datavorous/entrance-exam-dataset huggingface dataset:
If $n$ is a positive integer, then $(\sqrt{3}+1)^{2 n}-(\sqrt{3}-1)^{2 n}$ is

Options:
A
an irrational number

B
an odd positive integer

C
an even positive integer

D
a rational number other than positive integers

The correct answer is:
an irrational number

Solution:
$\begin{aligned}
& (\sqrt{3}+1)^{2 n}-(\sqrt{3}-1)^{2 n}=\left[(\sqrt{3}+1)^2\right]^n-\left[(\sqrt{3}-1)^2\right]^n=(4+2 \sqrt{3})^n-(4-2 \sqrt{3})^n \\
& =2^n\left[(2+\sqrt{3})^n-(2-\sqrt{3})^n\right] \\
& =2^n\left\{\left[{ }^n C_0 2^n+{ }^n C_1 2^{n-1} \sqrt{3}+{ }^n C_2 2^{n-2} 3+\cdots \cdot\right]-\left[{ }^n C_0 2^n-{ }^n C_1 2^{n-1} \sqrt{3}+{ }^n C_2 2^{n-2} 3-\cdots \cdot\right]\right\} \\
& =2^{n+1}\left[{ }^n C_1 2^{n-1} \sqrt{3}+{ }^n C_3 2^{n-3} 3 \sqrt{3}+\cdots \cdot\right]=2^{n+1} \sqrt{3} \text { (some integer) }
\end{aligned}$
Which is irrational

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2012 (Offline)
If n is an even positive integer greater than 1 and x>0, then the condition that the greatest term in the expansion of 1+xn may have the greatest coefficient also is

Options:
A n-1n<x<nn-1
B nn+1<x<n+1n
C nn+3<x<n+3n
D nn+2<x<n+2n

The correct answer is:
nn+2<x<n+2n

Solution:
If n is even, the greatest coefficient is nCn/2
â´ The greatest term =ânCn/2âxn/2
â´ânCn/2âxn/2>ânCn/2â1xn/2â1
and nCn/2âxn/2>ânCn/2+1xn/2+1
â n-n2+1n2x>1
and n-n2+1+1n2+1x<1
ân/2n/2+1<x<n2+1n2
Hence, nn+2<x<n+2n

Topic: Mathematics, Binomial Theorem, JEE Main
If ' $n$ ' is an integer, the domain of the function $\sqrt{\sin 2 x}$ is

Options:
A $\left[n \pi-\frac{\pi}{2}, n \pi\right]$
B $\left[n \pi, n \pi+\frac{\pi}{2}\right]$
C $[(2 n-1) \pi, 2 n \pi]$
D $[2 n \pi,(2 n+1) \pi]$

The correct answer is:
$\left[n \pi, n \pi+\frac{\pi}{2}\right]$

Solution:
According to question, as $\sqrt{\sin 2 x}$ can't be negative. So the option (b) is correct Domain of function $\sqrt{\sin 2 x}$ is $[n \pi, n \pi+\pi / 2]$.

Topic: Mathematics, Functions, JEE Main
If $n$ is an integer which leaves remainder one when divided by three, then $(1+\sqrt{3} i)^n+(1-\sqrt{3} i)^n$ equals

Options:
A $-2^{n+1}$
B $2^{n+1}$
C $-(-2)^n$
D $-2^n$

The correct answer is:
$-(-2)^n$

Solution:
$\begin{aligned} & \text { Now, }(1+\sqrt{3} i)^n+(1-\sqrt{3} i)^n \\ &= {\left[2\left(\frac{1+\sqrt{3} i}{2}\right)\right]^n+\left[2\left(\frac{1-\sqrt{3} i}{2}\right)\right]^n } \\ &=\left(-2 \omega^2\right)^n+(-2 \omega)^n \\ &=(-2)^n\left[\left(\omega^2\right)^{3 r+1}+(\omega)^{3 r+1}\right] \\ & {[\because n=3 r+1, \text { where } r \text { is an integer }] } \\ &=(-2)^n\left(\omega^2+\omega\right)=-(-2)^n\end{aligned}$

Topic: Mathematics, Complex Number, TS EAMCET, TS EAMCET 2009
If $n$ is an integer which leaves remainder one when divided by three, then $(1+\sqrt{3} i)^n+(1-\sqrt{3} i)^n$ equals

