| 1) D | 7) D | 13) B | 19) C | 25) B |
| 2) B | 8) A | 14) D | 20) C | 26) D |
| 3) C | 9) D | 15) D | 21) A | 27) B |
| 4) A | 10) A | 16) D | 22) A | 28) C |
| 5) B | 11) B | 17) B | 23) C | 29) D |
| 6) D | 12) B | 18) A | 24) B | 30) A |
If $x=1+\sqrt[5]{2}+\sqrt[5]{4}+\sqrt[5]{8}+\sqrt[3]{16}$, then the value of $\left(1+\frac{1}{x}\right)^{30}$ is
(a) 2
(b) 5
(c) 32
(d) 64
$x=1+2^{1 / 5}+2^{2 / 5}+2^{3 / 5}+2^{4 / 5}$
Let $a=2^{1 / 5}$
$\therefore x=1+a+a^2+a^3+a^4$
$=\frac{a^5-1 i}{a-1}=\frac{1}{2^{1 / 5}-1}$
$\begin{aligned} & \therefore \frac{1}{x}=2^{1 / 5}-1 \ & \quad\left(1+\frac{1}{x}\right)^{30}=2^6=64\end{aligned}$
Let $j$ be a number selected at random from ${\{1,2, \ldots, 2024}\}$. What is the probability that $j$ is divisible by 9 and 15 ?
(a) $\frac{1}{23}$
(b) $\frac{1}{46}$
(c) $\frac{1}{44}$
(d) $\frac{1}{253}$
$\operatorname{lcm}(9,15)=45$
$\lfloor \frac{2024}{45} \rfloor,=44.$
So the set of favourable integers has 44 elements.
So, probability = $\frac{44}{2024}=\frac{1}{46}$
Let $S_n$ be the set of all $n$-digit numbers whose digits are all 1 or 2 and there are no consecutive 2's. (Example: 112 is in $S_3$ but 221 is not in $S_3$ ). Then the number of elements in $S_{10}$ is
(a) $512$
(b) $256$
(c) $144$
(d) $89$
12th Fibonacci Number
$\sum_{k=0}^n\binom{n-k+1}{k}$
$=F_{n+2}$
=$144$
There are 30 True or False questions in an examination. A student knows the answer to 20 questions and guesses the answers to the remaining 10 questions at random. What is the probability that the student gets exactly 24 answers correct?
(a) $\frac{105}{2^9}$
(b) $\frac{105}{2^8}$
(c) $\frac{105}{2^{10}}$
(d) $\frac{4}{2^{10}}$
Binomial Distribution
$x$ be the R.V. of selecting unknown questions. (answer is unknown)
Then $x \sim \operatorname{Bin}(10,1 / 2)$
So, $P(x=4)=\binom{10}{4} \frac{1}{2^{10}}=\frac{105}{2^9}$
Let $T$ be a right-angled triangle in the plane whose side lengths are in a geometric progression. Let $n(T)$ denote the number of sides of $T$ that have integer lengths. Then the maximum value of $n(T)$ over all such $T$ is
(a) $0$
(b) $1$
(c) $2$
(d) $3$
$a, a r, a r^2$
Now, $a^2+a^2 r^2=a^2 r^4$
$\Rightarrow 1+r^2=r^4$
But $r$ has a positive solution
So, only $a$ can be an integer $\max (n(T))=1$
Let $x_1, x_2, \ldots, x_n$ be non-negative real numbers such that $\sum_{i=1}^n x_i=1$. What is the maximum possible value of $\sum_{i=1}^n \sqrt{x_i}$ ?
