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Mathematics Mock Test - 3

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Mathematics Mock Test - 3
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  • Question 1
    5 / -1
    The projection of a vector on another vector is 
    Solution

    Concept:

    Vector projection: The vector projection of a on b is a vector a1 which is either null or parallel to b. More exactly, a1 = 0 if θ = 90°, a1 and b have the same direction if 0 ≤ θ < 90 degrees,

    a1 and b have opposite directions if 90 degrees < θ ≤ 180 degrees.

     

    A projection of one vector on another vector is always scalar quantity.

  • Question 2
    5 / -1

    The solution of ∫logx/x2dx is

    Solution

    I = ∫logx/x2dx

    By integration by parts:

    I = ∫1/x2logxdx

    I = logx(- 1/x) - ∫1/x (- 1/x) dx

    I = - logx/x - 1/x + c
  • Question 3
    5 / -1

    The degree of the given equation is

    dydxx=(yxdydx)6

    Solution

    Concept:

    Order: The order of a differential equation is the order of the highest derivative appearing in it.

    Degree: The degree of a differential equation is the power of the highest derivative occurring in it after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

    Calculation:

    dydxx=(yxdydx)6, rearranging the given equation we get,

    dydxx=1(yxdydx)6

    (dydxx)×(yxdydx)6=1

    After multiplying the two terms the power of the highest derivative will be 7

    Hence the degree = 7

  • Question 4
    5 / -1
    The function y = f(x) has relative minima where:
    Solution

    Explanation:

    Maxima/ Minima of a function y = f(x):

    • Relative (Local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
    • Relative (Local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
    • At the points of relative (local) maxima or minima, f'(x) = 0.
    • At the points of relative (local) maxima, f''(x) < 0.
    • At the points of relative (local) minima, f''(x) > 0.

    From the above definition it is clear that at the points of relative minima of a function y = f(x), the value of f'(x) = 0 and f''(x) > 0.

  • Question 5
    5 / -1
    What is the principal value of cosec1(2)?
    Solution

    Concept:

    Principal value of cosec-1θ is  [π2, π2]0 

    cosec[cosec-1x] = x

    Calculation:

    Let, 

    y = cosec1(2)

    ⇒  cosec y = - √2

    ⇒  cosec y = cosec(π4)

    ∴ Principal value is π4.

    Additional InformationPrincipal Values of Inverse Trigonometric Functions:

    Function

    Domain

    Range of Principal Value

    sin-1 x

    [-1, 1]

    [-π/2, π/2]

    cos-1 x

    [-1, 1]

    [0, π]

    csc-1 x

    R - (-1, 1)

    [-π/2, π/2] - {0}

    sec-1 x

    R - (-1, 1)

    [0, π] - {π/2}

    tan-1 x

    R

    (-π/2, π/2)

    cot-1 x

    R

    (0, π)

  • Question 6
    5 / -1
    The line with direction ratios 1, 0, -1 is inclined with z - axis at an angle
    Solution

    Direction ratio of the given line are 1, 0, -1

    Direction cosine of the given line are

    112+02+(1)2,012+02+(1)2,112+02+(1)2

    i.e. 12,0,12,

    cosγ=12γ=cos1(12)=135=(π180×135)=3π4

  • Question 7
    5 / -1
    If [1x1] [123 456 325] [1 2 3] = 0 then the value of x is
    Solution

    Concept:

    For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. 

    Calculation:

    Consider,

    [1x1] [123 456 325]

    [1+4x+3 2+5x+2 3+6x+5]

    [4+4x 4+5x 8+6x]

    [1x1] [123 456 325][4+4x 4+5x 8+6x]

    Now multiply [1 2 3] both hand sides we have,

    [1x1] [123 456 325] [1 2 3] = [4+4x 4+5x 8+6x] [1 2 3]

    ⇒ 4 + 4x - 8 - 10x + 24 + 18x = 0

     20 + 12x = 0

     20 = - 12x

    ⇒ x = -20/12

    ⇒ x = - 5/3

    Hence, the correct answer is option 1).

  • Question 8
    5 / -1

    Let \(\rm f(x)=\left\{\begin{matrix}3x-4,&0\le x\le2\\ 2x+l, &2

    If f is continuous at x = 2, then what is the value of l?

    Solution

    Concept:

    Let y = f(x) be a function. Then,

    The function is continuous if it satisfies the following conditions:

    limxaf(x)=limxa+f(x)=f(a)

    Calculation:

    \(\rm f(x)=\left\{\begin{matrix}3x-4,&0\le x\le2\\ 2x+l, &2

    LHL = limx2f(x)

    ⇒ LHL = limh0f(2h)

    ⇒ LHL = limh03(2h)4

    ⇒ LHL = 2

    RHL = limx2+f(x)

    ⇒ RHL = limh0f(2+h)

    ⇒ RHL = limh02(2+h)+l

    RHL = 4 + l

    Since, f(x) is continuous at x = 2,

    LHL= RHL = f(2)

    4 + l = 3(2) - 4

    ⇒ 4 + l = 2

    ⇒ l = -2

    ∴ The value of l is -2.

