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Page 1
CBSE XI | Mathematics
Sample Paper – 2 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 2 Solution
SECTION – A
1. The sum of the infinite series will be:
2 3 4
1 1 1 1
...
3
3 3 3
? ? ? ?
=
1
1
3
1
2
1
3
?
?
OR
23
0.5 0.5555...
0.5 0.05 0.005 ...
5 5 5
...
10
10 10
5
10
1
1
10
5
9
?
? ? ? ? ?
? ? ? ? ?
?
?
?
2. Since A and B are disjoint sets, so their intersection is an empty or a null set. So the
statement is true.
3. cos cos sin sin
4 4 4 4
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
cos
44
cos ( )
2
sin( )
?? ?? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ??
? ? ? ? ??
???
? ? ? ? ?
??
??
? ? ? ?
Page 2
CBSE XI | Mathematics
Sample Paper – 2 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 2 Solution
SECTION – A
1. The sum of the infinite series will be:
2 3 4
1 1 1 1
...
3
3 3 3
? ? ? ?
=
1
1
3
1
2
1
3
?
?
OR
23
0.5 0.5555...
0.5 0.05 0.005 ...
5 5 5
...
10
10 10
5
10
1
1
10
5
9
?
? ? ? ? ?
? ? ? ? ?
?
?
?
2. Since A and B are disjoint sets, so their intersection is an empty or a null set. So the
statement is true.
3. cos cos sin sin
4 4 4 4
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
cos
44
cos ( )
2
sin( )
?? ?? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ??
? ? ? ? ??
???
? ? ? ? ?
??
??
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 2 Solution
4.
? ? ? ?
22
1 1 1 i 1 i 1 i 1 1
i
1 i 1 i 1 i 1 1 2 2
1i
???
? ? ? ? ? ?
? ? ? ?
?
1 1 1
11
Comparing with x iy, x ,y
22
1
y
2
Argument = tan tan tan 1
1
x4
2
? ? ?
? ? ?
??
??
?
? ? ? ? ?
??
??
??
SECTION – B
5.
22
4x + y = 100
22
2 2 2 2
xy
1
25 100
b 100,a 25;c b a 100 25 75 5 3
c 5 3 3
eccentricity,e
b 10 2
? ? ?
? ? ? ? ? ? ? ? ?
? ? ?
6. Probability of both the friends not having the same birthday is
365 364
365 365
?
So the probability of two friends having the same birthday is
365 364 364 1
= 1 1 1
365 365 365 365
? ? ? ? ? ?
OR
Let A be the event that the patient will have his teeth cleaned and B be the event that he
will have cavity filled.
P(A) = 0.44, P(B) = 0.24, P(A ? B) = 0.60
P(A n B) = P(A) + P(B) – P(A ? B)
= 0.44 + 0.24 – 0.60
= 0.08
7. Let the four parts be a - 2d, a - d, a + d, a + 2d
So a - 2d + a - d + a + d + a + 2d = 20
a = 5
22
22
[(a 2d)(a 2d)] 2
[(a d)(a d)] 3
(a 4d ) 2
3
(a d )
??
?
??
?
?
?
3a
2
- 12d
2
= 2a
2
- 2d
2
a
2
- 10 d
2
= 0
So, 25=10 d
2
5
d
10
?
?
Page 3
CBSE XI | Mathematics
Sample Paper – 2 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 2 Solution
SECTION – A
1. The sum of the infinite series will be:
2 3 4
1 1 1 1
...
3
3 3 3
? ? ? ?
=
1
1
3
1
2
1
3
?
?
OR
23
0.5 0.5555...
0.5 0.05 0.005 ...
5 5 5
...
10
10 10
5
10
1
1
10
5
9
?
? ? ? ? ?
? ? ? ? ?
?
?
?
2. Since A and B are disjoint sets, so their intersection is an empty or a null set. So the
statement is true.
3. cos cos sin sin
4 4 4 4
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
cos
44
cos ( )
2
sin( )
?? ?? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ??
? ? ? ? ??
???
? ? ? ? ?
??
??
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 2 Solution
4.
? ? ? ?
