Solve the following
Question: If ${ }^{\left(a^{2}-a\right)} C_{2}={ }^{\left(a^{2}-a\right)} C_{4}$, then $a=$ (a) 2 (b) 3 (c) 4 (d) none of these Solution: (b) 3 $a^{2}-a=2+4 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow a^{2}-a-6=0$ $\Rightarrow a^{2}-3 a+2 a-6=0$ $\Rightarrow a(a-3)+2(a-3)=0$ $\Rightarrow(a+2)(a-3)=0$ $\Rightarrow a=-2$ or $a=3$ But, $a=-2$ is not possible. $\therefore a=3$...
Read More →If f : R→ (−1, 1) is defined by
Question: If $f: R \rightarrow(-1,1)$ is defined by $f(x)=\frac{-x|x|}{1+x^{2}}$, then $f^{-1}(x)$ equals (a) $\sqrt{\frac{|x|}{1-|x|}}$ (b) $\operatorname{Sgn}(x) \sqrt{\frac{|x|}{1-|x|}}$ (c) $-\sqrt{\frac{x}{1-x}}$ (d) None of these Solution: (b) $-\operatorname{Sgn}$ $(x) \sqrt{\frac{|x|}{1-|x|}}$ We have, $f(x)=\frac{-x|x|}{1+x^{2}} \quad x \in(-1,1)$ Case - (I) When, $x0$, Then, $|x|=-x$ And $f(x)0$ Now, $f(x)=\frac{-x(-x)}{1+x^{2}}$ $\Rightarrow y=\frac{x^{2}}{1+x^{2}}$ $\Rightarrow \frac...
Read More →Into how many chapters was the famous treatise, 'The Elements' divided by Euclid?
Question: Into how many chapters was the famous treatise, 'The Elements' divided by Euclid?(a) 13(b) 12(c) 11(d) 9 Solution: The famous treatise, 'The Elements' by Euclid is divided into 13 chapters.Hence, the correct answer is option (a)....
Read More →In Indus Valley Civilisation (about BC 3000), the bricks used for construction work were having dimensions in the ratio of
Question: In Indus Valley Civilisation (about BC 3000), the bricks used for construction work were having dimensions in the ratio of(a) 5 : 3 : 2(b) 4 : 2 : 1(c) 4 : 3 : 2(d) 6 : 4 : 2 Solution: (b) 4 : 2 : 1...
Read More →Solve the following
Question: IfnCr+nCr+ 1=n+1Cx, thenx= (a) r (b)r 1 (c) n (d)r+ 1 Solution: (d)r + 1 $n_{C_{r}}+n_{C_{r+1}}=n+1_{C_{x}} \quad$ [Given] We have: ${ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{x} \quad\left[\because{ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}\right]$ $\Rightarrow{ }^{n+1} C_{r+1}={ }^{n+1} C_{x}$$\Rightarrow r+1=x \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$...
Read More →The number of interwoven isosceles triangles in a Sriyantra is
Question: The number of interwoven isosceles triangles in aSriyantrais(a) five(b) seven(c) nine(d) eleven Solution: (c) nine...
Read More →If 10 times the 10th term of an A.P. is equal to 15 times the 15th term,
Question: If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero. Solution: Here, let us take the first term of the A.P. asaand the common difference asd We are given that 10 times the 10thterm is equal to 15 times the 15thterm. We need to show that 25thterm is zero. So, let us first find the two terms. So, as we know, $a_{n}=a+(n-1) d$ For $10^{\text {th }}$ term $(n=10)$, $a_{10}=a+(10-1) d$ $=a+9 d$ For $15^{\text {th }}$ term $(n=15)$...
Read More →In ancient India, altars with combination of shapes like rectangles, triangles and trapeziums were used for
Question: In ancient India, altars with combination of shapes like rectangles, triangles and trapeziums were used for(a) household rituals(b) public rituals(c) both (a) and (b)(d) none of (a), (b) and (c) Solution: The construction of altars (or vedis) and fireplaces for performining vedic rituals resulted in the origin of the geometry of vedic period. Square and circular altars were used for household rituals whereas the altars with combination of shapes like rectangles, triangles and trapezium...
