datavorous/entrance-exam-dataset huggingface dataset:
If \(f(x+2 y, x-2 y)=x y\), then \(f(x, y)\) is equal to

Options:
A \(\frac{1}{4} x y\)
B \(\frac{1}{4}\left(x^2-y^2\right)\)
C \(\frac{1}{8}\left(x^2-y^2\right)\)
D \(\frac{1}{2}\left(x^2+y^2\right)\)

The correct answer is:
\(\frac{1}{8}\left(x^2-y^2\right)\)

Solution:
\(\begin{aligned}\text{Hint: }
& x+2 y=a \\
& \frac{x-2 y=b}{2 x=a+b} \\
& 4 y=a-b
\end{aligned}\)
\(f(a, b)=\left(\frac{a+b}{2}\right)\left(\frac{a-b}{4}\right)=\frac{a^2-b^2}{8}\)

Topic: Mathematics, Functions, WBJEE, WBJEE 2011
If \(f(x)=3 \sqrt[3]{x^2}-x^2\), then

Options:
A f has no extrema
B \(f\) is maximum at two points \(x=1\) and \(x=-1\)
C \(f\) is minimum at \(x=0\)
D f has maximum at \(x=1\) only

The correct answers are:
\(f\) is maximum at two points \(x=1\) and \(x=-1\), \(f\) is minimum at \(x=0\)

Solution:
Hint : \(f(x)=3(x)^{\frac{2}{3}}-x^2\)
Differentiate both sides wrt. \(x\)
\(=f^{\prime}(x)=3 \cdot \frac{2}{3} x^{-1 / 3}-2 x=2 x^{-1 / 3}-2 x=2\left(x^{-1 / 3}-x\right)=2\left(\frac{1-x^{4 / 3}}{x^{1 / 3}}\right)\)
\(\therefore f\) has maximum at two points \(x=-1 x=1\)
\(f\) has minimum at \(x=0\)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2023
If $I=\int_{0}^{2} e^{x^{4}}(x-a) d x=0$, then $\alpha$ lies in the interval

Options:
A (0,2)
B (-1,0)
C (2,3)
D (-2,-1)

The correct answer is:
(0,2)

Solution:
Given, $I=\int_{0}^{2} e^{x^{4}}(x-\alpha) d x=0$
$\Rightarrow \quad \int_{0}^{2} e^{x^{4}} x d x=\int_{0}^{2} e^{x^{4}} \alpha d x$
Here, we see that $\int_{0}^{2} e^{x^{4}} x d x$ gives us an area between two curves $e^{x^{4}}$ and $x$ from $x=0$ to $x=2$
Similarly. $\int_{0}^{2} e^{x^{4}} \alpha d x$ gives us a same area, so $a$ should lies between 0 to $2, i . e ., \alpha \in(0,2)$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2014
If $\mathrm{I}=\int_0^{\mathrm{I}} \frac{\mathrm{dx}}{1+\mathrm{x}^{\pi / 2}}$, then

Options:
A $\log _{\mathrm{e}} 2 < 1 < \pi / 4$
B $\log _{\mathrm{e}} 2>1$
C $\mathrm{I}=\pi / 4$
D $\mathrm{I}=\log _{\mathrm{e}} 2$

The correct answer is:
$\log _{\mathrm{e}} 2 < 1 < \pi / 4$

Solution:
Hints: $\mathrm{x}^2 < \mathrm{x}^{\frac{\pi}{2}} < \mathrm{x}, \quad 1+\mathrm{x}^2 < 1+\mathrm{x}^{\frac{\pi}{2}} < 1+\mathrm{x}$
$\frac{1}{1+x^2}>\frac{1}{1+x^{\frac{\pi}{2}}}>\frac{1}{1+x}$
$\frac{\pi}{4}>\mathrm{I}>(\log (1+\mathrm{x})), \quad \frac{\pi}{4}>\mathrm{I}>\log 2$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2010
If $I=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$ and $P=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2\end{array}\right)$ Then, the matrix $P^{3}+2 P^{2}$ is equal to