22
1 1 1 i 1 i 1 i 1 1
i
1 i 1 i 1 i 1 1 2 2
1i
???
? ? ? ? ? ?
? ? ? ?
?
1 1 1
11
Comparing with x iy, x ,y
22
1
y
2
Argument = tan tan tan 1
1
x4
2
? ? ?
? ? ?
??
??
?
? ? ? ? ?
??
??
??
SECTION – B
5.
22
4x + y = 100
22
2 2 2 2
xy
1
25 100
b 100,a 25;c b a 100 25 75 5 3
c 5 3 3
eccentricity,e
b 10 2
? ? ?
? ? ? ? ? ? ? ? ?
? ? ?
6. Probability of both the friends not having the same birthday is
365 364
365 365
?
So the probability of two friends having the same birthday is
365 364 364 1
= 1 1 1
365 365 365 365
? ? ? ? ? ?
OR
Let A be the event that the patient will have his teeth cleaned and B be the event that he
will have cavity filled.
P(A) = 0.44, P(B) = 0.24, P(A ? B) = 0.60
P(A n B) = P(A) + P(B) – P(A ? B)
= 0.44 + 0.24 – 0.60
= 0.08
7. Let the four parts be a - 2d, a - d, a + d, a + 2d
So a - 2d + a - d + a + d + a + 2d = 20
a = 5
22
22
[(a 2d)(a 2d)] 2
[(a d)(a d)] 3
(a 4d ) 2
3
(a d )
??
?
??
?
?
?
3a
2
- 12d
2
= 2a
2
- 2d
2
a
2
- 10 d
2
= 0
So, 25=10 d
2
5
d
10
?
?
CBSE XI | Mathematics
Sample Paper – 2 Solution
So A.P. is 5 - v10, 5 -
5
10
, 5 +
5
10
, 5 + v10
Or 5 + v10, 5 +
5
10
, 5 -
5
10
, 5 - v10
8. Sn = (pn + qn
2
)
S1 = (p.1 + q.1
2
)
= p + q = a1
S2 = (p.2 + q.2
2
)
= 2p + 4q = a1 + a2
a2 = S2 - S1 = p + 3q
d = a2 - a1 = (p + 3 q) - (p + q) = 2q
9. If we need to prove something false, one counter example is sufficient.
(i) (a, a) ?R, for all a ?N is not true
For example take 2 ?N. we have 2 ? 2
2
, therefore (2, 2) ?R.
(ii) (a, b) ?R, implies (b, a) ?R is also not true for example take a = 9, b = 3.
As 9 = 3
2
, we have (9, 3) ?R, but 3 ? 9
2
, therefore (3, 9) ?R.
10. Let y = |2x – 1|
Differentiating w.r.t. x we get
? ?
? ?
2
2
2
2
2
dy 2x 1 d d x
2x 1 ............... x
dx dx dx x 2x 1
4x 2x 1
1
....... x
2x 1 2
??
? ??
? ? ? ?
??
?
??
??
?
? ? ?
?
OR
y = x sin x log x
dy dlog x dx dsinx
xsinx sinxlog x xlog x
dx dx dx dx
dy xsinx
sinxlog x xlog xcosx
dx x
dy
sinx sinxlog x xlog xcosx
dx
? ? ?
? ? ?
? ? ?
11. Equation of the circle is x
2
+y
2
– 3x + 5y – 4 = 0
? Centre is (3/2, -5/2)
Which is the midpoint of diameter having end points (2, 1) & (x, y)……… say
So by mid point formula
2 3 1 5
&
2 2 2 2
xy ? ? ?
??
? x = 1, y = -6
Page 4
CBSE XI | Mathematics
Sample Paper – 2 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 2 Solution
SECTION – A
1. The sum of the infinite series will be:
2 3 4
1 1 1 1
...
3
3 3 3
? ? ? ?
=
1
1
3
1
2
1
3
?
?
OR
23
0.5 0.5555...
0.5 0.05 0.005 ...
5 5 5
...
10
10 10
5
10
1
1
10
5
9
?
? ? ? ? ?
? ? ? ? ?
?
?
?
2. Since A and B are disjoint sets, so their intersection is an empty or a null set. So the
statement is true.