(a) $1$
(b) $\sqrt{n}$
(c) $n^{3 / 4}$
(d) $n$
Cauchy Schwarz
$\left(\sum\left(\sqrt{\left.x_i-1\right)}\right)^2 \leq\left(2\left(\sqrt{x_i}\right)^2\right)\left(\sum 1^2\right)\right.$
$\therefore \sum \sqrt{x_1} \leqslant \sqrt{n}$
The precise interval on which the function $f(x)=\log _{1 / 2}\left(x^2-2 x-3\right)$ is monotonically decreasing, is
(a) $(-\infty,-1)$
(b) $(-\infty,1)$
(c) $(1, \infty)$
(d) $(3, \infty)$
$f(x)=-\log _2\left(x^2-2 x-3\right)$
$f^{\prime}(x)=-\frac{\ln^2}{\log _2(x-3)}-\frac{\ln^2}{\log _2(x+1)}$
in $(3, \infty)$
$f^{\prime}(x)<0 \quad \forall x$
So, in $(3, \infty), f(x)$ is monotonically decreasing.
The angle subtended at the origin by the common chord of the circles $x^2+y^2-6 x-6 y=0$ and $x^2+y^2=36$ is
(a) $\pi / 2$
(b) $\pi / 4$
(c) $\pi / 3$
(d) $2 \pi / 3$
The length of the chord
$=2 \sqrt{36-18}$
$=6 \sqrt{2}$
So, ABC is an isosceles triangle. $ (\frac{\pi}{2}) $
In $\triangle A B C, C D$ is the median and $B E$ is the altitude. Given that $\overline{C D}=\overline{B E}$, what is the value of $\angle A C D ?$
(a) $\pi / 3$
(b) $\pi / 4$
(c) $\pi / 5$
(d) $\pi / 6$
If the points $z_1$ and $z_2$ are on the circles $|z|=2$ and $|z|=3$, respectively, and the angle ineluded between these vectors is $60^{\circ}$, then the value of $\frac{\left|z_1+z_2\right|}{\left|z_1-z_2\right|}$ is
(a) $\sqrt{\frac{19}{7}}$
(b) $\sqrt{19}$
(c) $\sqrt{7}$
(d) $\sqrt{\frac{7}{19}}$
Let $n \geqslant 1$. The maximum possible number of primes in the set ${\{n+6, n+7, \ldots . n+34, n+35}\}$ is
(a) 7
(b) 8
(c) 12
(c) 13
Suppose 40 distinguishable balls are to be distributed into 4 different boxes such that each box gets exactly 10 balls. Out of these 40 balls, 10 are defective and 30 are non-defective. In how many ways can the balls be distributed such that all the defective balls go to the first two boxes?
(a) $\frac{40!}{(10!)^4}$
(b) $\frac{30!\cdot 20!}{(10!)^5}$
(c) $\frac{20!\cdot 20!}{(10!)^5}$
(d) $\frac{30!-10!}{(10!)^4}$
The number of elements in the set
${\{x: 0 \leqslant x \leqslant 2,\left|x-x^5\right|=\left|x^5-x^6\right|}\}$ is
(a) 2
(b) 3
(c) 4
(d) 5
For $0 \leq x \leq 1$
$x=0$ is two solutions.
$x=1$
if $0<x<1$ then,
$x-x^5=x^3-x^6$
$\Rightarrow x+x^6=2 x^5$
$\Rightarrow 1+x^3=2 x^4$
$\text { Consider }=x^5-2 x^4+1=f(x)$
$f^{\prime}(x)=5 x^4-8 x^3=x^3(5 x-8)=5 x^3\left(x-\frac{8}{5}\right)$
$\text { So, } f^{\prime}(x)<0 \text { for } x \in(0,1)$
$\text { So, Now sloution for } x \text { for (1) }$
$\text { in } x \in(0,1) \text {. }$
$\text { if } x \in(1,2) \text { then, }$
$x^5-x=x^6-x^5$
$x^6+x=2 x^5$
$x^5+1=2 x^4$
Consider,
$f(x)=x^5-2 x^4+1$
$f^{\prime}(x)=5 x^3\left(x-\frac{8}{5}\right) \quad \text { Here } x \in(1,2)$
So, at some point $f^{\prime}(x)$ changes sign for $x \in(1,2)$
$\text { Now, } f(1.5)<0$
$f(2)>0$
So, by $I VT$, there is a solution for $f$.
