Mathematics Class XI

Question 1 of 25

1. Question
1 point(s)
Sum of $ n $ terms of the series $ 1^3+3^3+5^3+… $ is______
- $ n^2(2n^2-1) $
- $ n^3(n-1) $
- $ n^3+8n+4 $
- $ 2n^4+3n^2 $
Correct

Incorrect

Hint

$Tn=\left( 2n-1 \right)^3=8n^3-1^3-3\cdot 2n\left( 2n-1 \right)$
$=8n^3-1-12n^2+6n$
$=8n^3-12n^2+6n-1$
$Sn=\Sigma~ Tn=8~\Sigma n^3-12~\Sigma~ n^2+6~\Sigma n-\Sigma 1$
$=8\left[ \frac{n\left( n+1 \right)}{2} \right]-2\left[ n\left( n+1 \right)\left( 2n+1 \right) \right]+3n\left( n+1 \right)-n$
$=n\left( n+1 \right)\left[ 2n^2+2n-4n-2+3 \right]-n$
$=n\left( n+1 \right)\left[ 2n^2-2n+1 \right]-n$
$=n\left( n+1 \right)2n\cdot \left( n-1 \right)+n\left( n+1 \right)-n$
$=n^2\left( 2n^2-1 \right)$
Question 2 of 25

2. Question
1 point(s)
Consider an infinite GP with first term $ a $ and common ratio $ r $. If the sum is $ 4 $ and the second term is $ \dfrac{3}{4} $, then______
- $ a=2, r=\dfrac{3}{8} $
- $ a=\dfrac{4}{7}, r=\dfrac{3}{7}$
- $ a=\dfrac{3}{2}, r=\dfrac{1}{2} $
- $ a=3, r=\dfrac{1}{4} $
Correct

Incorrect

Hint

$S_\infty =\frac{a}{1-r}\implies4=\frac{a}{1-r}$ –(i)
$T_2=ar=\dfrac{3}{4}$ —(ii)
From (i) and (ii)
$a=4-4\left( \dfrac{a}{4a} \right)$
$\implies a^2-4a+3=0\implies a=1,3$
Where $a=3\implies 3r=\dfrac{3}{4}\implies r=\dfrac{1}{4}$
$\therefore a=3,r=\dfrac{1}{4}$
Question 3 of 25

3. Question
1 point(s)
If $ a,b \text{ and } c $ are positive number in a GP, then the roots of the quadratic equation $ (\log_ea)x^2-(2\log_eb)x+\log_ec=0$ are _____
- $ -1 $ and $ \dfrac{\log_ec}{\log_ea} $
- $ 1 $ and $ -\dfrac{log_ec}{log_ea} $
- $ 1 $ and $ \log_ac $
- $ -1 $ and $ \log_ca $
Correct

Incorrect

Hint

$\left( \log_ea \right)x^2-2\left( \log_eb \right)x+\log_3c=0$
Put $x=1$
$\implies \log_ea-2\log_eb+\log_ec=0$
$\implies 2\log_eb=\log_ea+\log_ec$
$\implies b^2=ac\left( \therefore a,b,c\ \text{are in GP } \right)$
$1+\alpha =\dfrac{2\log_eb}{\log_ea}=\dfrac{\log_eb^2}{\log_ea}$
$\implies \alpha =\dfrac{\log_eac}{\log_ea}-1=\dfrac{\log_ec}{\log_ea}=\log_ac$
$\therefore$ Roots are $1$ and $\log_a c$
Question 4 of 25

4. Question
1 point(s)
The $ 5^{th} $ term of the series $ \dfrac{10}{9}, \dfrac{1}{3}, \sqrt{\dfrac{20}{3}}, \dfrac{2}{3}… $ is_____
- $ \dfrac{1}{3} $
- $ 1 $
- $ \dfrac{2}{5} $
- $ \sqrt{\dfrac{2}{3}} $
Correct

