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Mathematics Test - 11
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Mathematics Test - 11
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  • Question 1/10
    1 / -0

    State the nature of the given quadratic equation (x + 1) (x + 2) + 2 = 0

    Solutions
    let event A= minimun is 3 let event B= maximum is 6
    total ways =8C3 P(AB)=2.8C3=28!3!5!
    =2×3!8×7×6=128
    P(AB)=P(AB)P(B)
    =128s5C2sC3
    =2s8C3sC2sC3
    =255C2
    =2×25×4=15
    option 1 is correct
  • Question 2/10
    1 / -0

    If A&B are two events then P{(AB¯)(A¯B)}=

    Solutions


    P(AB)P(AB)
  • Question 3/10
    1 / -0

    The number of four digit odd numbers that can be formed using 1,2,3,4,5,6,7,8,9 (repetition of digits is allowed ) are

    Solutions
    For a number to be odd, the units digit should be either of 1,3,5,7 or 9
    Total number of ways of choosing units digit =5 since, it is given that the digit may be repeated. So, we have 9 choices for ten's digit Similarly, for rest two digits we can choose any of the 9 digits
    No. of ways of choosing first and second digit is also 9 for each.
    Total number of ways to form odd numbers is 5×9×9×9=5×93
    Hence, the number of 4 digit odd numbers can be formed is 5×93
  • Question 4/10
    1 / -0

    The number of nine digit numbers that can be formed with different digits is

    Solutions
    The first digit is one of (1,2,3,4,5,6,7,8,9) (i.e., all
    expect 0 ). So, the no. of ways =9 The remaining 8 digits are to be taken from the remaining 9 numbers. So, the no. of ways =9P8 So, total no. of ways =9×9P8 =9×9!
  • Question 5/10
    1 / -0

    What are the nature of  quadratic equation x+ 3x + 4 = 0 ?

    Solutions
    Using D=(3)2(4)(1)(4)

    D=916

    D=7

    D<0

    So the given quadratic equation have imaginary roots.
  • Question 6/10
    1 / -0

    If the first term of an AP is 5 and the common difference is −2, then the sum of the first 6 terms is

    Solutions
    Sum of an AP is:
    n2×[2a+(n1)d]
    a=5,d=2,n=6
    Sum =62×[(2×5)+(61)×(2)]
    =3×(1010)
    =0
  • Question 7/10
    1 / -0

    The value of sin4θcosθ is

    Solutions
    The value of sin4θcosθ is
    =2sin2θcos2θcosθ
    =4sinθcosθcos2θcosθ
    =4sinθcos2θ
    Hence, answer is option B.
  • Question 8/10
    1 / -0

    Find the value of sin1x+sin11x+cos1x+cos11x
    Solutions
    The value of sin1x+sin11x+cos1x+cos11x is defined only when
    x,1xϵ[1,1]
    which is possible only when x=±1 For which sin1x+sin11x+cos1x+cos11x=π
  • Question 9/10
    1 / -0

    esinxsin2xdx=

    Solutions
    esinx2sinxcosxdx
    =2esinxsinxd(sinx)
    Take sinx=t
    =2ettdt=2(tetet)
    =2et(t1)
    =2esinx(sinx1)
    esinxsin2xdx=2esinx(sinx1)+c
  • Question 10/10
    1 / -0

    (x+2)x+1dx=

    Solutions
    (x+2)x+1dx
    [(x+1)+1](x+1)dx
    (x+1)32dx+x+1dx
    =25(x+1)5/2+23(x+1)3/2+c
    =215[3(x+1)5/2+5(x+1)3/2]+c
    =215(x+1)3/2(3x+8)+c
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