$\text { So, for } 0 x \in(1,2)$
there is a solution.
Total 3 solutions
In a room with $n \geqslant 2$ people, each pair shakes hands between themselves with probability $\frac{2}{n^2}$ and independently of other pairs. If $p_n$ is the probability that the total number of handshakes is at most 1 , then $\lim _{n \rightarrow \infty} p_n$ is equal to
(a) 0
(b) 1
(c) $e^{-1}$
(d) $2e^{-1}$
The number of positive solutions to the equation
$$e^x \sin x=\log x+e^{\sqrt{x}}+2$$
(a) 0
(b) 1
(c) 2
(d) $\infty$
$e^x \sin x=\log x+e^{\sqrt{x}}+2$
$f(x)=e^x \sin x$
$f^{\prime}(x)=e^x \cos x+e^x \sin x$
Now, $-e^x<e^x \sin x<R^x$
Now, $\begin{aligned} g(x)=\log +e^{\sqrt{x}}+2\end{aligned}$ is an increasing graph for $x>0$. So infinitely many.
Let $n>1$ be the smallest composite integer that is coprime to $\frac{10000}{9900 \%}$. Then
(a) $n \leqslant 100$
(b) $100<n \leqslant 9900$
(c) $9900<n \leqslant 10000$
(d) $n>10000$
Let $P={\{(x, y): x+1 \geqslant y, x \geqslant-1, y \geqslant 2 x}\}$. Then the minimum value of $(x+y)$ where $(x, y)$ varies over the set $P$ is
(a) $-1$
(b) $-3$
(c) $3$
(d) $0$
Let $A={1, \ldots, 5}$ and $B={1, \ldots, 10}$. Then the number of ordered pairs $(f, g)$ of functions $f: A \rightarrow B$ and $g: B \rightarrow A$ satisfying $(g \circ f)(a)=a$ for all $a \in A$ is
(a) $\frac{10!}{5!} \times 5^5$
(b) $5^{10} \times 5!$
(c) $10!\times 5!$
(d) $\binom{10}{5} \times 10^5$
Let
$$S=\frac{1}{\sqrt{10000}}+\frac{1}{\sqrt{10001}}+\cdots+\frac{1}{\sqrt{160000}}$$
Then the largest positive integer not exceeding $S$ is
(a) $200$
(b) $400$
(c) $600$
(d) $800$
The real number $x$ satisfies
$$
\frac{|x|^2-|x|-2}{2|x|-|x|^2-2}>2
$$
if and only if $x$ belongs to
(a) $(-2,-1) \cup(1,2)$
(b) $(-2 / 3,0) \cup(0,2 / 3)$
(c) $(-1,-2 / 3) \cup(2 / 3,1)$
(d) $(-1,0) \cup(0,1)$
Consider points of the form $\left(n, n^k\right)$, where $n$ and $k$ are integers with $n \geq 0, k \geq 1$. How many such points are strictly inside the circle of radius 10 with centre at the origin?
