Solve the following

Question: The value ofnCr+ 2nCr 1+nCr 2, 2 rn, is ____________. Solution: nCr+ 2nCr1+nCr2 =nCr+nCr1+nCr1+nCr2 =n+1Cr+n+1Cr1 =n+2Crusing identity nCr+2nCr1+nCr2=n+2C2...

Two arithmetic progression have the same common difference.

Question: Two arithmetic progression have the same common difference. The difference between their 100th terms is 100, What is the difference between their 1000th terms? Solution: Here, we are given two A.P sequences which have the same common difference. Let us take the first term of one A.P. asaand of other A.P. asa Also, it is given that the difference between their 100thterms is 100. We need to find the difference between their 100thterms So, let us first find the 100thterms for both of them...

Solve the following

Question: IfnP4= 24.nC5, then the value ofnis ____________. Solution: Given :- ${ }^{n} P_{4}=24{ }^{n} C_{5}$ Since ${ }^{n} C_{5}=\frac{{ }^{n} P_{5}}{5 !} \quad\left(\because{ }^{n} C_{r}=\frac{{ }^{n} P_{r}}{r !}\right)$ $\therefore{ }^{n} P_{4}=\frac{24{ }^{n} P_{5}}{5 !}$ $\therefore \frac{{ }^{n} P_{5}}{{ }^{n} P_{4}}=\frac{5 !}{24}=\frac{5 \times 4 \times 3 \times 2}{24}$ i. e $\frac{n !}{(n-5) !} \times \frac{(n-4) !}{n !}=5$ i.en 4 = 5 i.en= 9...

Let A = {1, 2, 3, 4, 5, ..., 10} and f : A → A be an invertible function.

Question: Let $A=\{1,2,3,4,5, \ldots, 10\}$ and $f: A \rightarrow A$ be an invertible function. Then, $\sum_{r=1}^{10}\left(f^{-1} o f\right)(r)=$___________. Solution: Given:f:AAis an invertible function, whereA = {1, 2, 3, 4, 5, ..., 10} Since, $f$ is invertible Therefore, $f^{-1} o f(x)=x \quad \ldots(1)$ Now, $\sum_{r=1}^{10}\left(f^{-1} o f\right)(r)=f^{-1} o f(1)+f^{-1} o f(2)+f^{-1} o f(3)+\ldots .+f^{-1} o f(10)$ $\left(\because 1+2+3+\ldots+n=\frac{n(n+1)}{2}\right)$ $=1+2+3+\ldots+10 \...

Solve the following

Question: If189C35+189Cr=190Cr, thenr= ____________. Solution: Given :-189C35+189Cr=190Cr UsingnCr+nCr+1=n+1Cr+1 ⇒n= 89 andr= 36...

If two straight lines intersect in such a way that one of the angles formed measures 90°,

Question: If two straight lines intersect in such a way that one of the angles formed measures 90, show that each of the remaining angles measures 90. Solution: We know that if two lines intersect, then the vertically-opposite angles are equal. $\angle A O C=90^{\circ}$. Then, $\angle A O C=\angle B O D=90^{\circ}$ And let $\angle B O C=\angle A O D=x$ Also, we know that the sum of all angles around a point is $360^{\circ}$ $\therefore \angle A O C+\angle B O D+\angle A O D+\angle B O C=360^{\ci...

Solve the following

Question: IfnC12=nC6, thennC2= ____________. Solution: GivennC12=nC6 UsingnCr=nCnr ⇒n 6 = 12 n= 18 $\Rightarrow^{18} C_{2}=\frac{18 \times 17 \times 16 !}{16 ! \times 2}$ $=9 \times 17$ i.e18C2= 153...

Which term of the A.P. 3, 10, 17, ... will be 84 more than its 13th term?

Question: Which term of the A.P. 3, 10, 17, ... will be 84 more than its 13thterm? Solution: In the given problem, let us first find the 13thterm of the given A.P. A.P. is 3, 10, 17 Here, First term (a) = 3 Common difference of the A.P. $(d)=10-3=7$ Now, as we know, $a_{n}=a+(n-1) d$ So, for 13thterm (n= 13), $a_{13}=3+(13-1)(7)$ $=3+12(7)$ $=3+84$ $=87$ Let us take the term which is 84 more than the 13thterm asan. So, $a_{n}=84+a_{13}$ $=84+87$ $=171$ Also, $a_{n}=a+(n-1) d$ $171=3+(n-1) 7$ $17...

