datavorous/entrance-exam-dataset huggingface dataset:
The value of the integral I=â«12tt+t1+lnt dt is equal to
(. and . denotes the greatest integer and fractional part function respectively)

Options:
A 0
B 1
C 2
D 3

The correct answer is:
3

Solution:
As t=0, we get,

I=∫12tt1+ln⁡tdt
=∫12dtt
=tt12
=22–11=3

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral I=â«133dx1+x2+x3+x5 is equal to

Options:
A π2
B π3
C π12
D π6

The correct answer is:
π12

Solution:
Given integral is I=∫133dx1+x21+x3
Let, tan-1⁡x=θ
⇒dx=sec2⁡θdθ
∴I=∫π6π3dθ1+tan3⁡θ
=∫π6π3cos3⁡θsin3⁡θ+cos3⁡θdθ
Applying a+b-x property and adding, we get,
2I=∫π6π3cos3⁡θ+sin3⁡θsin3⁡θ+cos3⁡θdθ
2I=θπ6π3
⇒2I=π6
⇒I=π12

Topic: Mathematics, Definite Integration, JEE Main
The value of the integral ∫sin2xsec2x+21-x2tanxsin-1x1-x21+tan2xdx is (where, C is the constant of integration)

Options:
A sin-1xcos2x+C
B sin-1xsin2x+C
C cos-1xsin2x+C
D -sin-1xsin2x+C

The correct answer is:
sin-1xsin2x+C

Solution:
∫sin2x1-x2+2tanx1+tan2xsin-1xdx
=∫sin2x1-x2+sin⁡2xsin-1xdx
=sin2⁡xsin-1x+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral ∫sin⁡xx6xcosx-sin⁡xx2dx is

(where, c is an arbitrary constant)

Options:
A sinxx+c
B sinxx2+c
C sin7xx7+c
D sin7x7x7+c

The correct answer is:
sin7x7x7+c

Solution:
Put sinxx=t.
Then, I=â«t6dt=t77+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral ∫x 131-x3dx is equal to (where c is the constant of integration)

Options:
A 6x438+311x116+314x73+117x176+c
B 6x438-311x116+314x73-117x176+c
C 2x438-311x46-314x73-117x176+c
D 2x48-311x11-314x7-117x17+c

The correct answer is:
6x438-311x116+314x73-117x176+c

Solution:
Let, I=∫x131-x3dx
Let, x=t6 ⇒dx=6t5dt
I=∫t21-t336t5dt
I=6∫t71-3t3+3t6-t9dt
=6∫t7-3t10+3t13-t16dt
I=6t88-3t1111+3t1414-t1717+c
I=6x438-311x116+314x73-117x176+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the integral â«x2-4xx+6x-4x+1x-2x+1dx is equal to

Options:
A x322+x+c
B x22-43x32+x+c
C x32+x2+c
D 23x32+c

The correct answer is:
x22-43x32+x+c

Solution:
Ι=∫x-14x-12dx=∫x-12dx=∫x-2x+1dx
=x22-43x32+x+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the limit limθ→0tanπcos2θsin2πsin2θ is equal to :

Options:
A -12
B -14
C 0
D 14

The correct answer is:
-12

Solution:
We have to find l=limθ→0tanπcos2θsin2πsin2θ

Using sin2θ+cos2θ=1, we get

⇒l=limθ→0tanπ1-sin2θsin2πsin2θ

⇒l=limθ→0tanπ-πsin2θsin2πsin2θ

Using tanπ-θ=-tanθ, we get

⇒l=limθ→0-tanπsin2θsin2πsin2θ

⇒l=limθ→0-tanπsin2θπsin2θ2πsin2θsin2πsin2θ×12

Using, limx→0sinxx=1 & limx→0tanxx=1, we get

l=-12.