Options:
A P
B I-P
C $2I+P$
D $2I-P$

The correct answer is:
$2I+P$

Solution:
Given, $I=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
and $P=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2\end{array}\right)$
The characteristic equation of $P$ is
$$
|P-\lambda|=0
$$
$\Rightarrow\left|\begin{array}{ccc}1-\lambda & 0 & 0 \\ 0 & -1-\lambda & 0 \\ 0 & 0 & -2-\lambda\end{array}\right|=0$
$\Rightarrow \quad(1-\lambda)\{(1+\lambda)(2+\lambda)\}=0$
$\Rightarrow \quad\left(1-\lambda^{2}\right)(2+\lambda)=0$
$\Rightarrow \quad 2-2 \lambda^{2}+\lambda-\lambda^{3}=0$
$\Rightarrow \quad \lambda^{3}+2 \lambda^{2}-\lambda-2=0$
We know that, Caylay Hamilton theorem states that Every square matrix satisfy its characteristic equation'.
$\therefore$
$$
P^{3}+2 P^{2}-P-2 I=0
$$
$\Rightarrow$\(p^{3}+2 p^{2}=2 I+p\)

Topic: Mathematics, Matrices, WBJEE, WBJEE 2013
If $\mathrm{I}_1=\int_0^{3 \pi} \mathrm{f}\left(\cos ^2 \mathrm{x}\right) \mathrm{dx}$ and $\mathrm{I}_2=\int_0^\pi \mathrm{f}\left(\cos ^2 \mathrm{x}\right) \mathrm{dx}$, then

Options:
A $\mathrm{I}_1=\mathrm{I}_2$
B $3 \mathrm{I}_1=\mathrm{I}_2$
C $\mathrm{I}_1=3 \mathrm{I}_2$
D $\mathrm{I}_1=5 \mathrm{I}_2$

The correct answer is:
$\mathrm{I}_1=3 \mathrm{I}_2$

Solution:
Hints : $I_1=3 \int_0^\pi \mathrm{f}\left(\cos ^2 x\right) d x=3 I \quad$ [period is $\pi$ ]

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2010
If $\mathrm{I}_1=\int_0^{\pi / 4} \sin ^2 \mathrm{x} d \mathrm{x}$ and $\mathrm{I}_2=\int_0^{\pi / 4} \cos ^2 \mathrm{x} d \mathrm{x}$, then,

Options:
A $\mathrm{I}_1=\mathrm{I}_2$
B I $_1 < \mathrm{I}_2$
C $\mathrm{I}_1>\mathrm{I}_2$
D $\mathrm{I}_2=\mathrm{I}_1+\pi / 4$

The correct answer is:
I $_1 < \mathrm{I}_2$

Solution:
Hints : $I_1=\int_0^{\pi / 4} \sin ^2 x d x ; I_2=\int_0^{\pi / 4} \cos ^2 x d x$
$$
\begin{aligned}
& \text { In }\left(0, \frac{\pi}{4}\right), \cos ^2 x>\sin ^2 x \therefore \int_0^{\pi / 4} \cos ^2 x d x>\int_0^{\pi / 4} \sin ^2 x d x \\
& I_2>I_1 \text { i.e. } I_1 < I_2
\end{aligned}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2009
If $i=\sqrt{-1}$ and $n$ is a positive integer, then $i^n+i^{n+1}+i^{n+2}+i^{n+3}$ is euqal to

Options:
A 1
B $i$
C $i^n$
D 0

The correct answer is:
0

Solution:
$$
\text { Hints : } \dot{i}^n\left(1+\dot{i}+\dot{i}^2+\dot{i}^3\right)=\dot{i}^n(1+\dot{i}-1-\hat{i})=0
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2009
If $\mathrm{I}$ is the greatest of $I_1=\int_0^1 e^{-x} \cos ^2 x d x, I_2=\int_0^1 e^{-x^2} \cos ^2 x d x, I_3=\int_0^1 e^{-x^2} d x, I_4=\int_0^1 e^{-x^2 / 2} d x$, then

Options:
A $\mathrm{I}=\mathrm{I}_1$
B $\mathrm{I}=\mathrm{I}_2$
C $I=I_3$
D $\mathrm{I}=\mathrm{I}_4$

The correct answer is:
$\mathrm{I}=\mathrm{I}_4$

Solution:
$\mathrm{e}^{-\mathrm{x}} \cdot \cos ^2 \mathrm{x} < \mathrm{e}^{-\mathrm{x}^2} \cos ^2 \mathrm{x} < \mathrm{e}^{-\mathrm{x}^2} < \mathrm{e}^{-\frac{\mathrm{x}^2}{2}}$ for $0 < \mathrm{x} < 1$
$\therefore$ By domination Law, $\mathrm{I}_4$ is maximum
$\therefore \mathrm{I}=\mathrm{I}_4$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2022
If $\mathrm{I}=\lim _{\mathrm{x} \rightarrow 0} \sin \left(\frac{\mathrm{e}^{\mathrm{x}}-\mathrm{x}-1-\frac{\mathrm{x}^{2}}{2}}{\mathrm{x}^{2}}\right)$, then limit

