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NCERT Exemplar · Class 11 Mathematics Relations and Functions

42 questions · 42 still being checked

EXERCISE 2.3 21–30 (part 3 of 4)

  1. Exercise 21

    Let f(x)=x\displaystyle f(x)=\sqrt{x} and g(x)=x\displaystyle g(x)=x be two functions defined in the domain R+∪{0}\displaystyle \mathrm{R}^{+} \cup\{0\}. Find
    (i)
    (f+g)(x)\displaystyle (f+g)(x)
    (ii)
    (f−g)(x)\displaystyle (f-g)(x)
    (iii)
    (fg)(x)\displaystyle (f g)(x)
    (iv)
    (fg)(x)\displaystyle \left(\frac{f}{g}\right)(x)

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    NCERT’s answer
    (i)
    \(\displaystyle (f+g) x=\sqrt{x}+x\)
    (ii)
    \(\displaystyle (f-g) x=\sqrt{x}-x\)
    (iii)
    \(\displaystyle (f g) x=x^{\frac{3}{2}}\)
    (iv)
    \(\displaystyle \left(\frac{f}{g}\right) x=\frac{1}{\sqrt{x}}\)
    \[(f+g)(x) = f(x) + g(x) = \sqrt{x} + x \] \[(f-g)(x) = f(x) - g(x) = \sqrt{x} - x \] \[(fg)(x) = f(x)\,g(x) = x\sqrt{x} = x^{3/2} \] \[\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}, \quad x > 0 \] Since \(\displaystyle g(0) = 0\), \(\displaystyle 0\) is excluded from the quotient. Answer: \(\displaystyle \sqrt{x} + x\); \(\displaystyle \sqrt{x} - x\); \(\displaystyle x^{3/2}\); \(\displaystyle \dfrac{1}{\sqrt{x}}\) for \(\displaystyle x > 0\).

  2. Exercise 22

    Find the domain and Range of the function f(x)=1x−5\displaystyle f(x)=\frac{1}{\sqrt{x-5}}.

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    NCERT’s answer
    Domain of \(\displaystyle f=(5, \infty)\) and Range of \(\displaystyle f=\mathrm{R}^{+}\)
    The root needs a positive radicand, and the denominator cannot vanish: \[x-5>0 \Rightarrow x>5 \] Solve for \(\displaystyle x\) to find which \(\displaystyle y\) occur: \[y=\frac{1}{\sqrt{x-5}} \Rightarrow y>0 \] \[x=5+\frac{1}{y^{2}} \] Every \(\displaystyle y>0\) gives an \(\displaystyle x>5\), so every \(\displaystyle y>0\) is attained.Answer: Domain \(\displaystyle (5,\infty)\), Range \(\displaystyle (0,\infty)\).

  3. Exercise 23

    If f(x)=y=ax−bcx−a\displaystyle f(x)=y=\frac{a x-b}{c x-a}, then prove that f(y)=x\displaystyle f(y)=x.

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    \[f(y)=\frac{ay-b}{cy-a},\qquad y=\frac{ax-b}{cx-a} \] \[ay-b=\frac{a(ax-b)-b(cx-a)}{cx-a}=\frac{(a^{2}-bc)\,x}{cx-a} \] \[cy-a=\frac{c(ax-b)-a(cx-a)}{cx-a}=\frac{a^{2}-bc}{cx-a} \] \[f(y)=\frac{(a^{2}-bc)\,x}{a^{2}-bc}=x \qquad (a^{2}\neq bc) \]Answer: \(\displaystyle f(y)=x\)
  4. Choose the correct answers in Exercises from $\displaystyle 24$ to $\displaystyle 35$ (M.C.Q.)

    Exercise 24

    Let n( A)=m\displaystyle n(\mathrm{~A})=m, and n( B)=n\displaystyle n(\mathrm{~B})=n. Then the total number of non-empty relations that can be defined from A to B is
    (A)
    mn\displaystyle m^n
    (B)
    nm−1\displaystyle n^m-1
    (C)
    mn−1\displaystyle m n-1
    (D)
    2mn−1\displaystyle 2^{m n}-1

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    NCERT’s answer
    D
    (D) \(\displaystyle 2^{mn}-1\). A relation from A to B is a subset of \(\displaystyle A\times B\); the empty subset is excluded. \[n(A\times B)=mn \] \[\text{number of subsets}=2^{mn} \] \[\text{non-empty relations}=2^{mn}-1 \]

  5. Exercise 25

    If [x]2−5[x]+6=0\displaystyle [x]^2-5[x]+6=0, where [ . ]\displaystyle [\,.\,] denote the greatest integer function, then
    (A)
    x∈[3,4]\displaystyle x \in[3,4]
    (B)
    x∈(2,3]\displaystyle x \in(2,3]
    (C)
    x∈[2,3]\displaystyle x \in[2,3]
    (D)
    x∈[2,4)\displaystyle x \in[2,4)

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    NCERT’s answer
    D
    (D) \(\displaystyle x\in[2,4)\). Factorise in \(\displaystyle [x]\): \[[x]^{2}-5[x]+6=([x]-2)([x]-3)=0 \] \[[x]=2 \ \text{ or } \ [x]=3 \] \[2\le x<3 \ \text{ or } \ 3\le x<4 \] \[x\in[2,4) \]