Solve the following

Question: If ${ }^{18} C_{15}+2\left({ }^{18} C_{16}\right)+{ }^{17} C_{16}+1={ }^{n} C_{3}$, then $n=$ __________________ Solution: Given ${ }^{18} C_{15}+2\left({ }^{18} C_{16}\right)+{ }^{17} C_{16}+1={ }^{n} C_{3}$ L.H.S18C15+18C16+18C16+17C16 + 1 Since,nCr+1+nCr=n+1Cr+1and 1 =17C17 L.H.S reduces to, 19C16+18C16+17C16+17C17 =19C16+18C16+18C17 =19C16+19C17 i.e L.H.S =20C17and R.H.S =nC3= L.H.S (given) 20C17=nC3 or20C3=nC3 [∵20C17=20C2017=20C3] ⇒n= 20...

The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P.

Question: The $7^{\text {th }}$ term of an A.P. is 32 and its $13^{\text {th }}$ term is 62 . Find the A.P. Solution: Here, let us take the first term of the A.P. asaand the common difference of the A.P asd Now, as we know, $a_{n}=a+(n-1) d$ So, for $7^{\text {th }}$ term $(n=7)$, $a_{7}=a+(7-1) d$ $32=a+6 d$.........(1) Also, for 13thterm (n= 13), $a_{13}=a+(13-1) d$ $62=a+12 d$.........(2) Now, on subtracting (2) from (1), we get, $62-32=(a+12 d)-(a+6 d)$ $30=a+12 d-a-6 d$ $30=6 d$ $d=\frac{30...

If f : R → R be defined by f(x) = (2 – x5)1/5,

Question: If $f: R \rightarrow R$ be defined by $f(x)=\left(2-x^{5}\right)^{1 / 5}$, then $f \circ f(x)=$ Solution: Given: $f(x)=\left(2-x^{5}\right)^{1 / 5}$ $f o f(x)=f(f(x))$ $=f\left(\left(2-x^{5}\right)^{1 / 6}\right)$ $=\left[2-\left(\left(2-x^{5}\right)^{1 / 5}\right)^{5}\right]^{1 / 5}$ $=\left[2-\left(2-x^{5}\right)^{\frac{1}{5} \times 5}\right]^{1 / 5}$ $=\left[2-2+x^{5}\right]^{1 / 5}$ $=\left[x^{5}\right]^{1 / 5}$ $=x^{5 \times 1 / 5}$ $=x$ Hence, if $f: R \rightarrow R$ be defined b...

Let A = {1, 2, 3, 4} and f : A → A be given by f = {(1, 4), (2, 3), (3, 2), (4, 1)}.

Question: Let $A=\{1,2,3,4\}$ and $f: A \rightarrow A$ be given by $f=\{(1,4),(2,3),(3,2),(4,1)\}$. Then $f^{-1}=$ Solution: Given: A function $f: A \rightarrow A$ be given by $f=\{(1,4),(2,3),(3,2),(4,1)\}$ $f=\{(1,4),(2,3),(3,2),(4,1)\}$ $\Rightarrow f^{-1}=\{(4,1),(3,2),(2,3),(1,4)\}$ Thus, $f^{-1}=\{(4,1),(3,2),(2,3),(1,4)\}$ Hence, $f^{-1}=\{(4,1),(3,2),(2,3),(1,4)\}$....

The 24th term of an A.P. is twice its 10th term.

Question: The $24^{\text {th }}$ term of an A.P. is twice its $10^{\text {th }}$ term. Show that its $72^{\text {nd }}$ term is 4 times its $15^{\text {th }}$ term. Solution: Letabe the first term andd be the common difference. We know that, $n^{\text {th }}$ term $=a_{n}=a+(n-1) d$ According to the question, $a_{24}=2 a_{10}$ $\Rightarrow a+(24-1) d=2(a+(10-1) d)$ $\Rightarrow a+23 d=2 a+18 d$ $\Rightarrow 23 d-18 d=2 a-a$ $\Rightarrow 5 d=a$ $\Rightarrow a=5 d$......(1) Also, $a_{72}=a+(72-1) ...