Options:
A $-2^{n+1}$
B $2^{n+1}$
C $-(-2)^n$
D $-2^n$

The correct answer is:
$-(-2)^n$

Solution:
$\begin{aligned} & \text { Now, }(1+\sqrt{3} i)^n+(1-\sqrt{3} i)^n \\ &= {\left[2\left(\frac{1+\sqrt{3} i}{2}\right)\right]^n+\left[2\left(\frac{1-\sqrt{3} i}{2}\right)\right]^n } \\ &=\left(-2 \omega^2\right)^n+(-2 \omega)^n \\ &=(-2)^n\left[\left(\omega^2\right)^{3 r+1}+(\omega)^{3 r+1}\right] \\ & {[\because n=3 r+1, \text { where } r \text { is an integer }] } \\ &=(-2)^n\left(\omega^2+\omega\right)=-(-2)^n\end{aligned}$

Topic: Mathematics, Complex Number, AP EAMCET, AP EAMCET 2009
If $n$ is an integer which leaves remainder one when divided by three, then $(1+\sqrt{3} i)^n+(1-\sqrt{3} i)^n$ equals

Options:
A $-2^{n+1}$
B $2^{n+1}$
C $-(-2)^n$
D $-2^n$

The correct answer is:
$-(-2)^n$

Solution:
$\begin{aligned} & \text { Now, }(1+\sqrt{3} i)^n+(1-\sqrt{3} i)^n \\ &= {\left[2\left(\frac{1+\sqrt{3} i}{2}\right)\right]^n+\left[2\left(\frac{1-\sqrt{3} i}{2}\right)\right]^n } \\ &=\left(-2 \omega^2\right)^n+(-2 \omega)^n \\ &=(-2)^n\left[\left(\omega^2\right)^{3 r+1}+(\omega)^{3 r+1}\right] \\ & {[\because n=3 r+1, \text { where } r \text { is an integer }] } \\ &=(-2)^n\left(\omega^2+\omega\right)=-(-2)^n\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\mathrm{n}$ is an integer which leaves remainder one when divided by three, then $(1+\sqrt{3} i)^{n}+(1-\sqrt{3} i)^{n}$ equals

Options:
A $-2^{\mathrm{n}+1}$
B $2^{n+1}$
C $-(-2)^{\mathrm{n}}$
D $-2^{n}$

The correct answer is:
$-(-2)^{\mathrm{n}}$

Solution:
$$
\begin{aligned}
&(1+\sqrt{3} i)^{n}+(1-\sqrt{3} i)^{n} \\
=&\left[2\left(\frac{1+\sqrt{3} i}{2}\right)\right]^{n}+\left[2\left(\frac{1-\sqrt{3} i}{2}\right)\right]^{n} \\
=&\left(-2 \omega^{2}\right)^{n}+(-2 \omega)^{n} \\
=&(-2)^{n}\left[\left(\omega^{2}\right)^{3 r+1}+(\omega)^{3 r+1}\right] \\
&(\because n=3 r+1, \text { where } r \text { is an integer }) \\
=&(-2)^{n}\left(\omega^{2}+\omega\right)=-(-2)^{n}
\end{aligned}
$$

Topic: Mathematics, Complex Number, VITEEE, VITEEE 2009
If $n$ is even and the value of ${ }^n C_r$ is maximum, then $r=$

Options:
A $\frac{n}{2}$
B $\frac{n+1}{2}$
C $\frac{n-1}{2}$
D None of these

The correct answer is:
$\frac{n}{2}$

Solution:
Apply the formula for binomial coefficient series,
${ }^n C_0,{ }^n C_1,{ }^n C_2, \ldots,{ }^n C_r, \ldots$, ${ }^n C_{n-2},{ }^n C_{n-1},{ }^n C_n$
Here, we have to find the maximum value of value of ${ }^n C_r$
And maximum value will be the middle binomial coefficient.
Total number of terms $=n+1 ; r=0$ to $n$
If we compare, ${ }^n C_r$ is $(r+1)$ th term And middle term
$=\frac{(n+1+1)}{2}=\frac{n}{2}+1$
$\Rightarrow r+1=\frac{n}{2}+1$
$\Rightarrow \quad r=\frac{n}{2}$