Incorrect

Hint

$r=\dfrac{1}{3}\sqrt{\dfrac{20}{3}}\times \dfrac{9}{10}=\sqrt{\dfrac{3}{5}}$
$T_5=ar^4=\dfrac{10}{9}\left[ \sqrt{\dfrac{3}{5}} \right]^4$
$=\dfrac{10}{9}\times \dfrac{9}{25}=\dfrac{2}{5}$
Question 5 of 25

5. Question
1 point(s)
If $ a,b,c $ are in GP and $ 4a,5b,4c $ are in AP such that $ a+b+c=70 $, then the value of $ b $ is______
- $ 5 $
- $ 10 $
- $ 15 $
- $ 20 $
Correct

Incorrect

Hint

$a,b,C$ are in GP $\implies b^2=ac$
$4a,5b,4C$ are in AP $\implies 5b=\dfrac{4a+4c}{2}$
$\implies 5b=2a+2C$
$a+b+c=70$
$\implies 2a+2b+2C=140$
$\implies 5b+2b=140$
$\implies 7b=140$
$\implies b=\dfrac{140}{7}=20$
Question 6 of 25

6. Question
1 point(s)
If $ a^{\frac{1}{x}}=b^{\frac{1}{y}}=C^{\frac{1}{2}}=k$ and $ a,b,c $ are in GP, then $x,y,z $ are in ______
- AP
- GP
- HP
- AGP
Correct

Incorrect

Hint

Let $\text{[math]}$
$\therefore \log\ a=x \log k$
$\log b=y~\log k$
$\log c=z~\log k$
$\therefore b^2=ac$
$\implies 2\log\ b=\log a+\log c$
$\implies 2\left[ y~\log k \right]=x \log k+z \log k$
$\implies 2y=x+z$
$\therefore x,y,z$ are in AP
Question 7 of 25

7. Question
1 point(s)
The difference between two numbers is $ 48 $ and the difference between their AM and GM is $ 18 $. Then, the greatest of two number is______
- $ 96 $
- $ 60 $
- $ 54 $
- $ 49 $
Correct

Incorrect

Hint

Given $a-b=48$ –(i)
$\dfrac{a+b}{2}-\sqrt{ab}=18$
$\implies \left( \sqrt{a}-\sqrt{b} \right)^2=2\times 18=36$
$\implies \sqrt{a}-\sqrt{b}=6$
From $\left( i \right)\left( \sqrt{1}+\sqrt{b} \right)\left( \sqrt{a}-\sqrt{b} \right)=48$
$\therefore \sqrt{a}+\sqrt{b}=\dfrac{48}{6}=8$
$\therefore \dfrac{\sqrt{1}+\sqrt{b}=8\\\sqrt{a}-\sqrt{b}=6}{2\sqrt{a}=14}\implies \sqrt{a}=7\implies a=49$
$\therefore b=1$
$\therefore$ Greater number is $49$ .
Question 8 of 25

8. Question
1 point(s)
If $ f(x)=\cos^2x+\sec^2x $, then ______
- $ f(x) $ < $ 1 $
- $ f(x)=1 $
- $ 1 $ < $ f(x) $ < $2 $
- $ f(x)\geq 2 $
Correct

Incorrect

Hint

$\text{AM}=\dfrac{\cos^2x+\sec^2x}{2}$
$\text{GM}=\sqrt{\cos^2c\times \sec^2x}=1$
$\text{AM}\ge \text{GM}$
$\therefore \dfrac{\cos^2x+\sec^2x}{2}\ge 1$
$\implies \cos^2x+\sec^2x\ge 2$
$\therefore f\left( x \right)\ge 2$
Question 9 of 25

9. Question
1 point(s)
If $ a,b,c $ are in AP and $ |a| $ < $ 1 $, $|b| $< $1 $, $|c| $ < $ 1 $. If $ x=1+a+a^2+…\ ,y=1+b+b^2+…\ ,z=1+c+c^2+…\ , $ then $ x,y,z $ are in_______
- AP
- GP
- HP
- AGP
Correct

