The sum of first n odd natural numbers is
Question: The sum of firstnodd natural numbers is(a) 2n 1(b) 2n+ 1(c)n2(d)n2 1 Solution: In this problem, we need to find the sum of firstnodd natural numbers. So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2. So here, First term (a) = 1 Common difference (d) = 2 So, let us take the number of terms asn Now, as we know, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ So, fornterms, $S_{n}=\frac{n}{2}[2(1)+(n-1) 2]$ $=\frac{n}{2}[2+2 n-...
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Question: If $\tan ^{-1} x-\tan ^{-1} y=\frac{\pi}{4}$, then $x-y-x y=$____________________. Solution: $\tan ^{-1} x-\tan ^{-1} y=\frac{\pi}{4}$ $\Rightarrow \tan ^{-1}\left(\frac{x-y}{1+x y}\right)=\frac{\pi}{4}$ $\Rightarrow \frac{x-y}{1+x y}=\tan \frac{\pi}{4}$ $\Rightarrow \frac{x-y}{1+x y}=1$ $\Rightarrow x-y=1+x y$ $\Rightarrow x-y-x y=1$ If $\tan ^{-1} x-\tan ^{-1} y=\frac{\pi}{4}$, then $x-y-x y=__1__.$...
Read More →If the first term of an A.P. is a and nth term is b, then its common difference is
Question: If the first term of an A.P. is a and nth term isb, then its common difference is (a) $\frac{b-a}{n+1}$ (b) $\frac{b-a}{n-1}$ (C) $\frac{b-a}{n}$ (d) $\frac{b+a}{n-1}$ Solution: Here, we are given the first term of the A.P. asaand thenthterm (an) asb.So, let us take the common difference of the A.P. asd. Now, as we know, $a_{n}=a+(n-1) d$ On substituting the values given in the question, we get. $b=a+(n-1) d$ $(n-1) d=b-a$ $d=\frac{b-a}{n-1}$ Therefore, $d=\frac{b-a}{n-1}$ Hence the co...
Read More →The common difference of the A.P. 1/2b,1−6b/2b,1−12b/2b, ... is
Question: The common difference of the A.P. $\frac{1}{2 b}, \frac{1-6 b}{2 b}, \frac{1-12 b}{2 b}, \ldots$ is (a) 2b(b) 2b(c) 3(d) 3 Solution: Letabe the first term andd be the common difference. The given A.P. is $\frac{1}{2 b}, \frac{1-6 b}{2 b}, \frac{1-12 b}{2 b}, \ldots$ Common difference =d = Second term First term $=\frac{1-6 b}{2 b}-\frac{1}{2 b}$ $=\frac{-6 b}{2 b}=-3$ Hence, the correct option is (d)....
Read More →If a and b are coefficients of
Question: If $a$ and $b$ are coefficients of $x^{n}$ in the expansions of $(1+x)^{2 n}$ and $(1+x)^{2 n-1}$ respectively, then write the relation between a and $b$. Solution: Coefficient of $x^{n}$ in the expansion $(1+x)^{2 n}={ }^{2 n} C_{n}=a$ Coefficient of $x^{n}$ in the expansion $(1+x)^{2 n-1}={ }^{2 n-1} C_{n}=b$ Thus we have. ${ }^{2 n} C_{n}=\frac{2 n !}{n ! n !}=\frac{2 n(2 n-1) !}{n(n-1) ! n !} \quad \ldots(1)$ and ${ }^{2 n-1} C_{n}=\frac{(2 n-1) !}{n !(n-1) !}$$\ldots(2)$ Dividing ...
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Question: If $\tan ^{-1}-\frac{1}{\sqrt{3}}+\cot ^{-1} x=\frac{x}{2}$, then $x=$ Solution: Disclaimer:The solution is provided for the following question. If $\tan ^{-1}-\frac{1}{\sqrt{3}}+\cot ^{-1} x=\frac{\pi}{2}$, then $x=$__________________. Solution:We know $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$, for all $x \in \mathrm{R}$ $\therefore \tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=\frac{\pi}{2}$ ...(1) It is given that, $\tan ^{-1}\left(-\frac{1}{\...