  • Question 9
    5 / -1
    The number that exceeds its square by the greatest amount is 
    Solution

    Concept:

    • We have to first formulate the function of maximization as the given data.
    • Let the number is x, then,

    y = x - x2

    Calculation:

    • For the greatest value of y,

    dydx=12x

    d2ydx2=2(<0)

    • So, y will be the maximum for all the values of x where, 

    dydx=12x=0

     ⇒ 1 - 2x = 0

    ⇒  x = 1/2

    • So, the correct answer is option 3.
  • Question 10
    5 / -1
    If c is a point at which Rolle's theorem holds for the function f(x)=loge(x2+α7x) in the interval [3, 4], where α ∈ R, then the value of c is equal to 
    Solution

    Concept:

    By Rolle’s Theorem,

    f(3) = f(4)

    Also, f'(c) = 0

    Calculation:

    Given:

    f(x)=loge(x2+α7x)

    9+α21=16+α28

    ⇒ α = 12

    Now after putting the value of  α, differentiate it with respect to x,

    f(x)=7xx2+12×x2127x2=x212x(x2+12)

    As per Rolle’s Theorem,

    ∴ f'(c) = 0

    ⇒ c = 2√3 

    • So, the correct answer is option 2

     

  • Question 11
    5 / -1
    If the curve y = ax2 + bx + c, x ∈ R, passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are 
    Solution

    Concept:

    • The tangent line to this curve at origin is y = x,

     dydx|(0,0)=1

    • If the curve y = ax2 + bx + c, x ∈ R, passes through the point (1, 2) so,

    ⇒ a (1)2 + b × 1  + c = 2       ...(i)

    Calculation:

    Given: y = ax2 + bx + c, x ∈ R

    2ax+b|(0,0)=1

    ⇒ b = 1

    • The curve passes through the origin,

    ⇒ c = 0

    • By equation 1,

    ⇒ a (1)2 + 1 × 1  + 0 = 2  

    a = 1

    • So, the correct answer is option 3
  • Question 12
    5 / -1
    The mean of the numbers obtained on throwing a die having written 5 on four faces, 4 on three faces, 3 on two faces and 2 on one face is?
    Solution

    Calculation:

    Let X be the random variable representing a number on the die.

    The total number of observation is six.

    P(X = 5) = 46=23 

    P(X = 4) = 36=12

    P(X = 3) = 26=13

    P(X = 2) = 16

    Mean(X) = xipi

    Mean(X) = 5×23+4×12+3×13+2×16

    Mean(X) = 203

  • Question 13
    5 / -1
    A continuous random variable X has a probability density function f (x) = e-x, 0 < x < ∞, Then P{X > 1} is
    Solution

    Concept:-

    P{X > 1} is representing the probability for all the values of random variable X > 1.

    P(X>1)=1exdx

    Calculation:

    Given: f (x) = e-x, 0 < x < ∞

    P(X>1)=1exdx

    P(X>1)=ex1|1

    P(X>1)=ex1|1

    ⇒ P(X > 1) = - (e-∞ - e-1)

    ⇒  P(X > 1) = 1/e

    So, the correct answer is option 1

  • Question 14
    5 / -1

    A discrete random variable X has the probability functions as:

    X

    0

    1

    2

    3

    4

    5

    6

    7

    8

    f(x)

    K

    2k

    3k

    5k

    5k

    4k

    3k

    2k

    k


    The value of E(X) is:
    Solution

    Explanation:

    x

    F(x)

    xf(x)

    0

    k

    0

    1

    2k

    2k

    2

    3k

    6k

    3

    5k

    15k

    4

    5k

    20k

    5

    4k

    20k

    6

    3k

    18k

    7

    2k

    14k

    8

    k

    8k

    Total

    26k

    103k

     

    We know that for Fpr random variable X sum of Probaboility = 1

    ⇒ ∑Pi = 1

    ⇒ k + 2k + 3k + 5k + 5k + 4k + 3k + 2k + k + 1

    ⇒ 26k = 1

    ⇒ k = 1/26

    ∴ xf(x) = 103k = 103 × 1/26

    ∴ E(x) is 103/26

  • Question 15
    5 / -1
    A continuous random variable X has uncountable many values in the interval [a, b]. If C is a values in the interval [a, b], then P{ X = C }
    Solution

    Probability Density Function:

    It indicates the distribution of total probability to various random variables (R.V)

    f(x)=ddxF(x)

    f(x) : pdf (Probability density function)

    F(x) : PDF (probability Distribution function)

    For a continuous R.V., the probability of occurrence at a particular value of ‘x’ will be ‘zero’. [Area approaches to zero].

    Since the given R.V. is continuous and the period/interval is [a, b]:

    P[x = c] will be zero.

    Important Notes:

    Properties of pdf:

    1) axf(x)dx=1 (or) Total area = 1

    2) P[xx1]=F(x1)=x1f(x)dx

    3) P[x>x1]=1F(x1)=x1f(x)dx 

    4) P[x1<xx2]=x1x2f(x)dx

    For Discrete R.V., the Probabilities are defined at a particular value. Whereas for continuous R.V. probabilities are defined for a particular interval.
  • Question 16
    5 / -1
    Let R be a relation on the set of natural numbers defined by ‘xRy ⟺ x ≤ y’. Which one of the following is correct?
    Solution

    Concept:

    Properties of relation:

    Let R be a relation on Z, and let x, y, z ∈ Z.