Topic: Mathematics, Permutation Combination, JEE Main
If $n$ is even posirive integer, then the condition that the greatest term in the expansion of $(a+x)^{n}$ may also have the greatest coefficient, is

Options:
A $\frac{n}{n+2} < x < \frac{n+2}{n}$
B $\frac{n}{n+1} < x < \frac{n+1}{n}$
C $\frac{n+1}{n+2} < x < \frac{n+2}{n+1}$
D $\frac{n+2}{n+3} < x < \frac{n+3}{n+2}$

The correct answer is:
$\frac{n}{n+2} < x < \frac{n+2}{n}$

Solution:
For greatest term of $(1+x)^{n},$ we have
$\quad \frac{n}{2} < \frac{n+1}{1+x} < \frac{n}{2}+1$
$\Rightarrow \quad \frac{n}{2} < \frac{n+1}{1+x}$ and $\frac{n+1}{1+x} < \frac{n}{2}+1$
$\Rightarrow \quad 1+x < \frac{n+1}{n / 2}$ and $\frac{n+1}{\frac{n}{2}+1} < 1+x$
$\Rightarrow \quad x < \frac{n+1-n / 2}{n / 2}$
and $\quad \frac{n+1-(n / 2-1)}{\frac{n+2}{2}} < x$
$\Rightarrow \quad x < \frac{n+2}{n}$ and $\frac{n}{n+2} < x$
$\Rightarrow \quad \frac{n}{n+2} < x < \frac{n+2}{n}$

Topic: Mathematics, Binomial Theorem, JEE Main
If n is the degree of the polynomial, 25x3+1-5x3-18+ 25x3+1+5x3-18 and m is the coefficient of xn in it, then the ordered pair n, m is equal to

Options:
A 8, 5104
B 12, 8104
C 12, 204
D 24, 108

The correct answer is:
12, 204

Solution:
25x3+1-5x3-18+ 25x3+1+5x3-18

Rationalizing denominators of both the brackets,

=5x3+1+ 5x3-1 8+ 5x3+1 - 5x3-1 8

=28C0 5x3+18+8C25x3+165x3-12+8C45x3+145x3-14+8C65x3+125x3-16+8C85x3-18

=25x3-14+285x3+135x3-1+705x3+125x3-12+285x3+15x3-13+5x3+14

So, n=12 and

m=254+140.53+70.54+140.53+54=160000=204

Topic: Mathematics, Binomial Theorem, JEE Main
If n is the degree of the polynomial, 25x3+1-5x3-18+ 25x3+1+5x3-18 and m is the coefficient of xn in it, then the ordered pair n, m is equal to

Options:
A 8, 5104
B 12, 8104
C 12, 204
D 24, 108

The correct answer is:
12, 204

Solution:
25x3+1-5x3-18+ 25x3+1+5x3-18

Rationalizing denominators of both the brackets,

=5x3+1+ 5x3-1 8+ 5x3+1 - 5x3-1 8

=28C0 5x3+18+8C25x3+165x3-12+8C45x3+145x3-14+8C65x3+125x3-16+8C85x3-18

=25x3-14+285x3+135x3-1+705x3+125x3-12+285x3+15x3-13+5x3+14

So, n=12 and

m=254+140.53+70.54+140.53+54=160000=204

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2018 (15 Apr)
If $n$ is the degree of the polynomial,
$$
\left[\frac{1}{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}\right]^8+\left[\frac{1}{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}}\right]^8
$$
and $\mathrm{m}$ is the coefficient of $\mathrm{x}^{\mathrm{n}}$ in it, then the ordered pair $(\mathrm{n}, \mathrm{m})$ is equal to