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The value of the limit
$$
\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{6 / x^{2}}
$$
is

Options:
A e
B $\mathrm{e}^{-1}$
C $\mathrm{e}^{-1 / 6}$
D $\mathrm{e}^{6}$

The correct answer is:
e

Solution:
$\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{6 / x^{2}}$
$e^{\lim _{x \rightarrow 0} \frac{6}{x^{2}}\left(\frac{x-\sin x}{\sin }\right)}$
$\lim _{x \rightarrow 0} \frac{6}{x^{2}}\left(\frac{x-\left(x-\frac{x^{3}}{3}\right)}{x}\right)$
$=\mathrm{e}$

Topic: Mathematics, Limits, JEE Main
The value of the limit $\lim _{x \rightarrow-\infty}\left(\sqrt{4 x^{2}-x}+2 x\right)$ is

Options:
A $-\infty$
B $-\frac{1}{4}$
C 0
D $\frac{1}{4}$

The correct answer is:
$\frac{1}{4}$

Solution:
Rationalise
$\lim _{x \rightarrow-\infty}\left(\left(\sqrt{4 x^{2}-x}+2 x\right) \times \frac{\sqrt{4 x^{2}-x}-2 x}{\sqrt{4 x^{2}-x}-2 x}\right)$
$\lim _{x \rightarrow-\infty}\left(\frac{-x}{|x| \sqrt{4-\frac{1}{x}}-2 x}\right)$ at $x \rightarrow-\infty \quad|x|=-x$
$\lim _{\mathrm{x} \rightarrow-\infty}\left(\frac{-\mathrm{x}}{-\mathrm{x} \sqrt{4-\frac{1}{\mathrm{x}}}-2 \mathrm{x}}\right)=\frac{1}{2+2}=\frac{1}{4}$

Topic: Mathematics, Limits, JEE Main
The value of the numerically greatest term in the expansion of 4-3x7 when x=23 is equal to

Options:
A 71680
B 35840
C 10752
D 86016

The correct answer is:
86016

Solution:
We know that, Tr+1 has the greatest numerical value where,
r=n+1ax+a, where .=GIF
⇒r=7+1-3x-3x+4=83⋅23323+4=2
⇒T3 will be the term with greatest numerical value and
T3=7C2453x2=7C24532232=86016.

Topic: Mathematics, Binomial Theorem, JEE Main
The value of the parameter âa' such that the area bounded by y=a2x2+ax+1, coordinate axes and the line x=1 attains its least value, is equal to

Options:
A -14
B -12
C -34
D -1

The correct answer is:
-34

Solution:
a2x2+ax+1 is positive for all real values of x.
∴ Area=∫01a2x2+ax+1dx
=a23+a2+1
=162a2+3a+6
=162a2+32a+916+6-1816
=162a+342+398 , which is minimum for a=-34

Topic: Mathematics, Area Under Curves, JEE Main
The value of the resistance of a carbon resistor having the standard colour-coding as shown in the figure is n GΩ±5%. What is the value of n?

Options:

The correct answer is:
02.10

Solution:
From colour codes for resistors.

R=21×108 Ω±5%

R=2.1×109 Ω±5%

⇒R=2.1 GΩ±5%

n=2.1

Topic: Physics, Current Electricity, JEE Main
The value of the resistor, RS, needed in the DC voltage regulator circuit shown here, equals:

Options:
A Vi+VLn+1IL
B Vi-VLn IL
C Vi+VLn IL
D Vi-VLn+1IL

The correct answer is:
Vi-VLn+1IL

Solution:
Across zener diode VL=constant
∵Vi=VL+n+1IL RS
⇒Vi-VL=n+1IL RS
⇒RS=Vi-VLn+1IL

Topic: Physics, Semiconductors, JEE Main, JEE Main 2015 (11 Apr Online)
The value of the â«sin8x-cos8x1-2sin2xcos2xdx is equal to (where, C is the constant of integration)

Options:
A sin⁡2x2+C
B -sin⁡2x2+C
C cos⁡2x+C
D cos⁡x2+C

The correct answer is:
-sin⁡2x2+C

Solution:
Ι=∫sin4x+cos4xsin2x+cos2xsin2x-cos2x1-2sin2xcos2xdx
Ι=∫-cos⁡2xdx=-12sin⁡2x+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of the ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM. The correct one is

Options:
A
$\mathrm{d}^4$ (in strong ligand filed)

B
$\mathrm{d}^4$ (in weak ligand filed)

C
$\mathrm{d}^3$ (in weak as well as in strong fields)

D
$\mathrm{d}^5$ (in strong ligand filed)

The correct answer is:
$\mathrm{d}^4$ (in strong ligand filed)

Solution:
$\mathrm{d}^4$ in strong field, so unpaired electrons = 2.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2005
The value of the ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM. the correct one

Options:
A $d^4$ (in strong ligand field)
B $d^4$ (in weak ligand field)
C $d^3$ (in weak as well as in strong fields)
D $d^5$ (in strong ligand field)

The correct answer is:
$d^4$ (in strong ligand field)

Solution:
As we know,
Magnetic moment \(=\sqrt{n(n+2)}=2.84\)
so \(n=2\)
\(d^4\) (in strong ligand field) has two unpaired electron in octahedral field.
Hence, option (a) is correct.
All others doesnt have two unpaired electrons.