    1. Reflexive

     xRx

    1. Symmetric 

     xRy implies yRx

    1. Transitive

     xRy and yRz implies xRz

     

    Calculation:

    Given:

    ‘xRy ⟺ x ≤ y’

    R is a relation on the set of natural numbers (N) and x, y ∈ N

    ‘xRy ⟺ x ≤ y’

    i. x ≤ x or 5 ≤ 5 so its reflexive.

    ii. 3 ≤ 5 but 5 is not less than 3.  i.e., x ≤ y doesn’t imply y ≤ x so NOT symmetric

    iii. 3 < 5 and 5 < 7 that means 3 < 7 i.e., xRy and yRz implies xRz so transitive.

    Hence, option (2) is correct.

  • Question 17
    5 / -1
    if a function defined as  f : N → N f(x) = (x - 1)/2 Then which of the following is true regarding this function?
    Solution

    Given: f(x) = (x - 1)/2 = y

    • This given function is defined for the domain as N which is having the codomain as N

    Explanation:

    Definition of a function:

    • Note that the given expression is not a function by the definition of the function.
    • A single-valued function is a function for each point in the domain, that has a unique value in the range.
    • Also we know that the range is the subset of codomain.

    Here, for, x = 1 → y = 0 which is not the element of codomain N, so for this value of x in the domain we do not get any image or output which is violating the definition of the function.

    • So, the given expression is not a function actually so the given data is invalid.

     

  • Question 18
    5 / -1

    For the following Linear Programming Problem, the constraints are Graphically expressed as shown. What will be the optimal maximized value of objective function z?

    Where,
    Z = 2x + 4y

    x ≥ 0 and y ≥ 0

    Solution

    CONCEPT:

    • As shown in the given graph above, the shaded bounded region OABCO represents the common region of the above inequation. This region is the feasible region of the given LPP.
    • The coordinates of the vertices (corner point) of the shaded feasible region are O(0, 0), A(0, 2.5), B(3, 1), and C(4, 0).
    • The value of the objective function z will be a maximum value of among the values on the corner points of the feasible region.

    CALCULATION:

    • The value of the objective function on these corner points are given in the following table:
    Corner PointsCoordinatesObjective Function
    Z = 2x + 4y
    O(0, 0)0
    A(0, 2.5)

    10 (Max.)

    B(3, 1)10 (Max.)
    C(4, 0)8

     

    • Clearly, Z has maximized at two corner points A(0, 2.5) and B(3, 1).
    • Here, any point on the line segment joining points A and B will give the maximum value Z = 10 of the objective function.
    • The optimal maximized value of Z is 10 when x = 0 and y = 2.5 or when x = 3 and y = 1

    So, the correct answer is option 2.

  • Question 19
    5 / -1
    M is a square matrix of order ‘n’ and its determinant value is 5. If all the elements of M are multiplied by 2, its determinant value becomes 40. The value of ‘n’ is
    Solution

    Concept:

    If all the elements of matrix M be multiplied by some scalar k, new determinant value is if n-rowed square matrix is given by

    |k×M|=kn×|M|

    Data: 

    Scalarnumber(k)=2,|M|=5,|k×M|=40andOrderofM=n

    Calculation:

    kn×|M|=40,2n×5=40,2n=82n=23

    Therefore, the order of matrix is 3.

  • Question 20
    5 / -1

    Evaluate:

    tan-1(1) + cos-1(-1/2) + sin-1(-1/2) = ?

    Solution

    CONCEPT:

    • tan (π/4) = 1
    • sin-1 x + cos-1 x = π/2, x ∈ [-1, 1]

    CALCULATION:

    Let S = tan-1(1) + cos-1(-1/2) + sin-1(-1/2)

    As we know that, sin-1 x + cos-1 x = π/2, x ∈ [-1, 1]

    ⇒ S = tan-1(1) + π/2

    As we know that, tan(π/4) = 1

    ⇒ S = π/4 + π/2 = 3π/4

    Hence, correct option is 1. 

  • Question 21
    5 / -1
    The differential coefficient of f [log(x)] when f (x) = log x is 
    Solution

    Concept:

    • We will use the chain rule here for differentiating the given function.

    Calculation:

    Given: f (x) = log x;

    ⇒  f [log x] = log log x 

    ⇒  f' [log x] = 1logx.ddxlog x 

    f' [log x] = 1xlogx.

    So the correct answer is option 3.

  • Question 22
    5 / -1

    The system of equations

    x + 2y + z = 0, x – z = k, x + y = 0

    has infinite many solution then the possible value of k?

    Given: AdjA=[ 1 1 2  1 1 2 1 1 2]

    Solution

    Concept:

    Consider the system of m linear equations

    a11 x1 + a12 x2 + … + a1n xn = 0

    a21 x1 + a22 x2 + … + a2n xn = 0

    am1 x1 + am2 x2 + … + amn xn = 0

    The above equations contain the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrix.