Options:
A
$\left(12,(20)^4\right)$

B
$\left(8,5(10)^4\right)$

C
$\left(24,(10)^8\right)$

D
$\left(12,8(10)^4\right)$

The correct answer is:
$\left(12,8(10)^4\right)$

Solution:
$$
\left[\frac{1}{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}\right]^8+\left[\frac{1}{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}}\right]^8
$$
After rationalise the polynomial we get
$$
\begin{aligned}
&{\left[\frac{1}{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}} \times \frac{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}}{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}}\right]^8} \\
&+\left[\frac{1}{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}} \times \frac{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}\right]^8
\end{aligned}
$$
$$
\left[\frac{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}}{\left(5 x^3+1\right)-\left(5 x^3-1\right)}\right]^8+\left[\frac{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}{\left(5 x^3+1\right)-\left(5 x^3-1\right)}\right]^8
$$
$$
\begin{aligned}
&=\frac{1}{2^8}\left[\left(\sqrt{5 x^3+1}+\sqrt{5 x^3-1}\right)^8+\left(\sqrt{5 x^3+1}-\sqrt{5 x^3-1}\right)^8\right] \\
&\left.=\frac{1}{2^8}\left[\begin{array}{l}
\left.{ }^8 C_0\left(\sqrt{5 x^3+1}\right)^8+{ }^8 C_4\left(\sqrt{5 x^3+1}\right)^6\left(\sqrt{5 x^3-1}\right)^2\right)^4\left(\sqrt{5 x^3-1}\right)^4+ \\
{ }^8 C_6\left(\sqrt { 5 x ^ { 3 } + 1 } { } ^ { 2 } \left(\sqrt{\left.5 x^3-\right)^1}{ }^6+{ }^8 C_8\left(\sqrt{5 x^3-1}{ }^8\right.\right.\right.
\end{array}\right]\right) \\
&\frac{1}{2^8}\left[\begin{array}{l}
{ }^8 C_0\left(5 x^3+1\right)^4+{ }^8 C_2\left(5 x^3+1\right)^3\left(5 x^3-1\right)+8_{C_4} \\
\left(5 x^3+1\right)^2\left(5 x^3-1\right)^2+ \\
{ }^8 C_6\left(5 x^3+1\right)\left(5 x^3-1\right)^3+8_{C_8}\left(5 x^3-1\right)^4
\end{array}\right] \\
&
\end{aligned}
$$
So, the degree of polynomial is 12 ,
Now, coefficient of
$$
\begin{aligned}
&x^{12}=\left[{ }^8 \mathrm{C}_0 5^4+{ }^8 \mathrm{C}_2 5^4+{ }^8 \mathrm{C}_4 5^4+{ }^8 \mathrm{C}_6 5^4+{ }^8 \mathrm{C}_8 5^4\right] \\
&=5^4 \times \frac{2^8}{2}=5^4 \times 2^4 \times \frac{2^2}{2} \\
&=10^4 \times 2^3=8(10)^4
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If n is the number of irrational terms in the expansion of 31/4+51/860, then n-1 is divisible by :

Options:
A 26
B 30
C 8
D 7

The correct answer is:
26

Solution:
31/4+51/860

=Cr6031/460-r.51/8r

=Cr60360-r4.5r8

For rational terms.

r8=k; 0≤r≤60

0≤8k≤60

0≤k≤608

0≤k≤7.5

k=0,1,2,3,4,5,6,7

60-8k4 is always divisible by 4 for all value of k

Total rational terms =8

Total terms =61

irrational terms =53

n-1=53-1=52

52 is divisible by 26.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (16 Mar Shift 1)
If N is the number of positive integral solutions of the equation x1x2x3x4=770, then the value of N is

Options:
A 250
B 252
C 254
D 256

The correct answer is:
256

Solution:
Given, x 1 . x 2 . x 3 . x 4 =770=2.5.7.11
Let positive integer x 1 = 2 a 1 .5 b 1 . 7 c 1 .11 d 1
Let positive integer x 2 = 2 a 2 .5 b 2 .7 c 2 .11 d 2
Let positive integer x 3 = 2 a 3 .5 b 3 .7 c 3 .11 d 3
Let positive integer x 4 = 2 a 4 .5 b 4 .7 c 4 .11 d 4
Now, x 1 x 2 x 3 x 4 = 2 a 1 + a 2 + a 3 + a 4 .5 b 1 + b 2 + b 3 + b 4 .7 c 1 + c 2 + c 3 + c 4 .11 d 1 + d 2 + d 3 + d 4
As per given condition a 1 + a 2 + a 3 + a 4 =1, which can be in 4 ways ( 1, 0, 0, 0 ), ( 0, 1, 0, 0 )( 0, 0, 1, 0 ) and ( 0, 0, 0, 1 ), similarly for other powers.
Hence total number of ways =4×4×4×4=256

