datavorous/entrance-exam-dataset huggingface dataset:
If $\lim _{x \rightarrow 0}\left[1+x \log \left(1+b^2\right)\right]^{1 / x}=2 b \sin ^2 \theta b>0$ and $\theta \in(-\pi, \pi]$, then the value of $\theta$ is

Options:
A
$\pm \frac{\pi}{4}$

B
$\pm \frac{\pi}{3}$

C
$\pm \frac{\pi}{6}$

D
$\pm \frac{\pi}{2}$

The correct answer is:
$\pm \frac{\pi}{2}$

Solution:
Here, $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+b^2\right)\right\}^{1 / x}$
$$
\begin{aligned}
& \text { Given, } \quad\left[1^{\infty} \text { from }\right] \\
& \Rightarrow \quad e^{\lim _{x \rightarrow 0}\left\{x \log \left(1+b^2\right)\right\} \cdot \frac{1}{x}} \\
& \Rightarrow \quad e^{\log \left(1+b^2\right)}=(1+b)^2 \\
& \lim _{x \rightarrow 0}\left\{1+x \log \left(1+b^2\right)\right\}^{1 / x}=2 b \sin ^2 \theta \\
& \Rightarrow \quad\left(1+b^2\right)=2 b \sin ^2 \theta \\
& \therefore \quad \sin ^2 \theta=\frac{1+b^2}{2 b} \quad \ldots \text { (ii) }
\end{aligned}
$$
By $A M \geq G M$,
$$
\frac{b+\frac{1}{b}}{2} \geq\left(b \cdot \frac{1}{b}\right)^{1 / 2} \Rightarrow \frac{b^2+1}{2 b} \geq 1
$$
From Eqs. (ii) and (iii), we get $\sin ^2 \theta=1$
$$
\Rightarrow \quad \theta=\pm \frac{\pi}{2} \text { as } \theta \in(-\pi, \pi]
$$

Topic: Mathematics, Limits, JEE Main
If limx→03+αsinx+βcosx+loge(1-x)3tan2x=13, then 2α-β is equal to :

Options:
A 2
B 7
C 5
D 1

The correct answer is:
5

Solution:
Given,

limx→03+αsinx+βcosx+loge(1-x)3tan2x=13

⇒limx→03+αx-x33!+…+β1-x22!+x44!…+-x-x22-x33…3tan2x=13

⇒limx→0(3+β)+(α-1)x+-12-β2x2+….3x2×x2tan2x=13

The coefficient of all the powers of x except x2 is 0.

⇒β+3=0,α-1=0 and -12-β23=13

⇒β=-3,α=1

⇒2α-β=2+3=5

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (27 Jan Shift 2)
If limx→0aex-bcosx+ce-xxsinx=2, then a+b+c is equal to ________.

Options:

The correct answer is:
4

Solution:
Given limx→0aex-bcosx+ce-xxsinx=2

⇒limx→0a1+x+x22!…-b1-x22!+…+c1-x+x22!xsinxxx=2

⇒limx→0a1+x+x22!…-b1-x22!+…+c1-x+x22!x2=2 ∵limx→0sinxx=1

For the given limit to exist numerator should be 0, when x→0.

Then, a-b+c=0 ...(i)

Applying L'Hospital's rule, we get

limx→0a1+2x2!…-b-2x2!+…+c-1+2x2!2x=2

Again, for the given limit to exist numerator should be 0, when x→0.

Then, a-c=0 ...ii

Applying, L'Hospital's rule, we get

limx→0a1+…-b-1+…+c12=2

⇒a+b+c2=2

⇒a+b+c=4

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (16 Mar Shift 1)
If limxâ0((a-n)nx-tan x)sin nxx2=0, where n is a non zero real number, then a is equal to

Options:
A 0
B n + 1 n
C n
D n + 1 n

The correct answer is:
n + 1 n

Solution:
limx→0((a-n)nx-tan x)sin nxx2=0
⇒ lim x → 0 a - n n - tan x x sin nx x = 0
⇒ a - n n - 1 n = 0
⇒ a - n n - 1 = 0 ∵ n ≠ 0
∴ a = n + 1 n

Topic: Mathematics, Limits, JEE Main
If limx→0ax2ex−bloge1+x+cxe−xx2sinx=1, then 16a2+b2+c2 is equal to ______.