    A=[a11a12a1na21a22a2nam1am2amn]

    We can write,

    AX = B

    ⇒ X = A-1B

    • A is the coefficient matrix of the given system of equations.
    • For the system of the equation has an infinite number of solutions,

    ⇒ Adj(A). B = O and |A| = 0

    Calculation:

    Given : AdjA=[ 1 1 2  1 1 2 1 1 2]

    Also, A=[ 1 2 1  1 0 1 1 1 0]x=[ x  y z]B=[ 0  k 0]

    ⇒ |A| = 1 - 2 + 1 = 0

    • Now, we will find adj (A)

    ⇒ Adj(A).B = [ 1 1 2  1 1 2 1 1 2][ 0  k 0]

    Adj(A).B=[ k  k k]

    • For the system of the equation has an infinite number of solutions,

    ⇒ Adj(A).b = O Which is only possible when,

    ⇒ k = 0

    ⇒ Adj(A).B = O

    • ∴ The system equation has infinitely many solutions for k = 0.
    • So, the correct answer is option 2.
  • Question 23
    5 / -1
     The points (5, -2), (8, -3) and (a, -12) are collinear if the value of a is 
    Solution

    Concept:

    If the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

    Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area (A) of ΔABC is given as; 

    A=12|x1y11x2y21x3y31|

    Calculation

    Here, we have to find the value of a for  which the points (5, -2), (8, -3) and (a, -12) are collinear

    Let,

    x1 = 5, y1 = -2,

    x2 = 8, y2 = -3,

    x3 = a, y3 = -12.

    As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a Δ ABC then area of Δ ABC =   A=12|x1y11x2y21x3y31|

    A=12|521831a121|

    2A = 5 (-3 + 12) + 2(8 - a) + 1(-96 + 3a)

    2A = 45 + 16 - 2a - 96 + 3a

    2A = a - 35  

    ⇒ A = (a - 35)/2

    ∵ The given points are collinear.

    As we know that, if the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

    ⇒ (a - 35)/2 = 0

    ⇒ a = 35

    Hence, option D is the correct answer.

    Alternate Method 

    Concept:

    Three or more points are collinear if the slope of any two pairs of points is the same.

    The slope of a line passing through the distinct points (x1, y1) and (x2, y2) is y2y1x2x1

    Calculation:

    Let, A = (5, -2), B = (8, -3), C = (a, -12)

    Now, the slope of AB = Slope of BC = Slope of AC      (∵ points are collinear)

     3(2)85=12(3)a813=9a8

    ⇒ a - 8= 27

    ⇒ a = 27 + 8 = 35

    Hence, option (4) is correct.

  • Question 24
    5 / -1

    In the figure, a = 2i, b= i^ + j + k^^, c = 2k^  and vector x satisfies the equation x - w = v. Then x is

     

    Solution

    Concept:

    Triangle Law of Vector Addition: Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

    ⇒ R=a+b

    Calculation:

    Given that, 

    a = 2i, b= i^ + j + k^^, c = 2k^

    In Δ ABD and Δ ADC using triangle law

    w = b  a

    v = b  c

    According to question, x - w = v

     x = v + w

     x = b  c + b  a

     x = a + 2b  c

     x = 2i + 2(i + j + k^)  2k^

     x = 2j

    Hence, option (1) is correct.

  • Question 25
    5 / -1

    The two lines x+34=y63=z2andx+24=y1=z71  are?

    Take (B1×B2)=(1,4,8) where B1 and B2 are representing the directions of the first and second lines respectively.

    Solution

    Concept:

    • The shortest distance between the lines xx1a1=yy1b1=zz1c1  and xx2a2=yy2b2=zz2c2 is given by,

    d=|(A2A1).(B1×B2)^|

    d=|(A2A1).(B1×B2)|B1×B2||

    Where (B1×B2)^ is the unit vector perpendicular to B1 and B2

    • Here.  B1 and B2 are representing the direction of the first and second lines A1 and A2 are representing the points on the first and second lines
    • If B1B2 ⇒ So the lines are not parallel as the direction of both the lines is not the same.
    • If the two non-parallel lines are intersecting then d ≠ 0 ⇒ The lines are skew parallel lines.

    Calculation:

    Given: (B1×B2)=(1,4,8)

    |(B1×B2)|=1+16+64=9

    B1=(4,3,2)B2=(4,1,1)

     A1=(3,6,0) is a point on the first line and A2=(2,0,7)

    B1B2

    • So the lines are not parallel as the direction of both the lines is not the same.

    A2A1=(2(3),06,70)=(1,6,7)

    (B1×B2)=(1,4,8)

    d=|(A2A1).(B1×B2)|B1×B2||

    d=(2+3).1+(06)(4)+(70)81+16+64=819=9

    ⇒ d ≠ 0

    • So the lines are non-intersecting and non-parallel so the lines are skew parallel lines.
    • Hence, option 2 is correct.

    Note:

    • For the vector A=(1,1,9)
    • The above vector can also be written as, A=(1i^+1j^+9k^)
     
  • Question 26
    5 / -1
    Find limn(1n+1n+1+1n+2+...+13n)
    Solution

    Concept:

    Summation of infinite series bu integration:

    Step:1 Write the general form of series in terms of r and n.

    Step:2 Change  limxr=1n

    Step:3 Change f(rn)  f(x)

    Step:4 Change 1n  dx

    Step:5 Upper limit = limn (rn)at r = 1

    Step:6 Upper limit = limn (rn)at r = n

    Calculation:

    Let,

    I = limn(1n+1n+1+...+1n + 2n)

    The general form of the above series is written as

    I =limnr=02n1n+r

    I =limnr=02n1n11+rn

    This general series can be converted as integration, as discussed in the concept.

     I =0211+xdx

    We know that, ∫ (1/x)dx = ln x + C

     I =[log(1+x)]02

    Taking the limit of integration, we will get

    ⇒ I = log 3

  • Question 27
    5 / -1
    If a matrix  A is such that 3A3 + 2A2 + 5A + I = O, then what is A-1 equal to?
    Solution

    Concept:

    Let A be the square matrix.

