Jee Main Jan 30 Shift 1 Mathematics
Question 1
In an arithmetic progression if sum of 20 terms is 790 and sum of 10 terms is 145 , then $S _{15}-S _5$ is (when $S _n$ denotes sum of $n$ terms)
(1) 400
(2) 395
(3) 385
(4) 405
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Answer: (2)
Solution:
$S _{20}=\frac{20}{2}[2 a+19 d]=790$
$2 a+19 d=79$
$S _{10}=\frac{10}{2}[2 a+9 d]=145$
$2 a+9 d=29$
from (1) and (2) $a=-8, \quad d=5$
$S _{15}-S _5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d]$
$=\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20]$
$=405-10$
$=395$
Question 2
If the foot of perpendicular from $(1,2,3)$ to the line $\frac{x+1}{2}=\frac{y-2}{5}=\frac{z-4}{1}$ is $(\alpha, \beta, \gamma)$ then find $\alpha+\beta+\gamma$
(1) 6
(2) 5.8
(3) 4.8
(4) 5
Show Answer
Answer: (2)
Solution:
$(\alpha-1) \times 2+(\beta-2) \times 5+(\gamma-3) \times 1=0$
$2 \alpha+5 \beta+\gamma-15=0$
Also, $P$ lie on line
$\Rightarrow \alpha+1=2 \lambda$
$\beta-2=5 \lambda$
$$ \begin{aligned} & \gamma-4=\lambda \\ \Rightarrow & 2(2 \lambda-1)+5(5 \lambda+2)+\lambda+4-15=0 \\ \Rightarrow & 4 \lambda+25 \lambda+\lambda-2+10+4-15=0 \\ & 30 \lambda-3=0 \\ \Rightarrow & \lambda=\frac{1}{10} \\ \Rightarrow & \alpha+\beta+\gamma=(2 \lambda-1)+(5 \lambda+2)+(\lambda+4) \\ & 8 \lambda+5=\frac{8}{10}+5=5.8 \end{aligned} $$
Question 3
$\lim _{n \rightarrow \infty} \sum _{k=1}^{n} \frac{n^{3}}{\left(n^{2}+k^{2}\right)\left(n^{2}+3 k^{2}\right)}$
(1) $\frac{\pi}{2 \sqrt{3}}-\frac{\pi}{8}$
(2) $\frac{\pi}{2 \sqrt{3}}+\frac{\pi}{8}$
(3) $\frac{\pi}{2}-\frac{\pi}{\sqrt{3}}$
(4) $\frac{\pi}{\sqrt{3}}-\frac{\pi}{4}$
Show Answer
Answer: (1)
Solution:
$\lim _{n \rightarrow \infty} \sum _{k=1}^{n} \frac{n^{3}}{n^{4}\left(1+\frac{k^{2}}{n^{2}}\right)\left(1+\frac{3 k^{2}}{n^{2}}\right)}$
$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum _{k=1}^{n} \frac{1}{\left(1+\frac{k^{2}}{n^{2}}\right)\left(1+\frac{3 k^{2}}{n^{2}}\right)}$
$=\int _0^{1} \frac{d x}{3\left(1+x^{2}\right)\left(\frac{1}{3}+x^{2}\right)}$
$=\int _0^{1} \frac{1}{3} \times \frac{3}{2} \frac{\left(x^{2}+1\right)-\left(x^{2}+\frac{1}{3}\right)}{\left(1+x^{2}\right)\left(x^{2}+\frac{1}{3}\right)} d x$
$=\frac{1}{2} \int _0^{1}\left[\frac{1}{x^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}}-\frac{1}{1+x^{2}}\right] d x$
$=\frac{1}{2}\left[\sqrt{3} \tan ^{-1}(\sqrt{3} x)\right] _0^{1}-\frac{1}{2}\left(\tan ^{-1} x\right) _0^{1}$
$=\frac{\sqrt{3}}{2}\left(\frac{\pi}{3}\right)-\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{2 \sqrt{3}}-\frac{\pi}{8}$
Question 4
