Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\log (\operatorname{cosec} x-\cot x)$ Solution: Let $y=\log (\operatorname{cosec} x-\cot x)$ On differentiating $y$ with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}[\log (\operatorname{cosec} x-\cot x)]$ We know $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$ $\Rightarrow \frac{d y}{d x}=\frac{1}{\operatorname{cosec} x-\cot x} \frac{d}{d x}(\operatorname{cosec} x-\cot x)$ [using chain rule] $\...
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Question: Factorise:81 49x2 Solution: We have: $81-49 x^{2}=(9)^{2}-(7 x)^{2}$ $=(9+7 x)(9-7 x)$ $\therefore 81-49 x^{2}=(9+7 x)(9-7 x)$...
Read More →Calculate the mean of the following data
Question: Calculate the mean of the following data Solution: Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class. Now, we first find the class markxi, of each class and then proceed as follows Therefore, $\bar{x}($ mean $)=\frac{\Sigma f_{i} x_{i}}{\Sigma_{j}}=\frac{362}{28}=12.93$ Hence, mean of the given data is $12.93$....
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Question: Factorise:4a2 9 Solution: We have: $4 a^{2}-9=(2 a)^{2}-(3)^{2}$ $\quad=(2 a+3)(2 a-3)$ $\therefore 4 a^{2}-9=(2 a+3)(2 a-3)$...
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Question: Factorise:x2 36 Solution: We have: $x^{2}-36=(x)^{2}-(6)^{2}$ $=(x+6)(x-6)$ $\therefore x^{2}-36=(x+6)(x-6)$...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $\mathrm{x}$ : $\frac{e^{x} \log x}{x^{2}}$ Solution: Let $y=\frac{e^{x} \log x}{x^{2}}$ On differentiating y with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left(\frac{e^{x} \log x}{x^{2}}\right)$ Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule) $\Rightarrow \frac{d y}{d x}=\frac{\left(x^{2}\right) \frac{d}{d x}\left(e^{x} \...
Read More →Calculate the mean of the scores
Question: Calculate the mean of the scores of 20 students in a mathematics test Solution: We first, find the class mark of each class and then proceed as follows Therefore, $\operatorname{mean}(\bar{x})=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{700}{20}=35$ Hence, the mean of scores of 20 students in mathematics test is 35....
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Question: Factorise:x2x(a+ 2b) + 2ab Solution: We have: $x^{2}-x(a+2 b)+2 a b=x^{2}-a x-2 b x+2 a b$ $=x^{2}-2 b x-a x+2 a b$ $=\left(x^{2}-2 b x\right)-(a x-2 a b)$ $=x(x-2 b)-a(x-2 b)$ $=(x-2 b)(x-a)$ $\therefore x^{2}-x(a+2 b)+2 a b=(x-2 b)(x-a)$...
Read More →Find the mean of the distribution
Question: Find the mean of the distribution Solution: We first, find the class mark xi,of each class and then proceed as follows. Therefore, $\operatorname{mean}(\bar{x})=\frac{\Sigma f_{j} x_{i}}{\Sigma f_{i}}=\frac{412.5}{75}=5.5$ Hence, mean of the given distribution is 5.5....
Read More →Factorise:
Question: Factorise:ab(x2+y2) xy(a2+b2) Solution: We have: $a b\left(x^{2}+y^{2}\right)-x y\left(a^{2}+b^{2}\right)=a b x^{2}+a b y^{2}-a^{2} x y-b^{2} x y$ $=a b x^{2}-a^{2} x y+a b y^{2}-b^{2} x y$ $=a x(b x-a y)+b y(a y-b x)$ $=a x(b x-a y)-b y(b x-a y)$ $=(b x-a y)(a x-b y)$ $\therefore a b\left(x^{2}+y^{2}\right)-x y\left(a^{2}+b^{2}\right)=(b x-a y)(a x-b y)$...
Read More →A bag contains slips numbered from 1 to 100 .
Question: A bag contains slips numbered from 1 to 100 . If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since, this situation has only two possible outcomes, so the 1 probability of each is $\frac{1}{2}$. Justify. Solution: We know that, between 1 to 100 half numbers are even and half numbers are odd i.e., 50 numbers (2, 4, 6, 8. 96, 98,100) are even and 50 numbers (1,3, 5, 7. . , 97, 99) are odd. So, both events are equally likely. So, probabi...
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Question: Factorise:x3 3x2+x 3 Solution: We have: $x^{3}-3 x^{2}+x-3=\left(x^{3}-3 x^{2}\right)+(x-3)$ $=x^{2}(x-3)+1(x-3)$ $=(x-3)\left(x^{2}+1\right)$ $\therefore x^{3}-3 x^{2}+x-3=(x-3)\left(x^{2}+1\right)$...
Read More →If I toss a coin 3 times and get head each time,
Question: If I toss a coin 3 times and get head each time, should I expect a tail to have a higher chance in the 4th toss? Give reason in support of your answer. Solution: No, let we toss a coin, then we get head or tail, both are equaly likely events i.e., probability of each event is $\frac{1}{2}$. So, no question of expecting a tail to have a higher chance in 4th toss....
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\log \left(x+\sqrt{x^{2}+1}\right)$ Solution: Let $y=\log \left(x+\sqrt{x^{2}+1}\right)$ On differentiating $y$ with respect to $x$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\log \left(\mathrm{x}+\sqrt{\mathrm{x}^{2}+1}\right)\right]$ We know $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}+\sqrt{\mathrm{x...
Read More →Sushma tosses a coin 3 times and gets tail each time.
