FULL TEST – I Paper-2 [ANSWERS, HINTS & SOLUTIONS]

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    ANSWERS, HINTS & SOLUTIONS
    FULL TEST –I
    (Paper-2)
    Q.
    No.
    PHYSICS CHEMISTRY MATHEMATICS
    1. C D B
    2. C A B
    3. D C A
    4. B B C
    5. B B D
    6. D C D
    7. B D C
    8. C B A
    9. A C C
    10. D B C
    11. B A D
    12. D C A
    13. B A C
    14. A D A
    15. A C A
    16. B B C
    17. B A B
    18. A B A
    19. B C A
    1.
    (A) (s), (B) (q),
    (C) (p), (D) (r)
    (A p, q, r), (B q),
    (C p, r, S), (D p)
    (A) (q), (B) (r),
    (C) (s), (D) (p)
    2.
    (A) (q); (B) (r);
    (C) (s); (D) (p)
    (A p, q, s), (B r, s),
    (C q), (D r, s)
    (A) (s), (B) (p),
    (C) (q), (D) (r)
    3.
    (A) (p), (B) (r),
    (C) (s), (D) (q)
    (A p, r), (B q, p),
    (C q), (D q, s)
    (A) (s), (B) (r),
    (C) (p), (D) (q)
    ALL INDIA TEST SERIES
    FIITJEE
    JEE(Advanced)-2014
    From Classroom/Integrated School Programs
    7 in Top 20, 23 in Top 100, 54 in Top 300, 106 in Top 500 All India Ranks & 2314 Students
    from Classroom /Integrated School Programs & 3723 Students from All Programs have been Awarded a Rank in JEE (Advanced), 2013

    Page 1

  • AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/14
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    2
    P
    P
    h
    h
    y
    y
    s
    s
    i
    i
    c
    c
    s
    s
    PART – I
    2. T sin = mg/2
    T = T cos
    =
    cot
    2 2tan
    T sin
    T cos
    T
    T
    3.
    2
    1
    mv pt
    2
    (P = const)
    v =
    2Pt
    m
    a =
    dv 2P 1
    dt m
    2 t
    F = ma =
    mP 1
    2t v
    5.
    1 2 3 4 5
    F F F F F F
    2 5 2 4
    F F and F F
    1 3 2 1
    F F 2F cos30 2F cos60
    F
    3
    =
    2
    2
    Gm
    4a
    ; F
    2
    =
    2
    2
    Gm
    3a
    ; F
    1
    =
    2
    2
    Gm
    a
    m
    m
    m
    m
    m
    m
    F
    1
    F
    2
    F
    3
    F
    4
    F
    5
    F =
    2
    2
    Gm 5 1
    4
    a
    3
    = m
    2
    a
    =
    3
    Gm 5 1
    a 4
    3
    T = 2
    3
    4 3a
    Gm 5 3 4
    6.
    V
    P
    P
    Q
    R
    S
    7. a =
    v g a m g a
    m
    = 20 m/s
    2
    t =
    2h
    a
    = 1 sec
    10. Time period becomes 2
    R
    g
    . We can’t neglect the roundness of earth for the pendulum of
    infinite length.

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    3
    12. The ve sign is
    d
    dt
    signifies the direction of induced emf.
    14. process AB U = constant
    P RT
    M
    and U t
    P = const
    Process BC isochoric
    Process CA isothermal
    15. Q = Q
    AB
    + Q
    BC
    + Q
    CA
    Q = 5U
    0
    + 3U
    0
    +
    0
    10U
    ln2.5
    3
    16. W
    AB
    = Q
    AB
    U
    AB
    = 5U
    0
    (3U
    0
    ) = 2U
    0
    17. For lens L
    1
    , ray must move parallel to the axis after refraction
    11 w
    1
    x R
    x = 10 cm
    18. For lens L
    2
    , image must form at centre of curvature of the curved surface after refraction through
    plane part.
    2
    2
    R x
    0
    x = 8 cm

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    4
    C
    C
    h
    h
    e
    e
    m
    m
    i
    i
    s
    s
    t
    t
    r
    r
    y
    y PART – II
    1.
    Br
    CH
    3
    H
    O
    Mg
    Ether
    