Options:

The correct answer is:
81

Solution:
Given:

limx→0ax2ex-bloge1+x+cxe-xx2sinx=1

⇒limx→0ax2ex-bloge1+x+cxe-xx2·x·sinxx=1

⇒limx→0ax21+x1!+x22!+...-bx-x22+x33+...+cx1-x1!+x22!+...x2×x=1

Coefficient of x=0

⇒-b+c=0 ...iii

Coefficient of x2=0

⇒a+b2-c=0 ...ii

⇒a-c2=0

⇒a=c2

Coefficient of x3=1

⇒a-b3+c2=1 ...iii

⇒c2-c3+c2=1

⇒2c3=1

⇒c=32=b

⇒a=34

⇒16a2+b2+c2=16916+94×2

⇒16a2+b2+c2=81

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (31 Jan Shift 2)
If limx→0ax-e4x-1axe4x-1 exists and is equal to b, then the value of a-2b is ___ .

Options:

The correct answer is:
5

Solution:
limx→0ax-e4x-1axe4x-1 00

=limx→0ax-e4x-1ax·4x Use limx→0e4x-14x=1

Apply L'Hospital Rule

=limx→0a-4e4x8ax ( a-40 form )

limit exists only when a-4=0⇒a=4

=limx→04-4e4x32x

=limx→01-e4x8x 00

=limx→0-e4x·48=-12⇒b=-12

a-2b=4-2-12

=5

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If $\lim _{x \rightarrow 0} k x \operatorname{cosec} x=\lim _{x \rightarrow 0} x \operatorname{coseckx}$, then $k=$

Options:
A $1$
B $-1$
C $\pm 1$
D $\pm 2$

The correct answer is:
$\pm 1$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} k x \operatorname{cosec} x=\lim _{x \rightarrow 0} x \operatorname{cosec} k x \\ & \Rightarrow k \cdot \lim _{x \rightarrow 0} \frac{x}{\sin x}=\frac{1}{k} \lim _{x \rightarrow 0} \frac{k x}{\sin k x} \Rightarrow k=\frac{1}{k} \Rightarrow k= \pm 1\end{aligned}$

Topic: Mathematics, Limits, JEE Main
If limx→0sin-1x-tan-1x3x3 is equal to L, then the value of (6L+1) is

Options:
A 16
B 12
C 6
D 2

The correct answer is:
2

Solution:
Given limx→0sin-1x-tan-1x3x3

⇒limx→0x+x33!…-x-x33…3x3

⇒limx→01+3x23!…-1-3x23…9x2 (Use L'Hospital rule 00 form)

⇒limx→06x3!+…--6x3+…18x (Use L'Hospital rule 00 form)

⇒limx→063!+…--63+…18=1+218=16

Given that, limx→0sin-1x-tan-1x3x3=L

∴ 16=L⇒6L=1

So, 6L+1=2

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (18 Mar Shift 1)
If limx→0⁡sin-1⁡x-tan-1⁡xx3 equals L, then the value of ( 4L+1 ) is

Options:

The correct answer is:
3

Solution:
Let L= limxâ0â¡sin-1â¡x-tan-1â¡xx300form
=limxâ0â¡11-x2-11+x23x2Lâ² Hospital rule
=13limxâ0â¡1x21+x2-1-x21+x21-x2
=13limxâ0â¡1x21+x22-1-x21+x21-x2. 11+x2+1-x2
= 1 3 lim xâ0 ( 3+ x 2 ) ( 1+ x 2 ) 1â x 2 1 [ ( 1+ x 2 )+ 1â x 2 ]
=1332=12

Topic: Mathematics, Limits, JEE Main
If limx→0⁡sin⁡2x-asin⁡xx33=L exists finitely, then the absolute value of L is equal to

Options:

The correct answer is:
27

Solution:
L=limx→027sin⁡x2cos⁡x-ax⋅x2=limx→0sin⁡xx2cos⁡x-ax2×27
This limit to exists finitely
limx→0⁡2cos⁡x-ax2= finite
∴ It must be 00 form
∴2cos⁡0-a=0⇒a=2
⇒L=limx→0⁡sin⁡xx×2cos⁡x-1x2×27
⇒L=1×2×-12×27=-27

Topic: Mathematics, Limits, JEE Main
If $\alpha=\lim _{x \rightarrow 0^{+}}\left(\frac{\mathrm{e}^{\sqrt{\tan x}}-\mathrm{e}^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)$ and $\beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $a x^2+b x-\sqrt{\mathrm{e}}=0$, then $12 \log _{\mathrm{e}}(\mathrm{a}+\mathrm{b})$ is equal to__________

Options:

The correct answer is:
6

Solution:
$\begin{aligned} & \alpha=\lim _{x \rightarrow 0^{+}} e^{\sqrt{x}} \frac{\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{\sqrt{\tan x}-\sqrt{x}} \\ & =1 \\ & \beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x} \\ & =e^{1 / 2} \\ & x^2-(1+\sqrt{e})+\sqrt{e}=0 \\ & a x^2+b x-\sqrt{e}=0 \\ & \text { On comparing } \\ & a=-1, b=\sqrt{e}+1 \\ & 12 \ln (a+b)=12 \times \frac{1}{2}=6\end{aligned}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (08 Apr Shift 2)
If $\lim _{x \rightarrow 0} \frac{\log (3+x)-\log (3-x)}{x}=k$, the value of $k$ is

Options:
A
$-\frac{2}{3}$

B
0

C
$-\frac{1}{3}$

D
$\frac{2}{3}$

The correct answer is:
$\frac{2}{3}$

Solution:
$\operatorname{Lim}_{x \rightarrow 0} \frac{\log (3+x)-\log (3-x)}{x}=K$ (by L'Hospital rule)
$\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{1}{3+x}-\frac{-1}{3-x}}{1}=K \quad \therefore \frac{2}{3}=\mathrm{K}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2003
If limx→∞⁡1+ax-4x22x=e3, then a is equal to

Options:
A 2
B 32
C 12
D 23

The correct answer is:
32

Solution:
limx→∞1+ax-4x22x (1∞ form)
=elimx→∞⁡1+ax-4x2-12x
=elimx→∞⁡2a-8x=e2a
∴ e2a=e3

⇒2a=3

⇒ a=32

Topic: Mathematics, Limits, JEE Main, JEE Main 2016 (09 Apr Online)
If limx→1sin3x2-4x+1-x2+12x3-7x2+ax+b=-2, then the value of a-b is equal to

Options:

The correct answer is:
11

Solution:
Given limx→1sin3x2−4x+1−x2+12x3−7x2+ax+b=−2 ...i

Putting x=1 in limit we get

⇒limx→1sin0-1+12-7+a+b=-2 ⇒limx→10a+b-5=-2

For limit to exist a+b-5 should be zero

so a+b-5=0 ⇒ a+b=5 ...ii

Now Using L-Hospital in equation i we get

limx→1cos3x2-4x+16x-4-2x6x2-14x+a=-2

Again putting x=1 we get

⇒limx→1cos0×6-4-26-14+a=-2

⇒limx→10a-8=-2

Again for limit to exist a-8=0⇒a=8

So from equation ii we get

8+b=5⇒b=-3

So a-b=8--3=11

Topic: Mathematics, Limits, JEE Main, JEE Main 2022 (28 Jun Shift 2)
If limx→1x2-ax+bx-1=5, then a+b is equal to:

Options:
A 1
B 5
C –4
D –7

The correct answer is:
–7

Solution:
Given L=limx→1⁡x2-ax+bx-1=5

As x→1, the denominator becomes zero, hence for finite limit, the numerator must also approach to zero as x→1

⇒1-a+b=0

⇒b=a-1 …(1)

Now L=limx→1x2-ax+bx-1 00 form

By using L'Hospital Rule,

⇒L=limx→12x-a1

⇒L=limx→12x-a1=5

⇒2-a=5 ⇒a=-3

From the equation 1, we get

⇒b=a-1=-4

⇒a+b=-7.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (10 Apr Shift 2)
If $\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2 x}=e^2$, then the values of $a$ and $b$, are

Options:
A
$a \in \underline{\underline{R}}, \mathrm{~b} \in \underline{\underline{=}}$

B
$a=1, b \in \underline{R}$

C
$a \in \underline{\underline{R}}, b=2$

D
$a=1$ and $b=2$

The correct answer is:
$a=1, b \in \underline{R}$

Solution:

Topic: Mathematics, Limits, JEE Main, JEE Main 2004
If limx→1x+x2+x3+...+xn-nx-1=820, n∈N then the value of n is equal to....

Options:

The correct answer is:
40

Solution:
limx→1x+x2+x3+….+xn-nx-1=820 00form

Applying L'Hospital rule

limx→11+2x+3x2+…..+nxn-11=820

⇒1+2+3+……+n=820

⇒n(n+1)2=820

⇒n2+n=1640

⇒n2+n-1640=0

⇒n=40, -41

Since n is natural number ⇒n=40.