Topic: Mathematics, Permutation Combination, JEE Main
If $n={ }^m C_2$, then the value of ${ }^n C_2$ is given by

Options:
A
$3\left({ }^{m+1} C_4\right)$

B
${ }^{m-1} C_4$

C
${ }^{m+1} C_4$

D
$2\left({ }^{m+2} C_4\right)$

The correct answer is:
$3\left({ }^{m+1} C_4\right)$

Solution:
$n={ }^m \mathrm{C}_2=\frac{m(m-1)}{2}$
Since $m$ and $(m-1)$ are two consecutive natural numbers, therefore their product is an even natural number. So $\frac{m(m-1)}{2}$ is also a natural number.
Now $\frac{m(m-1)}{2}=\frac{m^2-m}{2}$
$$
\begin{aligned}
\therefore & \frac{m(m-1)}{2} \mathrm{C}_2=\frac{\left(\frac{m^2-m}{2}\right)\left(\frac{m^2-m}{2}-1\right)}{2} \\
& =\frac{m(m-1)\left(m^2-m-2\right)}{8} \\
& =\frac{m(m-1)\left[m^2-2 m+m-2\right]}{8} \\
& =\frac{m(m-1)[m(m-2)+1(m-2)]}{8} \\
& =\frac{m(m-1)(m-2)(m+1)}{8}
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
If $n={ }^m C_2$, then the value of ${ }^n C_2$ is given by

Options:
A
$3\left({ }^{m+1} C_4\right)$

B
${ }^{m-1} C_4$

C
${ }^{m+1} C_4$

D
$2\left({ }^{m+2} C_4\right)$

The correct answer is:
$3\left({ }^{m+1} C_4\right)$

Solution:
$n={ }^m \mathrm{C}_2=\frac{m(m-1)}{2}$
Since $m$ and $(m-1)$ are two consecutive natural numbers, therefore their product is an even natural number. So $\frac{m(m-1)}{2}$ is also a natural number.
Now $\frac{m(m-1)}{2}=\frac{m^2-m}{2}$
$$
\begin{aligned}
\therefore & \frac{m(m-1)}{2} \mathrm{C}_2=\frac{\left(\frac{m^2-m}{2}\right)\left(\frac{m^2-m}{2}-1\right)}{2} \\
& =\frac{m(m-1)\left(m^2-m-2\right)}{8} \\
& =\frac{m(m-1)\left[m^2-2 m+m-2\right]}{8} \\
& =\frac{m(m-1)[m(m-2)+1(m-2)]}{8} \\
& =\frac{m(m-1)(m-2)(m+1)}{8}
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2012 (19 May Online)
If n objects are arranged in a row, then the number of ways of selecting three objects such that no two of them are next to each other is

Options:
A n-3C3
B n-3C2
C n-2C2
D n-2C3

The correct answer is:
n-2C3

Solution:
Let \(x\) be the number of objects to the left of the first object chosen, \(y\) the number of objects between the first and the second, \(\mathrm{z}\) the number of objects between the second and the third and \(u\) the number of objects to the right of the third objects. Then, \(x, u \geq 0 ; y, z \geq 1\) and \(x+y+z+u=n-3\). Let \(y_{1}=\) \(\mathrm{y}-1\) and \(\mathrm{z}_{1}=\mathrm{z}-1\). Then, \(\mathrm{y}_{1} \geq 0, \mathrm{z}_{1} \geq 0\) such that \(\mathrm{x}+\mathrm{y}_{1}+\mathrm{z}_{1}+\mathrm{u}=\mathrm{n}-5\). The total number of non-negative integral solutions of this equation is \({ }^{\mathrm{n}-5+4-1} \mathrm{C}_{4-1}={ }^{\mathrm{n}-2} \mathrm{C}_{3} .\)

Topic: Mathematics, Permutation Combination, JEE Main
If ${ }^n P_5=9 \times{ }^{n-1} P_4$, then the value of $n$ is

