A square piece of tin of side 18 cm is to made into a box without top,
Question: A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? Solution: Let the side of the square to be cut off bexcm. Then, the length and the breadth of the box will be (18 2x) cm each and the height of the box isxcm. Therefore, the volumeV(x) of the box is given by, $V(x)=x(18-2 x)^{2}$ $\b...
Read More →what is the probability of the event ‘not A’.
Question: If $\frac{2}{11}$ is the probability of an event, what is the probability of the event 'not $A$ '. Solution: It is given that $P(A)=\frac{2}{11}$. Accordingly, $P($ not $A)=1-P(A)=1-\frac{2}{11}=\frac{9}{11}$...
Read More →Three coins are tossed once. Find the probability of getting
Question: Three coins are tossed once. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most two tails. Solution: When three coins are tossed once, the sample space is given by S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} $\therefore$ Accordingly,$n(S)=8$ It is known that the probability of an event A is given by $\mathrm{P}(\mathrm{A})=\frac{\text { Number of outcomes favo...
Read More →Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Question: Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum. Solution: Let one number bex. Then, the other number is (16 x). Let the sum of the cubes of these numbers be denoted by S(x). Then, $S(x)=x^{3}+(16-x)^{3}$ $\therefore S^{\prime}(x)=3 x^{2}-3(16-x)^{2}, S^{\prime \prime}(x)=6 x+6(16-x)$' Now, $S^{\prime}(x)=0 \Rightarrow 3 x^{2}-3(16-x)^{2}=0$ $\Rightarrow x^{2}-(16-x)^{2}=0$ $\Rightarrow x^{2}-256-x^{2}+32 x=0$ $\Rightarrow x=\frac{256}{32}=8$ Now, $S^{\p...
Read More →If α and β are the zeros of the quadratic polynomial f(x)
Question: If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-2 x+3$, find a polynomial whose roots are $(i) \alpha+2, \beta+2$ (ii) $\frac{\alpha-1}{\alpha+1}, \frac{\beta-1}{\beta+1}$. Solution: (i) Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-2 x+3$...
Read More →If (27)x = 9/3x, find x
Question: If $(27)^{x}=9 / 3^{x}$, find $x$ Solution: We have, $(27)^{x}=9 / 3^{x}$ $\left(3^{3 x}\right)^{x}=9 / 3^{x}$ $3^{3 x}=9 / 3^{x}$ $3^{3 x}=3^{2} / 3^{x}$ $3^{3 x}=3^{2-x}$ 3x = 2 x [On equating exponents] 3x + x = 2 4x = 2 x = 2/4 x = 1/2 Here the value of x is...
Read More →If α and β are the zeros of the quadratic polynomial f(x)
Question: If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $1(x)=x^{2}-p(x+1)-c$, show that $(\alpha+1)(\beta+1)=1-c$. Solution: Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-p(x+1)-c$ Then $x^{2}-p(x+1)-c$ $x^{2}-p x-p-c$ $\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$ $=\frac{-(-p)}{1}$ = p $\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$ $=\frac{-p-c}{1}$ $=-p-c$ We have to prove...
Read More →Find two positive numbers x and y such that their sum is
Question: Find two positive numbersxandysuch that their sum is 35 and the productx2y5is a maximum Solution: Let one number bex. Then, the other number isy= (35 x). Let $P(x)=x^{2} y^{5}$. Then, we have: When $x=35, f^{\prime}(x)=f(x)=0$ and $y=35-35=0$. This will make the product $x^{2} y^{5}$ equal to 0 . When $x=0 \quad y=35-0=35$ and the nroduct $x^{2} v^{2}$ will be 0 x= 0 andx= 35 cannot be the possible values ofx. Whenx= 10, we have: $\begin{aligned} P^{\prime \prime}(x) =7(35-10)^{3}(6 \t...
Read More →If 3x = 5y = (75)z, Show that
Question: If $3^{x}=5^{y}=(75)^{z}$, Show that $z=\frac{x y}{2 x+y}$ Solution: $3^{x}=k$ $3=k^{1 / x}$ $5^{y}=k$ $5=k^{1 / y}$ $75^{z}=k$ $75=k^{1 / z}$ $3^{1} \times 5^{2}=75^{1}$ $k^{1 / x} \times k^{2 / y}=k^{1 / z}$ $1 / x+2 / y=1 / z$ $\frac{y+2 x}{x y}=\frac{1}{z}$ $z=\frac{x y}{2 x+y}$...