Topic: Mathematics, Limits, JEE Main, JEE Main 2020 (02 Sep Shift 1)
If $\lim _{x \rightarrow 2} \frac{\tan \left(x-2\left\{x^2+k+2 x-2 k\right\}\right.}{x^2-4 x+4}=5$, then $\mathrm{k}$ is equal to:

Options:
A
0

B
1

C
2

D
3

The correct answer is:
3

Solution:
$$
\begin{aligned}
\text { } & \lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+(k-2) x-2 k\right\}}{x^2-4 x+4}=5 \\
\Rightarrow & \lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+k x-2 x-2 k\right\}}{(x-2)^2}=5 \\
\Rightarrow & \lim _{x \rightarrow 2} \frac{\tan (x-2)\{x(x-2)+k(x-2)\}}{(x-2) \times(x-2)}=5 \\
\Rightarrow \lim _{x \rightarrow 2}\left(\frac{\tan (x-2)}{(x-2)}\right) \times \lim _{x \rightarrow 2}\left(\frac{(k+x)(x-2)}{(x-2)}\right)=5 \\
\Rightarrow & 1 \times \lim _{x \rightarrow 2}(k+x)=5 \\
&\left\{\because \lim _{h \rightarrow 0} \frac{\tan h}{h}=1\right\} \\
& \text { or } k+2=5 \\
\Rightarrow k=3
\end{aligned}
$$

Topic: Mathematics, Limits, JEE Main, JEE Main 2014 (11 Apr Online)
If $\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80$, where $n$ is a positive integer, then $n=$

Options:
A $3$
B $5$
C $2$
D None of these

The correct answer is:
$5$

Solution:
$\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=n \cdot 2^{n-1} \Rightarrow n \cdot 2^{n-1}=80 \Rightarrow n=5$ (Using L-Hospital's rule)

Topic: Mathematics, Limits, JEE Main
If $\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80$, where $n$ is a positive integer, then $n=$

Options:
A $3$
B $5$
C $2$
D None of these

The correct answer is:
$5$

Solution:
$\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=n \cdot 2^{n-1} \Rightarrow n \cdot 2^{n-1}=80 \Rightarrow n=5$ (using L-Hospital's rule)

Topic: Mathematics, Limits, JEE Main
If α=limx→π/4tan3x-tanxcosx+π4 and β=limx→0cosxcotx are the roots of the equation, ax2+bx-4=0, then the ordered pair a, b is :

Options:
A -1, 3
B 1, -3
C 1, 3
D -1, -3

The correct answer is:
1, 3

Solution:
If, β=limx→0cosxcotx (1∞ form)

β=e lim x→0cotxcosx-1

β=e limx→0-cosxsinx1-1+2sin2x2

β=e limx→0 -cosx2sinx2.cosx21-1+2sin2x2

β=e limx→0 -cosx.sinx2cosx2

β=e0

β=1
α=limx→π/4tan3x-tanxcosx+π4

=limx→π/4tanx(tanx+1)(tanx-1)cosx+π4

α=limx→π/4tanx×limx→π/4tanx+1×limx→π/4(tanx-1)cosx+π4

α=1×1+1×limx→π/4(tanx-1)cosx+π4

=2limx→π/4tanx-1cosx+π4

=2limx→π/4sinx-cosxcosx.cosx+π4

=-22limx→π/412cosx-12sinxcosx.cosx+π4

α=-22limx→π/4cosx+π4cosx.cosx+π4

α=-4

So, the equation whose roots are α and β is

x2+3x-4=0

Hence, a=1, b=3

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (31 Aug Shift 2)
If $\lim _{x \rightarrow a} \frac{x^9+a^9}{x+a}=9$, then $a=$

Options:
A $9^{1 / 8}$
B $\pm 2$
C $\pm 3$
D None of these

The correct answer is:
$9^{1 / 8}$

Solution:
$\lim _{x \rightarrow a} \frac{x^9+a^9}{x+a}=9 \Rightarrow \frac{2 a^9}{2 a}=9 \Rightarrow a^8=9 \Rightarrow a=9^{1 / 8}$

Topic: Mathematics, Limits, JEE Main
If \(\lim _{x \rightarrow 0} \frac{[(a-n) n x-\tan x] \sin n x}{x^2}=0,(n \neq 0)\)
then the minimum possible positive value of \(a\) is

Options:
A 0
B -2
C 2
D 1

The correct answer is:
2

Solution:
\(\begin{aligned}
& \lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0, n \neq 0 \\
& \Rightarrow \quad \lim _{x \rightarrow 0}\left(\frac{(a-n) n x}{x}-\frac{\tan x}{x}\right) \frac{\sin (n x)}{x}=0 \\
& \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin n x}{n x}=n\right] \\
& \Rightarrow \quad n\left(a n-n^2-1\right)=0 \Rightarrow \quad a=n+\frac{1}{n}, n \neq 0 \\
& \because \quad \frac{n+\frac{1}{n}}{2} \geq \sqrt{n \frac{1}{n}} \text { (by AM } \geq \text { GM) } \\
& \Rightarrow \quad \frac{a}{2} \geq 1 \\
& \Rightarrow \quad a \geq 2
\end{aligned}\)
\(\therefore\) Minimum possible positive value of \(a\) is 2. Hence, option (c) is correct.

