If one angle of a triangle is greater than the sum of the other two,
Question: If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled. Solution: Let $\mathrm{ABC}$ be a triangle and let $\angle C\angle A+\angle B$. Then, we have: $2 \angle C\angle A+\angle B+\angle C \quad[$ Adding $\angle C$ to both sides $]$ $\Rightarrow 2 \angle C180^{\circ}\left[\because \angle A+\angle B+\angle C=180^{\circ}\right]$ $\Rightarrow \angle C\mathbf{9 0}^{\circ}$ Since one of the angles of the triangle is greater than $90^{\ci...
Read More →There are 25 trees at equal distances of 5 metres in a line with a well,
Question: There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener water all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to eater all the trees. Solution: In the given problem, there are 25 trees in a line with a well such that the distance between two trees is 5 meters an...
Read More →If each angle of a triangle is less than the sum of the other two,
Question: If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled. Solution: Let ABC be the triangle. Let $\angle A\angle B+\angle C$ Then, $2 \angle A\angle A+\angle B+\angle C \quad$ [Adding $\angle A$ to both sides] $\Rightarrow 2 \angle A180^{\circ} \quad\left[\because \angle A+\angle B+\angle C=180^{\circ}\right]$ $\Rightarrow \angle A\mathbf{9 0}^{\circ}$ Also, let $\angle B\angle A+\angle C$ Then, $2 \angle B\angle A+\angle B+\angle C \qua...
Read More →Find the coefficient of
Question: Find the coefficient ofa4in the product (1 + 2a)4(2 a)5using binomial theorem. Solution: $(1+2 a)^{4}(2-a)^{5}$ $\begin{aligned}=\left[{ }^{4} C_{0}(2 a)^{0}+{ }^{4} C_{1}(2 a)^{1}+{ }^{4} C_{2}(2 a)^{2}+{ }^{4} C_{3}(2 a)^{3}+{ }^{4} C_{4}(2 a)^{4}\right] \times \\ \left[{ }^{5} C_{0}(2)^{5}(-a)^{0}+{ }^{5} C_{1}(2)^{4}(-a)^{1}+{ }^{5} C_{2}(2)^{3}(-a)^{2}+{ }^{5} C_{3}(2)^{2}(-a)^{3}+{ }^{5} C_{4}(2)^{1}(-a)^{4}+{ }^{5} C_{5}(2)^{0}(-a)^{5}\right] \end{aligned}$ $=\left[1+8 a+24 a^{2...
Read More →A man arranges to pay off a dept of Rs 3600 by 40 annual instalments which form an
Question: A man arranges to pay off a dept of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of all debt unpaid, find the value of the first instalment. Solution: In the given problem, Total amount of debt to be paid in 40 installments After 30 installments onethird of his debt is left unpaid. This means that he paid two third of the debt in 30 installments. So, Amount he paid in 30 installments $=\frac{2}{3}(3600)...
Read More →Find a, if the coefficients of
Question: Find a, if the coefficients ofx2andx3in the expansion of (3 +ax)9are equal. Solution: $(3+a x)^{9}$ $={ }^{9} C_{0} \cdot 3^{9} \cdot(a x)^{0}+{ }^{9} C_{1} \cdot 3^{8} \cdot(a x)^{1}+{ }^{9} C_{2} \cdot 3^{7} \cdot(a x)^{2}+{ }^{9} C_{3} \cdot 3^{6} \cdot(a x)^{3}+\ldots$ We have Coefficient of $x^{2}=$ Coefficient of $x^{3}$ ${ }^{9} C_{2} \times 3^{7} a^{2}={ }^{9} C_{3} \times 3^{6} a^{3}$ $\Rightarrow a=\frac{{ }^{9} C_{2}}{{ }^{9} C_{3}} \times 3$ $=\frac{9 ! \times 3 ! \times 6 ...
