Mathematics — Class 11 practice questions

Worked examples for Mathematics — Class 11

1. Question 1

   If $\sin x = \dfrac{3}{5}$ and $0 < x < \dfrac{\pi}{2}$, then ${\sin\left(\dfrac{3\pi}{2} - x\right) =}$

   1. A. $-\dfrac{3}{5}$

      The sign is right, but the function was not changed. At an odd multiple of $\frac{\pi}{2}$, sine becomes cosine. Sine stays sine only for $\pi \pm x$ and $2\pi - x$. Writing $-\sin x$ gives $-\frac{3}{5}$ instead of $-\cos x = -\frac{4}{5}$.

   2. B. $\dfrac{4}{5}$

      This is $\sin\left(\frac{\pi}{2} - x\right) = \cos x$. The $\frac{3\pi}{2}$ was treated as $\frac{\pi}{2}$, which puts the angle in quadrant I. But $\frac{3\pi}{2} - x$ lies in quadrant III, where sine is negative, so the answer needs a minus sign.

   3. C. $-\dfrac{4}{5}$Correct answer

      $\frac{3\pi}{2} - x$ is an odd multiple of $\frac{\pi}{2}$ minus an acute angle. So sine changes to cosine. The angle lies in quadrant III, where the original function, sine, is negative. Hence $\sin\left(\frac{3\pi}{2} - x\right) = -\cos x = -\sqrt{1 - \frac{9}{25}} = -\frac{4}{5}$.

   4. D. $-\dfrac{8}{5}$

      Sine was split over the difference: $\sin\frac{3\pi}{2} - \sin x = -1 - \frac{3}{5}$. Sine is not linear, so $\sin(A - B) \neq \sin A - \sin B$. A value of magnitude greater than $1$ cannot be a sine at all.

   Explanation

   For an allied angle, look at the multiple of $\frac{\pi}{2}$. An odd multiple changes the function to its co-function. The sign is the sign of the original function in the quadrant where the allied angle lies, taking $x$ as acute. Here sine becomes cosine with a negative sign, and $\cos x = \frac{4}{5}$ comes from $\sin^{2} x + \cos^{2} x = 1$ since $x$ is acute.

2. Question 2

   The value of $\dfrac{\log_{3} 64}{\log_{3} 4}$ is

   1. A. $3$Correct answer

      By the change-of-base rule, $\frac{\log_{3} 64}{\log_{3} 4} = \log_{4} 64$. Since $4^{3} = 64$, the value is $3$.

   2. B. $\log_{3} 16$

      This uses the false rule $\frac{\log a}{\log b} = \log \frac{a}{b}$. The true rule is $\log a - \log b = \log \frac{a}{b}$, which is a difference of logs and not a quotient. A quotient of two logs to the same base is a change of base: $\frac{\log_{3} 64}{\log_{3} 4} = \log_{4} 64$.

   3. C. $16$

      The arguments were divided, $\frac{64}{4} = 16$, and the logarithms were dropped. The expression divides two logarithms, not the numbers inside them. Here $\log_{3} 64 = 6\log_{3} 2$ and $\log_{3} 4 = 2\log_{3} 2$, so the ratio is $3$.

   4. D. $\dfrac{1}{3}$

      The change of base was applied upside down, giving $\log_{64} 4 = \frac{1}{3}$. In $\frac{\log_{c} a}{\log_{c} b} = \log_{b} a$, the numerator's argument is the new argument and the denominator's argument is the new base.

   Explanation

   The change-of-base law says $\log_{b} a = \frac{\log_{c} a}{\log_{c} b}$ for any admissible base $c$. Read from right to left, a quotient of two logarithms to the same base is a single logarithm, so the expression is $\log_{4} 64$. Since $64 = 4^{3}$, its value is $3$.

3. Question 3

   If\ $\quad z = (4 + 3i)(2 - i)$,\ where $i = \sqrt{-1}$, then ${|z| =}$

   1. A. $5\sqrt{5}$Correct answer

      The modulus of a product is the product of the moduli: $|4 + 3i| = \sqrt{16 + 9} = 5$ and $|2 - i| = \sqrt{4 + 1} = \sqrt{5}$, so $|z| = 5\sqrt{5}$. Expanding gives the same result: $z = 8 - 4i + 6i - 3i^{2} = 11 + 2i$, so $|z| = \sqrt{121 + 4} = \sqrt{125} = 5\sqrt{5}$.

   2. B. $\sqrt{21}$

      This comes from squaring the whole imaginary term, $(3i)^{2} = -9$ and $(-i)^{2} = -1$, which gives $\sqrt{16 - 9} = \sqrt{7}$ and $\sqrt{4 - 1} = \sqrt{3}$. In $|a + ib| = \sqrt{a^{2} + b^{2}}$, $b$ is the real coefficient of $i$, so both terms under the root are positive.