Topic: Mathematics, Limits, JEE Main
If \( \lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{(x-1)^{2}}=2 \), then \( (a, b, c) \)

Options:
A \( (2,-4,2) \)
B \( (2,4,2) \)
C \( (2,4,-2) \)
D \( (2,-4,-2) \)

The correct answer is:
\( (2,-4,2) \)

Solution:
Given, limx→1⁡ax2+bx+cx-12=2
This limit will exist, if
ax2+bx+c=2x-12
⇒ ax2+bx+c=2x2-4x+2
⇒ a=2, b=-4, c=2

Topic: Mathematics, Limits, JEE Main
If limx→∞x2-x+1-ax=b, then the ordered pair (a, b) is:

Options:
A 1,-12
B -1,12
C -1,-12
D 1,12

The correct answer is:
1,-12

Solution:
Given that:limx→∞ x2−x+1−ax=b

⇒limx→∞ x2−x+1−ax×x2−x+1+axx2−x+1+ax=b

⇒limx→∞ x2−x+1−(ax)2x2−x+1+ax=b

⇒ Limit exist only If a2=1, because x2 must not be there in the numerator

∴a=1,−1

⇒limx→∞ −x+1x2−x+1+ax=b

⇒limx→∞ −1+1x1−1x+1x2+a=b

⇒−11+a=b

Here, a≠−1

So a=1

b=−12

(a,b)≡1,−12

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (27 Aug Shift 2)
If $\lim _{x \rightarrow} x \sin \left(\frac{1}{x}\right)=A$ and $\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=B$,

then which one of the following is correct?

Options:
A $A=1$ and $B=0$
B $\quad A=0$ and $B=1$
C $A=0$ and $B=0$
D $A=1$ and $B=1$

The correct answer is:
$A=1$ and $B=0$

Solution:
As given,

$$

\mathrm{A}=\lim _{\mathrm{x} \rightarrow \infty} \mathrm{x} \sin \left(\frac{1}{\mathrm{x}}\right)=\lim _{\mathrm{x} \rightarrow \infty} \frac{\sin \left(\frac{1}{\mathrm{x}}\right)}{\left(\frac{1}{\mathrm{x}}\right)}

$$

Let $\mathrm{t}=\frac{1}{\mathrm{x}}$ when $\mathrm{x} \rightarrow \alpha, \mathrm{t} \rightarrow 0$

$$

\begin{aligned}

\Rightarrow \mathrm{A} &=\lim _{t \rightarrow \infty} \frac{\sin \mathrm{t}}{\mathrm{t}}=1 \\

&\left[\because \lim _{\mathrm{t} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}}=1\right]

\end{aligned}

$$

and $\mathrm{B}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{x} \sin \left(\frac{1}{\mathrm{x}}\right)$

$$

\Rightarrow \mathrm{B}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{x} \cdot \lim _{\mathrm{x} \rightarrow 0} \sin \left(\frac{1}{\mathrm{x}}\right)

$$

$$

\Rightarrow \mathrm{B}=0

$$

$\therefore \quad A=1$ and $B=0$ is correct

Topic: Mathematics, Limits, JEE Main
If linear density of a rod of length 3 m varies as λ=2+x, then the position of the centre of gravity of the rod is

Options:
A 73m
B 127m
C 107m
D 97m

The correct answer is:
127m

Solution:
Let rod is placed along x-axis. Mass of element PQ of length dx situated at x=x is
dm=λ dx=2+x dx
The CM of the element has coordinates (x, 0, 0).
Therefore, x-coordinates of CM of the rod will be
xCM=∫03xdm∫03dm
=∫03x2+xdx∫032+xdx
=∫032x+x2dx∫032+xdx
=2x22+x33032x+x2203
=32+3332×3+322=9+96+9/2
=18×221=127m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
If lines $4 x+3 y=1, y=x+5$ and $5 y+b x=3$ are concurrent, then $b$ equals