Incorrect

Hint

$x=1+a+a^2+...=\dfrac{1}{1-a}$
$y=a+b+b^2+...=\dfrac{1}{1-b}$
$z=1+c+c^2+...=\dfrac{1}{1-c}$
As $a, b, c$ are in AP
$1-a,1-b,1-c$ are in AP
$\implies \dfrac{1}{1-a}, \dfrac{1}{1-b},\dfrac{1}{1-c}$ are in HP
$\therefore x,y,z$ are in HP
Question 10 of 25

10. Question
1 point(s)
If $ y=3^{x-1}+3^{-x-1} $, then the least value of $ y $ is______
- $ 2 $
- $ 6 $
- $ \dfrac{2}{3} $
- None of these
Correct

Incorrect

Hint

$y=\dfrac{3^x}{3}+\dfrac{1}{3.3^x}=\dfrac{1}{3}\left( 3^x +3^{-x}\right)$
$\therefore \text{AM}\ge \text{GM}$
$\therefore \dfrac{3^x+3^{-x}}{2}\ge \sqrt{3^x\cdot 3^{-x}}$
$\implies 3^x+3^{-x}\ge 2$
$\implies 3^{x-1}+3^{-n-1}\ge \dfrac{2}{3}$
$\implies y\ge \dfrac{2}{3}$
$\therefore$ least value of $y$ is $\dfrac{2}{3}$
Question 11 of 25

11. Question
1 point(s)
The value of $\dfrac{1}{2!}+\dfrac{2}{3!}+…\dfrac{999}{1000!} $ is______
- $ \dfrac{1000!-1}{1000!} $
- $ \dfrac{1000!+1}{1000!} $
- $ \dfrac{999!-1}{999!} $
- $ \dfrac{999!+1}{999!} $
Correct

Incorrect

Hint

$\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{999}{1000!}$
$=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+...+\dfrac{1000-1}{1000!}$
$=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+...+\dfrac{1}{999!}-\dfrac{1}{1000!}$
$=1-\dfrac{1}{1000!}=\dfrac{1000!-1}{1000!}$
Question 12 of 25

12. Question
1 point(s)
If $ S_n=\dfrac{1}{_13}+\dfrac{1+2}{_13+2^3}+…+\dfrac{1+2+…n}{1^3+2^3+…+n^3} $ where $ n=1,2,3,\ … $, then $ S_n $ is not greater than_____
- $ \dfrac{1}{2}$
- $ 1 $
- $ 2 $
- $ \dfrac{1}{4}$
Correct

Incorrect

Hint

$Tn=\dfrac{1+2+...+n}{1^3+2^3+...+n^3}=\dfrac{\Sigma n}{\Sigma n^3}$
$=\dfrac{n\left( n+1 \right)/2}{\left\{ n\left( n+1 \right)/2 \right\}^2}=\dfrac{2}{n\left( n+1 \right)}=2\left[ \dfrac{1}{n}-\dfrac{1}{n+1} \right]$
$Sn=\Sigma\ Tn=2\left( \dfrac{1}{1}-\dfrac{1}{2} \right)+2\left( \dfrac{1}{2}-\dfrac{1}{3} \right)+...+2\left( \dfrac{1}{n}-\dfrac{1}{n+1} \right)$
$=2\left( 1-\dfrac{1}{n+1} \right)=2-\dfrac{2}{n+1}\lt 2$
$\therefore Sn$ is not greater than $2$ .
Question 13 of 25

13. Question
1 point(s)
The sum of the series $ \left( 1+2 \right)+\left( 1+2+2^2 \right)+\left( 1+2+2^2+2^3 \right)+… $ up to $ n $ terms is ______
- $ 2^{n+1}-n-4 $
- $ 2(2^n-1)-n $
- $ 2^{n+1}-n $
- $ 2^{n+1}-n $
Correct

