The function
neither injective nor surjective.
invertible
injective but not surjective.
surjective but not injective
D.
surjective but not injective
If, for a positive integer n, the quadratic equation,
x(x + 1) + (x + 1) (x + 2) + .....
+ (x + n -1 ) (x + n) = 10n
has two consecutive integral solutions, then n is equal to :
11
12
9
10
A.
11
Let a, b, c ∈ R. If f(x) = ax^{2} + bx + c is such
that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, ∀x,y ∈ R,then is equal to
225
330
165
190
B.
330
f(x) = ax^{2} + bx + c
f(1) = a + b + c = 3
Now f(x + y) = f(x) + f(y) + xy
put y = 1
f(x + 1) = f(x) + f(1) + x
f(x + 1) = f(x) + x + 3
Now
f(2) = 7
f(3) = 12
Now
Sn = 3 + 7 + 12 + ..... t^{n} ...(1)
Sn = 3 + 7 + ......t^{n – 1} + t^{n} ...(2)
On subtracting (2) from (1)
tn = 3 + 4 + 5 + ....... upto n terms
For any three positive real numbers a, b and c, (25a^{2} + b^{2}) + 25(c^{2} – 3ac) = 15b(3a + c), Then
b , c and a are in G.P
b, c and a are in A.P
a, b and c are in A.P
a, b and c are in G.P
B.
b, c and a are in A.P
225a^{2} + 9b^{2} + 25c^{2} – 75ac – 45ab – 15bc = 0
(15a)^{2} + (3b)^{2} + (5c)^{2} – (15a)(3b) – (3b)(5c) – (15a) (5c) = 0
1/2[(15a – 3b)^{2} + (3b – 5c)^{2} + (5c – 15a)^{2}] = 0
15a = 3b , 3b = 5c , 5c = 15a
5a = b , 3b = 5c , c = 3a
a/1 = b/5 = c/3
a = λ, b = 5λ, c = 3λ
a, c, b are in AP
The value of
(^{21}C_{1} – ^{10}C_{1}) + (^{21}C_{2} – ^{10}C_{2}) + (^{21}C_{3} – ^{10}C_{3}) + (^{21}C_{4} – ^{10}C_{4}) + .... +
(^{21}C_{10} – ^{10}C_{10}) is
2^{20} – 2^{10}
2^{21} – 2^{11}
2^{21} – 2^{10}
2^{20} – 2^{9}
A.
2^{20} – 2^{10}
^{(21}C_{1} + ^{21}C_{2} .........+ ^{21}C_{10})– (^{10}C_{1} + ^{10}C_{2} ......... ^{10}C_{10})
=1/2[(^{21}C_{1} + ....+ ^{21}C_{10}) + (^{21}C_{11} +.....^{21}C_{20})]
– (2^{10} – 1)
=1/2[2^{21} – 2] – (2^{10} – 1)
= (2^{20} – 1) – (2^{10} – 1)
= 2^{20} – 2^{10}
Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector.Then the maximum area (in sq. m) of the flower-bed, is
30
12.5
10
25
D.
25
Total length = r + r + rθ = 20
hyperbola passes through the point P(√2, √3) and has foci at (± 2, 0). Then the tangent to this hyperbola at P also passes through the point
C.
Equation of hyperbola is
foci is (±2, 0) hence ae = 2, ⇒ a^{2}e^{2} = 4
b^{2} = a^{2}(e^{2} – 1)
∴ a^{2} + b^{2} = 4
Hyperbola passes through √2,√3
The following statement
(p → q ) → [(~p → q) → q] is
a fallacy
a tautology
equivalent to ~ p → q
equivalent to p → ~q
B.
a tautology
(p → q) → [(~p → q) →q]
(p → q) → ((p → q) → q)
(p → q) → ((~p → ~q) → q)
(p → q) → ((~p→ q) → (~q→ q))
(p→ q) v (p → q) which is tautology
If 5(tan^{2}x – cos^{2}x) = 2cos ^{2}x + 9, then the value of cos^{4}x is
-7/9
-3/5
1/3
2/9
A.
-7/9
⇒5(1 – t – t^{2}) = t(4t + 7)
⇒ 9t^{2} + 12t – 5 = 0
⇒ 9t^{2} + 15t – 3t – 5 = 0
⇒ (3t – 1) (3t + 5) = 0
⇒ t = t/3 as t≠-5/3.
cos2x = 2(1/3)-1 = -1/3
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