    A-1A = I and A-1.I = A-1

    Calculation:

    Given, a matrix A is such that 3A3 + 2A2 + 5A + I = 0

    Consider, 3A+ 2A2 + 5A + I = 0

    Pre-Multiplying the above polynomial by A-1.

    A-1(3A+ 2A2 + 5A + I) = A-1(0)

    ⇒ 3A-1A+ 2A-1A2 + 5A-1A + A-1.I = 0

    ⇒ 3A-1A.A2 + 2A-1A.A + 5A-1A + A-1.l = 0

    We know that A-1A = I and A-1.I = A-1

    ⇒ 3IA2+ 2IA + 5I + A-1 = 0

    ⇒ 3A2 + 2A + 5I + A-1 = 0

    ⇒ A-1 = - (3A2 + 2A + 5I)

    ∴ A-1 = - (3A2 + 2A + 5I)

  • Question 28
    5 / -1
    Let A and B be two events such that P(AB)=16, P(A ∩ B) = 14and P(A̅) = 14, where A̅ stands for complement of event A. Then, events A and B are:
    Solution

    Concept:

    Independent Events

    Let E and F be two events associated with a sample space s. If the probability of occurrence of one of them is not affected by the occurrence of the other, then we say that the two events are independent. thus, two events E and F will be independent, if 

         (a) P(F | E)= P(F) , provided P(E) ≠ 0

         (b) P(E | F) =P(E) , provided P(F) ≠ 0

    Using the multiplication theorem on probability, we have 

         (c) P(E ⋂ F) = P(E) P(F)

    There events A, B and C are said to be mutually independent if all the following condition hold:

    • P(A ⋂ B) = P(A) P(B)
    • P(A ⋂ C) = P(A) P(C)
    • P(B ⋂ C) = P(B) P(C)
    • P(A ⋂ B ⋂ C) = P(A) P(B) P(C)

    Calculation:

    Given:

    P(A ∩ B) = 14 and P(A̅) = 14,

    and P(AB)=16

    ⇒ 1 - P(A ∪ B) = 1/6 {Such that p(A) + p(A¯) = 1)}

    ⇒ 1 - P(A) - P(B) + P(A ∩ B) = 1/6

    Such that P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

    ⇒ P(A¯) - P(B) + 1/4 =1/6

    ⇒ P(B) = 1/4 + 1/4 - 1/6

    ⇒ P(B) = 1/3 and P(A) = 3/4 

    Now  P(A ∩ B) = 1/4 = 3/4 × 1/3 = P(A) P(B)

    Hence, events A and B are independent events but not equally likely.

  • Question 29
    5 / -1

    Maximize Z = 2X1 + 3X2

    Subject to

    2X1 + X2 ≤ 6

    X1 – X2 ≥ 3

    X1, X2 ≥ 0

    The solution to the above LPP is
    Solution

    Concept:

    • In order to find the maximum value of the objective function, the constraints of the objective function are drawn and the region formed by the constraints is the feasible region.

    Draw the constraints to find the feasible region:

    • To draw the inequalities, first, draw the equation form of the inequalities.
    • Convert all the constraints to equality and plot on the graph. Put the value of (x1, x2) obtained from the corner points of the feasible region and put it in the objective function.
    • Now check the region which we have to choose depending on the sign of inequality.
    • To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
    • If it is satisfying the inequality then take the region containing t (0,0) else the opposite side of (0,0).

    Calculation:

    Given:

    • Following cases are observed by the region formed by the constraints

    Calculation:

    Given:

    • The objective function to maximize is,

    Z = 2X1 + 3X2

    • Which is subjected to the constraints,

    2X1 + X2 ≤ 6

    X1 – X2 ≥ 3

    X1, X2 ≥ 0

    • If the given constraints are drawn graphically then,

    • The feasible region is a point here, therefore the given LPP has an optimal solution and it will occur at point (3, 0).
  • Question 30
    5 / -1
    Assuming that the sums and products given below are defined, which of the following is not true for matrices?
    Solution

    A+B=B+A

    If AB=ACA1AB=A1ACB=C

    If AB=0 then A=0 or B=0 satisfies but always need not be true.

    A=[0001],B=[1000]

    Here both A and B are not null matrices but AB is a null matrix.

    (AB)T=BTAT
  • Question 31
    5 / -1
    What is the area bounded by the curves |y| = 1 – x2?
    Solution

    Calculation:

    |y|={y,y<0y,y0

    For y ≥ 0

    y = 1- x2

    For y < 0

    -y = 1- x2

    ⇒ y = x2 – 1

    So area under the curve = 4 × Area under the region OABO (symmetry)

    =4×01(1x2)dx

    =4×[xx33]01

    =4×(113)=83square units

  • Question 32
    5 / -1
    If I1=exdxex+ex and I2=dxe2x+1,then what is I1 + I2 equal to?
    Solution

    Formula used:

    xndx=xn+1n+1+C

    ∫ dx = x + C

    Calculation:

    I1=exdxex+ex       

    I1=exdxex+1ex

    I1=e2xdxe2x+1        ----(1)

     I2=dxe2x+1,          ----(2)

    Add equation (1) & (2)