Incorrect

Hint

$Tn=1+2+2^2+...+2^n=1\left( \dfrac{2^{n+1}-1}{2-1} \right)$
$=2^{n+1}-1$
$\therefore \Sigma\ Tn=Sn=\Sigma\left( 2^{n+1}-1 \right)$
$=\Sigma\ 2^{n+1}-\Sigma\ 1$
$=\left( 2^2+2^3+...+2^{n+1} \right)-n$
$=2\left( 2+2^2+...+2^n \right)-n$
$=4\left( \dfrac{2^n-1}{2-1} \right)-n$
$=2^{n+2}-n-y$
Question 14 of 25

14. Question
1 point(s)
If $ S_n=1^3+2^3+…+n^3 $ and $ T_n=1+2+…+n $, then
- $ S_n=T_n $
- $ S_n=T_n^4 $
- $ S_n=T_n^2 $
- $ S_n=T_n^3 $
Correct

Incorrect

Hint

$Tn=1+2+3+...+n$
$Sn=1_3+1^3+3^3+...+n^3$
$\therefore Tn=\Sigma\ n=\frac{n\left( n+1 \right)}{2}$
$Sn=\Sigma\ n^3=\left[ \frac{n\left( n+1 \right)}{2} \right]^2=Tn^2$
$\therefore Sn=Tn^2$
Question 15 of 25

15. Question
1 point(s)
The sum of $ n $ terms of the series $ \dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+… $ is______
- $ \sqrt{2n+1} $
- $ \sqrt{2n+1}-1 $
- $ \dfrac{1}{2}\sqrt{2n+1} $
- $ \dfrac{1}{2}(\sqrt{2n+1}-1) $
Correct

Incorrect

Hint

$Sn=\dfrac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3+}\sqrt{5}}+...$
$=\dfrac{1}{2}\left[ \left( \sqrt{3}-1 \right)+\left( \sqrt{5}-\sqrt{3} \right)+...+\left( \sqrt{2n+1}-\sqrt{2n-1} \right) \right]$
$=\dfrac{1}{2}\left[ \sqrt{2n+1}-1 \right]$
Question 16 of 25

16. Question
1 point(s)
If the sum of first $ n $ natural number is $ \dfrac{1}{5} $ times the sum of their squares, then the value of $ n $ is_____
- $ 5 $
- $ 6 $
- $ 7 $
- $ 8 $
Correct

Incorrect

Hint

$\Sigma\ n=\dfrac{1}{5}\Sigma\ n^2$
$\implies \dfrac{n\left( n+1 \right)}{2}=\dfrac{1}{5}\left[ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right]$
$\implies 2n+1=15$
$n=7$
Question 17 of 25

17. Question
1 point(s)
The sum of the series $ 1^3+2^3+…+15^3 $ is______
- $ 22000 $
- $ 10000 $
- $ 14400 $
- $ 15000 $
Correct

Incorrect

Hint

$S=1^3+2^3+3^3+...+15^3$
$=\sum_{n=1}^{15}n^3$
$=\left[ \dfrac{15\left( 15+1 \right)}{2} \right]^2$
$=120^2=14400$
Question 18 of 25

18. Question
1 point(s)
$ \dfrac{2}{1}\cdot\dfrac{1}{3}+\dfrac{3}{2}\cdot \dfrac{1}{9}+\dfrac{4}{3}\cdot \dfrac{1}{27}+… $ is equal to _______
- $ \dfrac{1}{2}+ln\left( \dfrac{3}{2} \right) $
- $ \ln\dfrac{3}{2} $
- $\ln 6 $
- $ \ln \dfrac{2}{3} $
Correct

Incorrect

Hint

$Tn=\left( \dfrac{n+1}{n} \right)\dfrac{1}{3n}=\dfrac{1}{3^n}+\dfrac{1}{n3^n}$
$\therefore S=\sum_{n=1}^{\infty }Tn=\sum_{n=1}^{\infty }\dfrac{1}{3^n}+\sum_{n=1}^{\infty }\dfrac{1}{n3^n}$
$=\dfrac{\dfrac{1}{3}}{1-\frac{1}{3}}+\left\{ -\ln\left( 1-\dfrac{1}{3} \right) \right\}$
$=\dfrac{1}{2}-\ln\dfrac{2}{3}$
$\implies S=\dfrac{1}{2}+\ln\dfrac{3}{2}$
Question 19 of 25