Read More →A man saved Rs. 32 during the first year, Rs 36 in the second year and
Question: A man saved Rs. 32 during the first year, Rs 36 in the second year and in this way he increases his saving by Rs 4 every year. Find in what time his saving will be Rs. 200. Solution: Here, we are given that the total saving of a man is Rs 200. In the first year he saved Rs 32 and every year he saved Rs 4 more than the previous year. So, the first installment = 32. Second installment = 36 Third installment = So, these installments will form an A.P. with the common difference (d) = 4 The...
Read More →The value of tan
Question: The value of $\tan \left\{\cos ^{-1} \frac{1}{5 \sqrt{2}}-\sin ^{-1} \frac{4}{\sqrt{17}}\right\}$ is (a) $\frac{\sqrt{29}}{3}$ (b) $\frac{29}{3}$ (c) $\frac{\sqrt{3}}{29}$ (d) $\frac{3}{29}$ Solution: (d) $\frac{3}{29}$ Let, $\cos ^{-1} \frac{1}{5 \sqrt{2}}=y$ and $\sin ^{-1} \frac{4}{\sqrt{17}}=z$ $\therefore \cos y=\frac{1}{5 \sqrt{2}} \Rightarrow \sin y=\frac{7}{5 \sqrt{2}} \Rightarrow \tan y=7$ $\sin z=\frac{4}{\sqrt{17}} \Rightarrow \cos z=\frac{1}{\sqrt{17}} \Rightarrow \tan z=4$...
Read More →If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.
Question: If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled. Solution: LetABCbe a triangle. Then, $\angle A=\angle B+\angle C$ $\therefore \angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle] $\Rightarrow \angle B+\angle C+\angle B+\angle C=180^{\circ}$ $\Rightarrow 2 \angle B+\angle C=180^{\circ}$ $\Rightarrow \angle B+\angle C=90^{\circ}$ $\Rightarrow \angle A=\mathbf{9 0}^{\circ} \quad[\because \angle A=\angle B+...
Read More →If in the expansion of
Question: If in the expansion of $(1+x)^{n}$, the coefficients of pth and qth terms are equal, prove that $p+q=n+2$, where $p \neq q$. Solution: Coefficients of the $p$ th and $q$ th terms are ${ }^{n} C_{p-1}$ and ${ }^{n} C_{q-1}$ respectively. Thus, we have : ${ }^{n} C_{p-1}={ }^{n} C_{q-1}$ $\Rightarrow p-1=q-1$ or, $p-1+q-1=n \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{s} \Rightarrow r=s\right.$ or, $\left.r+s=n\right]$ $\Rightarrow p=q$ or, $p+q=n+2$ If $p \neq q$, then $p+q=n+2$ Hence pr...
Read More →A man saved Rs 16500 in ten years. In each year after the first he saved
Question: A man saved Rs 16500 in ten years. In each year after the first he saved Rs 100 more than he did in the preceding year. How much did he save in the first year? Solution: Here, we are given that the total saving of a man is Rs 16500 and every year he saved Rs 100 more than the previous year. So, let us take the first installment asa. Second installment = Third installment = So, these installments will form an A.P. with the common difference (d) = 100 The sum of his savings every year Nu...
Read More →If
Question: If $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)=, then $x^{2}=$ (a) $\sin 2 \alpha$ (b) $\sin a$ (c) $\cos 2 a$ (d) $\cos \alpha$ Solution: (a) $\sin 2 \alpha$ $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)=\alpha$ $\Rightarrow \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}=\tan \alpha$ $\Rightarrow \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}} \times \...
Read More →Let there be an A.P. with first term 'a', common difference 'd'.
Question: Let there be an A.P. with first term 'a', common difference 'd'. Ifandenotes innthterm andSnthe sum of firstnterms, find.(i)nand Sn, ifa= 5,d= 3 andan= 50.(ii) n and a, if an= 4, d = 2 and Sn= 14.(iii) d, if a = 3, n = 8 and Sn= 192.(iv) a, if an= 28, Sn= 144 and n= 9.(v) n and d, if a = 8, an= 62 and Sn= 210(vi) n and an, if a= 2, d = 8 and Sn= 90. Solution: (i) Here, we have an A.P. whosenthterm (an), first term (a) and common difference (d) are given. We need to find the number of t...