Options:
A does not exist
B exists and equals 1
C exists and equals 0
D exists and equals $\frac{1}{2}$

The correct answer is:
exists and equals 0

Solution:
$I=\lim _{x \rightarrow 0} \sin \left(\frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}\right)=\lim _{x \rightarrow 0} \frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}=\lim _{x \rightarrow 0} \frac{e^{x}-1-0-x}{2 x}=\lim _{x \rightarrow 0} \frac{e^{x}-1}{2}=0$

Topic: Mathematics, Limits, WBJEE, WBJEE 2021
If \(I=\int \frac{x^2 d x}{(x \sin x+\cos x)^2}=f(x)+\tan x+c\), then \(f(x)\) is

Options:
A \(\frac{\sin x}{x \sin x+\cos x}\)
B \(\frac{1}{(x \sin x+\cos x)^2}\)
C \(\frac{-x}{\cos x(x \sin x+\cos x)}\)
D \(\frac{1}{\sin x(x \cos x+\sin x)}\)

The correct answer is:
\(\frac{-x}{\cos x(x \sin x+\cos x)}\)

Solution:
Hint : \(I=\int \frac{x^2}{(x \sin x+\cos x)^2} d x=\int \frac{x}{(x \sin x+\cos x)^2} \times \frac{x}{\cos x} d x\)
\(I=-\frac{1}{(x \sin x+\cos x)} \cdot \frac{x}{\cos x}+\int \frac{1}{(x \sin x+\cos x)} \cdot \frac{(\cos x+x \sin x)}{\cos ^2 x} d x\)
\(\begin{aligned} & I=-\frac{1}{(x \sin x+\cos x)} \cdot \frac{x}{\cos x}+\int \sec ^2 x d x \\ & I=-\frac{x}{(x \sin x+\cos x) \cos x}+\tan x+C \\ & \therefore f(x)=\frac{-x}{\cos x(x \sin x+C)}\end{aligned}\)

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2023
If in a $\triangle A B C, A D, B E$ and $C F$ are the altitudes and $R$ is the circumradius, then the radius of the circumcircle of $\Delta D E F$ is

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

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

Solution:
Let circumradius of $\Delta$ DEF be $R'$ We know. $\angle F D E=180^{\circ}-2 A$ and $F E=R \sin 2 A$

Now, by sine rule in $\triangle D E F$ $2 R^{\prime}=\frac{E F}{\sin \angle F D E}=\frac{A \sin 2 A}{\sin \left(180^{\circ}-2 A\right)}$
$\therefore$
$R^{\prime}=\frac{R}{2}$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2016
If in a triangle \(A B C, \sin A, \sin B, \sin C\) are in A.P., then

Options:
A the altitudes are in A.P.
B the altitudes are in H.P.
C the angles are in A.P.
D the angles are in H.P.

The correct answer is:
the altitudes are in H.P.

Solution:
Hints: \(\frac{1}{2} \mathrm{ap}_1=\frac{1}{2} \mathrm{bp}_2=\frac{1}{2} \mathrm{cp}_3=\Delta\)
\(\mathrm{a}=\frac{2 \Delta}{\mathrm{p}_1}\left|\mathrm{~b}=\frac{2 \Delta}{\mathrm{p}_2}\right| \mathrm{c}=\frac{2 \Delta}{\mathrm{p} 3}\)
H.P.

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2011
If in case of a radio isotope the value of half-life \(\left(T_{1/2}\right)\) and decay constant \((\lambda)\) are identical in magnitude, then their value should be

Options:
A \(0.693 / 2\)
B \((0.693)^{1 / 2}\)
C \((0.693)^2\)
D 0.693

The correct answer is:
\((0.693)^{1 / 2}\)

Solution:
Hint: For a radio decay \(T_{1 / 2}=\frac{0.693}{\lambda}\)
If \(T_{1 / 2}=\lambda=x\) then \(\mathrm{x}=\frac{0.693}{\mathrm{x}}\)
\(\Rightarrow \mathrm{x}^2=0.693, \quad \Rightarrow \mathrm{x}=\mathrm{T}_{1 / 2}=\lambda=(0.693)^{1 / 2}\)

