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NCERT Exemplar · Class 11 Mathematics Mathematical Reasoning

37 questions · 37 still being checked

This chapter is from the older syllabus and is not in the current NCERT textbook, so you may skip it.

EXERCISE 14.3 11–20 (part 2 of 4)

  1. Exercise 11

    Identify the Quantifiers in the following statements.
    (i)
    There exists a triangle which is not equilateral.
    (ii)
    For all real numbers x\displaystyle x and y,xy=yx\displaystyle y, x y=y x.
    (iii)
    There exists a real number which is not a rational number.
    (iv)
    For every natural number x,x+1\displaystyle x, x+1 is also a natural number.
    (v)
    For all real numbers x\displaystyle x with x>3,x2\displaystyle x>3, x^2 is greater than 9.
    (vi)
    There exists a triangle which is not an isosceles triangle.
    (vii)
    For all negative integers x,x3\displaystyle x, x^3 is also a negative integers.
    (viii)
    There exists a statement in above statements which is not true.
    (ix)
    There exists a even prime number other than 2.
    (x)
    There exists a real number x\displaystyle x such that x2+1=0\displaystyle x^2+1=0.

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    NCERT’s answer
    (i)
    There exists
    (ii)
    For all
    (iii)
    There exists
    (iv)
    For every
    (v)
    For all
    (vi)
    There exists
    (vii)
    For all
    (viii)
    There exists
    (ix)
    There exists
    (x)
    There exists
    \[\exists:\ \text{(i), (iii), (vi), (viii), (ix), (x)} \qquad \forall:\ \text{(ii), (iv), (v), (vii)} \]
    (i)
    There exists
    (ii)
    For all
    (iii)
    There exists
    (iv)
    For every
    (v)
    For all
    (vi)
    There exists
    (vii)
    For all
    (viii)
    There exists
    (ix)
    There exists
    (x)
    There exists
    Answer: "There exists" in (i), (iii), (vi), (viii), (ix), (x); "For all / For every" in (ii), (iv), (v), (vii).

  2. Exercise 12

    Prove by direct method that for any integer 'n\displaystyle n', n3−n\displaystyle n^3-n is always even. [Hint: Two cases (i) n\displaystyle n is even, (ii) n\displaystyle n is odd.]

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    \[n^3 - n = n(n^2 - 1) \]Case $\displaystyle 1$: \(\displaystyle n\) even, \(\displaystyle n = 2k\).\[n^3 - n = 2k(4k^2 - 1) = 2\,[\,k(4k^2 - 1)\,] \]Case $\displaystyle 2$: \(\displaystyle n\) odd, \(\displaystyle n = 2k + 1\).\[n^2 - 1 = 4k^2 + 4k = 4k(k+1) \]\[n^3 - n = (2k+1)\cdot 4k(k+1) = 2\,[\,2k(k+1)(2k+1)\,] \]In both cases \(\displaystyle n^3 - n = 2m\) with \(\displaystyle m\) an integer.Answer: \(\displaystyle n^3 - n\) is even for every integer \(\displaystyle n\).

  3. Exercise 13

    Check the validity of the following statement.
    (i)
    p:125\displaystyle p: 125 is divisible by 5\displaystyle 5 and 7.
    (ii)
    q:131\displaystyle q: 131 is a multiple of 3\displaystyle 3 or 11.

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    (i) \(\displaystyle p\) is a conjunction, so both parts must hold.\[125 = 5^3 \quad\Rightarrow\quad 5 \mid 125 \]\[125 = 7\cdot 17 + 6 \quad\Rightarrow\quad 7 \nmid 125 \]\(\displaystyle p\) is false.(ii) \(\displaystyle q\) is a disjunction, so at least one part must hold.\[131 = 3\cdot 43 + 2 \quad\Rightarrow\quad 3 \nmid 131 \]\[131 = 11\cdot 11 + 10 \quad\Rightarrow\quad 11 \nmid 131 \]Both parts fail, so \(\displaystyle q\) is false.Answer: neither \(\displaystyle p\) nor \(\displaystyle q\) is valid.