Topic: Chemistry, Coordination Compounds, JEE Main
The value of the sum $1 \times 2 \times 3+2 \times 3 \times 4+3 \times 4 \times 5+\ldots$ upto $n$ terms is equal to

Options:
A $\frac{1}{6} n^2\left(2 n^2+1\right)$
B $\frac{1}{6}\left(n^2-1\right)(2 n-1)(2 n+3)$
C $\frac{1}{8}\left(n^2+1\right)\left(n^2+5\right)$
D $\frac{1}{4} n(n+1)(n+2)(n+3)$

The correct answer is:
$\frac{1}{4} n(n+1)(n+2)(n+3)$

Solution:
Let given series be
$S=1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\ldots n$ terms
Now, general term
$$
\begin{aligned}
& T_n=\{1+(n-1) \cdot 1\}\{2+(n-1) \cdot 1\}\{3+(n-1) \cdot 1\} \\
& =(1+n-1)(2+n-1)(3+n-1) \\
& =n(n+1)(n+2) \\
& =n\left(n^2+3 n+2\right) \\
& =n^3+3 n^2+2 n
\end{aligned}
$$
Now,
$$
\begin{aligned}
& \text { Sum of the series } \mathrm{S}=\sum T_n=\sum\left(n^3+3 n^2+2 n\right) \\
& =\sum n^3+3 \sum n^2+2 \sum n \\
& =\frac{n^2(n+1)^2}{4}+\frac{3 n(n+1)(2 n+1)}{6}+\frac{2 n(n+1)}{2} \\
& =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+(2 n+1)+2\right]
\end{aligned}
$$
$\begin{aligned} & =\frac{n(n+1)}{2}\left[\frac{n^2+n+4 n+2+4}{2}\right] \\ & =\frac{n(n+1)\left(n^2+5 n+6\right)}{4} \\ & \therefore \mathrm{S}=\frac{n(n+1)(n+2)(n+3)}{4}\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main
The value of the sum of the series \( 3^{n} C_{0}-8^{n} C_{1}+13^{n} C_{2}-18^{n} C_{3}+\ldots \ldots \) upto \( (n+1) \) terms where \( n \geq 2 \), is

Options:
A \( 3^{n} \)
B \( 5^{n} \)
C \( 0 \)
D None of these

The correct answer is:
\( 0 \)

Solution:
3·nC0 - 8·nC1 + 13·nC2 - 18·nC3 +........
= 3 ( nC0 - nC1 + nC2 - nC3 +........)+5 (- nC1 + 2·nC2 - 3·nC3 + ......)
∵(1-x)n=nC0-nC1x+nC2x2-nC3x3+....... ...i
and - n(1-x)n-1 = -nC1 + 2·nC2x - 3·nC3x2 +..... ......(ii)
Putting x = 1 in Eqs. i & ii, we get,
nC0 - nC1 + nC2 - nC3 +.........= 0 ....(iii)
and - nC1 + 2·nC2 - 3·nC3 +........ = 0 .....(iv)
Now, by 3×iii+5×iv, we get,
3·nC0 - 8·nC1 + 13·nC2 - 18·nC3 +......= 0

Topic: Mathematics, Binomial Theorem, JEE Main
The value of $\int \frac{1}{x^4} d x$ is

Options:
A $\frac{1}{-3 x^3}+c$
B $\frac{1}{3 x^3}+c$
C $\frac{1}{-4 x^3}+c$
D $-\frac{1}{3 x^2}+c$

The correct answer is:
$\frac{1}{-3 x^3}+c$

Solution:
$\int x^{-4} d x=\frac{x^{-3}}{-3}+c=-\frac{1}{3 x^3}+c$.