3. cos cos sin sin
4 4 4 4
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
cos
44
cos ( )
2
sin( )
?? ?? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ??
? ? ? ? ??
???
? ? ? ? ?
??
??
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 2 Solution
4.
? ? ? ?
22
1 1 1 i 1 i 1 i 1 1
i
1 i 1 i 1 i 1 1 2 2
1i
???
? ? ? ? ? ?
? ? ? ?
?
1 1 1
11
Comparing with x iy, x ,y
22
1
y
2
Argument = tan tan tan 1
1
x4
2
? ? ?
? ? ?
??
??
?
? ? ? ? ?
??
??
??
SECTION – B
5.
22
4x + y = 100
22
2 2 2 2
xy
1
25 100
b 100,a 25;c b a 100 25 75 5 3
c 5 3 3
eccentricity,e
b 10 2
? ? ?
? ? ? ? ? ? ? ? ?
? ? ?
6. Probability of both the friends not having the same birthday is
365 364
365 365
?
So the probability of two friends having the same birthday is
365 364 364 1
= 1 1 1
365 365 365 365
? ? ? ? ? ?
OR
Let A be the event that the patient will have his teeth cleaned and B be the event that he
will have cavity filled.
P(A) = 0.44, P(B) = 0.24, P(A ? B) = 0.60
P(A n B) = P(A) + P(B) – P(A ? B)
= 0.44 + 0.24 – 0.60
= 0.08
7. Let the four parts be a - 2d, a - d, a + d, a + 2d
So a - 2d + a - d + a + d + a + 2d = 20
a = 5
22
22
[(a 2d)(a 2d)] 2
[(a d)(a d)] 3
(a 4d ) 2
3
(a d )
??
?
??
?
?
?
3a
2
- 12d
2
= 2a
2
- 2d
2
a
2
- 10 d
2
= 0
So, 25=10 d
2
5
d
10
?
?
CBSE XI | Mathematics
Sample Paper – 2 Solution
So A.P. is 5 - v10, 5 -
5
10
, 5 +
5
10
, 5 + v10
Or 5 + v10, 5 +
5
10
, 5 -
5
10
, 5 - v10
8. Sn = (pn + qn
2
)
S1 = (p.1 + q.1
2
)
= p + q = a1
S2 = (p.2 + q.2
2
)
= 2p + 4q = a1 + a2
a2 = S2 - S1 = p + 3q
d = a2 - a1 = (p + 3 q) - (p + q) = 2q
9. If we need to prove something false, one counter example is sufficient.
(i) (a, a) ?R, for all a ?N is not true
For example take 2 ?N. we have 2 ? 2
2
, therefore (2, 2) ?R.
(ii) (a, b) ?R, implies (b, a) ?R is also not true for example take a = 9, b = 3.
As 9 = 3
2
, we have (9, 3) ?R, but 3 ? 9
2
, therefore (3, 9) ?R.
10. Let y = |2x – 1|
Differentiating w.r.t. x we get
? ?
? ?
2
2
2
2
2
dy 2x 1 d d x
2x 1 ............... x
dx dx dx x 2x 1
4x 2x 1
1
....... x
2x 1 2
??
? ??
? ? ? ?
??
?
??
??
?
? ? ?
?
OR
y = x sin x log x
dy dlog x dx dsinx
xsinx sinxlog x xlog x
dx dx dx dx
dy xsinx
sinxlog x xlog xcosx
dx x
dy
sinx sinxlog x xlog xcosx
dx
? ? ?
? ? ?
? ? ?
11. Equation of the circle is x
2
+y
2
– 3x + 5y – 4 = 0
? Centre is (3/2, -5/2)
Which is the midpoint of diameter having end points (2, 1) & (x, y)……… say
So by mid point formula
2 3 1 5
&
2 2 2 2
xy ? ? ?
??
? x = 1, y = -6
CBSE XI | Mathematics
Sample Paper – 2 Solution
12. Since the centre of ellipse is at the origin & foci lie on y-axis, its equation is
? ?