Read More →If ax = by = cz and b2
Question: If $a^{x}=b^{y}=c^{2}$ and $b^{2}=a c$, then show that $y=\frac{2 z x}{z+x}$ Solution: Let $a^{x}=b^{y}=c^{z}=k$ $a=k^{1 / x}, b=k^{1 / y}, c=k^{1 / z}$ Now $b^{2}=a c$ $\left(k^{1 / y}\right)^{2}=k^{1 / x} \times k^{1 / z}$ $k^{2 / y}=k^{1 / x+1 / z}$ $2 / y=1 / x+1 / z$ $2 y=\frac{x+z}{x z}$ $y=\frac{2 x z}{x+z}$...
Read More →If α and β are the zeros of a quadratic polynomial such that α + β = 24 and α − β = 8,
Question: If $\alpha$ and $\beta$ are the zeros of a quadratic polynomial such that $\alpha+\beta=24$ and $\alpha-\beta=8$, find a quadratic polynomial have $\alpha$ and $\beta$ as its zeros. Solution: Given, $\alpha+\beta=24 \ldots \ldots$ (i) $\alpha-\beta=8 \ldots \ldots$ (ii) By subtracting equation $(i i)$ from $(i)$ we get $\alpha=\frac{32}{2}$ $\alpha=16$ Substituting $\alpha=16$ in equation $(i)$ we get, $\alpha+\beta=24$ $16+\beta=24$ $\beta=24-16$ $\beta=8$ Let S and P denote respectiv...
Read More →If 2x = 3y = 6−z, show that
Question: If $2^{x}=3^{y}=6$, show that $1 / x+1 / y+1 / z=0$ Solution: $2^{x}=3^{y}=6^{-z}$ $2^{x}=k$ $2=k^{1 / x}$ $3^{y}=k$ $3=k^{1 / y}$ $6^{-z}=k$ $k=1 / 6^{z}$ $6=k^{-1 / z}$ $2 \times 3=6$ $k^{1 / x} \times k^{1 / y}=k^{-1 / z}$ $1 / x+1 / y=-1 / z$ [by equating exponents] $1 / x+1 / y+1 / z=0$...
Read More →A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up.
Question: A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up.From the sample space calculate how many different amounts of money you canhave after four tosses and the probability of having each of these amounts. Solution: Since the coin is tossed four times, there can be a maximum of 4 heads or tails. When 4 heads turns up, Rel+Rel+Rel+Rel = Rs 4 is the gain. When 3 heads and 1 tail turn up, Re 1 + Re 1 + Re 1 Rs 1.50 = Rs 3 Rs 1.50...
Read More →If α and β are the zeros of the quadratic polynomial f(x)
Question: If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $1(x)=x^{2}-3 x-2$, find a quadratic polynomial whose zeros are $\frac{1}{2 \alpha+\beta}$ and $\frac{1}{2 \beta+\alpha}$. Solution: Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-3 x-2$ The roots are $\alpha$ and $\beta$ $\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$ $\alpha+\beta=-\left(\frac{-3}{1}\right)$ $\alpha+\beta=-(-3)$ $\alpha+\beta=3$ $\alph...
Read More →If 2x = 3y = 12z, show that
Question: If $2^{x}=3^{y}=12^{z}$, show that $1 / z=1 / y+2 / x$ Solution: $2^{x}=3^{y}=(2 \times 3 \times 2)^{z}$ $2^{x}=3^{y}=\left(2^{2} \times 3\right)^{z}$ $2^{x}=3^{y}=\left(2^{2 z} \times 3^{z}\right)$ $2^{x}=3^{y}=12^{z}=k$ $2=k^{1 / x}$ $3=k^{1 / y}$ $12=k^{1 / z}$ $12=2 \times 3 \times 2$ $12=k^{1 / z}=k^{1 / y} \times k^{1 / x} \times k^{1 / x}$ $k^{1 / z}=k^{2 / x}+{ }^{1 / y}$ $1 / z=1 / y+2 / x$...
Read More →There are four men and six women on the city council.
Question: There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman? Solution: There are four men and six women on the city council. As one council member is to be selected for a committee at random, the sample space contains 10 (4 + 6) elements. Let A be the event in which the selected council member is a woman. Accordingly, $n(\mathrm{~A})=6$ $\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of o...
Read More →A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed.
Question: A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12 Solution: Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} Accordingly, $n(S)=12$ (i) Let A be the event in which ...