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2023
If in the expansion of $(a-2 b)^{\mathrm{n}}$, the sum of the 5 th and 6 th term is zero, then the value of $\frac{\mathrm{a}}{\mathrm{b}}$ is

Options:
A $\frac{\mathrm{n}-4}{5}$
B $\frac{2(n-4)}{5}$
C $\frac{5}{n-4}$
D $\frac{5}{2(n-4)}$

The correct answer is:
$\frac{2(n-4)}{5}$

Solution:
$$
\text { Hints: } \begin{gathered}
(a-2 b)^{\mathrm{n}}=\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(a)^{\mathrm{n}-\mathrm{r}}(-2 \mathrm{~b})^{\mathrm{r}} \\
\mathrm{t}_5+\mathrm{t}_6=0
\end{gathered}
$$
$$
\begin{aligned}
& \Rightarrow{ }^{\mathrm{n}} \mathrm{C}_4(\mathrm{a})^{\mathrm{n}-4}(-2 \mathrm{~b})^4+{ }^{\mathrm{n}} \mathrm{C}_5(\mathrm{a})^{\mathrm{n}-5}(-2 \mathrm{~b})^5=0 \Rightarrow \frac{\mathrm{n} !}{4 !(\mathrm{n}-4) !} \mathrm{a}^{\mathrm{n}-4}(-2 \mathrm{~b})^4=-\frac{\mathrm{n} !}{5 !(\mathrm{n}-5) !}(\mathrm{a})^{\mathrm{n}-5}(-2 \mathrm{~b})^5 \\
& \Rightarrow \frac{1}{(\mathrm{n}-4)} \times \mathrm{a}=\frac{-1}{5}(-2 \mathrm{~b}) \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{2(\mathrm{n}-4)}{5}
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2010
If \(I_n=\int_0^{\frac{\pi}{2}} \cos ^n x \cos n x d x\), then \(I_1, I_2, I_3 \ldots\) are in

Options:
A A.P.
B G.P.
C H.P
D no such relation

The correct answer is:
G.P.

Solution:
Hint : \(I_n=\int_0^{\pi / 2} \cos ^n x \cdot \cos n x d x\)
\(I_1=\int_0^{\pi / 2} \cos x \cdot \cos x d x=\int_0^{\pi / 2}\left(\frac{1+\cos 2 x}{2}\right) d x=\frac{\pi}{4}\)
\(I_2=\int_0^{\pi / 2} \cos ^2 x \cdot \cos 2 x d x=\int_0^{\pi / 2}-\sin ^2 x \cdot \cos 2 x d x\)
\(2 I_2=\int_0^{\pi / 2} \cos ^2 2 x=\int_0^{\pi / 2} \frac{1+\cos 4 x}{2} d x=\frac{\pi}{4}\)
\(\therefore \mathrm{I}_2=\frac{\pi}{8}\)
\(I_3=\int_0^{\pi / 2} \cos ^3 x \cdot \cos 3 x d x=\int_0^{\pi / 2}\left(\frac{3 \cos x+\cos 3 x}{4}\right) \cos 3 x d x\)
\(\begin{aligned}
& =\frac{3}{4} \int_0^{\pi / 2} \cos x \cdot \cos 3 x d x+\frac{1}{4} \int_0^{\pi / 2} \cos ^2 3 x d x \\
& =\frac{3}{8} \int_0^{\pi / 2}(\cos 2 x+\cos 4 x) d x+\frac{1}{4} \int_0^{\pi / 2} \frac{1+\cos 6 x}{2} d x \\
& =\frac{\pi}{16}
\end{aligned}\)
\(\therefore \mathrm{I}_1, \mathrm{I}_2, \mathrm{I}_3\) are in G.P.

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2023
If \(\int \frac{d x}{(x+1)(x-2)(x-3)}=\frac{1}{k} \log _e\left\{\frac{|x-3|^3|x+1|}{(x-2)^4}\right\}+c\), then the value of \(k\) is

