Get Answers to all your Questions

Filter By

Clear Filter

All Questions

A sum of money under compound interest doubles itself in 4 years. In how many years will it become 16 times itself?

Option: 1
12 years

Option: 2
16 years

Option: 3
8 years

Option: 4
None of these

16 years

View Full Answer(6)

Posted by

Debjani Behera

A certain loan amounts, under compound interest, compounded annually earns an interest of Rs.1980 in the second year and Rs.2178 in the third year. How much interest did it earn in the first year?
Option: 1
Rs.1600

Option: 2
Rs.1800

Option: 3
Rs.1900

Option: 4
None of these

xyz pqr abc year tt

View Full Answer(2)

Posted by

wgqefiqyg

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

If a, b and c are the greatest values of $^{19}C_{p},^{20}C_{q},^{21}C_{r}$ respectively, then:

Option: 1 $\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$
Option: 2 $\frac{a}{10}=\frac{b}{11}=\frac{c}{42}$
Option: 3 $\frac{a}{11}=\frac{b}{22}=\frac{c}{21}$
Option: 4 $\frac{a}{10}=\frac{b}{11}=\frac{c}{21}$

Binomial Coefficient of the middle term is greatest.

Now,

$\\^nC_{r}\;\text{ is max at middle term}\\\begin{array}{l}{a=^{19} C_{p}=^{19} C_{10}=^{19} C_{9}} \\ {b=^{20} C_{q}=^{20} C_{10}} \\ {c=^{21} C_{r}=^{21} C_{10}=^{21} C_{11}}\end{array}$

$\frac{a}{^{19}C_9}=\frac{b}{ ^{20} \mathrm{C}_{10}}=\frac{c}{^{21} \mathrm{C}_{11}}$

$\frac{a}{^{19}C_9}=\frac{b}{\frac{20}{10} \cdot ^{19} \mathrm{C}_9}=\frac{c}{\frac{21}{11} \cdot \frac{20}{10} ^{19} \mathrm{C}_{9}}$

$\\\frac{a}{1}=\frac{b}{2}=\frac{c}{42 / 11}\\\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$

Correct Option (1)

View Full Answer(1)

Posted by

Kuldeep Maurya

If the sum of the coefficients of all even powers of x in the product $(1+x+x^{2}+....+x^{2n})(1-x+x^{2}-x^{3}+....+x^{2m})$ is 61, then n is equal to _________.
Option: 1 30
Option: 260
Option: 315
Option: 4 45

$\text { Let } (1+x+x^{2}+....+x^{2n})(1-x+x^{2}-x^{3}+....+x^{2n})=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots$

$\\Put \,\, x=1 \\ (2 n+1).1 =a_{0} + a_{1} + a_{2}+ \ldots \ldots \\ Put\, x =-1 \\ 1.(2 n+1)= a_{0} - a_{1} + a_{2}+\ldots \ldots \\ From (i)+(ii) \\ 4 n+2 =2 (a_{0}+a_{2}+\ldots ) \\ So,\,\, 2 n+1=61 \\ \Rightarrow n=30$

View Full Answer(1)

Posted by

Kuldeep Maurya

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

If $\alpha$ and $\beta$ be the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of $\left ( x+\sqrt{x^{2}-1} \right )^{6}+\left ( x-\sqrt{x^{2}-1} \right )^{6},$ then :
Option: 1 $\alpha +\beta =30$
Option: 2 $\alpha -\beta =-132$
Option: 3 $\alpha +\beta =60$
Option: 4 $\alpha -\beta =60$

We know

$\mathrm{(x+y)^{n}+(x-y)^{n}=2\left[^{n} C_{0}\; x^{n}\; y^{0}+^{n} C_{2} \;x^{n-2}\; y^{2}+^{n} C_{4}\; x^{n-4} \;y^{4}+\ldots .\right]}$