    ⇒ I = I1 + I2

    ⇒ I = e2xdxe2x+1 + dxe2x+1

    I=[e2xe2x+1+1e2x+1]dx

    I=[e2x+1e2x+1]dx

    ⇒ I = ∫dx  

    Using the formula given above 

    ∴ I1 + I = x + C

  • Question 33
    5 / -1
    If f(x)=2ln(ex), what is the area bounded by f(x) for the interval [0, 2] on the x-axis?
    Solution

    Explanation:

    y=f(x)=2ln(ex)

    The area bounded by the curve y = f(x) and in the interval where x ∈ (0, 2) is given by:

    A=022ln(ex)dx

    A=022ln(ex/2)dx

    A=022×x2dx

    A=[x22]02=[20]=2

  • Question 34
    5 / -1
    Find the area of the parabola y= 4ax bounded by it's latus rectum.
    Solution

    Concept:

    Parabola:

    • The focus of the parabola y2 = 4ax is at (a, 0).
    • The latus rectum of the parabola y2 = 4ax cuts the parabola at (a, 2a) and (a, -2a).

     

    Area under a curve:

    • The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral |abf(x) dx|, for curves which are entirely on the same side of the x-axis in the given range.
    • If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.

     

    Calculation:

    Since the graph of the parabola y2 = 4ax is symmetrical about the x-axis, the required area is:

    2 × |0a4ax dx| 

    = 2 × 2√a 0ax dx 

    = 2 × 2√a [23x32]0a 

    83a×a32 = 8a23.

    Additional Information:

    The latus rectum is a line which passes through the focus and is parallel to the directrix. 

  • Question 35
    5 / -1

    Ram is a fruit seller who can purchase apples at the rate of Rs 80/kg and mango at the rate of Rs 120/kg. Both can be sold at a profit of Rs 10/kg and Rs 12/kg. The total kgs of fruit that can be bought for one time is limited to 20 kg. Ram has a budget of Rs 2000 to bring the fruits. Ram wants to minimize the purchase cost of the fruits. Let x be the number of kilograms of apples being bought and y be the number of kilograms of mangoes being bought. The objective function Z of such a problem will be:

    Solution

    Concept: 

    There are three main components of linear programming:

    • Decision Variables: These variables are the activities that share the available resources while also competing with one another. In other words, these are interrelated in terms of resource utilization and need to be solved simultaneously. They are assumed to be non-negative and continuous.  
    • The Objective Function: Each linear programming problem is aimed to have an objective to be measured in quantitative terms such as profits, sales, cost, time, or some other parameter, etc which needs to be maximized or minimized. The objective function will vary depending upon the business requirements or other factors.
    • Constraints: These represent real-life limitations such as money, time, labor, or other restrictions and the objective function needs to be maximized or minimized while satisfying the constraints. They are represented as linear equations or inequations in terms of the decision variables.
    • For example, suppose x and y are the decision variables. The objective function will be given by:

    Z = ax + by ….(1)

    Where a and b are constants and Z is the function to be maximized or minimized. There will be conditions x ≥ 0, and y ≥ 0, which indicates the non-negative constraints on the decision variable.

    • The equation looks very simple since there are various assumptions involved while forming a linear programming example. These are mentioned below:
    • Parameters such as resources available, profit contribution of unit decision variable and resource used by unit decision variable needs to be known.
    • Decision variables are continuous. Hence the outputs can be an integer or a fraction.
    • The contribution of each decision variable in the objective function is directly proportional to the objective function.

    Calculation:

    Given:

    • From the above problem, Ram wants to maximize the profit. x represents kilograms of apples bought and y represents kilograms of mangoes bought. These represent the decision variables for the problem.
    • Rs 10/kg is the profit that can be made by selling apples and Rs 12/kg can be made by selling the mangoes.   
    • Since Ram wants to have maximum profits for his shop, the objective function would be given by,

    Z = 10x + 12y

    • Hence, correct answer will be option 4.
  • Question 36
    5 / -1
    The function y = e-4x is a solution to the differential equation
    Solution

    Concept:

    Some useful formulas are:

    d(eax)dx=aeax

    Calculation:

    Given function is

    y = e-4x, differentiating it we get

    dydx=4e4x

    Differentiating it further we get,

    d2ydx2=16e4x

    Now, d2ydx2+3dydx4y=0

    =16e4x+3(4e4x)4e4x

    = 0

    So, d2ydx2+3dydx4y=0 

  • Question 37
    5 / -1
    The value of 13[tan1(xx2+1)+tan1(x2+1x)]dx is
    Solution

    Given:

    13[tan1(xx2+1)+tan1(x2+1x)]dx

    Formula used :

    tan1p+cot1p=π2        -----(1)

    Calculations:

    Using equation (1), we get

    ⇒ tan1p+tan11p=π2         -----(2)

    Applying equation (2) in the given equation

    ⇒ 13π2dx

    ⇒ π2131.dx

    ⇒ (π/2) [x]-13

    ⇒ π2[3(1)]

    ⇒ (π/2) × 4 = 2π 

    ∴ The value of 13[tan1(xx2+1)+tan1(x2+1x)]dx is 2π.

  • Question 38
    5 / -1
    Solved dydx=exy(exey)
    Solution

    Concept:

    The standard form of a linear equation of the first order is given by dydx+Py=Q where P, Q are arbitrary functions of x.

