Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Let f : [a, b] → R be such that f is differentiable in (a, b), f is continuous at x = a and x = b and moreover f(a) = 0 = f(b). Then
(A) there exists at least one point c in (a, b) such that f'(c) = f(c)
(B) f'(x) = f(x) does not hold at any point in (a, b)
(C) at every point of (a, b), f'(x) > f(x)
(D) at every point of (a, b), f'(x) < f(x)
The correct option (A) there exists at least one point c in (a, b) such that f'(c) = f(c) Explanation:Let h(x) = e–xf(x) h(a) = 0, h(b) = 0 h(x) is continuous and diff. by rolles theoremh'(c) = 0, c ∈ (a, b)e–xf(x) + (–e–x)f(x) = 0 e–cf'(c) = e–cf(c) f'(c) = f(c)
The correct option (A) there exists at least one point c in (a, b) such that f'(c) = f(c)
Explanation:
Let h(x) = e–xf(x) h(a) = 0,
h(b) = 0 h(x) is continuous and diff. by rolles theorem
h'(c) = 0, c ∈ (a, b)
e–xf(x) + (–e–x)f(x) = 0
e–cf'(c) = e–cf(c)
f'(c) = f(c)
See lessIf f : R → R be defined by f(x) = ex and g : R → R be defined by g(x) = x2. The mapping gof : R → R be defined by (gof)(x) = g[f(x)] ∀ x ∈ R, Then
(A) gof is bijective but f is not injective
(B) gof is injective and g is injective
(C) gof is injective but g is not bijective
(D) gof is surjective and g is surjective
The correct option (C) gof is injective but g is not bijectiveExplanation:f(x) = ex : R → R g(x) = x2 : R → R g(f(x)) = g(ex) = (ex)2 = e2x ∀ x ∈ R clearly g(f(x)) is injective and g(x) is not injective
The correct option (C) gof is injective but g is not bijective
Explanation:
f(x) = ex : R → R g(x) = x2 : R → R
g(f(x)) = g(ex) = (ex)2 = e2x ∀ x ∈ R
clearly g(f(x)) is injective and g(x) is not injective
See lessLet f : [a, b] → R be differentiable on [a, b] and k ∈ R. Let f(a) = 0 = f(b). Also let J(x) = f'(x) + kf(x). Then
(A) J(x) > 0 for all x ∈ [a, b]
(B) J(x) < 0 for all x ∈ [a, b]
(C) J(x) = 0 has at least one root in (a, b)
(D) J(x) = 0 through (a, b)
The correct option (C) J(x) = 0 has at least one root in (a, b) Explanation:Let g(x) = ekx f(x) f(a) = 0 = f(b) by rolles theorem g'(c) = 0, c ∈ (a, b) g'(x) = ekxf'(x) + kekxf(x) g'(c) = 0 ekc(f'(c) + kf(c) = 0 ⇒ f'(c) + kf(c) = 0 for at least one c in (a, b)
The correct option (C) J(x) = 0 has at least one root in (a, b)
Explanation:
Let g(x) = ekx f(x)
f(a) = 0 = f(b)
by rolles theorem
g'(c) = 0, c ∈ (a, b)
g'(x) = ekxf'(x) + kekxf(x)
g'(c) = 0
ekc(f'(c) + kf(c) = 0
⇒ f'(c) + kf(c) = 0 for at least one c in (a, b)
See lessIf (2 ≤ r ≤ n), then nCr + 2.nCr+ 1 + nCr + 2 is equal to
(A) 2. nCr + 2
(B) n+1Cr + 1
(C) n+2Cr + 2
(D) n+1C
The correct option (C) n+2Cr + 2 Explanation:nCr + nCr+1 + nCr+1 + nCr+2 = n+1Cr+1 + n+1Cr+2 = n+2Cr+2
The correct option (C) n+2Cr + 2
Explanation:
nCr + nCr+1 + nCr+1 + nCr+2 = n+1Cr+1 + n+1Cr+2 = n+2Cr+2
See lessA proton of mass ‘m’ moving with a speed v (<< c, velocity of light in vacuum) completes a circular orbit in time ‘T’ in a uniform magnetic field. If the speed of the proton is increased to 2v, what will be time needed to complete the circular orbit?
(A) √2T
(B) T
(C) T/√2
(D) T/2
The correct option (B) TExplanation:T = 2mπ/qBT is independent of v.
The correct option (B) T
Explanation:
T = 2mπ/qB
T is independent of v.
See lessHow the linear velocity ‘v’ of an electron in the Bohr orbit is related to its quantum number ‘n’?
(A) v ∝ 1/n
(B) v ∝ 1/n2
(C) v ∝ 1/√n
(D) v ∝ n
The correct option (A) v ∝ 1/nExplanation:v = ze2/2∈0nh
The correct option (A) v ∝ 1/n
Explanation:
v = ze2/2∈0nh
See lessLet ρ be a relation defined on N, the set of natural numbers, as ρ = {(x, y) ∈ N × N : 2x + y = 41} Then
(A) ρ is an equivalence relation
(B) ρ is only reflexive relation
(C) ρ is only symmetric relation
(D) ρ is not transitive
The correct option (D) ρ is not transitiveExplanation:ρ = {(x, y) ∈ N × N, 2x + y = 41} for reflexive relation x R x ⇒ 2x + x = 41 ⇒ x = 41/3 ∈ N for symmetric ⇒ x R y ⇒ 2x + y = 41 ≠ y R x (Not symmetric) for transitive xRy ⇒ 2x + y = 41 and yRz ⇒ 2y + z = 41, xRz (not transitive )
The correct option (D) ρ is not transitive
Explanation:
ρ = {(x, y) ∈ N × N, 2x + y = 41}
for reflexive relation x R x ⇒ 2x + x = 41 ⇒ x = 41/3 ∈ N
for symmetric ⇒ x R y ⇒ 2x + y = 41 ≠ y R x (Not symmetric)
for transitive xRy ⇒ 2x + y = 41 and yRz ⇒ 2y + z = 41, xRz (not transitive )
See lessThe intensity of a sound appears to an observer to be periodic. Which of the following can be the cause of it?
(A) The intensity of the source is periodic.
(B) The source is moving towards the observer.
(C) The observer is moving away from the source.
(D) The source is producing a sound composed of two nearby frequencies.
The correct option(A) The intensity of the source is periodic.(D) The source is producing a sound composed of two nearby frequencies.
The correct option
(A) The intensity of the source is periodic.
(D) The source is producing a sound composed of two nearby frequencies.
See lessThe number of selection of n objects from 2n objects of which n are identical and the rest are different is
(A) 2n
(B) 2n – 1
(C) 2n – 1
(D) 2n –1 + 1
The correct option (A) 2n Explanation:nC0 + nC1 + nC2 …… + nCn = 2n
The correct option (A) 2n
Explanation:
nC0 + nC1 + nC2 …… + nCn
= 2n
See lessThe correct order of reactivity for the addition reaction of the following carbonyl compounds with ethylmagnesium iodide is
(A) I > III > II > IV
(B) IV > III > II > I
(C) I > II > IV > III
(D) III > II > I > IV
The correct option (A) I > III > II > IVExplanation:Reactivity order for nucleophilic addition reaction isH–CHO > CH3–CHO > (CH3)2CO > [C(CH3)3]2COReactivit ∝ 1/Steric crowding
The correct option (A) I > III > II > IV
Explanation:
Reactivity order for nucleophilic addition reaction is
H–CHO > CH3–CHO > (CH3)2CO > [C(CH3)3]2CO
Reactivit ∝ 1/Steric crowding
See less