Options:
A 4
B 6
C 8
D 12

The correct answer is:
12

Solution:
Hint: \(\int \frac{d x}{(x+1)(x-2)(x-3)}=\int\left(\frac{A}{x+1}+\frac{B}{(x-2)}+\frac{C}{x-3}\right) d x\)
\(\begin{aligned}
& \frac{1}{(x+1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x+1)(x-3)+C(x+1)(x-2)}{(x+1)(x-2)(x-3)} \\
& \Rightarrow 1=A(x-2)(x-3)+B(x+1)(x-3)+C(x+1)(x-2) \\
& \text { for } x=-1 \\
& 1=A(-3)(-4) \Rightarrow A=\frac{1}{12} \\
& \text { for } x=2 \\
& 1=B(3)(-1) \Rightarrow B=-\frac{1}{3}
\end{aligned}\)
\(\text { for } x=3\)
\(\begin{aligned} & 1=C(4)(1) \Rightarrow C=\frac{1}{4} \\ & \therefore I=\frac{1}{12} \int \frac{1}{x+1} d x-\frac{1}{3} \int \frac{1}{x-2} d x+\frac{1}{4} \int \frac{1}{x-3} d x \\ & =\frac{1}{12} \ln |x+1|-\frac{1}{3} \ln |x-2|+\frac{1}{4} \ln |x-3|+C \\ & =\frac{1}{12}\left[\ln |x+1|-\ln |x-2|^4+\ln |x-3|^3\right]+C \\ & =\frac{1}{12} \ln \frac{(|x+1|)(|x-3|)^3}{(x-2)^4}-C \\ & \therefore k=12\end{aligned}\)

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2023
If $\vec{\alpha}$ is a unit vector, $\vec{\beta}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{\gamma}=\hat{\mathrm{i}}+\hat{\mathrm{k}}$, then the maximum value of $[\vec{\alpha} \vec{\beta} \vec{\gamma}]$ is

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

The correct answer is:
$\sqrt{6}$

Solution:
$[\vec{\alpha} \vec{\beta} \vec{\gamma}]=\vec{\alpha} \cdot(\vec{\beta} \times \vec{\gamma})=\vec{\alpha} \cdot\left|\begin{array}{ccc}\hat{i} & \hat{j} & k \\ 1 & 1 & -1 \\ 1 & 0 & 1\end{array}\right|$
$=\vec{\alpha} \cdot(\hat{i}+2 \hat{j}-\hat{k})$ is maximum $\Rightarrow$ angle between $\vec{\alpha} \& \hat{i}+2 \hat{j}-\hat{k}$ will be 0
$\therefore[\vec{\alpha} \vec{\beta} \vec{\gamma}]=|\vec{\alpha}||\hat{i}+2 \hat{j}-\hat{k}|=\sqrt{6}$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2022
If $\omega$ is an imaginary cube root of unity, then the value of $(2-\omega)\left(2-\omega^{2}\right)+2(3-\omega)\left(3-\omega^{2}\right)$ $+\ldots+(n-1)(n-\omega)\left(n-\omega^{2}\right)$ is

Options:
A $\frac{n^{2}}{4}(n+1)^{2}-n$
B $\frac{n^{2}}{4}(n+1)^{2}+n$
C $\frac{n^{2}}{4}(n+1)^{2}$
D $\frac{n^{2}}{4}(n+1)-n$

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

Solution:
Given,
$(2-\omega)\left(2-\omega^{2}\right)+2(3-\omega)\left(3 \omega^{2}\right)+\ldots+$ $\quad(n-1)(n-\omega)\left(n \omega^{2}\right)$
Now, $\begin{aligned} T_{n} &=(n-1)(n-\omega)\left(n-\omega^{2}\right) \\ &=(n-1)\left(n^{2}-n \omega-n \omega^{2}+\omega^{3}\right) \\ &=(n-1)\left(n^{2}+n+1\right)=n^{3}-1 \\ \therefore \quad S_{n} &=\Sigma T_{n} \\ &=\Sigma n^{3}-\Sigma 1=\frac{n^{2}(n+1)^{2}}{4}-n \end{aligned}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2016
If $\omega$ is an imaginary cube root of unity, then the value of the determinant $\left|\begin{array}{ccc}1+\omega & 0 & -\omega \\ 1+\omega^{2} & \omega & -\omega^{2} \\ \omega+\omega^{2} & \omega & -\omega^{2}\end{array}\right|$ is

Options:
A $-2 \omega$
B $-3 w^{2}$
C -1
D 0

The correct answer is:
$-3 w^{2}$

Solution:
$\left|\begin{array}{ccc}1+\omega & \omega^{2} & -\omega \\ 1+\omega^{2} & \omega & -\omega^{2} \\ \omega+\omega^{2} & \omega & -\omega^{2}\end{array}\right|$
$\left|\begin{array}{ccc}0 & \omega^{2} & -\omega \\ 0 & \omega & -\omega^{2} \\ -1+\omega & \omega & -\omega^{2}\end{array}\right|$
$=(-1+\omega)\left(-\omega^{4}+\omega^{2}\right)=(\omega-1)\left(\omega^{2}-c\right.$
$=\omega^{3}-\omega^{2}-\omega^{2}+\omega=-3 \omega^{2}$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2015
If it is assumed that ${ }_{92}^{235} U$ decays only by emitting $\alpha$ and $\beta$ particles, the possible product of the decay is :