19. Question
1 point(s)
The sum of infinite series $ \left( \dfrac{1}{3} \right)^2+\dfrac{1}{3}\left( \dfrac{1}{3} \right)^4+\dfrac{1}{5}\left( \dfrac{1}{3} \right)^6+… $ is______
- $ \dfrac{1}{4} \ln\ 2 $
- $ \dfrac{1}{2} \ln\ 2 $
- $ \dfrac{1}{6} \ln\ 2 $
- $ \dfrac{1}{4} \ln\ 2 $
Correct

Incorrect

Hint

$\left( \dfrac{1}{3} \right)^2+\dfrac{1}{3}\left( \dfrac{1}{3} \right)^4+\dfrac{1}{5}\left( \dfrac{1}{3} \right)^6+...$
$=\dfrac{1}{3}\left[ \dfrac{1}{3}+\left( \dfrac{1}{3} \right)^3\dfrac{1}{3}+\left( \dfrac{1}{3} \right)^5\dfrac{1}{5}+... \right]$
$=\dfrac{1}{3}\cdot \dfrac{1}{2}\ln\left( \dfrac{1+\dfrac{1}{3}}{1=\dfrac{1}{3}} \right)$
$=\dfrac{1}{6}\ \ln\ 2$
Question 20 of 25

20. Question
1 point(s)
The sum of $ n $ terms of the sseries $\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+… $ is equal to ______
- $ \dfrac{3^n\left( 2n+1 \right)+1}{2\left( 3^n \right)} $
- $ \dfrac{3^n\left( 2n+1 \right)-1}{2\left( 3^n \right)} $
- $ \dfrac{\left( 3^nn-1 \right)}{2\left( 3^n \right)} $
- $ \dfrac{3^n-1}{2\left( 3^n \right)} $
Correct

Incorrect

Hint

$\dfrac{4}{3}+\frac{10}{9}+\dfrac{28}{27}+...\left( n\ \text{ terms} \right)$
$=\left( 1+\dfrac{1}{3} \right)+\left( 1+\dfrac{1}{9} \right)+\left( 1+\dfrac{1}{27} \right)+...\left( n\ \text{ terms} \right)$
$=\left( 1+1+1...n\ \text{ terms} \right)+\left( \dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...n\ \text{ terms} \right)$
$=n+\dfrac{1}{3}\left[ \dfrac{1-\dfrac{1}{3^n}}{1-\dfrac{1}{3}} \right]$
$=\dfrac{3^n\left( 2n+1-1 \right)}{2\left( 3^n \right)}$
Question 21 of 25

21. Question
1 point(s)
The sum of infinite terms of the series $ \dfrac{1}{\left( 1+a \right)\left( 2+a \right)}+\dfrac{1}{\left( 2+a \right)\left( 3+a \right)}+\dfrac{1}{\left( 3+a \right)\left( 4+a \right)}+… $, where $ a $ is constant, is______
- $ \dfrac{1}{1+a} $
- $ \dfrac{2}{1+a} $
- $ \infty $
- None of these
Correct

Incorrect

Hint

$Tn=\sum_{n=1}^{\infty }\dfrac{1}{\left( n+1 \right)\left( n+1+a \right)}$
$=\sum_{n=1}^{\infty }\left\{ \dfrac{1}{n+a}-\dfrac{1}{n+1+a} \right\}$
$\therefore Sn=\Sigma\ Tn$
$=\left( \dfrac{1}{1+a}-\dfrac{1}{2+a} \right)+\left( \dfrac{1}{2+a}-\dfrac{1}{3+a} \right)+...+\left( \dfrac{1}{n+a}-\dfrac{1}{n+1+a} \right)$
$=\dfrac{1}{1+a}-\dfrac{1}{n+1+a}$
$\therefore \lim_{n \to \infty } Sn=\dfrac{1}{1+a}$
Question 22 of 25