   3. C. $125$

      This is $|z|^{2} = 11^{2} + 2^{2}$. The square root was never taken. The modulus is $\sqrt{125} = 5\sqrt{5}$.

   4. D. $5 + \sqrt{5}$

      The two moduli were added. The rule is $|z_1 z_2| = |z_1|\,|z_2|$, a product. Addition only gives an upper bound for $|z_1 + z_2|$, and that is a sum of the numbers, not a product.

   Explanation

   The modulus is multiplicative: $|z_1 z_2| = |z_1|\,|z_2|$, because the moduli multiply when two complex numbers are multiplied in polar form. Each modulus is $\sqrt{a^{2} + b^{2}}$, where $a$ and $b$ are the real and imaginary parts, both taken as real numbers. Expanding the product first to $11 + 2i$ gives the same value.

4. Question 4

   If\ $\quad f(x) = x^{2} + 5, \quad g(x) = x - 8$, then ${(f \circ g)(x) =}$

   1. A. $x^{2} - 16x + 69$Correct answer

      $(f \circ g)(x) = f(g(x)) = f(x - 8) = (x - 8)^{2} + 5 = x^{2} - 16x + 64 + 5 = x^{2} - 16x + 69$.

   2. B. $x^{2} - 3$

      This is $(g \circ f)(x) = g(f(x)) = (x^{2} + 5) - 8$. The order was reversed. In $(f \circ g)(x) = f(g(x))$, $g$ acts first and $f$ is applied to its output.

   3. C. $x^{2} + 69$

      Here $(x - 8)^{2}$ was expanded as $x^{2} + 64$ and the middle term $-2 \cdot 8 \cdot x = -16x$ was left out. The correct expansion is $(a - b)^{2} = a^{2} - 2ab + b^{2}$.

   4. D. $x^{3} - 8x^{2} + 5x - 40$

      This is the product $f(x) \cdot g(x) = (x^{2} + 5)(x - 8)$. Composition does not multiply the two functions. It puts $g(x)$ in place of $x$ inside $f$.

   Explanation

   A composite function $(f \circ g)(x)$ is defined as $f(g(x))$: first find $g(x)$, then use that value as the input of $f$. So you replace every $x$ in $f(x) = x^{2} + 5$ with $x - 8$ and expand the square in full.

5. Question 5

   If $\omega$ is a complex cube root of unity, then ${(1 - \omega + \omega^{2})^{5} + (1 + \omega - \omega^{2})^{5} =}$

   1. A. $-32$

      The minus sign was dropped from the collapsed brackets: $1 - \omega + \omega^{2} = (1 + \omega^{2}) - \omega = -2\omega$ was written as $2\omega$, and likewise $-2\omega^{2}$ as $2\omega^{2}$. That gives $(2\omega)^{5} + (2\omega^{2})^{5} = 32(\omega^{2} + \omega) = -32$. From $1 + \omega + \omega^{2} = 0$ the brackets are $-2\omega$ and $-2\omega^{2}$, so the sum is $(-2)^{5}(\omega^{5} + \omega^{10}) = -32(\omega^{2} + \omega) = 32$.

   2. B. $2$

      The power was applied to $\omega$ but not to the coefficient: $(-2\omega)^{5}$ was written as $-2\omega^{5}$. This gives $-2(\omega^{2} + \omega) = 2$. A power of a product raises every factor, so $(-2)^{5} = -32$ must appear as well.

   3. C. $10$

      The coefficient was multiplied by the index instead of raised to it: $(-2)^{5}$ was taken as $-2 \times 5 = -10$. This gives $-10(\omega^{5} + \omega^{10}) = -10(-1) = 10$. The fifth power of $-2$ is $-32$.

   4. D. $32$Correct answer

      $1 - \omega + \omega^{2} = -\omega - \omega = -2\omega$ and $1 + \omega - \omega^{2} = -2\omega^{2}$. So the sum is $(-2\omega)^{5} + (-2\omega^{2})^{5} = -32(\omega^{5} + \omega^{10}) = -32(\omega^{2} + \omega) = -32(-1) = 32$.

   Explanation

   Replace each pair of terms using $1 + \omega + \omega^{2} = 0$, which gives $1 + \omega^{2} = -\omega$ and $1 + \omega = -\omega^{2}$. Each bracket then becomes a single term, $-2\omega$ or $-2\omega^{2}$, and both factors of that term are raised to the fifth power. Reduce the exponents of $\omega$ mod $3$ using $\omega^{3} = 1$, so $\omega^{5} = \omega^{2}$ and $\omega^{10} = \omega$. Then $\omega^{2} + \omega = -1$ gives a real value.