Topic: Mathematics, Indefinite Integration, JEE Main
The value of ∫x-4x2x-2dx is equal to (where, C is the constant of integration)

Options:
A 2xx-2+C
B -2xx-2+C
C x-2x+C
D xx-2+C

The correct answer is:
-2xx-2+C

Solution:
∫x-4x2x-2dx
∫1x2-4x31x-2x2dx
Let 1x-2x2=t
-1x2+4x3dx=dt
∫-dtt=-2t+C
=-21x-2x2+C
=-2xx-2+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of â«x dx(x+3)x+1 is (where, c is the constant of integration)

Options:
A 2x+1+3Â Â tan-1x+1+c
B 2x+1+32Â Â tan-1x+12+c
C 2x+1-32 tan-1x+12+c
D 2x+1-3Â Â tan-1x+1+c

The correct answer is:
2x+1-32 tan-1x+12+c

Solution:
Substitute x+1=t, we get,

2∫t2-1t2+2 dt=2t-6∫1t2+2 dt
=2t-32tan-1t2+c
=2x+1-32tan-1x+12+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of $x$ for the maximum value of $\sqrt{3} \cos x+\sin x$ is

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
$\begin{aligned} & \text { Let } f(x)=\sqrt{3} \cos x+\sin x \\ & \Rightarrow f(x)=2\left(\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x\right)=2 \sin \left(x+\frac{\pi}{3}\right) \\ & \text { But }-1 \leq \sin \left(x+\frac{\pi}{3}\right) \leq 1 \\ & \text { Hence, } f(x) \text { is maximum, if } x+\frac{\pi}{3}=90^{\circ} \Rightarrow x=30^{\circ}\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of x for which sin⁡cot-1⁡1+x=cos⁡(tan-1⁡x) is

Options:
A 12
B 1
C 0
D -12

The correct answer is:
-12

Solution:
Let, cot-1⁡x+1=θ and tan-1⁡x=ϕ
Now, given equation becomes sinθ=cosϕ
⇒11+x+12=11+x2
⇒x+12+1=x2+1⇒x=-12

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of \( x \) for which \( \sin \left[\cot ^{-1}(1+x)\right]=\cos \left(\tan ^{-1} x\right) \)

Options:
A \( \frac{1}{2} \)
B \( 1 \)
C \( 0 \)
D \( -\frac{1}{2} \)

The correct answer is:
\( -\frac{1}{2} \)

Solution:
sin θ = 1 x + 1 2 + 1 2
cos Ï = 1 1 + x 2
⇒ 1 + x 2 = 1 + x + 1 2
⇒ x = ± x + 1
⇒ x = - x - 1 , x = - 1 2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of x for which the fourth term in the expansion of 525log5⁡4x+44+15log5⁡2x-1+738 is 336 can be equal to

Options:
A 12
B 1
C 2
D 3

The correct answer is:
1

Solution:
T4=C83515log5â¡4x+4451513log5â¡2x-1+73
=C835log5â¡4x+445log5â¡2x-1+7-1
=C83Ã4x+44Ã2x-1+7-1
Given, T4=336
â4x+442x-1+7-1=6
â4x+44=3â2x+42
â2x2-3â2x+2=0â2x=1,2
âx=0 or 1

Topic: Mathematics, Binomial Theorem, JEE Main
The value of $x$, if $\left|\begin{array}{ccc}-x & 1 & 0 \\ 1 & -x & 1 \\ 0 & 1 & -x\end{array}\right|=0$ is equal to

Options:
A $\pm \sqrt{6}$
B $\pm \sqrt{2}$
C $\pm \sqrt{3}$
D $\sqrt{2}, \sqrt{3}$

The correct answer is:
$\pm \sqrt{2}$

Solution:
We have $\Delta=\left|\begin{array}{ccc}-x & 1 & 0 \\ 1 & -x & 1 \\ 0 & 1 & -x\end{array}\right|$
$\Delta=-x\left(x^2-1\right)-1(-x-0)=0 \Rightarrow x= \pm \sqrt{2}$

Topic: Mathematics, Determinants, JEE Main
The value of $x$ in the partial redox equation $\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+x e \rightleftharpoons \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$ is

Options:
A $5$
B $3$
C $1$
D $0$

The correct answer is:
$5$

Solution:
\(\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \Leftrightarrow \mathrm{Mn}^{++}+4 \mathrm{H}_2 \mathrm{O}\).