  4. Exercise 14

    Prove the following statement by contradication method. p\displaystyle p : The sum of an irrational number and a rational number is irrational.

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    Let \(\displaystyle a\) be irrational and \(\displaystyle r = \dfrac{m}{n}\) rational (\(\displaystyle m, n\) integers, \(\displaystyle n \ne 0\)). Assume \(\displaystyle a + r\) is rational.\[a + r = \frac{p}{q} \quad (p, q \text{ integers},\ q \ne 0) \]\[a = \frac{p}{q} - \frac{m}{n} = \frac{pn - mq}{qn}, \qquad qn \ne 0 \]So \(\displaystyle a\) is a ratio of integers, i.e. rational, which contradicts \(\displaystyle a\) being irrational.Answer: the assumption is false, so \(\displaystyle a + r\) is irrational.

  5. Exercise 15

    Prove by direct method that for any real numbers x,y\displaystyle x, y if x=y\displaystyle x=y, then x2=y2\displaystyle x^2=y^2.

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    \[x = y \quad \text{(given)} \]\[x\cdot x = x\cdot y \quad \text{(multiply both sides by } x) \]\[x\cdot y = y\cdot y \quad \text{(multiply both sides of the given by } y) \]\[x^2 = y^2 \quad \text{(the two results share } xy) \]Answer: \(\displaystyle x = y \Rightarrow x^2 = y^2\).

  6. Exercise 16

    Using contrapositive method prove that if n2\displaystyle n^2 is an even integer, then n\displaystyle n is also an even integers.

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    Contrapositive: if \(\displaystyle n\) is odd, then \(\displaystyle n^2\) is odd. Take \(\displaystyle n\) odd.\[n = 2k + 1 \quad (k \in \mathbb{Z}) \]\[n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 \]\[n^2 = 2m + 1, \quad m = 2k^2 + 2k \in \mathbb{Z} \]So \(\displaystyle n^2\) is odd. The contrapositive holds, hence the original statement holds.Answer: if \(\displaystyle n^2\) is even, then \(\displaystyle n\) is even.
  7. Choose the correct answer out of the four options given against each of the Exercises $\displaystyle 17$ to $\displaystyle 36$ (M.C.Q.).

    Exercise 17

    Which of the following is a statement.
    (A)
    x\displaystyle x is a real number.
    (B)
    Switch off the fan.
    (C)
    6\displaystyle 6 is a natural number.
    (D)
    Let me go.

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    NCERT’s answer
    C
    (C) \(\displaystyle 6\) is a natural number.\[6 \in \mathbb{N} \quad \text{(true)} \]It is definitely true, so it is a statement. (A) depends on the value of \(\displaystyle x\), so it is neither true nor false; (B) and (D) are commands.

  8. Exercise 18

    Which of the following is not a statement
    (A)
    Smoking is injurious to health.
    (B)
    2+2=4\displaystyle 2+2=4
    (C)
    2\displaystyle 2 is the only even prime number.
    (D)
    Come here.

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    NCERT’s answer
    D
    (D) Come here.It is a command, so it is neither true nor false.\[2+2=4 \quad \text{(true)} \]The other three are definitely true, hence statements.

  9. Exercise 19

    The connective in the statement "2+7>9\displaystyle 2+7>9 or 2+7<9\displaystyle 2+7<9" is
    (A)
    and
    (B)
    or

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    NCERT’s answer
    B
    (B) or\[p:\ 2+7>9, \qquad q:\ 2+7<9 \] \[\text{statement} = p \vee q \]The signs \(\displaystyle >\) and \(\displaystyle <\) are relations inside the two clauses; the word joining them is "or".

  10. Exercise 20

    The connective in the statement "Earth revolves round the Sun and Moon is a satellite of earth" is
    (A)
    or
    (B)
    Earth
    (C)
    Sun
    (D)
    and

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    NCERT’s answer
    D
    (D) and\[p:\ \text{Earth revolves round the Sun} \] \[q:\ \text{Moon is a satellite of earth} \] \[\text{statement} = p \wedge q \]The word joining the two clauses is "and".