Options:
A 1
B 3
C 6
D 0

The correct answer is:
6

Solution:
Lines are concurrent, if $\left|\begin{array}{ccc}4 & 3 & -1 \\ 1 & -1 & 5 \\ b & 5 & -3\end{array}\right|=0$
$\begin{aligned}
& \Rightarrow 4(3-25)-3(-3-5 b)-1(5+b)=0 \\
& \Rightarrow-88+9+15 b-5-b=0 \Rightarrow-84+14 b=0 \\
& \therefore b=6
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
If liquids $\mathrm{A}$ and $\mathrm{B}$ form an ideal solution

Options:
A
the entropy of mixing is zero

B
the free energy of mixing is zero

C
the free energy as well as the entropy of mixing are each zero

D
the enthalpy of mixing is zero

The correct answer is:
the enthalpy of mixing is zero

Solution:
No Solution Available

Topic: Chemistry, Solutions, JEE Main, JEE Main 2003
If lnâ¡2x2-5, lnâ¡x2-1 and lnâ¡x2-3 are the first three terms of an arithmetic progression, then its fourth term is

Options:
A ln⁡8-ln⁡3
B ln⁡3-ln8
C ln⁡24
D 2ln⁡6

The correct answer is:
ln⁡8-ln⁡3

Solution:
Condition x2>3 for all logs to be defined
2ln⁡x2-1=ln⁡2x2-5+ln⁡x2-3
⇒x2-12=2x2-5x2-3
⇒x4+1-2x2=2x4-6x2-5x2+15
⇒x4-9x+14=0 ⇒x2=2,7
Since, x2>3 ⇒x2=7
So the numbers are →ln⁡9, ln⁡6, ln⁡4
⇒d=ln⁡6-ln⁡9=ln⁡23
⇒ fourth term =ln⁡4+ln⁡23=ln⁡83

Topic: Mathematics, Sequences and Series, JEE Main
If $\log (1+x)-\frac{2 x}{2+x}$ is increasing, then

Options:
A $0 < x < \infty$
B $-\infty < x < 0$
C $-\infty < x < \infty$
D $-1 < x < 2$

The correct answer is:
$-\infty < x < \infty$

Solution:
Let $f(x)=\log (1+x)-\frac{2 x}{2+x^2}$
$$
f^{\prime}(x)=\frac{1}{(1+x)}-\frac{2}{2+x}+\frac{2 x}{(2+x)^2}
$$
For increasing function, $f^{\prime}(x)>0$
$$
\begin{aligned}
\Rightarrow \quad(2+x)^2-2(2+x)(1+x)+2 x(1+x) & >0 \\
\Rightarrow 4+x^2+4 x-4-6 x-2 x^2+2 x+2 x^2 & >0 \\
x^2 & >0
\end{aligned}
$$
This shows $x$ lies between $-\infty$ to $\infty$.

Topic: Mathematics, Application of Derivatives, JEE Main
If $\log _3 2, \log _3\left(2^x-5\right)$ and $\log _3\left(2^x-\frac{7}{2}\right)$ are in A.P., then $x$ is equal to

Options:
A $1, \frac{1}{2}$
B $1, \frac{1}{3}$
C $1, \frac{3}{2}$
D None of these

The correct answer is:
None of these

Solution:
$\log _3 2, \log _3\left(2^x-5\right)$ and $\log _3\left(2^x-\frac{7}{2}\right)$ are in A.P.
$\Rightarrow \quad 2 \log _3\left(2^x-5\right)=\log _3$ $\left[(2)\left(2^x-\frac{7}{2}\right)\right]$
$\Rightarrow\left(2^x-5\right)^2=2^{x+1}-7 \Rightarrow 2^{2 x}-12 \cdot 2^x-32=0$
$\Rightarrow \quad x=2,3$
But $x=2$ does not hold, hence $x=3$.

Topic: Mathematics, Sequences and Series, JEE Main
If $\int \log \left(a^2+x^2\right) d x=h(x)+C$, then $\mathrm{h}(x)$ is equal to

Options:
A $x \log \left(a^2+x^2\right)+2 \tan ^{-1}\left(\frac{x}{a}\right)$
B $x^2 \log \left(a^2+x^2\right)+x+a \tan ^{-1}\left(\frac{x}{a}\right)$
C $x \log \left(a^2+x^2\right)-2 x+2 a \tan ^{-1}\left(\frac{x}{a}\right)$
D $x^2 \log \left(a^2+x^2\right)+2 x-a^2 \tan ^{-1}\left(\frac{x}{a}\right)$