Topic: Chemistry, Redox Reactions, JEE Main
The value of
$\left\{x \in \mathbb{R} \mid\left[\log (1.6)^{1-x^2}-(0.625)^{6(1+x)}\right] \in \mathbb{R}\right\}$ is

Options:
A $(-\infty,-1) \cup(7, \infty)$
B $(-1,5)$
C $(1,7)$
D $(-1,7)$

The correct answer is:
$(-\infty,-1) \cup(7, \infty)$

Solution:
$$
\left.x \in \mathbb{R} \mid \log \left[(1.6)^{1-x^2}-(0.625)^{6(1+x)}\right] \in \mathbb{R}\right\}
$$
Now, $\quad(1.6)^{1-x^2}>(0.625)^{6(1+x)}$
$$
\begin{array}{ll}
\Rightarrow & (1.6)^{1-x^2}>(0.625)^{6(1+x)} \\
& =\left(\frac{8}{5}\right)^{1-x^2}>\left(\frac{8}{5}\right)^{-6(1+x)} \\
\therefore & 1-x^2>-6(1+x) \\
\Rightarrow & x^2-6 x-7 < 0 \\
\Rightarrow \quad & (x-7)(x+1) < 0 \\
\Rightarrow \quad & x \in(-\infty,-1) \cup(7, \infty)
\end{array}
$$
Hence,
$$
\begin{gathered}
\left.x \in \mathbb{R}\left|\log \left[(1.6)^{1-x^2}-(0.625)^{6(1+x)}\right]\right| \in \mathbb{R}\right\} \\
=(-\infty,-1) \cup(7, \infty)
\end{gathered}
$$

Topic: Mathematics, Functions, JEE Main
The value of x satisfying the equation sinâ¡xcosâ¡x+2+tan2â¡x+cot2â¡x=3

Options:
A belongs to 0,π3
B belongs to π3,π2
C belongs to 3π4,π
D does not exist

The correct answer is:
does not exist

Solution:
sin⁡xcos⁡x+2+tan2⁡x+cot2⁡x=3
⇒sin⁡xcos⁡x+tan⁡x+cot⁡x2=3
⇒sin⁡xcos⁡x+tan⁡x+cot⁡x=3
⇒12sin⁡2x+tan⁡x+cot⁡x=3
⇒tan⁡x1+tan2⁡x+1+tan2⁡xtan⁡x=3
Let t=tan⁡x1+tan2⁡x
t+1t=3⇒t2-3t+1=0⇒D<0
So, no real values of t
Hence, x does not exists.

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of \(x\) that satisfies the equation \(\int_{\sqrt{2}}^x \frac{d t}{|t| \sqrt{t^2-1}}=\frac{\pi}{12}\) is

Options:
A 1
B 0
C \(-\sqrt{2}\)
D 2

The correct answer is:
2

Solution:
\(\begin{aligned}
& \int_{\sqrt{2}}^x \frac{d t}{|t| \sqrt{t^2-1}}=\frac{\pi}{12} \\
& \Rightarrow \quad\left[\sec ^{-1} t\right]_{\sqrt{2}}^x=\frac{\pi}{12} \Rightarrow \sec ^{-1} x-\sec ^{-1} \sqrt{2}=\frac{\pi}{12} \\
& \Rightarrow \quad \sec ^{-1} x=\frac{\pi}{12}+\frac{\pi}{4}=\frac{\pi}{3} \\
& \Rightarrow \quad x=\sec \left(\frac{\pi}{3}\right)=2
\end{aligned}\)

Topic: Mathematics, Definite Integration, JEE Main
The value of $x$, where $x>0$ and $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$ is

Options:
A $\sqrt{5}$
B $\frac{\sqrt{5}}{3}$
C $1$
D $2 / 3$

The correct answer is:
$\frac{\sqrt{5}}{3}$

Solution:
Given, $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$
$\begin{array}{r}\Rightarrow \tan \left(\tan ^{-1} \frac{\sqrt{1-x^2}}{x}\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{1+2^2}}\right) \\ \left(\because \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^2}}\right)\end{array}$
$\begin{aligned} \Rightarrow & & \frac{\sqrt{1-x^2}}{x} & =\frac{2}{\sqrt{5}} \\ & \Rightarrow & \sqrt{5} \sqrt{1-x^2} & =2 x \\ & \Rightarrow & 4 x^2 & =5\left(1-x^2\right) \\ & \Rightarrow & x^2 & =\frac{5}{9} \\ & \Rightarrow & x & =\frac{\sqrt{5}}{3}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of $x$ which satisfies the equation $\tan ^{-1} x=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$ is

