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Relations and Functions Test - 13

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Relations and Functions Test - 13
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  • Question 1
    1 / -0
    Let $$f_k(x)=\dfrac{1}{k}(sin^kx+cos^kx)$$ for $$k=1,2,3,....$$ Then for all $$x\in R$$, the value of $$f_4(x)-f_6(x)$$ is equal to:-
    Solution
    $$f_4(x)-f_6(x)$$

     $$=\dfrac{1}{4}(sin^4x+cos^4x)-\dfrac{1}{6}(sin^6x+cos^6x)$$

    $$=\dfrac{1}{4}\left(1-\dfrac{1}{2}sin^22x\right)-\dfrac{1}{6}\left(1-\dfrac{3}{4}sin^22x\right)=\dfrac{1}{12}$$
  • Question 2
    1 / -0
    Let  $$S = \{ 1,2,3 , \ldots , 100 \} .$$  The number of non-empty subsets  $$A$$  of  $$S$$  such that the product of elements in  $$A$$  is even is :-
    Solution
    $$\mathrm { S } = \{ 1,2,3 - \ldots - 100 \}$$

    = Total non empty subsets-subsets with product of element is odd

    $$= 2 ^ { 100 } - 1 - 1 \left[ \left( 2 ^ { 50 } - 1 \right) \right]$$

    $$= 2 ^ { 100 } - 2 ^ { 50 }$$

    $$= 2 ^ { 50 } \left( 2 ^ { 50 } - 1 \right)$$
  • Question 3
    1 / -0
    Let $$a, b, c,$$ $$\epsilon\ R$$. If $$f(x)=ax^2+bx+c$$ is such that $$a+b+c=3$$ and $$f(x+y)=f(x)+f(y)+xy, \forall \, x, y\,\epsilon\, R,$$ the $$\displaystyle \sum_{n=1}^{10}f(n)$$ is equal to.
    Solution
    Let $$a,b,c\ \epsilon$$ R
    $$f(1)=a+b+c=3$$
    $$f(2)=f(1+1)=f(1)+f(1)+1$$ ............. $$\because f(x+y)=f(x)+f(y)+xy$$
    $$\implies f(2)=2f(1)+1$$
    $$f(3)=f(2+1)=f(2)+f(1)+2$$
             $$=2f(1)+1+f(1)+2$$
    $$\implies f(3)=3f(1)+3$$
    $$f(4)=f(3+1)=f(3)+f(1)+3$$
             $$=3f(1)+3+f(1)+3$$
    $$\implies f(4)=4f(1)+6$$
    $$\displaystyle \sum_{n=1}^{10}f(n)=f(1)+f(2)+f(3)+--------+f(10)$$
                    $$=f(1)+2f(1)+1+3f(1)+3+4f(1)+6+5f(1)+10--------$$
                    $$=f(1)[1+2+3+4+-----+10]+(1+3+6+10+15+21+28+36+45)$$
                    $$=f(1)(55)+(165)$$
                    $$=3\times 55+165=330$$
    $$\therefore \displaystyle \sum_{n=1}^{10} f(n)=330$$
    Hence, option A is correct.
  • Question 4
    1 / -0
    Let $$f : (-1,1)\to R$$ be a be a function defined by $$f(x) = max \left\{ -|x|, -\sqrt{1-x^2}\right\}$$. If $$K$$ be the set of all points at which f is not differentiable, then $$K$$ has exactly :
    Solution
    $$f : (-1, 1) \to R$$
    $$f(x) = max \left\{ -|x|, -\sqrt{1-x^2}\right\}$$
    Non-derivable at 3 points in $$(1, 1)$$

  • Question 5
    1 / -0
    For every pair of continuous functions $$f, g : [0, 1]  \rightarrow R$$ such that $$\mathrm{m}\mathrm{a}\mathrm{x} \{f(x) : x \in [0, 1]\} = \mathrm{m}\mathrm{a}\mathrm{x} \{g(x) : x \in [0, 1]\}$$, the correct statement(s) is (are)
    Solution
    Let $$f(x)$$ and $$g(x)$$ achieve their maximum value at $$x_1$$ and $$x_2$$ respectively.
    $$h(x) = f(x) - g(x)$$
    $$h(x_1) = f(x_1) - g (x_1)  \geq 0$$
    $$h(x_2) = f(x_2) - g(x_2) \leq 0$$
    $$\Rightarrow h(c) = 0$$ where $$c\in [0, 1]$$  $$\Rightarrow f(c)= g(c)$$
    The correct options are therefore,
    A, D
  • Question 6
    1 / -0
    Let $$f(x) = \dfrac {ax + b}{cx + d}$$, then $$fof(x) = x$$, provided that
    Solution
    $$f(x) = \dfrac {ax + b}{cx + d}$$

    $$fof(x) = \dfrac {a\left \{\dfrac {ax + b}{cx + d}\right \} + b}{c  \left \{\dfrac {ax + b}{cx + d}\right \} + d} \Rightarrow \dfrac {a^{2}x + ab + bcx + bd}{acx + bc + cdx + d^{2}} = x$$

    $$\Rightarrow (ac + dc)x^{2} + (bc + d^{2} - bc - a^{2})x - ab - bd = 0, \forall x \epsilon R$$

    $$\Rightarrow (a + d)c = 0, d^{2} - a^{2} = 0$$ and $$(a + d)b = 0$$

    $$\Rightarrow a + d = 0$$

    $$\Rightarrow d=-a$$
  • Question 7
    1 / -0
    $$A$$ and $$B$$ are two sets having $$3$$ and $$4$$ elements respectively and having $$2$$ elements in common. The number of relations which can be defined from $$A$$ to $$B$$ is:
    Solution
    Given:
    $$n(A) = 3 = m$$
    $$n(B) = 4 = n$$

    Number of relation 
    $$= { 2 }^{ mn }$$
    $$= { 2 }^{ 3\times 4 }$$
    $$= { 2 }^{ 12 }$$.
  • Question 8
    1 / -0
    Let $$R$$ be a relation from a set $$A$$ to a set $$B$$, then:
    Solution
    $$R : $$$$A\rightarrow B$$
    then $$R$$ is a subset $$A\times B$$
     $$\therefore$$ $$R\quad \subseteq \quad A\times B$$
  • Question 9
    1 / -0
    If $$f(x)=3x+1, g(x)= x^{3}+2,$$ then $$(f+g)(0)-f(0)g(0)=$$
    Solution
    $$f(0)=3(0)+1=1\\g(0)=0^3+2=2$$

    $$\therefore (f+g)(0)-f(0)\cdot g(0)=f(0)+g(0)-f(0).g(0)$$ $$=1+2 -1.2=1$$
  • Question 10
    1 / -0
    If $$f(x)=x^3-x^2+x+1$$, then the value of $$\dfrac {f(1)+f(-1)}{2}$$ will be
    Solution
    Consider the given function.
    $$f(x)=x^3-x^2+x+1$$

    Put $$x=1$$
    $$f(1)=1-1+1+1$$
    $$f(1)=2$$

    Put $$x=-1$$
    $$f(-1)=-1-1-1+1$$
    $$f(-1)=-2$$

    Since,
    $$\dfrac{f(1)+f(-1)}{2}$$

    $$=\dfrac{2-2}{2}$$
    $$=0$$

    Hence, this is the answer.
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