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Relations and Functions Test - 21

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Relations and Functions Test - 21
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  • Question 1
    1 / -0
    If f(x)=2x1,g(x)=x2f(x)=2x-1, g(x)=x^{2} then (3f2g)(x)= (3f-2g)(x)=
    Solution
    3f(x)2g(x)3f\left ( x \right )-2g\left ( x \right )
    =3(2x1)2(x2)=3\left ( 2x-1 \right )-2\left ( x^{2} \right )
    =6x2x23=6x-2x^{2}-3
  • Question 2
    1 / -0
    If f(x)f(x) is a polynomial in x(>0)x(>0) satisfying the equation f(x)+f(1x)=f(x).f(1x) f(x)+f\left (\dfrac1x\right)=f(x).f\left (\dfrac1x\right) and f(2)=7f(2)=-7,  then f(3)f(3) is equal to
    Solution
    Whenever, f(x)+f(1x)=f(x).f(1x)f(x)+f\left (\dfrac{1}{x}\right)=f(x).f\left (\dfrac{1}{x}\right), then f(x)=1+xnf(x)=1+x^n or f(x)=1xnf(x)=1-x^n keeping in mind that nn must be greater than 00.
    So here, f(2)=7f(2)=-7.
    So we choose f(x)=1xnf(x)=1-x^n.
    So, f(2)=12nf(2)=1-2^n
    2n=8\Rightarrow 2^n=8
    n=3\Rightarrow n=3
    So, the function is f(x)=1x3f(x)=1-x^3.
    Now,f(3)=133f(3)=1-3^3
    f(3)=26\Rightarrow f(3)=-26
  • Question 3
    1 / -0
    If f(x+y)=f(x)f(y)f(x+y)=f(x)f(y)  and  f(5)=32f(5)=32  then  f(7)=f(7)=
    Solution
    f(x+y)=f(x).f(y)f(x+y)=f(x).f(y)
    Put x=1,y=1x=1, y=1
    f(x+y)=f(x).f(y)f(x+y)=f(x).f(y)
    f(2)=f(1)f(1)\Rightarrow f(2)=f(1)f(1)
    f(3)=f(2)f(1)=f(1)3\Rightarrow f(3)=f(2)f(1)=f(1)^3
    f(4)=f(3)f(1)=f(1)4\Rightarrow f(4)=f(3)f(1)=f(1)^4
    f(5)=f(4)f(1)=f(1)5\Rightarrow f(5)=f(4)f(1)=f(1)^5       
    Givenf(5)=32 f(5)=32
    f(1)=2\therefore f(1)=2
    And f(2)=4f(2)=4
    Now,
    f(7)=f(5)f(2)=32×4=128f(7)=f(5)f(2)=32\times4=128
  • Question 4
    1 / -0

    If a non-zero function ff satisfies the relation f(x+y)+f(xy)=2f(x).f(y)f(x+y)+f(x-y)=2f(x).f(y) for all x,yx, y in RR and f(0)0f(0) \neq 0; then f(10)f(10)=f(10)-f(-10)=

    Solution
    Given, f(x+y)+f(xy)=2f(x)f(y)f(x+y)+f(x-y)=2f(x)\cdot f(y)

    Put y=0,x=xy=0, x=x

    f(x)+f(x)=2f(x)f(0)f(x)+f(x)=2f(x)f(0)

    2f(x)=2f(x)f(0)2f(x)=2f(x)f(0)

    f(x)=0f(x)=0   that is not possible

    f(0)=1f(0)=1

    Take x=0x = 0 and y=10y = 10

    f(10)+f(10)=2f(10)f(0)f(10) + f(-10) = 2f(10)f(0)

    f(10)+f(10)=2f(10)f(10) + f(-10) = 2f(10)

    f(10)=f(10)f(-10) = f(10)

    f(10)f(10)=0\therefore f(10)-f(-10)=0
  • Question 5
    1 / -0
    If relation RR is defined by R={(x, y):2x2+3y26}\mathrm{R}=\{(\mathrm{x},\ \mathrm{y}):2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6\}, then the domain of R\mathrm{R} is
    Solution
     Domain == value of x\mathrm{x} in f=2x2+3y26f=2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6  then
    2x2+3y262\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6x23+y221\displaystyle \frac{\mathrm{x}^{2}}{3}+\frac{\mathrm{y}^{2}}{2}\leq 1 
    Since it is an equation of ellipse then value of xx   x[3,3] \Rightarrow  \mathrm{x}\in[-\sqrt{3},\sqrt{3}]
  • Question 6
    1 / -0
    The domain and range of relation R={(x,y)x,yNR=\{(x,y) | x, y \in N, x+2y=5}x+2y=5\} is?
    Solution
    The possible values of xx which satisfies the given relation is the domain of the relation
    The possible values of yy which satisfies the given relation is the range of the relation
    Given that x,yx,y are natural numbers and the relation is x+2y=5x+2y=5
    For x=1x=1 , the value of yy is 22
    For x=3x=3 , the value of yy is 11
    If x5x\geq 5, then yy is negative which does not belong to naturals.
    If x=2,4x=2,4 then yy becomes rational number which is not in naturals.
    Thus, only 11 and 33 gives yy as natural numbers 22 and 11.
    Therefore, the domain is {1,3}\{1,3\} and the range is {2,1}\{2,1\}
  • Question 7
    1 / -0
    If RR is the relation 'less than' from A={1,2,3,4,5}A=\{1, 2, 3, 4, 5\} to B={1,4}B=\{1, 4\}, the set of ordered pairs corresponding to RR, then the inverse of RR is
    Solution
     R\because  R is the relation 'less than' from AA to BB.

    So R={(1,4),(2,4),(3,4)}R=\{(1,4), (2,4), (3,4)\}

    R1={(4,1),(4,2),(4,3)}R^{-1}=\{(4,1), (4,2), (4, 3)\}
  • Question 8
    1 / -0
    If AA is the set of even natural numbers less than 88 and BB is the set of prime numbers less than 77, then the number of relations from AA to BB is
    Solution
    A={2,4,6},B={2,3,5}A = \{2, 4, 6\}, B = \{2,3,5\}
    n(A×B)=3×3=9n (A\times B) = 3\times 3 = 9
    Number of relations from AA to BB is 292^9
  • Question 9
    1 / -0
    The range of the function f(x)=sin(π[x])x2+1f(x)=\dfrac{\sin(\pi [x])}{x^{2}+1} (where [.][.] denotes greatest integer function) is
    Solution
    f(x)=sin(π[x])x2+1f(x)=\dfrac {\sin(\pi[x])}{x^2+1}
    \because [x][x] always gives integer value and sinnπ=0\sin n\pi =0
    f(x)=0\therefore f(x)=0
  • Question 10
    1 / -0
    If A={1,2,3}A =\{1,2,3\}, B={1,4,6,9}B=\{1,4,6,9\} and RR is a relation from AA to BB defined by 'xx is greater than yy'. The range of RR is
    Solution
    R={(x,y):xA,yBR=\left \{ ( x,y \right ): x\in A, y\in B and x>y}x> y\}
    R={(2,1),(3,1)}R= \left \{ \left ( 2,1 \right ),\left ( 3,1 \right ) \right \}
    Therefore, R={1}R=\left \{ 1 \right \}
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