6. Question 6

   Two fair dice are thrown. Given that the number on the first die is even, the probability that the sum of the numbers on the two dice is $8$ is

   1. A. $\dfrac{1}{12}$

      The three favourable outcomes $(2, 6)$, $(4, 4)$, $(6, 2)$ were counted correctly but divided by all $36$ outcomes. That gives $P(A \cap B)$, not $P(A|B)$. Once we know the first die is even, only the $18$ outcomes with an even first number are possible, so the denominator must be $18$.

   2. B. $\dfrac{3}{5}$

      This divides by the wrong event. It is $\frac{n(A \cap B)}{n(A)} = \frac{3}{5}$, the probability that the first die is even given that the sum is $8$, which is $P(B|A)$. The given event is the first die being even, so we divide by $n(B) = 18$.

   3. C. $\dfrac{1}{6}$Correct answer

      Let $A$: the sum is $8$ and $B$: the first die is even. Then $n(B) = 3 \times 6 = 18$, and $A \cap B = \{(2, 6), (4, 4), (6, 2)\}$, so $n(A \cap B) = 3$. Hence $P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{3}{18} = \frac{1}{6}$.

   4. D. $\dfrac{5}{36}$

      This is $P(A)$, from the five outcomes $(2,6), (3,5), (4,4), (5,3), (6,2)$ out of $36$. You get it by treating the events as independent, writing $P(A \cap B) = P(A)P(B)$ and cancelling $P(B)$. They are not independent: an even first die rules out $(3, 5)$ and $(5, 3)$, so the condition changes the probability.

   Explanation

   The phrase "given that" makes this conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Knowing that $B$ occurred shrinks the sample space to the outcomes in $B$. So we count the outcomes in both $A$ and $B$ and divide by the number of outcomes in $B$, not by all $36$.

7. Question 7

   Six boys and four girls are to be seated around a circular table. The number of ways in which they can be seated so that no two girls sit next to each other is

   1. A. $100800$

      This uses the row rule of $n + 1$ gaps at a round table: $5! \times {}^{7}P_{4} = 120 \times 840 = 100800$. Six people seated in a circle leave exactly $6$ gaps between neighbours, because a circle has no two ends to add extra slots.

   2. B. $259200$

      The boys were arranged as if in a row, in $6! = 720$ ways, giving $720 \times 360 = 259200$. Round a table only the relative order matters, so the $6$ rotations of one seating are the same seating and the boys can be seated in $(6 - 1)! = 5!$ ways.

   3. C. $1800$

      The gaps were only chosen, $\binom{6}{4} = 15$, and not filled in order: $5! \times 15 = 1800$. The girls are different people, so once $4$ gaps are chosen they can sit in them in $4!$ ways. That is why the count is ${}^{6}P_{4}$ and not $\binom{6}{4}$.

   4. D. $43200$Correct answer

      First seat the boys round the table in $(6 - 1)! = 5! = 120$ ways. They leave $6$ gaps, one between each pair of neighbours. The $4$ girls take $4$ of these gaps in order, in ${}^{6}P_{4} = 360$ ways. Total: $120 \times 360 = 43200$.

   Explanation

   For a "no two together" condition, first arrange the unrestricted people and then place the restricted ones in the gaps between them. A circular arrangement of $6$ distinct boys counts as $(6-1)!$, since rotations coincide, and it leaves exactly $6$ gaps, not $7$ as a row would. The $4$ distinct girls then fill $4$ of those gaps in order in ${}^{6}P_{4}$ ways, and the multiplication principle combines the two stages.

8. Question 8

   How many five-digit numbers divisible by $5$ can be formed using the digits $0, 1, 2, 3, 4, 5$, if no digit is repeated?

   1. A. $240$

      Both cases were counted as $5 \times 4 \times 3 \times 2 = 120$. When the units digit is $5$, the digit $0$ is still available and cannot go in the ten-thousands place, so that case has only $4 \times 4 \times 3 \times 2 = 96$ numbers, not $120$.

   2. B. $216$Correct answer

      The units digit must be $0$ or $5$. If it is $0$, the remaining $5$ digits fill the other four places in $5 \times 4 \times 3 \times 2 = 120$ ways. If it is $5$, the first place can take any of $1, 2, 3, 4$ ($4$ ways), and the other three places are filled from the $4$ digits left in $4 \times 3 \times 2 = 24$ ways, which gives $96$. The total is $120 + 96 = 216$.

   3. C. $120$

      Only numbers ending in $0$ were counted. A number is divisible by $5$ when it ends in $0$ or $5$, so the $96$ numbers ending in $5$ are missing.

   4. D. $96$

      Only numbers ending in $5$ were counted. Numbers ending in $0$ are also divisible by $5$, and there are $120$ of them.

   Explanation

   Divisibility by $5$ fixes the units digit as $0$ or $5$. The two cases differ because a five-digit number cannot begin with $0$, so they must be counted separately. With $0$ in the units place there is no restriction on the first place. With $5$ in the units place, $0$ is still available and has to be kept out of the first place. The two disjoint cases are then added.