Read More →For any a, b, x, y > 0, prove that:
Question: For anya,b,x,y 0, prove that: $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1} \frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}$ where = ax+by, =bx+ay Solution: Let $a=b \tan m$ and $x=y \tan n$ Then, $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\frac{2}{3} \tan ^{-1}\left(\f...
Read More →If the coefficients of 2nd, 3rd and 4th terms in the expansion
Question: If the coefficients of 2 nd, 3 rd and 4 th terms in the expansion of $(1+x)^{n}$ are in A.P., then find the value of $n$. Solution: Coefficients of the $2 \mathrm{nd}, 3 \mathrm{rd}$ and 4 th terms in the given expansion are: ${ }^{n} C_{1},{ }^{n} C_{2}$ and ${ }^{n} C_{3}$ We have: $2 \times{ }^{n} C_{2}={ }^{n} C_{1}+{ }^{n} C_{3}$ Dividing both sides by ${ }^{n} C_{2}$, we get : $2=\frac{{ }^{n} C_{1}}{{ }^{n} C_{2}}+\frac{{ }^{n} C_{3}}{{ }^{n} C_{2}}$ $\Rightarrow 2=\frac{2}{n-1}...
Read More →In a right-angled triangle, one of the acute angles measures 53°.
Question: In a right-angled triangle, one of the acute angles measures 53. Find the measure of each angle of the triangle. Solution: LetABCbe a triangle right-angled atB. Then, $\angle B=90^{\circ}$ and let $\angle A=53^{\circ}$. $\therefore \angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle] $\Rightarrow 53^{\circ}+90^{\circ}+\angle C=180^{\circ}$ $\Rightarrow \angle C=37^{\circ}$ Hence, $\angle A=53^{\circ}, \angle B=90^{\circ}$ and $\angle C=37^{\circ}$....
Read More →If the coefficients of 2nd, 3rd and 4th terms in the expansion
Question: If the coefficients of $2 \mathrm{nd}, 3 \mathrm{rd}$ and 4 th terms in the expansion of $(1+x)^{2 \mathrm{n}}$ are in A.P., show that $2 n^{2}-9 n+7=0$. Solution: Given: $(1+x)^{2 n}$ Thus, we have: $T_{2}=T_{1+1}$ $={ }^{2 n} C_{1} x^{1}$ $T_{3}=T_{2+1}$ $={ }^{2 n} C_{2} x^{2}$ $T_{4}=T_{3+1}$ $={ }^{2 n} C_{3} x^{3}$ We have coefficients of the 2 nd, 3 rd and 4 th terms in AP. $\therefore 2\left({ }^{2 n} C_{2}\right)={ }^{2 n} C_{1}+{ }^{2 n} C_{3}$ $\Rightarrow 2=\frac{{ }^{2 n} ...
Read More →Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest.
Question: Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles. Solution: Let the smallest angle of the triangle be $\angle C$ and let $\angle A=2 \angle C$ and $\angle B=3 \angle C$. Then, $\angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle] $\Rightarrow 2 \angle C+3 \angle C+\angle C=180^{\circ}$ $\Rightarrow 6 \angle=180^{\circ}$ $\Rightarrow \angle C=30^{\circ}$ $\therefore \angle A=2 \angle C$ $=2(30...
Read More →Prove that:
Question: Prove that: $\tan ^{-1} \frac{2 a b}{a^{2}-b^{2}}+\tan ^{-1} \frac{2 x y}{x^{2}-y^{2}}=\tan ^{-1} \frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}$ where $\alpha=a x-b y$ and $\beta=a y+b x$ Solution: We know $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right), \quad x y1$ $\therefore \tan ^{-1} \frac{2 a b}{a^{2}-b^{2}}+\tan ^{-1} \frac{2 x y}{x^{2}-y^{2}}=\tan ^{-1}\left(\frac{\frac{2 a b}{a^{2}-b^{2}}+\frac{2 x y}{x^{2}-y^{2}}}{1-\frac{2 a b}{a^{2}-b^{2}} \frac{2 x y}{x^{...