    CH
    2
    MgBr
    CH
    3
    H
    O
    OH
    H
    Rearrangment
    
    2.
    O
    H
    3
    CO
    O
    O
    3
    CH O
    
    O
    H
    3
    CO
    O
    O
    
    O
    COOCH
    3
    O
    H
    
    O
    H
    3
    COOC
    OH
    
    O
    O
    COOCH
    3
    3.
    o o o o
    A A B A B B B
    P P X P P 1 X P X
    o o o
    A A B
    P P P P B
    Thus
    o
    A
    P 120 Torr
    o o o
    A B B
    P P 75 P 45 Torr
    Hence C is correct answer.
    4.
    4
    o
    BaSO
    1000
    Conc normality
    4
    5
    o
    BaSO
    1000 1000 8 10
    Normality
    400
    = 2 × 10
    -4
    .
    4
    Normality
    Molarity 10 M Solubility
    2
    Ksp = S
    2
    = 10
    –8
    M
    2
    .
    5.
    av
    1
    T
    K
    99.9 2
    10 0.693 6.93
    T 10 ty
    K K
    Number of natural life times =
    6.693 1
    /
    K K
    = 6.93

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    5
    6.
    B
    O
    O
    B
    B
    O
    O
    BOH
    OH
    OH
    OH
    O
    7.
    2 2
    71
    145
    Ca OCl Cl.H O Cl
    
    Percentage
    71
    100 49
    145
    Hence D is correct answer.
    17.
    4 1
    I
    d c
    0.0033
    1.32 10 m min
    dt 25
    4 1
    II
    d c
    2.6 10 m min
    dt
    3 1
    III
    d c
    1.02 10 m min
    dt
    18. On comparing rates order w.r.t A = 2, and w.r.t. B = 1. Thus rate law = K[A]
    2
    [B]
    19.
    2
    dx
    K A B
    dt
    2
    dx / dt
    K 0.26
    A B

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    6
    M
    M
    a
    a
    t
    t
    h
    h
    e
    e
    m
    m
    a
    a
    t
    t
    i
    i
    c
    c
    s
    s PART – III
    1. Total number of lines made =
    9
    C
    2
    = 36
    Now, these 36 lines are 9 sets each with 4 parallel members. With 5 vertices number of lines
    made
    =
    5
    C
    2
    = 10
    Clearly, atleast 2 members belong to the same set
    So, atleast one pair is parallel
    2.
    2 2 2
    OA OB AB
    cos
    2OA OB
    =
    2
    2 2
    2 2
    OA OB
    OA OB
    3 OA OB 1
    2
    2OA OB 8 OA OB 4
    For maximum cos ,
    2 2
    3 OA OB 1 3 2 OA OB 1
    8 OA OB 2 8 OA OB 4
    =
    1
    2
    3
    3. S
    1
    = a
    1
    S
    2
    = a
    1
    + a
    2
    S
    3
    = a
    1
    + a
    2
    + a
    3
    S
    n
    = a
    1
    + a
    2
    + ….. + a
    n
    If we divide all S
    1
    , S
    2
    , ….. S
    n
    by 23 we get remainders 1, 2, 3, 4, ….., 22
    So, two of these give same remainders S
    p
    , S
    q
    S
    p
    – S
    q
    will be divisible by 23
    S
    p
    – S
    q
    = a
    p + 1
    +
    a
    p + 2
    + ….. + a
    q
    4. Using Cauchy we get,
    2 2 2 2
    2
    1 2 3 4
    1 2 3 4
    1 1 2 4
    z z z z 1 1 2 4 64
    z z z z
    5. Point of intersection is
    1 b
    ln
    2 a
    For C
    1
    ,
    dr
    ae
    d
    , now
    1
    d
    tan r
    dr
    =
    1
    ae a e 1
    1
    4
    For C
    2
    ,
    dr
    be
    d
    2
    1
    tan be e 1
    b
    2
    3
    4
    Angle of intersection is
    2 1
    3
    4 4 2