Question: Sushma tosses a coin 3 times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons. Solution: The outcome of next toss may or may not be tail, because on tossing a coin, we get head or tail so both are equally likely events....
Read More →If you toss a coin 6 times and it comes down
Question: If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1? Given reasons. Solution: No. if let we toss a coin, then we get head or tail, both are equally likely events So,probability is $\frac{1}{2}$. If we toss a coin 6 times, then probability will be same in each case. So, the 2 probability of getting a head is not $1 .$...
Read More →Factorise:
Question: Factorise:ab2+ (a 1)b 1 Solution: We have: $a b^{2}+(a-1) b-1=a b^{2}+b a-b-1$ $=\left(a b^{2}+b a\right)-(b+1)$ $=a b(b+1)-1(b+1)$ $=(b+1)(a b-1)$ $\therefore a b^{2}+(a-1) b-1=(b+1)(a b-1)$...
Read More →I toss three coins together.
Question: I toss three coins together. The possible outcomes are no heads, 1 head, 2 head and 3 heads. So, I say that probability of no heads is $\frac{1}{4}$. What is wrong with this conclusion? Solution: I toss three coins together [given] So, total number of outcomes = 23= 8 and possible outcomes are (HHH), (HTT), (THT), (TTH),(HHT), (THH), (HIH)and (TTT) Now, probability of getting no head $=\frac{1}{8}$ Hence, the given conclusion is wrong because the probability of no head is $\frac{1}{8}$...
Read More →A student says that, if you throw a die,
Question: A student says that, if you throw a die, it will show up 1 or not 1 . Therefore, the probability of getting 1 and the probability of getting not 1 . each is equal to $\frac{1}{2}$. Is this correct? Give reasons. Solution: No, this is not correct. Suppose we throw a die, then total number of outcomes = 6 Possible outcomes = 1 or 2 or 3 or 4 or 5 or 6 $\therefore \quad$ Probability of getting $1=\frac{1}{6}$ Now, probability of getting not $1=1$ - Probability of getting 1 $=1-\frac{1}{6}...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\tan \left(e^{\sin x}\right)$ Solution: Let $y=\tan \left(e^{\sin x}\right)$ On differentiating $y$ with respect to $x$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\tan \left(\mathrm{e}^{\sin \mathrm{x}}\right)\right]$ We know $\frac{d}{d x}(\tan x)=\sec ^{2} x$ $\Rightarrow \frac{d y}{d x}=\sec ^{2}\left(e^{\sin x}\right) \frac{d}{d x}\left(e^{\sin x}\right)$ [using chain rule] We have $\frac...
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Question: Factorise:(ax+by)2+ (bxay)2 Solution: We have: $(a x+b y)^{2}+(b x-a y)^{2}=\left(a^{2} x^{2}+b^{2} y^{2}+2 a x b y\right)+\left(b^{2} x^{2}+a^{2} y^{2}-2 b x a y\right)$ $=a^{2} x^{2}+a^{2} y^{2}+b^{2} y^{2}+b^{2} x^{2}+2 a x b y-2 b x a y$ $=a^{2}\left(x^{2}+y^{2}\right)+b^{2} x^{2}+b^{2} y^{2}+2 a x b y-2 a x b y$ $=a^{2}\left(x^{2}+y^{2}\right)+b^{2}\left(x^{2}+y^{2}\right)$ $=\left(x^{2}+y^{2}\right)\left(a^{2}+b^{2}\right)$ $\therefore(a x+b y)^{2}+(b x-a y)^{2}=\left(x^{2}+y^{2}...
Read More →When we toss a coin, there are two possible
Question: When we toss a coin, there are two possible outcomes-head or tail. Therefore, the probability of each outcome is $\frac{1}{2}$. Justify your answer Solution: Yes, probability of each outcome is $\frac{1}{2}$ because head and tail both are equally likely events....
Read More →Factorise:
Question: Factorise:(ax+by)2+ (bxay)2 Solution: We have: $(a x+b y)^{2}+(b x-a y)^{2}=\left(a^{2} x^{2}+b^{2} y^{2}+2 a x b y\right)+\left(b^{2} x^{2}+a^{2} y^{2}-2 b x a y\right)$ $=a^{2} x^{2}+a^{2} y^{2}+b^{2} y^{2}+b^{2} x^{2}+2 a x b y-2 b x a y$ $=a^{2}\left(x^{2}+y^{2}\right)+b^{2} x^{2}+b^{2} y^{2}+2 a x b y-2 a x b y$ $=a^{2}\left(x^{2}+y^{2}\right)+b^{2}\left(x^{2}+y^{2}\right)$ $=\left(x^{2}+y^{2}\right)\left(a^{2}+b^{2}\right)$ $\therefore(a x+b y)^{2}+(b x-a y)^{2}=\left(x^{2}+y^{2}...
Read More →Apoorv throws two dice once and computes
Question: Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why? Solution: Apoorv throws two dice once. So total number of outcomes = 36 Number of outcomes for getting product 36 = 1 (6 x 6) $\therefore \quad$ Probability for Apoorv $=\frac{1}{36}$ Also, Peehu throws one die, So, total number of outcomes $=6$ Number of outcomes for getting s...
Read More →Factorise:
Question: Factorise:y2xy(1 x) x3 Solution: We have: $y^{2}-x y(1-x)-x^{3}=y^{2}-x y+x^{2} y-x^{3}$ $=\left(y^{2}-x y\right)+\left(x^{2} y-x^{3}\right)$ $=y(y-x)+x^{2}(y-x)$ $=(y-x)\left(y+x^{2}\right)$ $\therefore y^{2}-x y(1-x)-x^{3}=(y-x)\left(y+x^{2}\right)$...
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