    The integrating factor of the linear equation is given by

     I.F.=epdx

    The solution of the linear equation is given by y(I.F.)=Q(I.F.)dx+C

    Calculation:

    Given equation is

    dydx=exy(exey)

    eydydx+exey=e2x    ---(1)

    Put ex = u, ey = v then

    eydydx = dvdx

    The equation reduces to the linear equation.

    dvdu+uv=u

    P = u, Q = eu

    I.F.=eudx = eu

    General solution is

    y(I.F.)=Q(I.F.)dx+C

    ⇒ veu=ueudu+C

    ⇒ veu=ueu  eu + C

    ⇒ v=u  1 + Ceu

     ey = ex  1 + Ceex

  • Question 39
    5 / -1

    The function f(x) = 8 logx - x2 + 3 attains its global minimum over the interval [1, e] at x = ________.

    (Here logx is the natural logarithm of x and  e2 = 7.39)

    Solution

    Calculation:

    f(x) = 8 logx - x2 + 3

    f ' (x) = 8/x - 2 x 

    For Stationary points  f ' (x) = 0

    8x2x=0  

    x = ± 2 

    x=  2[1  e]  

    f(x)  =  (8x22) |at  x=2  =4  

    f '' (x) < 0 

    f(x) is maximum at x = 2

    f(1) = 2

    f(e) = 8 - e2 + 3 = 3.61

    Minimum of f(x) in [1 , e] = min { f (1) , f (e) }

    Minimum of f(x) in [1 , e] = min { 2 , 3.61 }

    f(x) minimum  at x = 1

  • Question 40
    5 / -1
    For a point of inflection of y = f(x), which one of the following is correct?
    Solution

    Calculations:

    An Inflection Point of y = f(x)  is where a curve changes from Concave upward to Concave downward (or vice versa).

    The derivative of a function gives the slope.

    The second derivative tells us if the slope increases or decreases.

    • When the second derivative is positive, the function is concave upward.
    • When the second derivative is negative, the function is concave downward.
    • For the point of inflection,

    ⇒ d2ydx2 must be zero.

     

  • Question 41
    5 / -1
    The identity element for the binary operation * defined on Q - {0} (Where Q is a set of rational numbers) as: a * b = ab2, is:
    Solution

    Concept:

    An identity element is a special type of element of a set with respect to a binary operation on that set, which leaves any element of the set unchanged when combined with it.

     

    Calculation:

    It is given that a * b = ab2; a,b ∈ Q - {0}.

    Let the identity element for * be i, i.e. a * i = a.

    ∴ a * i = ai2

    ⇒ a = ai2

    ⇒ i = 2.

    Hence, the identity element for the operation * is i = 2.

  • Question 42
    5 / -1
    What is the cosine of angle between the planes x + y + z + 1 = 0 and 2x - 2y + 2z + 1 = 0?
    Solution

    Concept:

    The angle between two plane a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d= 0 is given by

    cosθ = a1a2+b1b2+c1c2(a12+b12+c12)×(a22+b22+c22)

    Calculation: 

    Let the angle between the given planes be θ 

    Given planes are, x + y + z + 1 = 0 and 2x - 2y + 2z + 1 = 0

    On comparing the above plane with the general form of planes 

    a1x + b1y + c1z + d= 0 and  a2x + b2y + c2z + d= 0, we get 

    a1 = 1, b= 1, c= 1, a2 = 2, b= -2, c= 2

    According to the concept used, we have

    cosθ = a1a2+b1b2+c1c2(a12+b12+c12)×(a22+b22+c22)

    ⇒ cosθ = (1)(2)+(1)(2)+(1)(2)(12+12+12)×(22+(2)2+22)

    ⇒ cosθ = (2)+(2)+(2)3×12

    ⇒ cosθ = 23×23

    ⇒ cosθ = 13

    ∴ The cosine of the angle between the planes is 13.

  • Question 43
    5 / -1
    If the position vector of a point P with respect to origin O is î + 3ĵ - 2k̂ and that of a point Q is 3î + ĵ - 2k̂, then what is the position vector of the bisector of the angle POQ? 
    Solution

    Concept:

     A triangle ABC is said to be an isosceles triangle if triangle ABC must have two sides of equal length.

    Calculations:

    Given, the position vector of a point P with respect to origin O is î + 3ĵ - 2k̂ and that of a point Q is 3î + ĵ - 2k̂.

    OP¯ = î + 3ĵ - 2k̂ and  OQ¯ = 3î + ĵ - 2k̂.

    ⇒ |OP| = 1+9+4=15

    ⇒ |OQ| = 9+1+4=15

    Here, |OP| = |OQ|

    POQ is isoscale.

    The position vector of the bisector of the angle POQ = 12(OP+OQ)

    ⇒The position vector of the bisector of the angle POQ = 12[(î+3ĵ2k̂)+(3î+ĵ2k̂)]

    ⇒The position vector of the bisector of the angle POQ = 12(4î+4ĵ4k̂)

    ⇒The position vector of the bisector of the angle POQ = 2î+2ĵ2k̂

  • Question 44
    5 / -1
    If x is a real number, then the given single valued function f(x) = x3 + x2 + x + 1 has
    Solution

    Concept:

    The value of the function at a maximum point is called the maximum value of the function and the value of the function at a minimum point is called the minimum value of the function.