Read More →Solve the following
Question: IfnC12=nC8, thenn= (a) 20 (b) 12 (c) 6 (d) 30 Solution: (a) 20 $n=12+8=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$...
Read More →In ancient India, the shapes of altars used for household rituals were
Question: In ancient India, the shapes of altars used for household rituals were(a) squares and rectangles(b) squares and circles(c) triangles and rectangles(d) trapeziums and pyramids Solution: (b) squares and circles...
Read More →Solve the following
Question: IfmC1=nC2, then (a) 2m=n (b) 2m=n(n+ 1) (c) 2m=n(n 1) (d) 2n=m(m 1) Solution: (c) 2m=n(n 1) ${ }^{m} C_{1}={ }^{n} C_{2}$ $\Rightarrow \frac{m !}{1 !(m-1) !}=\frac{n !}{2 !(n-2) !}$ $\Rightarrow \frac{m(m-1) !}{(m-1) !}=\frac{n(n-1)(n-2) !}{2(n-2) !}$ $\Rightarrow 2 m=n(n-1)$...
Read More →In the given figure, L and M are the mid- points of AB and BC respectively.
Question: In the given figure,LandMare the mid- points ofABandBCrespectively. (i) If $A B=B C$, prove that $A L=M C$. (ii) If $B L=B M$, prove that $A B=B C$. Hint (i) $A B=B C \Rightarrow \frac{1}{2} A B=\frac{1}{2} B C \Rightarrow A L=M C$. (ii) $B L=B M \Rightarrow 2 B L=2 B M \Rightarrow A B=B C$. Solution: (i) It is given that L is the mid-point of AB. ...........(1) $\therefore \mathrm{AL}=\mathrm{BL}=\frac{1}{2} \mathrm{AB}$ Also, $M$ is the mid-point of $B C$. $\therefore B M=M C=\frac{1...
Read More →If C (n, 12) = C (n, 8), then C (22, n) is equal to
Question: IfC(n, 12) =C(n, 8), thenC(22,n) is equal to (a) 231 (b) 210 (c) 252 (d) 303 Solution: (a) 231 ${ }^{n} C_{12}={ }^{n} C_{8}$ $\Rightarrow n=12+8=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ Now, ${ }^{22} C_{n}={ }^{22} C_{20}$ $=\frac{22}{2} \times \frac{21}{1}$ $=231$...
Read More →If C (n, 12) = C (n, 8), then C (22, n) is equal to
Question: IfC(n, 12) =C(n, 8), thenC(22,n) is equal to (a) 231 (b) 210 (c) 252 (d) 303 Solution: (a) 231 ${ }^{n} C_{12}={ }^{n} C_{8}$ $\Rightarrow n=12+8=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ Now, ${ }^{22} C_{n}={ }^{22} C_{20}$ $=\frac{22}{2} \times \frac{21}{1}$ $=231$...
Read More →If 9th term of an A.P is zero,
Question: If 9th term of an A.P is zero, prove that its 29th term is double the 19th term. Solution: In the given problem, the 9thterm of an A.P. is zero. Here, let us take the first term of the A.P asaand the common difference asd So, as we know, $a_{n}=a+(n-1) d$ We get, $a_{9}=a+(9-1) d$ $0=a+8 d$ $a=-8 d$ .......(1) Now, we need to prove that $29^{\text {th }}$ term is double of $19^{\text {th }}$ term. So, let us first find the two terms. For $19^{\text {th }}$ term $(n=19)$, $a_{19}=a+(19-...
Read More →Solve the following
Question: If20Cr+ 1=20Cr 1, thenris equal to (a) 10 (b) 11 (c) 19 (d) 12 Solution: (a) 10 $r+1+r-1=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow 2 r=20$ $\Rightarrow r=10$...
Read More →Solve the following
Question: If20Cr+ 1=20Cr 1, thenris equal to (a) 10 (b) 11 (c) 19 (d) 12 Solution: (a) 10 $r+1+r-1=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow 2 r=20$ $\Rightarrow r=10$...