Options:
A $3$
B $-3$
C $\frac{1}{3}$
D $-\frac{1}{3}$

The correct answer is:
$3$

Solution:
$\begin{aligned}
& \text { Given: } \tan ^{-1} \mathrm{x}=\sin ^{-1}\left(\frac{3}{\sqrt{10}} \right) ....(i) \\
& \text { Let } \mathrm{A}=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right) \\
& \Rightarrow \sin A=\frac{3}{\sqrt{10}} \\
& \Rightarrow \cos A=\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2} \\
& {\left[\sin ^2 A+\cos ^2 A=1\right]} \\
& \Rightarrow \cos A=\sqrt{1-\frac{9}{10}} \\
& \Rightarrow \cos A=\sqrt{\frac{10-9}{10}} \\
& \Rightarrow \cos A=\frac{1}{\sqrt{10}} \\
& \tan A=\frac{\sin A}{\cos A}=\frac{3 / \sqrt{10}}{1 / \sqrt{10}}=3 \\
& A=\tan ^{-1} 3 \\
& \therefore \sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)=\tan ^{-1} 3 ....(ii)
\end{aligned}$
From equations (ii) and (ii)
$\begin{gathered}
\tan ^{-1}(x)=\tan ^{-1}(3)\\ x=3
\end{gathered}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of $x$ which satisfies the equation $\tan ^{-1} x=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$ is

Options:
A $3$
B $-3$
C $\frac{1}{3}$
D $-\frac{1}{3}$

The correct answer is:
$3$

Solution:
Given,
$\begin{aligned}
& \quad \tan ^{-1} x=\sin ^{-1}\left[\frac{3}{\sqrt{10}}\right] \\
& \text { Given, } \\
& \Rightarrow x=\tan \left\{\sin ^{-1}\left[\frac{3}{\sqrt{10}}\right]\right\}=\tan \left\{\tan ^{-1} 3\right\} \\
& \Rightarrow x=3 .
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of x; âxâR which satisfy the equation x - 1 x 2 - 4 x + 3 + 2 x 2 + 3 x - 5 = 0 is

Options:

The correct answer is:
1

Solution:
x - 1 x 2 - 4 x + 3 + 2 x 2 + 3 x - 5 = 0
Case I: when x 2 - 4 x + 3 ≥ 0 , then x ∈ ( - ∞ , 1 ] ∪ [ 3 , ∞ )
x - 1 x 2 - 3 x - x + 3 + 2 x 2 + 5 x - 2 x - 5 = 0
x - 1 x 2 - 3 x - x + 3 + 2 x + 5 = 0
x - 1 x 2 - 2 x + 8 = 0 (the value of quadratic can't be zero since its D<0)
∴x-1=0
⇒ x = 1
Case II: when x 2 - 4 x + 3 < 0 , then x ∈ 1 , 3
x - 1 - x 2 + 4 x - 3 + 2 x + 5 = 0
x - 1 ≠ 0
∴ - x 2 + 6 x + 2 = 0
x 2 - 6 x - 2 = 0
x = 3 ± 1 1
both values will not satisfy
so x=1

Topic: Mathematics, Quadratic Equation, JEE Main
The value of \( x>1 \) satisfying the equation \( \int_{1}^{x} t \log t d t=\frac{1}{4} \), is

Options:
A \( \sqrt{e} \)
B \( e^{\frac{3}{2}} \)
C \( e^{2} \)
D \( 2 e-1 \)

The correct answer is:
\( \sqrt{e} \)

Solution:
Consider that, I=â«1xtlogâ¡tÂ dt
=logâ¡t⋅t221x-â«1x1tÂ ⋅t22dt
=x22logâ¡x-12t221x
=x22logâ¡x-12x22-12
âÂ Â Â Â Â 14=x22logâ¡x-14x2-1
âÂ Â Â 12x2logâ¡x-14x2=0 (as x>1 )
âÂ Â Â Â Â Â x22logâ¡x-1=0
âÂ Â Â 2logâ¡x-1=0
âÂ Â Â Â logâ¡x=12
âÂ Â Â Â Â x=e12
âÂ Â Â Â Â Â x=e