    Steps to find out the maximum and minimum value of any function f(x):

    • Differentiate the given function.
    • Let f '(x) = 0 and find critical numbers.
    • Then find the second derivative f ''(x).
    • The function f(x) is maximum when f ''(x) < 0.
    • The function f(x) is minimum when f ''(x) > 0.
    • To find the maximum and minimum values we need to apply those x values in the original function.

    Key Points

     Consider a quadratic equation ax2 + bx + c = 0.

    In this equation, we call the term b2 −4ac as the discriminant. The discriminant is important because it tells us how many roots a quadratic function has. Specifically, if

    • b2 −4ac < 0 There are no real roots.
    • b2 −4ac = 0 There is one real root.
    • b2 −4ac > 0 There are two real roots. 

    Calculation:

    f(x) = x3 + x2 + x + 1

    ⇒ f ' (x) = 3x2 + 2x + 1

    Since, f ' (x) = 0

    ⇒ 3x2 + 2x + 1 = 0

    Here, Discriminant = (2)2 - 4 × 3 × 1 = 4 - 12 = -8

    ⇒ x has no real value. Hence, there is no critical point.

    Hence, f(x) has neither maximum nor minimum value.

  • Question 45
    5 / -1
    Find the absolute minimum value of the function y=3x24 in the interval [1,5]
    Solution

    Concept:

    Following steps to finding the absolute minimum.

    Differentiate the function and equate it with zero to get the critical point 

    Find the value of the function at critical point c

    Find the function value at end of the interval [a, b]

    Absolute minimum value is {min f(a), f(b) and f(c)}

     

    Calculation:

    y=3x24, x ∈ [-1, 5]

    Differentiation with respect to x

    ⇒  y' = 6x

    For critical points, y' = 0

    ⇒ x = 0

    Now,

    ⇒ y(0) = -4

    ⇒ y(-1) = -1

    y(2)=3(2)24

    y(2)=8

    Min{4,1,8}=4

     Minimum value of function in given interval is -4

    Hence, option 1 is correct

  • Question 46
    5 / -1

    Find the solution set of y if the system of inequalities is:

    10x + 8y ≤ 80

    X < 4

    y ≥ 6
    Solution

    Step 1: draw lines for:

    10x + 8y = 80

    x = 4

    y = 6

    Step 2: Shade the solution region for each inequality

    Step 3: find the common region.

    So, x ϵ (0, 3.2)

    y ϵ [6, 10]
  • Question 47
    5 / -1
    If A is the identity matrix of order 3 and B is its transpose, then what is the value of the determinant of the matrix C = A + B?
    Solution

    Concept:

    1). Transpose of a Matrix:

         The new matrix obtained by interchanging the rows and columns of the original matrix is called

         the transpose of the matrix.

         For example: A=[abcxyz]⇒ AT=[axbycz]

         It is denoted by A' or AT.
    2). For the addition and subtraction of two matrices, the order of matrices should be equal.

    Calculation:

    Given:

    A=[100010001]and B=AT=[100010001]

    C = A + B

    ⇒ C=[100010001]+[100010001]

    ⇒ C=[200020002]

    ⇒ ∣C∣ = 2(4 - 0)

    ⇒ ∣C∣ = 8

    ∴ The determinant of matrix C is 8.

  • Question 48
    5 / -1
    What is the area under the curve f(x) = xex above the x-axis and between the lines x = 0 and x = 1?
    Solution

    Concept:

    The area under the curve f(x) above the x-axis and between the lines x = a and x = b is given by,

    A=abf(x)dx

    Calculation:

    We know that the area under the curve f(x) above the x-axis and between the lines x = a and x = b is given by:

    Area = abf(x)dx

    The area under the curve f(x) = xex above the x-axis and between the lines x = 0 and x = 1 is given by:

    Area = 01xexdx

    Area = (xexex)01

    Area = (1.e - e) - (0 - 1)

    Area = 1 square unit

    Hence, the area under the curve f(x) = xex above the x-axis and between the lines x = 0 and x = 1 is 1 square unit.

  • Question 49
    5 / -1
    Fine the value of sin (π3  sin1(12))
    Solution

    Concept:

    Inverse trigonometric function for negative argument

    sin-1 (-x)- sin-1 xcos-1 (-x)π - cos-1 x
    cosec-1 (-x)- cosec-1 xsec-1 (-x)π - sec-1 x
    tan-1 (-x)- tan-1 xcot-1 (-x)π - cot-1 x

     

    Values of trigonometric function

    Calculation:

    Given: sin (π3  sin1(12))

    sin (π3 + sin1(12))                            (∵ sin-1 (- x) = - sin-1 x)

    sin (π3 + π6)

    sin π2

    = 1

  • Question 50
    5 / -1

    A pair of fair dice is thrown, what is the probability of getting a six on both, given atleast one die gets six?

    Solution

    Concept:

    P(getting at least one six) = 1 - P(getting no six)

    If A and B are two mutually exclusive events, then the probability of getting B when A has happened,

    P(BA)=P(AB)P(A)

    Calculation:

    A = {getting a six a at least 1}= 1-{getting no six}

    P(A)=156×56=1136

    B = {getting a six on both}

    P(B)=16×16=136

    Probability of getting a six on both, given atleast one die gets six i.e. Probability of B given A

    P(BA)=P(AB)P(A)

    Also A∩B = B

    ∵ 

    P(BA)=P(B)P(A)=16×16156×56=136×3611=111

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