Options:
A ${ }_{89}^{225} A c$
B ${ }_{89}^{227} A c$
C ${ }_{89}^{230} A c$
D ${ }_{89}^{231} A c$

The correct answer is:
${ }_{89}^{227} A c$

Solution:
New mass no. $=235-2 \times 4=227$
New at. no. $=92-2 \times 2+1=92-4+1=89$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2009
If $\int_{\log _{e} 2}^{x}\left(e^{x}-1\right)^{-1} d x=\log _{e} \frac{3}{2}$ then the value of $x$ is

Options:
A 1
B $\mathrm{e}^{2}$
C $\log 4$
D $\frac{1}{\mathrm{e}}$

The correct answer is:
$\log 4$

Solution:
$\int_{\log _{e}{ }^{2}}^{x}\left(e^{x}-1\right)^{-1} d x=\log _{e}^{\frac{3}{2}} \int_{\log _{e}{ }^{2}}^{x} \frac{e^{-x}}{1-e^{-x}} d x=\log _{e}^{\frac{3}{2}}$
$e^{-x}=\frac{1}{4}$
$x=\ln 4$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2021
If $\lim _{x \rightarrow 0}\left(\frac{1+c x}{1-c x}\right)^{1 / x}=4$, then $\lim _{x \rightarrow 0}\left(\frac{1+2 c x}{1-2 c x}\right)^{1 / x}$ is

Options:
A 2
B 4
C 16
D 64

The correct answer is:
16

Solution:
Hint:
$\mathrm{e}^{2 \mathrm{c}}=4 \Rightarrow \mathrm{e}^{4 \mathrm{c}}=16$

Topic: Mathematics, Limits, WBJEE, WBJEE 2020
If $\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x}$ exists and is equal to 1 then the value of $a$ is

Options:
A 2
B 1
C 0
D -1

The correct answer is:
1

Solution:
$\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x}$
$$
=\lim _{x \rightarrow 0} \frac{2 a\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\ldots\right)-\left(2 x-\frac{(2 x)^{3}}{3 !}+\frac{(2 x)^{5}}{5 !}-\ldots\right)}{\left(x+\frac{x^{3}}{3}+\frac{2}{15} x^{5}+\ldots\right)^{3}}
$$
$$
=\lim _{x \rightarrow 0} \frac{\left(2 a-2 x+\left(-\frac{2 a}{3 !}+\frac{2^{3}}{3 !}\right) x^{3}+\left(\frac{2 a}{5 !}-\frac{2^{5}}{5 !}\right) x^{5}+\ldots\right.}{x^{3}\left(1+\frac{x^{2}}{3}+\frac{2}{15} x^{4}+\ldots\right)^{3}}
$$
since, it is given that given limit is exist, then
$$
2 a-2=0 \Rightarrow \alpha=1
$$

Topic: Mathematics, Limits, WBJEE, WBJEE 2014
If $\lim _{x \rightarrow 0} \frac{a x e^{x}-b \log (1+x)}{x^{2}}=3$ then the values of $a$ and $b$ are, respectively

Options:
A 2,2
B 1,2
C 2,1
D 2,0

The correct answer is:
2,2

Solution:
We have. $\lim _{x \rightarrow 0} \frac{\operatorname{ax} e^{x}-b \log (1+x)}{x^{2}}=3\left[\frac{0}{0}\right.$ form $]$
Using L' Hospital's rule, we get $\lim _{x \rightarrow 0} \frac{a e^{x}+a x e^{x}-\frac{b}{1+x}}{2 x}=3 \quad\left[\frac{0}{0}\right.$ form $]$
$\Rightarrow a-b=0 \Rightarrow a=b$
Using L' Hospital's rule, we get
$$
\lim _{x \rightarrow 0} \frac{a e^{x}+a e^{x}+a x e^{x}+\frac{b}{(1+x)^{2}}}{2}=3
$$
$\Rightarrow \quad \lim _{x \rightarrow 0} 2 a e^{x}+a x e^{x}+\frac{b}{(1+x)^{2}}=6$
$\Rightarrow$
$2 a+b=6$
$\Rightarrow \quad 3 a=6 \Rightarrow a=2$
On putting the value of $a$ we get $b=2$