22
22
2 2 2
2
22
22
22
2
2
xy
1............. (1)
ba
where b a 1 e
3
b a 1
4
7
ba
16
Sub in (1)
xy
1..............(2)
7
a
a
16
??
??
??
??
????
??
??
??
??
?
??
Also it passes through (6, 4)
? 576 + 112 = 7a
2
Substituting a
2
and b
2
in (1)
We get equation 16x
2
+ 7y
2
= 688.
OR
Comparing
22
xy
1
36 20
?? with
22
22
xy
1
ab
??
a
2
= 36 and b
2
= 20
b
2
= a
2
(1 – e
2
)
20 = 36(1 – e
2
)
36e
2
= 16
e
= 2/3
Distance between directrices =
2a 2 6
18
2
e
3
?
??
SECTION – C
13. Let A, B and C represent the set of students who received medals for Soccer, Basketball
and Cricket.
n(A) = 38, n(B) = 15, n(C) = 20, n(A ? B ? C) = 58, n(A ? B ? C) = 3
Using counting theorems,
n (A ? B ? C) = n(A) + n(B) + n(C) – n(A ? B) – n(B ? C) –n(A ? C) + n(A ? B ? C)
Substituting the values we get,
? 58 = 38 + 15 + 20 - n(A ? B) - n(B ? C) –n(A ? C) + 3
? n(A ? B) + n(B ? C) + n(A ? C) = 76 – 58 = 18
Now, each of n(A ? B), n(B ? C), n(A ? C) include the 3 students who received medals
for all three sports. Number of students who received medals in exactly two sports,
n(A ? B) – 3 + n(B ? C) –3 + n(A ? C) – 3= 18 – 3 – 3 – 3 = 9
Page 5
CBSE XI | Mathematics
Sample Paper – 2 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 2 Solution
SECTION – A
1. The sum of the infinite series will be:
2 3 4
1 1 1 1
...
3
3 3 3
? ? ? ?
=
1
1
3
1
2
1
3
?
?
OR
23
0.5 0.5555...
0.5 0.05 0.005 ...
5 5 5
...
10
10 10
5
10
1
1
10
5
9
?
? ? ? ? ?
? ? ? ? ?
?
?
?
2. Since A and B are disjoint sets, so their intersection is an empty or a null set. So the
statement is true.
3. cos cos sin sin
4 4 4 4
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
cos
44
cos ( )
2
sin( )
?? ?? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ??
? ? ? ? ??
???
? ? ? ? ?
??
??
? ? ? ?
CBSE XI | Mathematics
Sample Paper – 2 Solution
4.
? ? ? ?
22
1 1 1 i 1 i 1 i 1 1
i
1 i 1 i 1 i 1 1 2 2
1i
???
? ? ? ? ? ?
? ? ? ?
?
1 1 1
11
Comparing with x iy, x ,y
22
1
y
2
Argument = tan tan tan 1
1
x4
2
? ? ?
? ? ?
??
??
?
? ? ? ? ?
??
??
??
SECTION – B
5.
22
4x + y = 100
22
2 2 2 2
xy
1
25 100
b 100,a 25;c b a 100 25 75 5 3
c 5 3 3
eccentricity,e
b 10 2
? ? ?
? ? ? ? ? ? ? ? ?
? ? ?
6. Probability of both the friends not having the same birthday is
365 364
365 365
?
So the probability of two friends having the same birthday is
365 364 364 1
= 1 1 1
365 365 365 365
? ? ? ? ? ?
OR
Let A be the event that the patient will have his teeth cleaned and B be the event that he
will have cavity filled.
P(A) = 0.44, P(B) = 0.24, P(A ? B) = 0.60
P(A n B) = P(A) + P(B) – P(A ? B)
= 0.44 + 0.24 – 0.60
= 0.08
7. Let the four parts be a - 2d, a - d, a + d, a + 2d
So a - 2d + a - d + a + d + a + 2d = 20
a = 5
22
22
[(a 2d)(a 2d)] 2
[(a d)(a d)] 3
(a 4d ) 2
3
(a d )
??
?
??
?
?
?
3a
2
- 12d
2
= 2a
2
- 2d
2
a
2
- 10 d
2
= 0
So, 25=10 d
2
5
d
10
?