Read More →The common difference of the A.P. 13,1−3b3,1−6b3, ... is
Question: The common difference of the A.P. $\frac{1}{3}, \frac{1-3 b}{3}, \frac{1-6 b}{3}, \ldots$ is (a) $\frac{1}{3}$ (b) $-\frac{1}{3}$ (c) $-b$ (d) $b$ Solution: Letabe the first term andd be the common difference. The given A.P. is $\frac{1}{3}, \frac{1-3 b}{3}, \frac{1-6 b}{3}, \ldots$ Common difference =d = Second term First term $=\frac{1-3 b}{3}-\frac{1}{3}$ $=\frac{-3 b}{3}=-b$ Hence, the correct option is (c)....
Read More →If $a$ and $b$ denote respectively the coefficients of
Question: If $a$ and $b$ denote respectively the coefficients of $x^{m}$ and $x^{n}$ in the expansion of $(1+x)^{m+n}$, then write the relation between $a$ and $b$. Solution: Coefficient of $x^{m}$ in the given expansion $={ }^{m+n} C_{m}=a$ Coefficient of $x^{n}$ in the given expansion $={ }^{m+n} C_{n}=b$ $\therefore a=b \quad\left[\because{ }^{m+n} C_{m}={ }^{m+n} C_{n}\right]$...
Read More →Which term is independent of x, in the expansion of
Question: Which term is independent of $x$, in the expansion of $\left(x-\frac{1}{3 x^{2}}\right)^{9} ?$ Solution: Suppose $T_{\mathrm{r}+1}$ is the term in the given expression that is independent of $x$. Thus, we have : $T_{r+1}={ }^{9} C_{r} x^{9-r}\left(\frac{-1}{3 x^{2}}\right)^{r}$ $=(-1)^{r}{ }^{9} C_{r} \frac{1}{3^{r}} x^{9-r-2 r}$ For this term to be independent of $x$, we must have $9-3 r=0$ $\Rightarrow r=3$ Hence, the required term is the 4 th term....
Read More →Write the middle term in the expansion of
Question: Write the middle term in the expansion of $\left(\frac{2 x^{2}}{3}+\frac{3}{2 x^{2}}\right)$. Solution: Here, $n$, i. e., 10 , is an even number. $\therefore$ Middle term $=\left(\frac{10}{2}+1\right)$ th term $=6$ th term Thus, we have : $T_{6}=T_{5+1}$ $={ }^{10} C_{5}\left(\frac{2 x^{2}}{3}\right)^{10-5}\left(\frac{3}{2 x^{2}}\right)^{5}$ $=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2} \times \frac{2^{5}}{3^{5}} \times \frac{3^{5}}{2^{5}}$ $=252$...
Read More →Write the number of terms in the expansion of
Question: Write the number of terms in the expansion of $\left(1-3 x+3 x^{2}-x^{3}\right)^{8}$. Solution: The given expression is $\left(1-3 x+3 x^{2}-x^{3}\right)^{8}$. It can be written as $\left[(1-x)^{3}\right]^{8}$ i. e. $(1-x)^{24}$ Hence, the number of terms is $(24+1)$ i.e. 25...
Read More →Write the sum of the coefficients in the expansion of
Question: Write the sum of the coefficients in the expansion of $\left(1-3 x+x^{2}\right)^{111}$. Solution: To find the sum of coefficients, we plug 1 for each variable then, we get the sum of coefficients of the given expression. $\therefore$ Sum of coefficient $=\left(1-3 x+x^{2}\right)^{111}$ $=\left(1-3 \times 1+1^{2}\right)^{111}$ $=(1-3+1)^{111}$ $=(1-3+1)^{111}$ $=(-1)^{111}$ $=-1$...