    Page 6

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    7
    6. Let c, d, e be the three points where y = f(x) crosses x–axis. Then, f(c) = f(d) = f(e) = 0
    Assuming a < c < d < e < b. The function f satisfies Rolle’s theorem in two intervals (c, d) and
    (d, e). Since f and f are continuous and f(c) = f(d) = 0
    So, there exists, at least one point in the interval (c, d) and (d, e) such that derivative is zero
    Let, C
    1
    (c, d) such that f(C
    1
    ) = 0 and C
    2
    (d, e) such that f(C
    2
    ) = 0. Now the function f
    satisfies Rolle’s theorem since f, f are continuous and f(C
    1
    ) = f(C
    2
    ) = 0
    So, by Rolle’s theorem, there exists a number C
    3
    in between C
    1
    and C
    2
    such that f(C
    3
    ) = 0
    Minimum one root C
    3
    of the equation f(x) = 0 lies in the interval (a, b)
    7. Dividing the given differential equation by 3xy(y
    2
    – x
    2
    )
    2 2 2 2
    2 2 2 2
    y y 2x x 2y x
    dx dy 0
    3xy y x 3xy y x
    2 2 2 2
    dx xdx ydy dy
    0
    x y
    y x y x
    2 2
    2 2
    d y x
    1
    d ln xy 0
    2
    y x
    2 2 2 2
    d ln x y y x 0
    2 2 2 2
    ln x y y x c
    x
    2
    y
    2
    (y
    2
    – x
    2
    ) = c
    8. Let the circle be
    2
    2 2
    x y a
    . Let the point of intersection of tangents at P and Q be (h, k).
    Then equation of PQ, is
    2
    hx k y a 0
    . As it passes through
    a,0
    , so,
    2
    ha k a 0
    .
    2 2
    k a h a 0. D 0 k 4a h a 0
    i.e.
    2
    y 4a x a
    .
    9. Consider
    1 1
    2
    2 2
    0 0
    x f x dx f x 2 xf x x f x dx
    =
    2
    – 2
    2
    +
    2
    = 0
    However f(x) assumes only positive values i.e. in (0, 1)
    ( – x)
    2
    (f(x)) > 0 integral can’t be zero
    10. Differential equation can be written as, (p – x)(p 2 sin x)(2p + cos x) = 0 which has solution as
    (2y – x
    2
    – c)(y + 2 cos x – c)(2y + sin x – c) = 0
    11. Put x = –1 we get (–1 + 1) p(–1) + 1 = (–1)
    n + 1
    (n + 1)!
    n 1
    1
    n 1 !
    So,
    n 1
    1 x x 1 ..... x n x
    p x
    n 1 ! n 1 x 1
    Clearly,
    1, where 'n' is odd
    p n 1
    n
    , where 'n' is even
    n 2

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    8
    13. If e is the eccentricity then,
    2 2
    2
    2
    e
    Now, we know,
    2 2
    1 1
    a b
    2
    2 2
    1
    ab h
    2 2
    2
    a b
    ab h
    ,
    2 2
    2
    1
    ab h
    2
    2
    2 2
    2
    a b 4 ab h
    ab h
    2
    2
    2 2
    1 1
    a b 4 ab h
    For an ellipse
    2
    2
    2
    2
    a b a b 4 ab h
    So,
    2
    2
    2
    2 2
    2
    a b 4h
    e a b a b 4h
    2 ab h
    14. Put y = z = t = 0
    f(0)[f(x) + f(0)] = f(0)
    Put x = 0
    2f
    2
    (0) = f(0)
    f(0) = 0,
    1
    2
    If f(0) =
    1
    2
    f(x) +
    1
    2
    = 1
    f(x) =
    1
    2
    If f(0) = 0, z = t = 0
    f(x) f(y) = f(xy)
    Let, x = y = 1 f
    2
    (1) = f(1)
    f(1) = 0 or f(1) = 1
    We have f(0) = 0, f(1) = 0, y = 1
    f(x) = 0
    Also, f(0) = 0, f(1) = 1, x = 0, y = t = 1
    (f(0) + f(z)) (f(1) + f(1)) = f(–z) + f(z)
    2f(z) = f(–z) + f(z)
    f(z) = f(–z)
    15. If y = x in f(x) f(y) = f(xy)
    f(x
    2
    ) = f
    2
    (x) 0
    Put x = t, y = z
    [f(x) + f(y)]
    2
    = f(x
    2
    + y
    2
    )
    f(x
    2
    + y
    2
    ) = f
    2
    (x) + f
    2
    (y) + 2f(x)f(y) f
    2
    (x)
    f(x
    2
    + y
    2
    ) f(x
    2
    )
    f is non decreasing for positive x
    16. Put y = z = t = 1
    2(f(x) + 1) = f(x 1) + f(x + 1)
    f(2) = 4, f(z) = 9, f(1) = 1, f(0) = 0
    f(n) = n
    2
    (Possible function), if f(n – 1) = (n – 1)
    2
    2[f(n – 1) + 1] = f(n – 2) + f(n)

