CBSE
π/2
1
-1
– π/2
D.
– π/2
For real x, let f(x) = x^{3}+ 5x + 1, then
f is one–one but not onto R
f is onto R but not one–one
f is one–one and onto R
f is neither one–one nor onto R
C.
f is one–one and onto R
f(x) = x^{3}+ 5x + 1
f'(x )3x^{2} +5>0
⇒ f is one–one
therefore, f is cubic
⇒ f is onto
‘f’ is one–one and onto.
Let a, b, c be such that 0 (a +c) ≠ . If ,then the value of 'n' is
0
any even integer
any odd integer
any integer
C.
any odd integer
Let f(x) = (x + 1)2– 1, x ≥ – 1
Statement – 1: The set {x : f(x) = f–1(x)} = {0, –1}.
Statement – 2: f is a bijection.
Statement–1 is true, Statement–2 is true,Statement–2 is a correct explanation for statement–1
Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1.
Statement–1 is true, statement–2 is false.
Statement–1 is false, Statement–2 is true
A.
Statement–1 is true, Statement–2 is true,Statement–2 is a correct explanation for statement–1
(x + 1)^{2} – 1 = x
(x + 1)^{2}= x + 1
⇒ x = 0, −1
Since co–domain of function is not given.So if we assume function
(a) as onto then A is correct
(b) as not onto then none of the answer is correct.
Let f(x) = x|x| and g(x) = sinx
Statement 1 : gof is differentiable at x = 0 and its derivative is continuous atthat point
Statement 2: gof is twice differentiable at x = 0
Statement–1 is true, Statement–2 is true, Statement–2 is a correct explanation for statement–1
Statement–1 is true, Statement–2 is true;Statement–2 is not a correct explanation for statement–1.
Statement–1 is true, statement–2 is false.
Statement–1 is false, Statement–2 is true
C.
Statement–1 is true, statement–2 is false.
One ticket is selected at random from 50 tickets numbered 00, 01, 02, ... , 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals
1/14
1/7
5/14
1/50
A.
1/14
A = Events that sum of the digits on selected ticket is 8
= {08, 17, 26, 35, 44}
n(A) = 5
Event that product of digits is zero
= {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20,30, 40}
⇒ n(B) = 14
=P(A/B) = (5/14)
Statement 1: ~ (p ↔ ~ q) is equivalent to p ↔ q
Statement 2 : ~ (p ↔ ~ q) is a tautology
Statement–1 is true, Statement–2 is true, Statement–2 is a correct explanation for statement–1
Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1.
Statement–1 is true, statement–2 is false.
Statement–1 is false, Statement–2 is true
C.
Statement–1 is true, statement–2 is false.
p | q | ~q | (p ↔ ~q) | ~ (p ↔ ~q) | p ↔ q |
T | T | F | F | T | T |
T | F | T | T | F | F |
F | T | F | T | F | F |
F | F | T | F | T | T |
Given P(x) = x^{4}+ ax^{3} + cx + d such that x = 0 is the only real root of P′ (x) = 0. If P(–1) < P(1),then in the interval [–1, 1].
P(–1) is the minimum and P(1) is the maximum of P
P(–1) is not minimum but P(1) is the maximum of P
P(–1) is the minimum but P(1) is not the maximum of P
neither P(–1) is the minimum nor P(1) is the maximum of P
B.
P(–1) is not minimum but P(1) is the maximum of P
P(x) = x^{4}+ ax^{3}+ bx^{2}+ cx + d
P′(x) = 4x^{3}+ 3ax^{2}+ 2bx + c
As P′(x) = 0 has only root x = 0
⇒ c = 0
P′(x) = x(4x^{2}+ 3ax + 2b)
⇒ 4x^{3}+ 3ax + 2b = 0 has non real root.
and 4x^{2}+ 3ax + 2b > 0 ∀ x ∈ [−1, 1].
As P(−1) < P(1) ⇒ P(1) is the max. of P(x) in [−1, 1]
Statement 1: The variance of first n even natural numbers is
Statement 2: The sum of first n natural numbers is and the sum of squares of first n natural numbers is
Statement–1 is true, Statement–2 is true, Statement–2 is a correct explanation for statement–1
Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1.
Statement–1 is true, statement–2 is false.
Statement–1 is false, Statement–2 is true
D.
Statement–1 is false, Statement–2 is true
Let y be an implicit function of x defined by x^{2x} – 2x^{x}coty – 1 = 0. Then y′ (1) equals
-1
1
log 2
-log 2
A.
-1
When x = 1, y=π/2
=(x^{x}– cot y)^{2}= cosec^{2}y
x^{x} = cot y + |cosec y|
when x = 1, y=π/2
⇒ x^{x} = cot y + cosec y
diff. w.r.t. to x