Topic: Mathematics, Limits, WBJEE, WBJEE 2015
If $\log _{2}^{6}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)$, then the values of $x$ are

Options:
A $\frac{1}{4}, \frac{1}{3}$
B $\frac{1}{4},\frac{1}{2}$
C $-\frac{1}{4}, \frac{1}{2}$
D $\frac{1}{3},-\frac{1}{2}$

The correct answer is:
$\frac{1}{4},\frac{1}{2}$

Solution:
We have,
$\log _{2}^{6}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)$
$\Rightarrow \quad 1+\log _{2}^{3}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)$
$\Rightarrow \quad \frac{2^{\frac{1}{x}}+8}{3}=2^{1+\frac{1}{2 x}}$
$\Rightarrow 2^{\frac{1}{x}}+8=3 \cdot 2 \cdot 2^{\frac{1}{2 x}}$
$\Rightarrow 2^{\frac{1}{x}}+8=6 \cdot 2^{\frac{1}{2 x}}$
Let $y=2^{\frac{1}{2 x}}$
$\Rightarrow \quad y^{2}+8=6 y$
$\Rightarrow \quad y^{2}-6 y+8=0$
$\Rightarrow \quad(y-4)(y-2)=0$
$y=4,2$
$2^{\frac{1}{2 x}}=4$ and $2^{\frac{1}{2 x}}=2$
$\frac{1}{2 x}=2$ and $\frac{1}{2 x}=1$
$x=\frac{1}{4}, \frac{1}{2}$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2019
$\text {If } \log _{0.2}(x-1)>\log _{0.04}(x+5), \text { then }$

Options:
A $-1 < x < 4$
B $2 < x < 3$
C $1 < x < 4$
D $1 < x < 3$

The correct answer is:
$1 < x < 4$

Solution:
We have, $\log _{0.2}(x-1)>\log _{0.04}(x+5)$
$\Rightarrow \quad \log _{0.2}(x-1)>\log _{0.2^{2}}(x+5)$
$\Rightarrow \quad \log _{0.2}(x-1)>\frac{1}{2} \log _{0.2}(x+5)$
$\Rightarrow \quad 2 \log _{0.2}(x-1)>\log _{0.2}(x+5)$
$\Rightarrow \quad \log _{0.2}(x-1)^{2}>\log _{0.2}(x+5)$
$\Rightarrow \quad(x-1)^{2} < x+5$
$\left[\because \log _{a} x>\log _{a} y \Rightarrow x < y,\right.$ if $\left.0 < a < 1\right]$
$\Rightarrow \quad x^{2}-2 x+1 < x+5$
$\Rightarrow \quad x^{2}-3 x-4 < 0$
$\Rightarrow \quad x^{2}-4 x+x-4 < 0$
$\Rightarrow x(x-4)+1(x-4) < 0$
$\Rightarrow \quad(x-4)(x+1) < 0$
$\Rightarrow \quad x \in(-1,4)$
But $\quad x>1$
$\Rightarrow \quad x \in(1,4)$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2015
If $\log _{0.3}(x-1) < \log _{0.09}(x-1),$ then $x$ lies in
the interval

Options:
A $(2, \infty)$
B (1,2)
C (-2,-1)
D None of these

The correct answer is:
$(2, \infty)$

Solution:
Given, $\quad \log _{0.3}(x-1) < \log _{0.09}(x-1)$
$\Rightarrow \quad \log _{0.3}(x-1) < \log _{0.3^{2}}(x-1)$
$\Rightarrow \quad \log _{0.3}(x-1)^{2} < \log _{0.3}(x-1)$
$\Rightarrow \quad(x-1)^{2}>x-1 \quad[\because 0.3 < 1]$
$\Rightarrow x^{2}+1-2 x-x+1>0$
$\Rightarrow \quad x^{2}-3 x+2>0$
$\Rightarrow \quad(x-1)(x-2)>0$
$\Rightarrow \quad x < 1, x>2 \Rightarrow x>2 \quad[\because x \neq 1]$
$\therefore \quad(2, \infty)$
Hence, x les in the interval $(2, \infty)$

Topic: Mathematics, Functions, WBJEE, WBJEE 2016
If $\left(\log _{5} x\right)\left(\log _{x} 3 x\right)\left(\log _{3 x} y\right)=\log _{x} x^{3},$ then $y$ equals