Read More →Solve the following
Question: If15C3r=15Cr+ 3, thenris equal to (a) 5 (b) 4 (c) 3 (d) 2 Solution: (c) 3 $3 r+r+3=15 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow 4 r+3=15$ $\Rightarrow 4 r=12$ $\Rightarrow r=3$...
Read More →Solve the following
Question: If15C3r=15Cr+ 3, thenris equal to (a) 5 (b) 4 (c) 3 (d) 2 Solution: (c) 3 $3 r+r+3=15 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow 4 r+3=15$ $\Rightarrow 4 r=12$ $\Rightarrow r=3$...
Read More →Solve the following
Question: If20Cr=20Cr+ 4, thenrC3is equal to (a) 54 (b) 56 (c) 58 (d) none of these Solution: (b) 56 $r+r+4=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow 2 r+4=20$ $\Rightarrow 2 r=16$ $\Rightarrow r=8$ Now, ${ }^{r} C_{3}={ }^{8} C_{3}$ ${ }^{8} C_{3}=\frac{8 !}{3 ! 5 !}=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56$...
Read More →Solve the following
Question: If20Cr=20Cr+ 4, thenrC3is equal to (a) 54 (b) 56 (c) 58 (d) none of these Solution: (b) 56 $r+r+4=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow 2 r+4=20$ $\Rightarrow 2 r=16$ $\Rightarrow r=8$ Now, ${ }^{r} C_{3}={ }^{8} C_{3}$ ${ }^{8} C_{3}=\frac{8 !}{3 ! 5 !}=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56$...
Read More →The 6th and 17th terms of an A.P.
Question: The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term. Solution: In the given problem, we are given $6^{\text {th }}$ and $17^{\text {th }}$ term of an A.P. We need to find the $40^{\text {th }}$ term Here, $a_{6}=19$ $a_{17}=41$ Now, we will find $a_{6}$ and $a_{17}$ using the formula $a_{n}=a+(n-1) d$ So, $a_{6}=a+(6-1) d$ $19=a+5 d$ .......(1) Also, $a_{17}=a+(17-1) d$ $41=a+16 d$.......(2) So, to solve foraandd On subtracting (1) from (2), we get $a+16 d-...
Read More →Solve the following
Question: If20Cr=20Cr10, then18Cris equal to (a) 4896 (b) 816 (c) 1632 (d) nont of these Solution: (b) 816 $r+r-10=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow 2 r-10=20$ $\Rightarrow 2 r=30$ $\Rightarrow r=15$ Now, ${ }^{18} C_{r}={ }^{18} C_{15}$ $\therefore{ }^{18} C_{15}={ }^{18} C_{3}$ $\therefore{ }^{18} C_{3}=\frac{18}{3} \times \frac{17}{2} \times 16=816$...
Read More →Solve the following
Question: If20Cr=20Cr10, then18Cris equal to (a) 4896 (b) 816 (c) 1632 (d) nont of these Solution: (b) 816 $r+r-10=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow 2 r-10=20$ $\Rightarrow 2 r=30$ $\Rightarrow r=15$ Now, ${ }^{18} C_{r}={ }^{18} C_{15}$ $\therefore{ }^{18} C_{15}={ }^{18} C_{3}$ $\therefore{ }^{18} C_{3}=\frac{18}{3} \times \frac{17}{2} \times 16=816$...
Read More →Let f : [2, ∞)→X be defined by f(x)
Question: Let $f:[2, \infty) \rightarrow X$ be defined by $f(x)=4 x-x^{2}$. Then, $f$ is invertible if $X=$ (a) $[2, \infty)$ (b) $(-\infty, 2]$ (c) $(-\infty, 4]$ (d) $[4, \infty)$ Solution: Sincefis invertible, range off= co domain off=XSo, we need to find the range offto findX.For finding the range, let $f(x)=y$ $\Rightarrow 4 x-x^{2}=y$ $\Rightarrow x^{2}-4 x=-y$ $\Rightarrow x^{2}-4 x+4=4-y$ $\Rightarrow(x-2)^{2}=4-y$ $\Rightarrow x-2=\pm \sqrt{4-y}$ $\Rightarrow x=2 \pm \sqrt{4-y}$ This is...
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