(a) $11$
(b) $12$
(c) $15$
(d) $17$
Let $n>1$, and let us arrange the expansion of $\left(x^{1 / 2}+\frac{1}{2 x^{1 / 4}}\right)^n$ in decreasing powers of $x$. Suppose the first three coefficients are in arithmetic progression. Then, the number of terms where $x$ appears with an integer power, is
(a) $3$
(b) $2$
(c) $1$
(d) $0$
The limit
$$
\lim _{n \rightarrow \infty} \frac{2 \log 2+3 \log 3+\cdots+n \log n}{n^2 \log n}
$$
is equals to
(a) $0$
(b) $1 / 4$
(c) $1 / 2$
(d) $1$
$\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{1 \log 1+2 \log \alpha+\cdots+n \log n}{n \log n}\right]$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{r=1}^n$ $\frac{r \log r}{n \log n}$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{\gamma=1}^n .$ $\frac{\left(\frac{\gamma}{n}\right)\left[\log \left(\frac{\gamma}{n}\right)+\log n\right]}{\log n}$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{\gamma=1}^n .$ $\frac{\left(\frac{\gamma}{n}\right) \log \left(\frac{\gamma}{n}\right)}{\log (n)}$ $+\lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{r=1}^n\left(\frac{r}{n}\right)$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{\log (n)}$ $\frac{1}{n} \cdot \sum_{r=1}^n\left(\frac{r}{n}\right) \log \left(\frac{r}{n}\right)$ $+\lim _{n \rightarrow \infty} \frac{n(n+1)}{2 n^2}$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{\log (n)}$ $× \lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{\gamma=1}^n\left(\frac{\gamma}{n}\right) \log \left(\frac{\gamma}{n}\right)+\frac{1}{2}$
$=\lim _{x \rightarrow \infty} \frac{1}{\log (x)} \times \int_0^0 x \log (x)+\frac{1}{2}$
$=0 \times 0+\frac{1}{2}$
$=\frac{1}{2}$
Let $p<q$ be prime numbers such that $p^2+q^2+7 p q$ is a perfect square. Then, the largest possible value of $q$ is:
(a) $7$
(b) $11$
(c) $23$
(d) $29$
$p^2+q^2+r p q=x^2 $ with $x>p+q \in \mathbb{N}$
$\Rightarrow \quad(p+q)^2+5 p q=x^2$
$\Rightarrow \quad 5 p q=x^2-(p+q)^2$
$\Rightarrow \quad 5 p q=(x+p+q)(x-p-q)$
Considering $x+p+q>x-p-q$ & $q>p$,
Case 1:
$x+p+q=5 p ; 5 q$
$x-p-q=q \text p$
$\Rightarrow 2 p+2 q=5 p-q$ $\text { (or) } 5 q-p$
$\Rightarrow p=q,$ absurd
Case 2:
$x+p+q=q$
$x-p-q=5 p$
$\quad \Rightarrow \quad 2 p+2 q=q-5 p$
$\quad \Rightarrow \quad q=7 p$ not possible
Case 3:
$x+p+q=5 p q$
$x-p-q=1$
$\quad \Rightarrow$ $2 p+2 q=5 p q-1$
$\quad \Rightarrow$ $(5 q-2)(5 p-2)=7 $,
but $5 q-2>7$, not possible
Case 4:
$x+p+q=p q$
$x-p-q=5$
$\Rightarrow 2 p+2 q=p q-5$
$\quad \Rightarrow \quad(p-2)(q-2)=9$
$\Rightarrow \quad q-2=q, \quad p-2=1$
$\Rightarrow \boxed {\quad q=11, p=3}$
The set of all real numbers $x$ for which $3^{2^{1-x^2}}$ is an integer has
(a) $3$ elements
(b) $15$ elements
(c) $24$ elements
(d) infinitely many elements.
The maximum possible value of $1-x^2$ is 1 $(when =0)$ and the minimum value is $(-\infty)$. So,
$-\infty<1-x^2 \leqslant 1$
and it wan take any value in the intervene So,
$0<2^{1-x^2} \leqslant 2$
Thus,
$1=3^0<3^{2^{1-2}} \leq 3^2=9$
Hence, $3^{2^{1-x^2}}$ an take all the putegors 2 to 9 and each of which has two corresponding valet for $x$ except when $x=0$. Thus, the number of such $x=2 \times 8-1=15$
Let $a, b, c$ be three complex numbers. The equation
$$a z+b \bar{z}+c=0$$
represents a straight line on the complex plane if and only if
(a) $a = b$
(b) $\tilde{a} c=b \vec{c}$
(c) $|a|=|b| \neq 0$
(d) $|a|=|b| \neq 0$ and $\bar{a} c=b \bar{c}$
In the adjoining figure, $C$ is the centre of the circle drawn, $A, F, E$ lie on the circle and $B C D F$ is a rectangle. If $\frac{D E}{A B}=2$, then $\frac{F E}{F A}$ equals
(a) $\sqrt{\frac{3}{2}}$
(b) $\sqrt{2}$
(c) $\sqrt{\frac{5}{2}}$
(d) $\sqrt{3}$
For every increasing function $b:[1, \infty) \rightarrow[1, \infty)$ such that
$$
\int_1^{\infty} \frac{\mathrm{d} x}{b(x)}<\infty
$$
we must have
(a) $\sum_{k=1}^{\infty} \frac{\sqrt{\log k}}{b(k)}<\infty$
(b) $\sum_{k=3}^{\infty} \frac{\log k}{b(\log k)}<\infty$
(c) $\sum_{k=1}^{\infty} \frac{e^k}{b\left(e^k\right)}<\infty$
(d) $\sum_{k=3}^{\infty} \frac{1}{\sqrt{b(\log k)}}<\infty$
Consider the following two statements:
(I) There exists a differentiable function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $g\left(x^3+x^5\right)=e^x-100$.