Now,

$\left ( x+\sqrt{x^{2}-1} \right )^{6}+\left ( x-\sqrt{x^{2}-1} \right )^{6}$

$={2\left[^{6} \mathrm{C}_{0} \mathrm{x}^{6}+^{6} \mathrm{C}_{2} \mathrm{x}^{4}\left(\mathrm{x}^{2}-1\right)+^{6} \mathrm{C}_{4} \mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{2}+^{6} \mathrm{C}_{6}\left(\mathrm{x}^{2}-1\right)^{3}\right]} \\ {\quad=2\left[\mathrm{x}^{6}+15\left(\mathrm{x}^{6}-\mathrm{x}^{4}\right)+15 \mathrm{x}^{2}\left(\mathrm{x}^{4}-2 \mathrm{x}^{2}+1\right)+\left(\mathrm{x}^{6}-1-3 \mathrm{x}^{4}+3 \mathrm{x}^{2}\right)\right]} \\ {\quad=2\left(32 \mathrm{x}^{6}-48 \mathrm{x}^{4}+18 \mathrm{x}^{2}-1\right)} \\ {\quad\alpha=-96 \text { and } \beta=36} \\ {\therefore \quad \alpha-\beta=-132}$ Correct Option (2)

View Full Answer(1)

Posted by

vishal kumar

The number of ordered pairs (r,k) for which $6\cdot ^{35}C_{r}=(k^{2}-3)\cdot ^{36}C_{r+1},$ where k is an integer, is :
Option: 1 $4$
Option: 2 6
Option: 3 2
Option: 4 3

As we have learnt

$^{n} C_{r}=\frac{n}{r} \cdot^{n-1} C_{r-1}$

Now,

$^{36}C_{r+1} (k^{2}-3)={^{35}C_{r}}\times {6}$

$\frac{36}{r+1}. ^{35}C_{r} (k^{2}-3)={^{35}C_{r}}\times {6}$

${k^{2}-3=\frac{r+1}{6} \Rightarrow k^{2}=3+\frac{r+1}{6}}$

r should be less than or equal to 35

Hence for k to be an integer, r can be 5 and 35

For r=5 we get k = -2, 2

For r=35 we get k=-3,3

We get 4 ordered pair (5,-2), (5,2), (35,-3), (35, 3)

Correct Option (1)

View Full Answer(1)

Posted by

Ritika Jonwal

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+....+x^{10}$ is :
Option: 1 $420$
Option: 2 $330$
Option: 3 $210$
Option: 4 $120$

Binomial Theorem

$(x+y)^{n}=^{n} C_{0} x^{n}+^{n} C_{1} x^{n-1} y+^{n} C_{2} x^{n-2} y^{2}+\cdots+^{n} C_{n} y^{n}\;\;where,\;n\in \mathbb{N}$

Now,

Given series S is a GP, with a = (1+x)¹⁰ , r = x/(1+x), n = 11

So, S = $\frac{(1+x)^{10} \left ( \frac{x}{1+x}^{11}-1 \right )}{(\frac{x}{1+x})-1}$

= (1+x)¹¹- x¹¹

Hence coefficient of x⁷ is ¹¹C₇= 330

Correct option (2)

View Full Answer(1)

Posted by

Ritika Jonwal

$\sum_{k=0}^{20}\left({ }^{20} \mathrm{C}_{\mathrm{k}}\right)^{2}$ is equal to :
Option: 1 ${ }^{40} \mathrm{C}_{21}$
Option: 2 ${ }^{41} \mathrm{C}_{20}$
Option: 3 ${ }^{40} C_{20}$
Option: 4 ${ }^{40} C_{19}$

$s= \sum_{k= 0}^{20}\left (\, ^{20}C_{k} \right )^{2}= \, ^{20}C_{0}\: ^{2}+ \, ^{20}C_{1}\: ^{2}+ \, ^{20}C_{2}\: ^{2}+\cdots ^{20}C_{20}\, ^{2}$
it is same as coefficient of $x^{20}$ in the expansion of $\left ( 1+x \right )^{20}\left ( x+1 \right )^{20}$
$= coeff\cdot of\, x^{20}\pi\left ( 1+x \right )^{40}$
$= \, ^{40}C_{20}$
option (3)