Topic: Mathematics, Definite Integration, JEE Main
The values of ∫-π2π2sin2⁡x1+2xdx is

Options:
A π4
B π8
C π2
D 4π

The correct answer is:
π4

Solution:
I= ∫-π2π2sin2x1+2x dx ...(i)
As ∫abf(x)dx=∫abf(a+b−x)dx
I=∫-π2π2sin2⁡-x1+2-xdx
I= ∫-π2π22xsin2⁡x1+2xdx ...(ii)
By adding Equations (i) and (ii),
2I= ∫-π2π21+2xsin2⁡x1+2xdx
2I=2∫0π2sin2⁡xdx

I=∫0π2sin2⁡xdx ...iii

I=∫0π2sin2⁡π2-xdx
I=∫0π2cos2⁡xdx ...iv

On adding iii & iv, we get
2I= ∫0π2sin2⁡x+cos2⁡xdx
2I= ∫0π21 dx

2I= x0π2
2I=π2-0
I=π2×12=π4

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2018 (08 Apr)
The values of a and b, for which the system of equations

2x+3y+6z=8

x+2y+az=5

3x+5y+9z=b

has no solution, are :

Options:
A a=3,b≠13
B a≠3,b≠13
C a≠3,b=3
D a=3,b=13

The correct answer is:
a=3,b≠13

Solution:
The given equation is

2x+3y+6z=8

x+2y+az=5

3x+5y+9z=b

So, D=23612a359

=218-5a-39-3a+65-6

=36-10a-27+9a+30-36

=3-a

Dz=23812535 b

=22b-10-3b-15+85-6

=4b-20-3b+15+40-48

=b-13

The system of equation has no solution if D=0 & Dz≠0

So, If a=3, b≠13, then the system of equation has no solution.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The values of ' $a$ ' for which one root of the equation $x^2-(a+1) x+a^2+a-8=0$ exceeds 2 and the other is lesser than 2 , are given by :

Options:
A
$3 < a < 10$
B
$a \geq 10$

C
$-2 < a < 3$
D
$a \leq-2$

The correct answer is:
$-2 < a < 3$

Solution:
$$
x^2-(a+1) x+a^2+a-8=0
$$
Since roots are different, therefore $\mathrm{D}>0$
$$
\begin{aligned}
& \Rightarrow(a+1)^2-4\left(a^2+a-8\right)>0 \\
& \Rightarrow(a-3)(3 a+1) < 0
\end{aligned}
$$
There are two cases arises.
Case I. $a-3>0$ and $3 a+1 < 0$
$\Rightarrow a>3$ and $a < -\frac{11}{3}$
Hence, no solution in this case
Case II : $a-3 < 0$ and $3 a+11>0$
$\Rightarrow a < 3$ and $a>-\frac{11}{3}$
$\therefore \quad-\frac{11}{3} < a < 3 \Rightarrow-2 < a < 3$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2013 (09 Apr Online)
The values of $a$, for which the points $A, B, C$ with position vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $a \hat{i}-3 \hat{j}+\hat{k}$ respectively are the vertices of a right-angled triangle with $C=\frac{\pi}{2}$ are

Options:
A
2 and 1

B
−2 and −1

C
−2 and 1

D
2 and −1

The correct answer is:
2 and 1

Solution:
$\Rightarrow \overrightarrow{B A}=\hat{i}-2 \hat{j}+6 \hat{k}$
$\overrightarrow{C A}=(2-a) \hat{i}+2 \hat{j}$
$\overrightarrow{C B}=(1-a) \hat{i}-6 \hat{k}$
$\overrightarrow{C A} \cdot \overrightarrow{C B}=0 \Rightarrow(2-a)(1-a)=0$
$\Rightarrow a=2,1$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2006
The values of a for which the two points $(1, a, 1)$ and $(-3,0, a)$ lie on the opposite sides of the plane $3 x+4 y-12 z+13=0$, satisfy

Options:
A
$0 < a < \frac{1}{3}$
B
$-1 < a < 0$
C
$a < -1$ or $a < \frac{1}{3}$
D
$a=0$