The correct answer is:
$x \log \left(a^2+x^2\right)-2 x+2 a \tan ^{-1}\left(\frac{x}{a}\right)$

Solution:
Let $l=\int \log \left(a^2+x^2\right) d x$
By using method of integration by parts, we get
$=\log \left(a^2+x^2\right) \int d x-\int\left\{\frac{d\left(\log \left(a^2+x^2\right)\right.}{d x} \int d x\right\} d x+C$
$\begin{aligned}
& =\log \left(a^2+x^2\right) \int \frac{2 x}{x^2+a^2} \cdot x d x+C \\
& =x \log \left(a^2+x^2\right)-2 \int \frac{x^2}{x^2+a^2} d x+C \\
& =x \log \left(a^2+x^2\right)-2 \int \frac{x^2+a^2}{x^2+a^2} d x+2 \int \frac{a^2}{x^2+a^2} d x \\
& =x \log \left(a^2+x^2\right)-\frac{2 a^2}{a} \tan ^{-1} \frac{x}{a}+C \\
& l=x \log \left(a^2+x^2\right)-2 x+2 a \tan ^{-1} \frac{x}{a}+C
\end{aligned}$
Comparing the above equation with $\int \log \left(x^2+a^2\right) d x=$ $h(x)+\mathrm{C}$, we get
$h(x)=x \log \left(a^2+x^2\right)-2 x+2 a \tan ^{-1}\left(\frac{x}{a}\right)$

Topic: Mathematics, Indefinite Integration, JEE Main
If $\log _e y=3 \sin ^{-1} x$, then $\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}$ at $x=\frac{1}{2}$ is equal to

Options:
A $3 e^{\pi / 6}$
B $9 e^{\pi / 2}$
C $3 e^{\pi / 2}$
D $9 e^{\pi / 6}$

The correct answer is:
$9 e^{\pi / 2}$

Solution:
$\begin{aligned} & \ln (y)=3 \sin ^{-1} x \\ & \frac{1}{y} \cdot y^{\prime}=3\left(\frac{1}{\sqrt{1-x^2}}\right) \\ & \Rightarrow y^{\prime}=\frac{3 y}{\sqrt{1-x^2}} \text { at } x=\frac{1}{2} \\ & \Rightarrow y^{\prime}=\frac{3 e^{3\left(\frac{\pi}{6}\right)}}{\frac{\sqrt{3}}{2}}=2 \sqrt{3} e^{\frac{\pi}{2}} \\ & \Rightarrow y^{\prime \prime}=3\left(\frac{\sqrt{1-x^2} y^{\prime}-y \frac{1}{2 \sqrt{1-x^2}}(-2 x)}{\left(1-x^2\right)}\right)\end{aligned}$
$\begin{aligned} & \Rightarrow\left(1-x^2\right) y^{\prime \prime}=3\left(3 y+\frac{x y}{\sqrt{1-x^2}}\right) \\ & \downarrow \text { at } x=\frac{1}{2}, y=e^{3 \sin ^{-1}\left(\frac{1}{2}\right)}=\mathrm{e}^{3\left(\frac{\pi}{6}\right)}=\mathrm{e}^{\frac{\pi}{2}}\end{aligned}$
$\begin{aligned} & \left.\left(1-x^2\right) y^{\prime \prime}\right|_{\text {at } x=\frac{1}{2}}=3\left(3 \mathrm{e}^{\frac{\pi}{2}}+\frac{\frac{1}{2}\left(e^{\frac{\pi}{2}}\right)}{\frac{\sqrt{3}}{2}}\right) \\ & =3 \mathrm{e}^{\frac{\pi}{2}}\left(3+\frac{1}{\sqrt{3}}\right) \\ & \left(1-x^2\right) y^{\prime \prime}-\left.x y^{\prime}\right|_{\text {atx } x=\frac{1}{2}} \\ & =3 \mathrm{e}^{\frac{\pi}{2}}\left(3+\frac{1}{\sqrt{3}}\right)-\frac{1}{2}\left(2 \sqrt{3} \mathrm{e}^{\frac{\pi}{2}}\right)=9 \mathrm{e}^{\frac{\pi}{2}}\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2024 (09 Apr Shift 2)
If log2â¡52x+1, log4â¡21-x+1 and 1 are in arithmetic progression, then x is equal to

Options:
A log⁡5log⁡2
B log2⁡0.6
C 1-log⁡5log⁡2
D log⁡2log⁡5

The correct answer is:
1-log⁡5log⁡2

Solution:
log2â¡5â2x+1, 12log2â¡21-x+1, log2â¡2 are in A.P.
â2Ã12log2â¡21-x+1=log2â¡5â2x+1+log22
â21-x+1=5â2x+12
Let, 2x=y

â2y+1=5y+12
â2+y=10y2+2y
â10y2+y-2=0ây=25,-12
â2x=25 or -12(not possible)

â2x=25
âx=log225=1-log2â¡5=1-logâ¡5logâ¡2

Topic: Mathematics, Sequences and Series, JEE Main
If log32,log32x-5,log32x-72 are in an arithmetic progression, then the value of x is equal to _____.