Options:
A 125
B 25
C $5 / 3$
D 243

The correct answer is:
125

Solution:
We have,
$\begin{aligned} & \log _{5} x \cdot \log _{x} 3 x \cdot \log _{3 x} y=\log _{x} x^{3} \\ \Rightarrow & \frac{\log x}{\log 5} \times \frac{\log 3 x}{\log x} \times \frac{\log y}{\log 3 x}=3 \log _{x} x \end{aligned}$
$\left[\because \log _{b} a=\frac{\log a}{\log b}\right.$ and $\left.\log a^{m}=m \log a\right]$
$\frac{\log y}{\log 5}=3$
$\left[\because \log _{a} a=1\right]$
$\begin{array}{ll}\Rightarrow & \log y=3 \log 5 \\ \Rightarrow & \log y=\log 5^{3}\end{array}$
$y=5^{3}=125$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2017
If $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$, then

Options:
A $4 < x \leq 5$
B $x < -4$ or $x>4$
C $-1 \leq x \leq 5$
D $x < -1$ or $x>5$

The correct answer is:
$-1 \leq x \leq 5$

Solution:
Given, \(\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)\) if and only if
\(\begin{array}{l}
x^{2}-16 \leq 4 x-11 \\
\Rightarrow x^{2}-4 x-5 \leq 0 \\
x^{2}-5 x+x-5 \leq 0 \\
\Rightarrow x(x-5)+1(x-5) \leq 0 \\
(x-5)(x+1) \leq 0
\end{array}\)
Sign scheme of \(x^{2}-4 x-5 \leq 0\)

\(-1 \leq x \leq 5\)

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2012
If \(\log _3 x+\log _3 y=2+\log _3 2\) and \(\log _3(x+y)=2\), then

Options:
A \(x=1, y=8\)
B \(x=8, y=1\)
C \(x=3, y=6\)
D \(x=9, y=3\)

The correct answer is:
\(x=3, y=6\)

Solution:
\(\begin{aligned}\text {Hints: }
& \log _3 \mathrm{x}+\log _3 \mathrm{y}=2+\log _3 2 \\
& \Rightarrow \mathrm{x} \cdot \mathrm{y}=18 \\
& \log (\mathrm{x}+\mathrm{y})=2 \Rightarrow \mathrm{x}+\mathrm{y}=9
\end{aligned}\)
we will get \(x=3\) and \(y=6\)

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2011
If \(\log _7 2=\lambda\), then the value of \(\log _{49}(28)\) is

Options:
A \((2 \lambda+1)\)
B \((2 \lambda+3)\)
C \(\frac{1}{2}(2 \lambda+1)\)
D \(2(2 \lambda+1)\)

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

Solution:
Hints: \(\log _{49} 28=\log _{72} 4 \times 7\)
\(=\frac{1}{2}\left[2 \log _7 2+\log _7 7\right]=\frac{1}{2}[2 \lambda+1]\)

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2011
If $M=\int_{0}^{n / 2} \frac{\cos x}{x+2} d x, N=\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{(x+1)^{2}} d x,$ then
the value of $M-N$ is

Options:
A $\pi$
B $\frac{\pi}{4}$
C $\frac{2}{\pi-4}$
D $\frac{2}{\pi+4}$

The correct answer is:
$\frac{2}{\pi+4}$

Solution:
Given, $M=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(x+2)} d x$
$\begin{aligned} \text { and } \quad N &=\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{(x+1)^{2}} d x \\ &=\int_{0}^{\pi / 4} \frac{1}{2} \cdot \frac{\sin 2 x}{(x+1)^{2}} d x \end{aligned}$
Put $2 x=t \Rightarrow d x=\frac{a}{2}$
$\begin{aligned} \therefore \quad N &=\int_{0}^{\frac{\pi}{2}} \frac{\sin t}{4(t / 2+1)^{2}} d t \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sin t}{(t+2)^{2}} d t \end{aligned}$
$\therefore M-N=\int_{0}^{\frac{\pi}{2}}(\cos x) \cdot \frac{1}{(x+2)} d x$
$-\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(x+2)^{2}} d x$
$=\left(\frac{\sin x}{x+2}\right)_{0}^{\pi 2}-\int_{0}^{\pi 2}-\frac{\sin x}{(x+2)^{2}} d x$
$-\frac{\sin \pi / 2}{\pi / 2+2}=\frac{1}{\frac{\pi+4}{2}}=\frac{2}{\pi+4}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2018