – 4)(x – 5) + 2(x – 3)(x – 5) – (x – 3)(x – 4) = 0
? – x
2
 + 9x – 20 + 2x
2
 – 16x + 30 – x
2
 + 7x – 12 = 0
? – 32 + 30 = 0 ?  –2 = 0
Which is not possible, hence no value of x satisfies the given condition.
8. (b) Let 
Now,  and a < 0
 Discriminant of ax
2
 + 2bxy + cy
2
 is negative and a < 0.
 [See Quadratics]
.
9. (a) We can write  as, 
= 
Page 5


= 
2. (a) We must have f (x) = k (x – a)
2
 where k is a constant. Now in order
that the given determinant [Say, D(x)] be divisible by f(x) we must
show that both D(x) and D'(x) vanish at x = a. Now
Also
 = 
which clearly gives D' (a) = 0, since first and third row become
identical.
3. (a)
 atleast two of A, B, C are equal.
Hence the triangle is isosceles or equilateral.
4. (a) Since, –1 < x < 0
? [x] = –1
0 < y < 1     ?  [y] = 0
1 < z < 2      ?  [z] = 1
? Given determinant =  = 1 = [z]
5. (b) We have
 = 
[Interchanging rows and columns]
=   [Taking –1 common from each row]
= –?
 is purely imaginary.
6. (d) Given system of equations can be written in matrix form as AX =
B where
Since, system is consistent and has infinitely many solutions
?  (adj. A) B = 0
?  
?   – 6 – 9 + b = 0 ? b = 15
and  6(10 – a) + 9(a – 6) – 2(b) = 0
?  60 – 6a + 9a – 54 – 30 = 0
?  3a = 24 ? a = 8
Hence, a = 8, b = 15.
7. (d) The given determinant  vanishes, i.e.,
Expanding along C
1
, we get
(x – 4)(x – 5)
2
 – (x – 5)(x – 4)
2
 – {(x – 3)(x – 5)
2
– (x – 5)(x – 3)
2
} + (x –
3)(x – 4)
2 
–
 
(x – 4)(x – 3)
2
 = 0
? (x – 4)(x – 5)(x – 5 – x + 4)– (x – 3)(x – 5)(x – 5 – x + 3)+(x –
3)(x – 4) (x – 4 –x + 3) = 0
? – (x – 4)(x – 5) + 2(x – 3)(x – 5) – (x – 3)(x – 4) = 0
? – x
2
 + 9x – 20 + 2x
2
 – 16x + 30 – x
2
 + 7x – 12 = 0
? – 32 + 30 = 0 ?  –2 = 0
Which is not possible, hence no value of x satisfies the given condition.
8. (b) Let 
Now,  and a < 0
 Discriminant of ax
2
 + 2bxy + cy
2
 is negative and a < 0.
 [See Quadratics]
.
9. (a) We can write  as, 
= 
= 
It is clear from here that ? cannot exceed 1.
10. (c) Applying , we get
.
For , x does not lie on 
For tan x = 1, x = , hence only one root.
11. (c) The given system of equations are :
p
3
x + (p +1)
3 
y = (p +2)
3
...(1)
px + (p +1)y = (p +2) ....(2)
x +y = 1 ....(3)
This system is consistent, if values of x and y from first two equation
satisfy the third equation.
which