Read More →The coefficients of 5th, 6th and 7th terms in the expansion
Question: The coefficients of 5th, 6th and 7th terms in the expansion of (1 +x)nare in A.P., findn. Solution: Coefficients of the 5 th, 6 th and 7 th terms in the given expansion are ${ }^{n} C_{4},{ }^{n} C_{5}$ and ${ }^{n} C_{6}$ These coefficients are in $A P$. Thus, we have $2^{n} C_{5}={ }^{n} C_{4}+{ }^{n} C_{6}$ On dividing both sides by ${ }^{n} C_{5}$, we get: $2=\frac{{ }^{n} C_{4}}{{ }^{n} C_{5}}+\frac{{ }^{n} C_{6}}{{ }^{n} C_{5}}$ $\Rightarrow 2=\frac{5}{n-4}+\frac{n-5}{6}$ $\Right...
Read More →Prove that
Question: Prove that $2 \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$ Solution: $\mathrm{LHS}=2 \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)$ $=\cos ^{-1}\left\{\frac{1-\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}{1+\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}\right\}$ $\left[\because 2 \tan ^{-1}(x)=\cos ^{-1}\left\{\frac{1-x^{2}}{1+x^{2}}\right\}\righ...
Read More →Prove that the term independent of x in the expansion
Question: Prove that the term independent of $x$ in the expansion of $\left(x+\frac{1}{x}\right)^{2 n}$ is $\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{n !} .2^{n} .$ Solution: Given: $\left(x+\frac{1}{x}\right)^{2 n}$ Suppose the term independent of $x$ is the $(r+1)$ th term. $\therefore T_{r+1}={ }^{2 n} C_{r} x^{2 n-r} \frac{1}{x^{r}}$ $={ }^{2 n} C_{r} x^{2 n-2 r}$ For this term to be independent of $x$, we must have: $2 n-2 r=0$ $\Rightarrow n=r$ $\therefore$ Required coefficient $={ }^{2 n} C_...
Read More →Prove that the coefficient of (r + 1)th term in the expansion
Question: Prove that the coefficient of (r+ 1)th term in the expansion of (1 +x)n+ 1is equal to the sum of the coefficients ofrth and (r+ 1)th terms in the expansion of (1 +x)n. Solution: Coefficient of the $(r+1)$ th term in $(1+x)^{n+1}$ is ${ }^{n+1} C_{r}$ Sum of the coefficients of the $r$ th and $(r+1)$ th terms in $(1+x)^{n}={ }^{n} C_{r-1}+{ }^{n} C_{r}$ $={ }^{n+1} C_{r} \quad\left[\because{ }^{n} C_{r+1}+{ }^{n} C_{r}={ }^{n+1} C_{r+1}\right]$ Hence proved....
Read More →In an A.P., the sum of first n terms is
Question: In an A.P., the sum of first $n$ terms is $\frac{3 n^{2}}{2}+\frac{13}{n} n$. Find its $25^{\text {th }}$ term. Solution: Here, the sum of firstnterms is given by the expression, $S_{n}=\frac{3 n^{2}}{2}+\frac{13}{2} n$ We need to find the 25thterm of the A.P. So we know that thenthterm of an A.P. is given by, $a_{n}=S_{n}-S_{n-1}$ So $a_{25}=S_{25}-S_{24} \ldots \ldots$ (1) So, using the expression given for the sum ofnterms, we find the sum of 25 terms (S25) and the sum of 24 terms (...
Read More →If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion
Question: If the coefficients of (2r+ 1)th term and (r+ 2)th term in the expansion of (1 +x)43are equal, findr. Solution: Given : $-(1+x)^{43}$ We know that the coefficient of the $r$ th term in the expansion of $(1+x)^{n}$ is ${ }^{n} C_{r-1}$ Therefore, the coefficients of the $(2 r+1)$ th and $(r+2)$ th term $s$ in the given expression are ${ }^{43} C_{2 r+1-1}$ and ${ }^{43} C_{r+2-1}$ For these coefficients to be equal, we must have: $\Rightarrow 2 r=r+1 \quad$ or, $2 r+r+1=43 \quad\left[\b...
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