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Wbjee

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  1. Asked: 3 years agoIn: Wbjee

    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)

    1ae94
    Added an answer about 3 years ago

    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)

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  2. Asked: 3 years agoIn: Wbjee

    If 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

    abb26
    Added an answer about 3 years ago

    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

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  3. Asked: 3 years agoIn: Wbjee

    Let 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)

    6666b
    Added an answer about 3 years ago

    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)

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  4. Asked: 3 years agoIn: Wbjee

    If (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

    54105
    Added an answer about 3 years ago

    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

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  5. Asked: 3 years agoIn: Wbjee

    A 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

    ab845
    Added an answer about 3 years ago

    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.

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  6. Asked: 3 years agoIn: Wbjee

    How 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

    01dbb
    Added an answer about 3 years ago

    The correct option (A) v ∝ 1/nExplanation:v = ze2/2∈0nh

    The correct option (A) v ∝ 1/n

    Explanation:

    v = ze2/2∈0nh

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  7. Asked: 3 years agoIn: Wbjee

    Let 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

    42b2a
    Added an answer about 3 years ago

    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 ) 

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  8. Asked: 3 years agoIn: Wbjee

    The 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.

    96373
    Added an answer about 3 years ago

    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.

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  9. Asked: 3 years agoIn: Wbjee

    The 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

    ec0c3
    Added an answer about 3 years ago

    The correct option (A) 2n  Explanation:nC0 + nC1 + nC2 …… + nCn = 2n

    The correct option (A) 2n  

    Explanation:

    nC0 + nC1 + nC2 …… + nCn 

    = 2n

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  10. Asked: 3 years agoIn: Wbjee

    The 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

    325c9
    Added an answer about 3 years ago

    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

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