If $f: R ightarrow R$ is defined by $f(x)=6-(x-9)^{3}$, then $f^{-1}(x)=$

Question: If $f: R \rightarrow R$ is defined by $f(x)=6-(x-9)^{3}$, then $f^{-1}(x)=$___________. Solution: Given: A function $f: R \rightarrow R$ is defined by $f(x)=6-(x-9)^{3}$ $f(x)=6-(x-9)^{3}$ $\Rightarrow y=6-(x-9)^{3}$ $\Rightarrow(x-9)^{3}=6-y$ $\Rightarrow x-9=(6-y)^{\frac{1}{3}}$ $\Rightarrow x=9+(6-y)^{\frac{1}{3}}$ Thus, $f^{-1}(x)=9+(6-x)^{\frac{1}{3}}$ Hence, if $f: R \rightarrow R$ is defined by $f(x)=6-(x-9)^{3}$, then $f^{-1}(x)=9+(6-x)^{\frac{1}{3}}$....

Two adjacent angles on a straight line are in the ratio 5 : 4.

Question: Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles. Solution: Let the two adjacent angles be 5xand 4x, respectively. Then, $5 x+4 x=180^{\circ}$ $\Rightarrow 9 x=180^{\circ}$ $\Rightarrow x=20^{\circ}$ Hence, the two angles are $5 \times 20^{\circ}=100^{\circ}$ and $4 \times 20^{\circ}=80^{\circ}$....

The 9th term of an A.P. is equal to 6 times its second term.

Question: The $9^{\text {th }}$ term of an A.P. is equal to 6 times its second term. If its $5^{\text {th }}$ term is 22 , find the A.P. Solution: Letabe the first term anddbe the common difference.We know that,nthterm =an=a+ (n 1)dAccording to the question,a9= 6a2⇒a+ (9 1)d= 6(a+ (2 1)d)⇒a+ 8d= 6a+ 6d⇒ 8d6d= 6aa⇒ 2d= 5a $\Rightarrow a=\frac{2}{5} d$ Also,a5= 22⇒a+ (5 1)d=22⇒a+ 4d =22 ....(2) On substituting the values of (1) in (2), we get $\frac{2}{5} d+4 d=22$ $\Rightarrow 2 d+20 d=22 \times ...

Question: If $f: R \rightarrow R$ is defined by $f(x)=6-(x-9)^{3}$, then $f^{-1}(x)=$ Solution: Given: A function $f: R \rightarrow R$ is defined by $f(x)=6-(x-9)^{3}$ $f(x)=6-(x-9)^{3}$ $\Rightarrow y=6-(x-9)^{3}$ $\Rightarrow(x-9)^{3}=6-y$ $\Rightarrow x-9=(6-y)^{\frac{1}{3}}$ $\Rightarrow x=9+(6-y)^{\frac{1}{3}}$ Thus, $f^{-1}(x)=9+(6-x)^{\frac{1}{3}}$ Hence, if $f: R \rightarrow R$ is defined by $f(x)=6-(x-9)^{3}$, then $f^{-1}(x)=9+(6-x)^{\frac{1}{3}}$....

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O.

Question: In the adjoining figure, three coplanar linesAB,CDandEFintersect at a pointO. Find the value ofx. Also, find AOD, COEand AOE. Solution: We know that if two lines intersect, then the vertically-opposite angles are equal. $\therefore \angle D O F=\angle C O E=5 x^{\circ}$ $\angle A O D=\angle B O C=2 x^{\circ}$ and $\angle A O E=\angle B O F=3 x^{\circ}$ Since, AOB is a straight line, we have: $\angle A O E+\angle C O E+\angle B O C=180^{\circ}$ $\Rightarrow 3 x+5 x+2 x=180^{\circ}$ $\Ri...

Solve the following

Question: If ${ }^{18} C_{15}+2\left({ }^{18} C_{16}\right)+{ }^{17} C_{16}+1={ }^{n} C_{3}$, then $n=$ __________________ Solution: Given ${ }^{18} C_{15}+2\left({ }^{18} C_{16}\right)+{ }^{17} C_{16}+1={ }^{n} C_{3}$ L.H.S ${ }^{18} \underline{\underline{C}}_{15} \pm \underline{18} \underline{\underline{C}}_{16}+{ }^{18} C_{16}+{ }^{17} C_{16}+1$ Since, ${ }^{n} C_{r+1}-{ }^{n} C_{r}={ }^{n+1} C_{r+1}$ and $1={ }^{17} C_{17}$ L.H.S reduces to, ${ }^{19} C_{16}+{ }^{18} C_{16}+{ }^{17} \underli...