(II) There exists a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $g\left(e^x\right)=x^3+x^5 $
Then
(a) Only (I) is correct
(b) Only (II) is correct
(c) Both (I) and (II) are correct.
(d) Neither (I) nor (II) is correct.
(I)
$g\left(x^3+x^5\right)=e^x-100$
differentiating both sides, we get
$g^{\prime}\left(x^3+x^5\right)\left(3 x^2+5 x^4\right)=e^x$
Note that the LHS is well-defined if $g(x)$ is differentiable substitute
$x=0 \Rightarrow 0=1$, which is absurd.
Hence, such a fraction $g(x)$ cannot be differentiable (I) is incorrect.
(II)
$g\left(e^x\right)=x^3+x^5$
$g(0)=\lim _{x \rightarrow-\infty} g\left(e^x\right)$ because $\lim _{x \rightarrow-\infty} e^x=0$ and that $g$ is continuous.
But notice that $\operatorname{Lim}_{x \rightarrow-\infty} x^3+x^5=-\infty$
$\Rightarrow \quad \lim _{x \rightarrow-\infty} g\left(e^x\right)=-\infty=g(0)$, which is absurd if we assume "g" as continuous (bounded on compact sets)
(II) is incorrect
Neither (I) nor (II) is correct
Let $\psi: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with $\int_{-1}^1 \psi(x) \mathrm{d} x=1$.
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. Then
$$
\lim_ {\varepsilon \rightarrow 0} \frac{1}{\varepsilon} \int_{1-\varepsilon}^{1+\varepsilon} f(y) \psi\left(\frac{1-y}{\varepsilon}\right) \mathrm{dy}
$$
equals
(a) $f(1)$
(b) $f(1) \psi(0)$
(c) $f^{\prime}(1) \psi(0)$
(d) $f(1) \psi(1)$
$\lim_{\varepsilon \mapsto 0} \frac{1}{\varepsilon} \int_{1-\varepsilon}^{1+\varepsilon} f(y) \psi\left(\frac{1-y}{\varepsilon}\right) d y$
Substitute $\frac{1-y}{\varepsilon}=t$ $\Rightarrow 1-y=\varepsilon t$ $\Rightarrow 1-\varepsilon t=y$
$1-y=\varepsilon t$
$-d y=\varepsilon d t$
=$\lim_{\varepsilon \rightarrow 0} \int_{-1}^{1} f(1-\varepsilon t) \psi(t)$
=$f(1)$
$\left(x^{1 / 2}+\frac{1}{2 x^{1 / 4}}\right)^n$
General Term, $\binom{n}{r}\left(x^{1 / 2}\right)^{n-r}\left(\frac{1}{2 x^{1 / a}}\right)^r$
$\binom{n}{r}x \frac{2 n-3 r}{4}$
Put, $r=0,1,2$
$\binom{n}{0},\binom{n}{1}\left(\frac{1}{2}\right),\binom{n}{2}\left(\frac{1}{2}\right)^2$ are in A.P.
So, $n=1$ or 8
$\therefore n=8$.
Now simple calculation, gives (A)
Attend trial class for join live classes