    Page 8

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    9
    f(n) = n
    2
    Now, for
    p
    x
    q
    (rational number)
    2
    p
    f f q f pq
    q
    2 2 2
    p
    f q p q
    q
    2
    p p
    f
    q q
    (True for rational number)
    Now, if x R, lets prove for positive x since if it is proved the function is even and will follow for
    negative x
    Assume for x > 0, f(x) < x
    2
    So, now a rational number ‘r’ between
    f x
    and x
    f x
    < r < x
    f(x) < r
    2
    < x
    2
    [f(r) = r
    2
    , f is non decreasing]
    f(r) = r
    2
    f(x) [contradiction]
    f(x) < x
    2
    (impossible)
    Similarly we can prove contradiction f(x) > x
    2
    So, only possibility f(x) = x
    2
    substituting f(x) = x
    2
    we get
    (f(x) + f(z)) (f(y) + f(t)) = f(xy zt) + f(xt + yz)
    (x
    2
    + z
    2
    )(y
    2
    + t
    2
    ) = (xy – zt)
    2
    + (xt + yz)
    2
    [Lagrange identity]
    So, f(x) =
    1
    2
    , f(x) = 0, f(x) = x
    2
    are the required solutions
    17. am
    3
    + m(2a – h) + k = 0
    am
    3
    + m(2a – x
    1
    ) = 0
    am
    2
    = x
    1
    – 2a {m = 0 (one possible value)}
    2
    1
    x 2a
    m
    a
    , if x (0, 2a)
    m
    2
    = (–) number, so non real roots
    18. Let P(h, k) be the point am
    3
    + m(2a – h) + k = 0
    Since, m
    1
    , m
    1
    , m
    1
    are the possible roots
    3m
    1
    = 0 m
    1
    = 0
    If m
    1
    = 0 is the root then k = 0
    am
    3
    + m(2a – h) = 0
    2
    h 2a
    m 0
    a
    h = 2a
    (2a, 0) is the only point
    19.
    8a
    3 / 2
    1
    2a
    4
    A x 2a dx
    27a
    =
    8a
    5/ 2
    2a
    2 2
    x 2a
    5
    3 3a
    =
    5/ 2
    2
    2 2 2 2
    6a 36a 6a
    5 5
    3 3a 3 3a
    =
    2
    48 2a
    5
    8a
    8a
    3 / 2
    0
    1
    0
    2
    A 4axdx 2 a x
    3
    =
    2
    4 64
    a 8a 8a 2a
    3 3
    Area = 2(A
    2
    – A
    1
    ) =
    2
    4 3
    2 16 2a
    3 5
    =
    2
    2
    11 352 2a
    32 2a
    15 15

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    10
    SECTION – B
    1. (A) |A – xI| = 0
    There exists a non zero matrix X such that AX = xX
    [(Adj A)A]X = x(adj A)X
    |A|IX = x(adj A)X
    |A|X = x(adj A)X
    A X
    adj A X
    x
    A X
    adj A X
    2
    (B)
    A' I A I ' A I
    A' I 0
    if and only if
    A I
    = 0
    A' I 0
    if and only if
    A I
    = 0
    or
    is the root of
    A' xI 0
    if and only if is the root of |A – yI| = 0
    e
    –i
    is the required solution
    (C)
    11 12 1n
    22 2n
    nn
    A A .......... A
    0 A ..... A
    0 | |
    B I
    | | |
    | | |
    | | A
    = 0
    (|A
    11
    | – )(|A
    22
    | – ) ….. (|A
    nn
    | – ) – 0
    Clearly the elements of principal diagonal become the roots
    (D) Let, AX = X [X is a non zero matrix]
    X'
    AX =
    X'
    X =
    X' AX'IX
    X'AX
    and
    X'IX
    are both real
    Also,
    X'X 0
    , X 0
    X'AX/ XIX
    is real so can have real values
    2. (A) z = cos + i sin ,
    1
    z 2cos
    z
    1
    z 2isin
    z
    p
    p
    1
    z 2cosp
    z
    Now,
    4 2
    4 2
    1 1
    2isin 2cos z z
    z z
    =
    6 4 2
    6 4 2
    1 1 1
    z 2 z z 4
    z z z
    then, 2 cos 6 – 2.2. cos 4 – 2 cos 2 + 4
    2
    4
    i
    4
    sin
    4
    2
    2
    cos
    2
    = 2(cos 6 – 2 cos 4 – cos 2 + 2)
    = 2

    Page 10

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