Solve the following

Question: IfnCr 1= 36,nCr= 84 andnCr+ 1= 126, thenr= ____________. Solution: $n C_{r-1}=36,{ }^{n} C_{r}=84$ and ${ }^{n} C_{r+1}=126$ here $\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{84}{36}$ i. e $\frac{n !}{r !(n-r) !} \times \frac{(r-1) !(n-r+1) !}{n !}=\frac{7}{3}$ i. e $\frac{(r-1) !(n-r+1)(n-r) !}{r(r-1) !(n-r) !}=\frac{7}{3}$ i. e $\frac{n-r+1}{r}=\frac{7}{3}$ i.e $3 n-3 r+3=7 r$ i. e $3 n+3=10 r$ ...$(1)$ also, $\frac{{ }^{n} C_{r+1}}{{ }^{n} C_{r}}=\frac{126}{84}$ $\frac{n !}{(r+1) !(...

The 19th term of an A.P. is equal to three times its sixth term.

Question: The $19^{\text {th }}$ term of an A.P. is equal to three times its sixth term. If its $9^{\text {th }}$ term is 19 , find the A.P. Solution: Letabe the first term anddbe the common difference.We know that,nthterm =an=a+ (n 1)dAccording to the question,a19= 3a6⇒a+ (19 1)d= 3(a+ (6 1)d)⇒a+ 18d= 3a+ 15d⇒18d15d= 3aa⇒ 3d= 2a⇒a =32$=\frac{3}{2} d$ ....(1)Also,a9= 19⇒a+ (9 1)d=19⇒a+ 8d=19 ....(2)On substituting the values of (1) in (2), we get32$\frac{3}{2} d$ + 8d=19⇒ 3d+ 16d=19 2⇒19d= 38⇒d=...

Find the number of all three digit natural numbers which are divisible by 9.

Question: Find the number of all three digit natural numbers which are divisible by 9. Solution: First three-digit number that is divisible by 9 is 108.Next number is 108 + 9 = 117.And the last three-digit number that is divisible by 9 is 999.Thus, the progression will be 108, 117, .... , 999.All are three digit numbers which are divisible by 9, and thus forms an A.P. having first term a 108 and the common difference as 9.We know that,nthterm =an=a+ (n 1)dAccording to the question,999= 108 +(n 1...

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O,

Question: In the adjoining figure, three coplanar linesAB,CDandEFintersect at a pointO, forming angles as shown. Find the values ofx,y,zandt. Solution: We know that if two lines intersect, then the vertically opposite angles are equal. $\therefore \angle B O D=\angle A O C=90^{\circ}$ Hence, $t=90^{\circ}$ Also, $\angle D O F=\angle C O E=50^{\circ}$ Hence, $z=50^{\circ}$ Since, AOB is a straight line, we have: $\angle A O C+\angle C O E+\angle B O E=180^{\circ}$ $\Rightarrow 90+50+y=180^{\circ}...

Solve the following

Question: If 2 nC5= 9 n 2C5, thenn= ____________. Solution: $2 \times{ }^{n} C_{5}=9 \times{ }^{n-2} C_{5}$ i. e $2 \times \frac{n !}{5 !(n-5) !}=9 \times \frac{(n-2) !}{5 !(n-2-5) !}$ i. $\mathrm{e} \frac{2 \times n !}{5 !(n-5) !}=\frac{9 \times(n-2) !}{5 !(n-7) !}$ i. e $\frac{2 \times n(n-1)(n-2) !}{5 !(n-5)(n-6)(n-7) !}=\frac{9 \times(n-2) !}{5 !(n-7) !}$ i. e $2 \times[n(n-1)]=9(n-5)(n-6)$ i. e $2 n^{2}-2 n=9\left[n^{2}-11 n+30\right]$ $2 n^{2}-2 n=9 n^{2}-99 n+270$ i. e $7 n^{2}-97 n+270$ ...

The 17th term of an A.P. is 5 more than twice its 8th term.

Question: The 17thterm of an A.P. is 5 more than twice its 8thterm. If the 11thterm of the A.P. is 43. find thenthterm. Solution: Letabe the first term anddbe the common difference.We know that,nthterm =an=a+ (n 1)dAccording to the question,a17= 5 + 2a8⇒a+ (17 1)d= 5 + 2(a+ (8 1)d)⇒a+ 16d= 5 + 2a+ 14d⇒16d14d= 5 + 2aa⇒ 2d= 5 +a⇒a =2d5 ....(1)Also,a11= 43⇒a+ (11 1)d=43⇒a+ 10d=43 ....(2)On substituting the values of (1) in (2), we get2d5+ 10d=43⇒12d=5 + 43⇒12d= 48⇒d= 4⇒a= 2 45 [From (1)]⇒a= 3an=a+ ...