Read More →The common difference of the A.P. is 12q,1−2q2q,1−4q2q, ... is
Question: The common difference of the A.P. is $\frac{1}{2 q}, \frac{1-2 q}{2 q}, \frac{1-4 q}{2 q}, \ldots$ is (a) 1(b) 1(c)q(d) 2q Solution: Letabe the first term andd be the common difference. The given A.P. is $\frac{1}{2 q}, \frac{1-2 q}{2 q}, \frac{1-4 q}{2 q}, \ldots$ Common difference =d = Second term First term $=\frac{1-2 q}{2 q}-\frac{1}{2 q}$ $=\frac{-2 q}{2 q}=-1$ Hence, the correct option is (a)....
Read More →Write the number of terms in the expansion of
Question: Write the number of terms in the expansion of $(2+\sqrt{3} x)^{10}+(2-\sqrt{3} x)^{10}$. Solution: Number of terms in the expansion $(x+y)^{n}+(x-y)^{n}$ where $n$ is even $=\left(\frac{n}{2}+1\right)$ Thus, we have: Number of terms in the given expansion $=\left(\frac{10}{2}+1\right)=6$...
Read More →The sum of first 20 odd natural numbers is
Question: The sum of first 20 odd natural numbers is (a) 100 (b) 210 (c) 400 (d) 420 Solution: Letabe the first term andd be the common difference. We know that, sum of first $n$ terms $=S_{n}=\frac{n}{2}[2 a+(n-1) d]$ The given series is 1 + 3 + 5 + ...... First term =a= 1.Common difference =d = 3 1= 2 $\therefore S_{20}=\frac{20}{2}[2 \times 1+(20-1) 2]$ $=10(2+19 \times 2)$ $=10(40)$ $=400$ Hence, the correct option is (c)....
Read More →The value of
Question: The value of $\tan ^{-1} 2+\tan ^{-1} 3$ is___________________. Solution: We know $\tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$, if $x y1$ $\therefore \tan ^{-1} 2+\tan ^{-1} 3$ $=\pi+\tan ^{-1}\left(\frac{2+3}{1-2 \times 3}\right)$ $=\pi+\tan ^{-1}(-1)$ $=\pi-\frac{\pi}{4}$ $=\frac{3 \pi}{4}$ The value of $\tan ^{-1} 2+\tan ^{-1} 3$ is $\frac{3 \pi}{4}$...
Read More →If Sn denote the sum of n terms of an A.P. with first term a and
Question: If $S_{n}$ denote the sum of $n$ terms of an A.P. with first term a and common difference $d$ such that $\frac{S x}{S k x}$ is independent of $x$, then (a)d=a(b)d= 2a(c)a= 2d(d)d= a Solution: Here, we are given an A.P. withaas the first term anddas the common difference. The sum ofnterms of the A.P. is given bySn. We need to find the relation between $a$ and $d$ such that $\frac{S_{x}}{S_{k 1}}$ is independent of So, let us first find the values ofSxandSkxusing the following formula fo...
Read More →If the sums of n terms of two arithmetic progressions are in the ratio
Question: If the sums of $n$ terms of two arithmetic progressions are in the ratio $\frac{3 n+5}{5 n-7}$, then their $n^{\text {th }}$ terms are in the ratio (a) $\frac{3 n-1}{5 n-1}$ (b) $\frac{3 n+1}{5 n+1}$ (c) $\frac{5 n+1}{3 n+1}$ (d) $\frac{5 n-1}{3 n-1}$ Solution: In the given problem, the ratio of the sum ofnterms of two A.Ps is given by the expression, $\frac{S_{z}}{S_{n}^{\dagger}}=\frac{3 n+5}{5 n+7}$ ....(1) We need to find the ratio of theirnthterms. Here we use the following formul...
Read More →if the
Question: If $x0$, then $\tan ^{-1} x+\tan ^{-1} \frac{1}{x}$ is equal to____________________. Solution: We know $\tan ^{-1} \frac{1}{x}= \begin{cases}\cot ^{-1} x, \text { for } x0 \\ -\pi+\cot ^{-1} x, \text { for } x0\end{cases}$ $\therefore \tan ^{-1} x+\tan ^{-1} \frac{1}{x}$ $=\tan ^{-1} x+\cot ^{-1} x-\pi$ $(x0)$ $=\frac{\pi}{2}-\pi$ $\left(\tan ^{-1} \mathrm{x}+\cot ^{-1} \mathrm{x}=\frac{\pi}{2}, \forall x \in \mathbf{R}\right)$ $=-\frac{\pi}{2}$ If $x0$, then $\tan ^{-1} x+\tan ^{-1} \...