22. Question
1 point(s)
If $ 0$ < $ y $ <$ 2^{\frac{1}{3}} $ and $ x(y^3-1)=1 $, then $ \dfrac{2}{x}+\dfrac{2}{3x^3}+\dfrac{2}{5x^5}+… $ is equal to______
- $ \log\left( \dfrac{y^3}{2-y^3} \right) $
- $ \log\left( \dfrac{y^3}{1-y^3} \right) $
- $ \log\left( \dfrac{2y^3}{1-y^3} \right) $
- $ \log\left( \dfrac{2y^3}{1-2y^3} \right) $
Correct

Incorrect

Hint

$x\left( y^3-1 \right)=1 \implies x=\frac{1}{y^3-1}=\dfrac{1}{k}$
$\implies k=\dfrac{1}{x}$
then $\dfrac{2}{x}+\dfrac{2}{3x^3}+\dfrac{2}{5x^5}+...$
$=2k+\dfrac{2}{3}k^3+\dfrac{2}{5}k^5+...$
$=\log\dfrac{1+k}{1-k}$
$=\log\left\{ \dfrac{1+y^3-1}{1-y^3+1} \right\}=log\left( \dfrac{y^3}{2-y^3} \right)$
Question 23 of 25

23. Question
1 point(s)
The value of $ \dfrac{2}{3!}+\dfrac{4}{5!}+\dfrac{6}{7!}+… $ is________
- $ e^{\frac{1}{2}} $
- $ e^{-1} $
- $ e $
- $ e^{-\frac{1}{3}} $
Correct

Incorrect

Hint

$Tn=\dfrac{2n}{\left( 2n+1 \right)!}=\dfrac{1}{2n!}-\frac{1}{\left( 2n+1 \right)!}$
$\Sigma\ Tn=\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-\dfrac{1}{5}+...$
$=1-\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-\dfrac{1}{5!}+...$
$=e^{-1}$
Question 24 of 25

24. Question
1 point(s)
The value of $ 1-\log\ 2+\dfrac{\left( \log\ 2 \right)^2}{2!}-\dfrac{\left( \log\ 2 \right)^3}{3!}+… $ is_______
- $ \log\ 3 $
- $ \log\ 2 $
- $ \dfrac{1}{2} $
- $ 0 $
Correct

Incorrect

Hint

$1-\log\ 2+\dfrac{\left( \log\ 2 \right)^2}{2!}-\dfrac{\left( log\ 2 \right)^3}{3!}+...$
$=e^{-log\ 2}=\dfrac{1}{2}$
Question 25 of 25

25. Question
1 point(s)
$ \dfrac{1}{2}-\dfrac{1}{2\cdot 2^2}+\dfrac{1}{3\cdot 2^3}-\dfrac{1}{4\cdot 2^4}+… $ is equal to________
- $ \dfrac{1}{4} $
- $ \ln \dfrac{3}{4} $
- $ \ln \dfrac{3}{2} $
- $ \ln \dfrac{2}{3} $
Correct

Incorrect

Hint

$\ln\left( 1+x \right)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+...$
Put $x=\dfrac{1}{2}$
$\therefore ln\left( 1+\dfrac{1}{2} \right)=\dfrac{1}{2}-\dfrac{1}{2\cdot 2^2}+\dfrac{1}{3\cdot 2^3}-...$
$\implies \left[ \dfrac{1}{2}-\dfrac{1}{2\cdot 2^2}+\dfrac{1}{3\cdot 2^3}-\dfrac{1}{4\cdot 2^4}+... \right]=\ln\ \dfrac{3}{2}$

Mathematics Class XI

Practice Questions 4-Sequence ans Series-Class XI

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