The correct answer is:
$a=0$

Solution:
Given equation of plane is
$$
3 x+4 y-12 z+13=0
$$
$(1, a, 1)$ and $(-3,0, a)$ satisfy the equation of plane.
$\therefore$ We have
$3+4(a)-12+13=0$ and
$3(-3)-12(a)+13=0$
$\Rightarrow 4+4 a=0$ and $4-12 a=0$
$\Rightarrow a=-1$ and $a=\frac{1}{3}$
Since, $(1, a, 1)$ and $(-3,0, a)$ lie on the opposite sides of the plane $\therefore a=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (07 May Online)
The values of a function $f(x)$ at different values of $x$ are as follows
Then, the approximate area (in square units) bounded by the curve $y=f(x)$ and $x$-axis between $x=0$ and 5 , using the Trapezoidal rule, is

Options:
A 50
B 75
C 52.5
D 62.5

The correct answer is:
52.5

Solution:
$h=$ difference of two values of $x$
Take value of $f(x)$ as $\left(y_0, y_1, y_2, \ldots, y_5\right)$
Then by Trapezoidal rule
Now, $\int_{x_0}^{x_0+n h} f(x) d x$
$=\frac{h}{2}\left[\left(y_0+y_5\right)+2\left(y_1+y_2+y_3+y_4\right)\right]$
$=\frac{1}{2}[(2+27)+2(3+6+11+18)]$
$=\frac{1}{2}[29+2 \cdot 38]=\frac{1}{2}(29+76)$
$=\frac{1}{2} \times 105=52.5$
Approximate area $=52.5 \mathrm{sq}$ unit

Topic: Mathematics, Area Under Curves, JEE Main
The values of λ and μ for which the system of linear equations x+y+z=2, x+2 y+3 z=5, x+3y+λz=μ has infinitely many solutions, are respectively

Options:
A 6 and 8
B 5 and 7
C 5 and 8
D 4 and 9

The correct answer is:
5 and 8

Solution:
For infinite number of solutions, all determinants D, D1, D2 and D3 must be zero and planes represented by these equations should not be parallel. So, we have,

D=11112313λ=0 ⇒λ=5

D1=211523μ3λ=0

D2=1211531μλ=0

And D3=11212513μ=0 ⇒μ=8

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The values of λ and μ such that the system of equations x+y+z=6, 3x+5y+5z=26 and x+2y+λz=μ has no solution, are:

Options:
A λ=3,μ=5
B λ=3,μ≠10
C λ≠2,μ=10
D λ=2,μ≠10

The correct answer is:
λ=2,μ≠10

Solution:
For the system of equations a1x+b1y+c1z=d1, a2x+b2y+c2z=d2 and a3x+b3y+c3z=d3,

We have D=a1b1c1a2b2c2a3b3c3, Dx=d1b1c1d2b2c2d3b3c3, Dy=a1d1c1a2d2c2a3d3c3 and Dz=a1b1d1a2b2d2a3b3d3.

And, the system has no solution, if D=0 and atleast one of Dx, Dy & Dz is non-zero.

Thus, for the given system of equations,

x+y+z=6, 3x+5y+5z=26 and x+2y+λz=μ

We have D=11135512λ=0

⇒15λ-10-13λ-5+16-5=0

⇒5λ-10-3λ+5+1=0

⇒2λ=4

⇒λ=2.

And, Dz=116352612μ≠0

⇒15μ-52-13μ-26+66-5≠0

⇒5μ-52-3μ+26+6≠0

⇒2μ≠20

⇒μ≠10.

∴ For no solution λ=2 and μ≠10.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The values of conductivity of some materials at 298.15K in Sm−1 are 2.1×103, 1.0×10−16, 1.2×10,3.91, 1.5×10−2, 1×10−7, 1.0×103. The number of conductors among the materials is _____.

Options:

The correct answer is:
4

Solution:
Molar conductivity is the conductance property of a solution containing one mole of the electrolyte, or it is a function of the ionic strength of a solution or the concentration of salt.

Λm = K / C

Where K is the specific conductivity and

C is the concentration in mole per litre.

Conductivity Sm−1

2.1×1031.2×103.911×103 conductors at 298.15K

1×10−16 Insulator at 298.15K

1.5×10−21×10−7 Semiconductor at 298.15K

Therefore number of conductors is 4.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (31 Jan Shift 2)