?
CBSE XI | Mathematics
Sample Paper – 2 Solution
So A.P. is 5 - v10, 5 -
5
10
, 5 +
5
10
, 5 + v10
Or 5 + v10, 5 +
5
10
, 5 -
5
10
, 5 - v10
8. Sn = (pn + qn
2
)
S1 = (p.1 + q.1
2
)
= p + q = a1
S2 = (p.2 + q.2
2
)
= 2p + 4q = a1 + a2
a2 = S2 - S1 = p + 3q
d = a2 - a1 = (p + 3 q) - (p + q) = 2q
9. If we need to prove something false, one counter example is sufficient.
(i) (a, a) ?R, for all a ?N is not true
For example take 2 ?N. we have 2 ? 2
2
, therefore (2, 2) ?R.
(ii) (a, b) ?R, implies (b, a) ?R is also not true for example take a = 9, b = 3.
As 9 = 3
2
, we have (9, 3) ?R, but 3 ? 9
2
, therefore (3, 9) ?R.
10. Let y = |2x – 1|
Differentiating w.r.t. x we get
? ?
? ?
2
2
2
2
2
dy 2x 1 d d x
2x 1 ............... x
dx dx dx x 2x 1
4x 2x 1
1
....... x
2x 1 2
??
? ??
? ? ? ?
??
?
??
??
?
? ? ?
?
OR
y = x sin x log x
dy dlog x dx dsinx
xsinx sinxlog x xlog x
dx dx dx dx
dy xsinx
sinxlog x xlog xcosx
dx x
dy
sinx sinxlog x xlog xcosx
dx
? ? ?
? ? ?
? ? ?
11. Equation of the circle is x
2
+y
2
– 3x + 5y – 4 = 0
? Centre is (3/2, -5/2)
Which is the midpoint of diameter having end points (2, 1) & (x, y)……… say
So by mid point formula
2 3 1 5
&
2 2 2 2
xy ? ? ?
??
? x = 1, y = -6
CBSE XI | Mathematics
Sample Paper – 2 Solution
12. Since the centre of ellipse is at the origin & foci lie on y-axis, its equation is
? ?
22
22
2 2 2
2
22
22
22
2
2
xy
1............. (1)
ba
where b a 1 e
3
b a 1
4
7
ba
16
Sub in (1)
xy
1..............(2)
7
a
a
16
??
??
??
??
????
??
??
??
??
?
??
Also it passes through (6, 4)
? 576 + 112 = 7a
2
Substituting a
2
and b
2
in (1)
We get equation 16x
2
+ 7y
2
= 688.
OR
Comparing
22
xy
1
36 20
?? with
22
22
xy
1
ab
??
a
2
= 36 and b
2
= 20
b
2
= a
2
(1 – e
2
)
20 = 36(1 – e
2
)
36e
2
= 16
e
= 2/3
Distance between directrices =
2a 2 6
18
2
e
3
?
??
SECTION – C
13. Let A, B and C represent the set of students who received medals for Soccer, Basketball
and Cricket.
n(A) = 38, n(B) = 15, n(C) = 20, n(A ? B ? C) = 58, n(A ? B ? C) = 3
Using counting theorems,
n (A ? B ? C) = n(A) + n(B) + n(C) – n(A ? B) – n(B ? C) –n(A ? C) + n(A ? B ? C)
Substituting the values we get,
? 58 = 38 + 15 + 20 - n(A ? B) - n(B ? C) –n(A ? C) + 3
? n(A ? B) + n(B ? C) + n(A ? C) = 76 – 58 = 18
Now, each of n(A ? B), n(B ? C), n(A ? C) include the 3 students who received medals
for all three sports. Number of students who received medals in exactly two sports,
n(A ? B) – 3 + n(B ? C) –3 + n(A ? C) – 3= 18 – 3 – 3 – 3 = 9
CBSE XI | Mathematics
Sample Paper – 2 Solution
14. cos6 cos3 (2 ) ? ? ?
? ?
? ?
??
? ? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
??
??
? ? ? ? ? ? ? ? ? ? ? ? ? ?
????
??