Options:
A $6$
B $8$
C $5$
D $9^{\prime}$

The correct answer is:
$9^{\prime}$

Solution:
$\frac{{ }^n P_5}{n^{n-1} P_4}=9 \Rightarrow \frac{n!}{(n-5)!} \times \frac{(n-5)!}{(n-1)!}=9 \Rightarrow n=9$

Topic: Mathematics, Permutation Combination, JEE Main
If ${ }^n P_r=30240$ and ${ }^n C_r=252$, then the ordered pair $(n, r)$ is equal to

Options:
A $(12,6)$
B $(10,5)$
C $(9,4)$
D $(16,7)$

The correct answer is:
$(10,5)$

Solution:
Given that, ${ }^n P_r=30240$ and ${ }^n C_r=252$
$$
\begin{aligned}
& \Rightarrow \frac{n !}{(n-r) !}=30240 \text { and } \frac{n !}{(n-r) ! r !}=252 \\
& \Rightarrow \quad r !=\frac{30240}{252}=120 \\
& \Rightarrow \quad r=5 \\
& \therefore \quad \frac{n !}{(n-5) !}=30240 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4)=30240 \\
& \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4) \\
& \quad=10(10-1)(10-2)(10-3)(10-4) \\
& \Rightarrow \quad n=10
\end{aligned}
$$
Hence, required ordered pair is $(10,5)$.

Topic: Mathematics, Permutation Combination, JEE Main
If nR and nV denote the number of photons emitted by a red bulb and violet bulb of equal power in a given time, then

Options:
A nR=nV
B nR>nV
C nR<nV
D nR ≤ nV

The correct answer is:
nR>nV

Solution:
Energy possessed by a photon is given by
E=hf=hcÎ»
If power of each bulb is P then energy given out in time t equal to Pt. Let the number of photons be n, then
n=PtE=Pt(hc/Î»)=PtÎ»hc
For red light, nR=PtÎ»Rhc
For violet light, nV=PtÎ»Vhc
â´ nRnV=Î»RÎ»V
As Î»R>Î»V
So, nR>nV

Topic: Physics, Dual Nature of Matter, JEE Main
If $n^{\text {th }}$ terms of two A.P.'s are $3 n+8$ and $7 n+15$, then the ratio of their $12^{\text {th }}$ terms will be

Options:
A $4 / 9$
B $7 / 16$
C $3 / 7$
D $8 / 15$

The correct answer is:
$4 / 9$

Solution:
Given: \(\mathrm{n}^{\text {th }}\) terms of two A.P.'s are \(3 \mathrm{n}+8\) and \(7 \mathrm{n}+15\).
Here, we have to find the ratio between their 12th terms.
So, \(12^{\text {th }}\) term in \(1^{\text {st }} \mathrm{AP}\) :
\(\begin{aligned}
& \left(T_{12}\right)_1=3 \times 12+8 \\
& \Rightarrow\left(T_{12}\right)_1=36+8=44 \\
& \Rightarrow\left(T_{12}\right)_1=44 \quad--(1)
\end{aligned}\)
Similarly, \(12^{\text {th }}\) term of \(2^{\text {nd }}\) AP:
\(\begin{aligned}
& \Rightarrow\left(T_{12}\right)_2=7 \times 12+15 \\
& \Rightarrow\left(T_{12}\right)_2=84+15=99 \\
& \Rightarrow\left(T_{12}\right)_2=99 \text {----(2) }
\end{aligned}\)
\(\therefore\) The required ratio \(=44 / 99=4 / 9=4: 9\)

Topic: Mathematics, Sequences and Series, JEE Main
If natality and mortality of a population are nearly same it will be termed as

Options:
A young population
B stable population
C declining population
D growing population

The correct answer is:
stable population

Solution:
Natality means the number of births per 1000 in a particular area for a particular time and mortality is the number of deaths per 1000 in a particular area for a particular time. In short, natality is birth rate and mortality is death rate.
If the natality and mortality remains the same then the population is neither growing nor reducing and it remains stable.
Declining population is the one where the old individuals outnumber young individuals and having more of death rate than birth rate.
Growing population is where the birth rate is higher than death rate.