Read More →In the adjoining figure, explain how one can find the breadth of the river without crossing it.
Question: In the adjoining figure, explain how one can find the breadth of the river without crossing it. Solution: Let AB be the breadth of the river.M is any point situated on the bank of the river.Let O be the mid point of BM. Moving along perpendicular to point such that A,O and N are in a straight line. Then MN is the required breadth of the river. In $\triangle \mathrm{OBA}$ and $\triangle \mathrm{OMN}$, we have: $\mathrm{OB}=\mathrm{OM} \quad(\mathrm{O}$ is midpoint $)$ $\angle \mathrm{OB...
Read More →The common difference of an A.P., the sum of whose n terms is Sn, is
Question: The common difference of an A.P., the sum of whose $n$ terms is $S_{n}$, is (a) $S_{n}-2 S_{n-1}+S_{n-2}$ (b) $S_{n}-2 S_{n-1}-S_{n-2}$ (c) $S_{n}-S_{n-2}$ (d) $S_{n}-S_{n-1}$ Solution: Here, we are given an A.P. the sum of whosenterms isSn. So, to calculate the common difference of the A.P, we find two consecutive terms of the A.P. Now, thenthterm of the A.P will be given by the following formula, $a_{n}=S_{n}-S_{n-1}$ Next, we find the (n 1)thterm using the same formula, $a_{n-1}=S_{...
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Question: If $\sin ^{-1} x=\frac{\pi}{5}$ for some $x \in(-1,1)$, then the value of $\cos ^{-1} x$ is____________________. Solution: Given: $\sin ^{-1} x=\frac{\pi}{5}, \forall x \in(-1,1)$ We know $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \frac{\pi}{5}+\cos ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \cos ^{-1} x=\frac{\pi}{2}-\frac{\pi}{5}=\frac{3 \pi}{10}$ Thus, the value of $\cos ^{-1} x$ is $\frac{3 \pi}{10}$. If $\sin ^{-1} x=\frac{\pi}{5}$ for some $x \in(-1,1)$, then the value of $\...
Read More →The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D,
Question: The image of an object placed at a pointAbefore a plane mirrorLMis seen at the pointBby an observer atD, as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror. Solution: Given: An object is placed at a pointA, the image of the object is seen at the pointB,an observer is at pointD, andLMisa plane mirror.To Prove:The image is as far behind the mirror as the object is in front of the mirror, i.e.BT=AT.Proof: $\because L M$ is a pl...
Read More →The range of
Question: The range of $\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x$ is_______________________. Solution: Domain of the given function $=[-1,1] \cap \mathbf{R}=[-1,1]$ Now, For $-1 \leq x \leq 1$ $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ and $-\frac{\pi}{4} \leq \tan ^{-1} x \leq \frac{\pi}{4}$ $\therefore \frac{\pi}{2}-\frac{\pi}{4} \leq \sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x \leq \frac{\pi}{2}+\frac{\pi}{4}$ $\Rightarrow \frac{\pi}{4} \leq \sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x \leq \frac{3 \pi}{4}...
Read More →The nth term of an A.P., the sum of whose n terms is Sn, is
Question: The $n$th term of an A.P., the sum of whose $n$ terms is $\mathrm{S}_{n}$, is (a) $S_{n}+S_{n-1}$ (b) $S_{n}-S_{n-1}$ (c) $S_{n}+S_{n+1}$ (d) $S_{n}-S_{n+1}$ Solution: A.P. we use following formula, $a_{n}=S_{n}-S_{n-1}$ So, the $n^{\text {th }}$ term of the A.P. is given by $a_{n}=S_{n}-S_{n-1}$. Therefore, the correct option is (b)....
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