  6. Exercise 26

    Range of f(x)=11−2cos⁡x\displaystyle f(x)=\frac{1}{1-2 \cos x} is
    (A)
    [13,1]\displaystyle \left[\frac{1}{3}, 1\right]
    (B)
    [−1,13]\displaystyle \left[-1, \frac{1}{3}\right]
    (C)
    (−∞,−1]∪[13,∞)\displaystyle (-\infty,-1] \cup\left[\frac{1}{3}, \infty\right)
    (D)
    [−13,1]\displaystyle \left[-\frac{1}{3}, 1\right]

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    (C) \(\displaystyle \left(-\infty,-1\right]\cup\left[\frac13,\infty\right)\). Put \(\displaystyle t=1-2\cos x\); \(\displaystyle t\neq0\) for \(\displaystyle f\) to exist. \[-1\le\cos x\le1 \Rightarrow -1\le t\le 3 \] \[t\in[-1,0)\Rightarrow \frac1t\in(-\infty,-1] \] \[t\in(0,3]\Rightarrow \frac1t\in\left[\frac13,\infty\right) \]

  7. Exercise 27

    Let f(x)=1+x2\displaystyle f(x)=\sqrt{1+x^2}, then
    (A)
    f(xy)=f(x).f(y)\displaystyle f(x y)=f(x) . f(y)
    (B)
    f(xy)≥f(x).f(y)\displaystyle f(x y) \geq f(x) . f(y)
    (C)
    f(xy)≤f(x).f(y)\displaystyle f(x y) \leq f(x) . f(y)
    (D)
    None of these
    [Hint : find f(xy)=1+x2y2,f(x).f(y)=1+x2y2+x2+y2\displaystyle f(x y)=\sqrt{1+x^2 y^2}, f(x) . f(y)=\sqrt{1+x^2 y^2+x^2+y^2} ]

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    NCERT’s answer
    C
    (C) \(\displaystyle f(xy)\le f(x)\,f(y)\). \[f(xy)=\sqrt{1+x^{2}y^{2}} \] \[f(x)f(y)=\sqrt{(1+x^{2})(1+y^{2})}=\sqrt{1+x^{2}y^{2}+x^{2}+y^{2}} \] \[x^{2}+y^{2}\ge0 \Rightarrow f(x)f(y)\ge f(xy) \]

  8. Exercise 28

    Domain of a2−x2(a>0)\displaystyle \sqrt{a^2-x^2}(a>0) is
    (A)
    (−a,a)\displaystyle (-a, a)
    (B)
    [−a,a]\displaystyle [-a, a]
    (C)
    [0,a]\displaystyle [0, a]
    (D)
    (−a,0]\displaystyle (-a, 0]

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    NCERT’s answer
    B
    (B) \(\displaystyle [-a,a]\). The radicand must be non-negative. \[a^{2}-x^{2}\ge0 \Rightarrow x^{2}\le a^{2} \] \[|x|\le a \Rightarrow -a\le x\le a \]

  9. Exercise 29

    If f(x)=ax+b\displaystyle f(x)=a x+b, where a\displaystyle a and b\displaystyle b are integers, f(−1)=−5\displaystyle f(-1)=-5 and f(3)=3\displaystyle f(3)=3, then a\displaystyle a and b\displaystyle b are equal to
    (A)
    a=−3,b=−1\displaystyle a=-3, b=-1
    (B)
    a=2,b=−3\displaystyle a=2, b=-3
    (C)
    a=0,b=2\displaystyle a=0, b=2
    (D)
    a=2,b=3\displaystyle a=2, b=3

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    NCERT’s answer
    B
    (B) \(\displaystyle a=2,\ b=-3\). Substitute the two given values and subtract.\[f(-1)=-a+b=-5 \] \[f(3)=3a+b=3 \] \[(3a+b)-(-a+b)=3-(-5) \] \[4a=8 \Rightarrow a=2 \] \[b=3-3a=-3 \]

  10. Exercise 30

    The domain of the function f\displaystyle f defined by f(x)=4−x+1x2−1\displaystyle f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}} is equal to
    (A)
    (−∞,−1)∪(1,4]\displaystyle (-\infty,-1) \cup(1,4]
    (B)
    (−∞,−1]∪(1,4]\displaystyle (-\infty,-1] \cup(1,4]
    (C)
    (−∞,−1)∪[1,4]\displaystyle (-\infty,-1) \cup[1,4]
    (D)
    (−∞,−1)∪[1,4)\displaystyle (-\infty,-1) \cup[1,4)

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    NCERT’s answer
    A
    (A) \(\displaystyle (-\infty,-1)\cup(1,4]\). The first root needs a non-negative radicand; the second sits in a denominator, so its radicand must be strictly positive.\[4-x\ge 0 \Rightarrow x\le 4 \] \[x^2-1>0 \Rightarrow x<-1 \ \text{or}\ x>1 \] \[\text{Domain}=(-\infty,-1)\cup(1,4] \]