Topic: Biology, Organisms and Populations, NEET
If non-zero numbers $a, b, c$ are in H.P., then the straight line $\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0$ always passes through a fixed point. That point is

Options:
A
$(-1,2)$

B
$(-1,-2)$

C
$(1,-2)$

D
$\left(1,-\frac{1}{2}\right)$

The correct answer is:
$(1,-2)$

Solution:
a, b, c are in H.P.
$$
\begin{aligned}
& \Rightarrow \frac{2}{b}-\frac{1}{a}-\frac{1}{c}=0 \\
& \frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0 \\
& \Rightarrow \frac{x}{-1} \quad-\frac{1}{-1} \quad \therefore x=1, y=-2
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2005
If non-zero real numbers b and c are such that min fx>max gx, where fx=x2+2bx+2c2 and gx=-x2-2cx+b2, x∈R; then cb lies in the interval

Options:
A 2 , ∞
B [ 1 2 , 1 2 )
C 0 , 1 2
D 1 2 , 2

The correct answer is:
2 , ∞

Solution:
Given fx=x2+2bx+2c2 and gx=-x2-2cx+b2

For finding the minimum value of f(x),

f'x=2x+2b=0

⇒x=-b

Also, f"x=2

⇒f"-b=2>0

So, fx has minimum value at x=-b

Hence, fmin=b2-2b2+2c2

⇒fmin=2c2-b2

Similarly, for finding the maximum value of g(x)

g'x=-2x-2c=0

⇒x=-c

Also, g"x=-2

g"-c=-2<0

So, gx has maximum value at x=-c.

Hence, gmax=-c2+2c2+b2

⇒gmax=c2+b2

Given fmin>gmax⇒2c2-b2>c2+b2⇒c2-2b2>0 ⇒c-2bc+2b>0

⇒cb-2cb+2>0

Using wavy curve method, we get cb<-2 or cb>2

∴cb∈2, ∞.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2014 (19 Apr Online)
If number of aldose, ketose, furanose and pyranose units present in maltose are p, q, r, s respectively. Find the value of p+q+r+s

Options:

The correct answer is:
4

Solution:
Monomer of maltose is α -D-glucopyranose. It is disaciharide, so 2 aldose and 2 pyranose forms are present in maltose.
Hence, p = 2, q = 0, r = 0 and s = 2
So, sum of p + q + r + s = 2 + 0 + 0 + 2 = 4.

Topic: Chemistry, Biomolecules, JEE Main
If number of lone pairs on central atom in XeF2,XeF4 and XeF6 are x, y, z. What is the sum of x + y + z here?

Options:

The correct answer is:
6.00

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
If number of terms in the expansion of $(x-2 y+3 z)^n$ are 45, then $n=$

Options:
A $7$
B $8$
C $9$
D None of these

The correct answer is:
$8$

Solution:
$\frac{(n+1)(n+2)}{2}=45$ or $n^2+3 n-88=0 \Rightarrow n=8$

Topic: Mathematics, Binomial Theorem, JEE Main
If numerical aperture of a microscope is increased then its

Options:
A Resolving power remains constant
B Resolving power becomes zero
C Limit of resolution is decreased
D Limit of resolution is increased

The correct answer is:
Limit of resolution is decreased

Solution:
d= 1.22λ 2nsinβ ∝ 1 NA
Limit of resolution (d) is decreases if NA of microscope increased

Topic: Physics, Ray Optics, MHT CET, MHT CET 2018
If numerical aperture of a microscope is increased then its

Options:
A Resolving power remains constant
B Resolving power becomes zero
C Limit of resolution is decreased
D Limit of resolution is increased

The correct answer is:
Limit of resolution is decreased

Solution:
d= 1.22λ 2nsinβ ∝ 1 NA
Limit of resolution (d) is decreases if NA of microscope increased

Topic: Physics, Wave Optics, NEET
If numerically greatest term in the expansion of 3-5x11, where x=15, is 729Î», then the value of Î»150 is

Options:

The correct answer is:
9.9

Solution:
3-5x11=3111-53x11
Now, m=-5x311+1-5x3+1=13×1243=3
⇒ The greatest terms in the expansion are T3 and T4
Greatest term T3=T2+1
=31111C2-132
=39×11C2=11×102×33×36
=55×27×729
⇒λ=55×27
⇒λ150=9.9