Read More →Show that
Question: Show that (i) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$ (ii) $\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}$ (iii) $\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}$ (iv) $\left(\frac{x^{a^{2}}+b^{2}}{x^{a b}}\right)^{a+b}\left(\frac{x^{b^{2}}+c^{2}}{x^{b c}}\right)^{b+c}\left(\frac{x^{c^{2}}+a^{2}}{x^{a c}}\right)^{a+c}=2\left(a^{3}+b^{3}+c^{3}\right)$ (v) ...
Read More →A card is selected from a pack of 52 cards.
Question: A card is selected from a pack of 52 cards. (a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades. (c) Calculate the probability that the card is (i) an ace (ii) black card. Solution: (a)When a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e., the sample space contains 52 elements. Therefore, there are 52 points in the sample space. (b)Let A be the event in which the card drawn is an ace of...
Read More →If α and β are the zeros of the quadratic polynomial f(x)
Question: If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $l(x)=x^{2}-1$, find the quadratic polynomial whose zeros are $\frac{2 \alpha}{\beta}$ and $\frac{2 \beta}{\alpha}$. Solution: Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-1$ The roots are $\alpha$ and $\beta$ $\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$ $\alpha+\beta=\frac{0}{1}$ $\alpha+\beta=0$ $\alpha \beta=\frac{\text { Constant term }}{\text {...
Read More →A die is thrown, find the probability of following events:
Question: A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to 3 will appear, (iii) A number less than or equal to one will appear, (iv) A number more than 6 will appear, (v) A number less than 6 will appear. Solution: The sample space of the given experiment is given by S = {1, 2, 3, 4, 5, 6} (i) Let A be the event of the occurrence of a prime number. Accordingly, A = {2, 3, 5} $\therefore \mathrm{P}(\mathrm{A})=\frac{...
Read More →Find two positive numbers
Question: Find two positive numbers $x$ and $y$ such that $x+y=60$ and $x y^{3}$ is maximum. Solution: The two numbers arexandysuch thatx+y= 60. $\Rightarrow y=60-x$ Let $f(x)=x y^{3}$ $\begin{aligned} \Rightarrow f(x) =x(60-x)^{3} \\ \therefore f^{\prime}(x) =(60-x)^{3}-3 x(60-x)^{2} \\ =(60-x)^{2}[60-x-3 x] \\ =(60-x)^{2}(60-4 x) \end{aligned}$ And, $f^{\prime \prime}(x)=-2(60-x)(60-4 x)-4(60-x)^{2}$ $=-2(60-x)[60-4 x+2(60-x)]$ $=-2(60-x)(180-6 x)$ $=-12(60-x)(30-x)$ Now, $f^{\prime}(x)=0 \Rig...
Read More →A coin is tossed twice, what is the probability that at least one tail occurs?
Question: A coin is tossed twice, what is the probability that at least one tail occurs? Solution: When a coin is tossed twice, the sample space is given by S = {HH, HT, TH, TT} Let A be the event of the occurrence of at least one tail. Accordingly, A = {HT, TH, TT} $\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of outcomes favourable to } \mathrm{A}}{\text { Total number of possible outcomes }}$ $=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}$ $=\frac{3}{4}$...
Read More →Which of the following can not be valid assignment of probabilities for outcomes of sample space
Question: Which of the following can not be valid assignment of probabilities for outcomes of sample space $S=\left\{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\right\}$ Solution: Here, each of the numbersp(i) is positive and less than 1. Sum of probabilities $=p\left(\omega_{1}\right)+p\left(\omega_{2}\right)+p\left(\omega_{3}\right)+p\left(\omega_{4}\right)+p\left(\omega_{5}\right)+p\left(\omega_{6}\right)+p\left(\omega_{7}\right)$ $=0.1+0.01+0.05+0.03+0....
Read More →If one zero of the quadratic polynomial f(x)
Question: If one zero of the quadratic polynomial $1(x)=4 x^{2}-8 k x-9$ is negative of the other, find the value of $k$. Solution: Since $\alpha$ and $-\alpha$ are the zeros of the quadratic polynomial $f(x)=4 x^{2}-8 k x-9$ $\alpha-\alpha=0$ $\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}=0$ $\frac{-8 k}{4}=0$ $-8 k=0 \times 4$ $-8 k=0$ $k=\frac{0}{-8}$ $k=0$ Hence, the Value of $k$ is 0 ....
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