? ? ? ? ? ? ? ?
33
3
3
2 2 2
2
6 4 2 2
6 4 2
4cos (2 ) 3cos(2 ) Using,cos3 4cos 3cos
4 cos(2 ) 3cos(2 )
4 2cos 1 3 2cos 1 Using cos2 =2cos 1
4 8cos 3 4cos 1 3 2cos 1 1 3 2cos 1
32cos 48cos 24cos 4
? ?
??
? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
2
6 4 2
6 4 2
6 4 2
6cos 3
32cos 48cos 18cos 1
2cos6 2 32cos 48cos 18cos 1
64cos 96cos 36cos 2
OR
Using sin3? = 3 sin? – 4 sin³? and cos3? = 4 cos³? – 3 cos? in L.H.S.
33
sin3 cos3
L.H.S
sin cos
3sin 4sin 4cos 3cos
sin cos
sin (3 4sin² ) cos (4cos² 3)
sin cos
= 3 - 4 sin² - 4 cos² + 3
= 6 - 4 sin² - 4 cos²
= 6 - 4(sin² + cos² )
= 6 - 4
= 2
= R.H.S.
??
??
??
? ? ? ? ? ?
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15. Five children could be arranged in 5! ways.
(i) If x and y have to sit together, then taking x and y as 1 unit there are 4 ways of
arranging them and the two can interchange places, so 2! x 4! = 48 ways
(ii) Number of ways in which the two boys x and y never sit together,
Total ways - ways in which x and y are together = 5! - 2! x 4! = 72 ways
OR
There are 9 places in all i.e. 1, 2, 3, 4, 5, 6, 7, 8, and 9.
Out of these, 5 are odd and 4 are even.
The 5 odd places i.e. 1, 3, 5, 7, and 9 have to be occupied by the 5 men as the women do
not have to sit on these.
This they can do in
5
P5 ways = 5!
Now the 4 even places i.e., 2, 4, 6, 8 have to be occupied by the 4 women
This they can do in
4
P4 ways = 4!
Together the men and women can be seated in 5! x 4! = 120 x 24 = 2880 ways
16. Let P(x, y, z) be the required point.
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FAQs on Sample Solution Paper 2 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce
| 1. What is the quadratic formula and how is it used to solve quadratic equations? |  |
Ans. The quadratic formula is a mathematical formula used to solve quadratic equations of the form ax^2 + bx + c = 0. It is given by x = (-b ± √(b^2 - 4ac)) / (2a). To use the quadratic formula, we substitute the values of a, b, and c into the formula and solve for x. The ± sign indicates that there are two possible solutions, one with a plus sign and another with a minus sign.
| 2. How do you determine the number of solutions to a quadratic equation without solving it? | |
Ans. The number of solutions to a quadratic equation can be determined by analyzing the discriminant, which is the expression under the square root in the quadratic formula. If the discriminant is positive, the equation has two distinct real solutions. If the discriminant is zero, the equation has one real solution (a repeated root). And if the discriminant is negative, the equation has no real solutions (only complex solutions).
| 3. Can a quadratic equation have more than two solutions? | |
Ans. No, a quadratic equation can have at most two solutions. This is because a quadratic equation represents a parabola, which is a U-shaped curve. A parabola can intersect the x-axis at most twice. Therefore, a quadratic equation can have two distinct real solutions, one repeated real solution, or no real solutions at all.
| 4. How can the graph of a quadratic equation help in solving it? | |
Ans. The graph of a quadratic equation can provide useful information to solve it. By plotting the quadratic equation on a coordinate plane, we can visually determine the x-intercepts (where the graph intersects the x-axis), which correspond to the solutions of the equation. Additionally, the shape of the graph can indicate whether the equation has two real solutions, one repeated solution, or no real solutions.
| 5. Can the quadratic formula be used to solve equations that are not quadratic? | |
Ans. No, the quadratic formula can only be used to solve quadratic equations. It is specifically derived for equations of the form ax^2 + bx + c = 0, where a, b, and c are constants. Equations that are not quadratic, such as linear equations (with no x^2 term) or higher-degree polynomial equations, require different methods to solve them.
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