Options:

The correct answer is:
3

Solution:
As log32,log32x-5,log32x-72 are in arithmetic progression, 2log32x-5=log32+log32x-72

⇒log32x-52=log322x-72

Let 2x=t

So, log3t-52=log32t-72

⇒t-52=2t-7

⇒t2-12t+32=0

⇒t-4t-8=0

⇒2x=4 or 2x=8

Here, x≠2, as log34-5 does not exist.

So, x=3

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (27 Jul Shift 1)
If log10⁡x3-y3x3+y3=2 , then dydx=

Options:
A xy
B -yx
C -xy
D yx

The correct answer is:
yx

Solution:
Given, log10â¡x3-y3x3+y3=2
âx3-y3x3+y3=102=100
Applying componendo and dividendo, we get,
x3y3=-10199âxy=-1019913......(i)
Hence, y=-9910113x
Differentiating, we get,
dydx=-9910113=yx (from (i))

Topic: Mathematics, Differentiation, JEE Main
If logcosx⁡sinx≥2 and 0≤x≤3π, then the value of sinx lies in the interval

Options:
A 5-12, 1
B 0,5-12
C 0,12
D None of these

The correct answer is:
0,5-12

Solution:
Since, sinx>0,ââcosx>0,ââcosxâ 1
So, sinxâ¤cos2â¡x
âsin2â¡x+sinx-1â¤0
âsinx+122-54â¤0
âââ|sinx+12|ââ¤ââ52
â´ sinx+12â[â52,52]
(As, sin x>0)
â´â0<sinxâ¤5â12

Topic: Mathematics, Trigonometric Equations, JEE Main
If logea,logeb,logec are in an A.P. and logea-loge2b,loge2b-loge3c,loge3c-logea are also in an A.P., then a:b:c is equal to

Options:
A 9: 6: 4
B 16: 4: 1
C 25: 10: 4
D 6: 3: 2

The correct answer is:
9: 6: 4

Solution:
Given: loga, logb, logc are in A.P.

⇒2logb=loga+logc

⇒b2=ac ...1

And loga-log2b, log2b-log3c, log3c-loga are in A.P

⇒2log2b3c=loga2b+log3ca

⇒2b3c2=a2b×3ca

⇒2b3c2=3c2b

⇒2b=3c

⇒4b2=9c2 & 6b=9c ...2

Now solving the equation 1 & 2 we get,

4ac=9c2

⇒4a=9c ...3

Then from equation 2 & 3 we get,

4a=6b=9c=k

⇒a=k4, b=k6 & c=k9

⇒a:b:c=14:16:19=9:6:4

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (29 Jan Shift 2)
If ∫logt+1+t21+t2dt=12gt2+c, where c is a constant, then g2, is equal to

Options:
A 15log2+5
B 2log2+5
C log2+5
D 12log(2+5)

The correct answer is:
log2+5

Solution:
Put, logt+1+t2=z
⇒1t+1+t21+2t21+t2dt=dz
⇒t+1+t2 t+1+t2 11+t2dt=dz
⇒11+t2dt=dz
∴I=∫zdz=z22+c=12logt+1+t22+c, where c is the constant of integration.
∴gt=logt+1+t2
⇒g2=log2+ 1+4
Hence, log2+5, is the answer.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2015 (11 Apr Online)
If \( \log (x+y)-2 x y=0 \). Then \( y^{\prime}(0)= \)

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

The correct answer is:
\( 1 \)

Solution:
Given, logx+y-2xy=0 .........1

Differentiate equation 1 w.r.t x

1x+y1+dydx-2xdydx+y=0

1x+y-2xdydx+1x+y-2y=0

⇒dydx=-1-2xy-2y21-2xy-2x2

Hence,y'0=-1-0-2y021-0-0 ........2

Now put x=0 in equation 1

log0+y=0⇒y0=1

from equation 2

y'0=1

Topic: Mathematics, Differentiation, JEE Main