Topic: Mathematics, Binomial Theorem, JEE Main
If OB is the semi-minor axis of an ellipse, F1 and F2 are its focii and the angle between F1B and F2B is a right angle, then the square of the eccentricity of the ellipse is

Options:
A 1 4
B 1 2
C 1 2
D 1 2 2

The correct answer is:
1 2

Solution:
In standard ellipse x2a2+y2b2=1

Semi-major axis OA=a

Semi-minor axis OB=b

Focus F1≡-ae,0

F2≡ae,0

∠F1BF2=90°

OF1=OF2⇒∠F1BO=∠F2BO=45°

⇒aeb=tan45°=1

⇒b=ae

Also, b2=a21-e2

⇒a2e2=a21-e2

⇒2e2=1

⇒e2=12

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2014 (09 Apr Online)
If O is the origin and OP, OQ are distinct tangents to the circle x2+y2+2gx+2fy+c=0, then the circumcentre of the triangle OPQ is

Options:
A -g, -f
B g, f
C -f, -g
D None of these

The correct answer is:
None of these

Solution:
Tangents drawn from the point O, meet the cirle at P & Q and C be the centre of given circle. Then points O, P, C and Q are concyclic. That means any circle passing through O, P and Q also passes through C and OC is the diameter for this circle. Hence mid point of OC is the circumcentre of triangle OPQ.
Coordinates of circumcenter =-g2,-f2

Topic: Mathematics, Circle, JEE Main
If on combining two charged bodies, the current does not flow then

Options:
A charge is equal on both
B capacitance is equal on both
C potential is equal on both
D resistance is equal on both

The correct answer is:
potential is equal on both

Solution:
No Solution available

Topic: Physics, Electrostatics, JEE Main
If one end of a diameter of the ellipse $4 x^2+y^2=16$ is $(\sqrt{3}, 2)$, then the other end is:

Options:
A $(-\sqrt{3}, 2)$
B $(\sqrt{3},-2)$
C $(-\sqrt{3},-2)$
D $(0,0)$

The correct answer is:
$(-\sqrt{3},-2)$

Solution:
Correct option is \(\mathrm{C}\)
Let \(P(h, k)\)
Since every diameter of an ellipse passes through the centre and is bisected by it, therefore the coordinates of the other end are \((-\sqrt{3},-2)\)
as, centre \((0,0)\) so, \(\frac{h+\sqrt{3}}{2}=0, \frac{k+2}{2}=0\)
So \((h, k)=(-\sqrt{3},-2)\)
So, \((h, k) \equiv(-\sqrt{3},-2)\)

Topic: Mathematics, Ellipse, JEE Main
If one end of a focal chord AB of the parabola y2=8x is at A12,-2, then the equation of the tangent to it at B is:

Options:
A 2x+y-24=0
B x-2y+8=0
C x+2y+8=0
D 2x-y-24=0

The correct answer is:
x-2y+8=0

Solution:
Let 12,-2 is 2t2,4t⇒t=-12

Parameter of other end of focal chord is t=2

⇒ Point is 8,8

⇒ Equation of tangent is 8y-4x+8=0

⇒2y-x=8

Topic: Mathematics, Parabola, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If one end of a focal chord of the parabola, y2=16x is at 1,4, then the length of this focal chord is

Options:
A 24
B 25
C 22
D 20

The correct answer is:
25

Solution:
The given parabola y2=16x is of the form y2=4ax, hence a=4 and the focus of the parabola is a, 0=4, 0.

The equation of a line joining the points x1, y1 and x2, y2 is y-y1=y2-y1x2-x1x-x1.

Therefore, equation of chord joining P1, 4 to focus S4, 0 is

y-0=-43x-4

⇒3y=-4x+16

⇒4x +3y-16=0

To find the points where this line will cut the parabola put x=y216 from the parabola into the line, to get

4y216+ 3y-16=0

⇒y2+ 12y-64=0

⇒y+16y-4=0

⇒y=-16 or y=4

And x=y216

⇒x=16 or x=1

Thus, the points are 1, 4 and 16,-16, but 1, 4 is the given point P.

⇒Q=16,-16

The distance between the points x1, y1 & x2, y2 is x1-x22+y1-y22

So, the length of focal chord PQ is

PQ=16-12+-16-42

=225+400=625=25 units.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (09 Apr Shift 1)