The value of maximum area possible of a $\triangle A B C$ such that $A(0,0)$ and $B(x, y)$ and $C(-x, y)$ such that $y=-2 x^{2}+54 x$ is (in sq. unit)
(1) 5800
(2) 5832
(3) 5942
(4) 6008
Show Answer
Answer: (2)
Solution:
Area of $\Delta$
$=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ x & y & 1 \\ -x & y & 1\end{array}\right|$
$\Rightarrow\left|\frac{1}{2}(x y+x y)\right|=|x y|$
Area $(\Delta)=|x y|=\left|x\left(-2 x^{2}+54 x\right)\right|$
$\frac{d(\Delta)}{d x}=\left|\left(-6 x^{2}+108 x\right)\right| \Rightarrow \frac{d \Delta}{d x}=0$ at $x=0$ and 18
$\Rightarrow$ at $x=0$, minima
and at $x=18$ maxima
Area $(\Delta)=\left|18\left(-2(18)^{2}+54 \times 18\right)\right|=5832$
Question 5
The range of $r$ for which circles $(x+1)^{2}+(y+2)^{2}=$ $r^{2}$ and $x^{2}+y^{2}-4 x-4 y+4=0$ coincide at two distinct points
(1) $3<r<7$
(2) $5<r<9$
(3) $\frac{1}{2}<r<4$
(4) $0<r<3$
Show Answer
Answer: (1)
Solution:
If two circles intersect at two distinct points
$\Rightarrow\left|r _1-r _2\right|<C _1 C _2<r _1+r _2$
$|r-2|<\sqrt{9+16}<r+2$
$|r-2|<5 \quad$ and $r+2>5$
$-5<r-2<5 \quad r>3$
$-3<r<7$
From (1) and (2)
$3<r<7$
Question 6
An ellipse whose length of minor axis is equal to half of length between foci, then eccentricity is
(1) $\frac{7}{2}$
(2) $\sqrt{17}$
(3) $\frac{2}{\sqrt{5}}$
(4) $\frac{3}{\sqrt{7}}$
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Answer: (3)
Solution:
$\because a e=2 b$
$\therefore \frac{4 b^{2}}{a^{2}}=e^{2}$
Or $4\left(1-e^{2}\right)=e^{2}$
$\therefore 4=5 e^{2} \Rightarrow e=\frac{2}{\sqrt{5}}$
Question 7
If $g^{\prime}\left(\frac{3}{2}\right)=g^{\prime}\left(\frac{1}{2}\right)$ and
$f(x)=\frac{1}{2}[g(x)+g(2-x)]$ and $f^{\prime}\left(\frac{3}{2}\right)=f^{\prime}\left(\frac{1}{2}\right)$ then
(1) $f^{\prime \prime}(x)=0$ has exactly one root in $(0,1)$
(2) $f^{\prime \prime}(x)=0$ has no root in $(0,1)$
(3) $f^{\prime \prime}(x)=0$ has at least two roots in $(0,2)$
(4) $f^{\prime \prime}(x)=0$ has 3 roots in $(0,2)$
Show Answer
Answer: (3)
Solution:
$f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0$
Also $f^{\prime}\left(\frac{1}{2}\right)=\frac{g^{\prime}\left(\frac{1}{2}\right)-g^{\prime}\left(\frac{3}{2}\right)}{2}=0, f^{\prime}(1)=0$
$\Rightarrow f^{\prime}\left(\frac{3}{2}\right)=f^{\prime}\left(\frac{1}{2}\right)=0$
$\Rightarrow$ roots in $\left(\frac{1}{2}, 1\right)$ and $\left(1, \frac{3}{2}\right)$
$\Rightarrow f^{\prime \prime}(x)$ is zero at least twice in $\left(\frac{1}{2}, \frac{3}{2}\right)$
Question 8
The domain of $y=\cos ^{-1}\left|\frac{2-|x|}{4}\right|+(\log (3-x))^{-1}$ is $[-\alpha, \beta)-{\gamma}$, then value of $\alpha+\beta+\gamma=$ ?
(1) 9
(2) 12
(3) 11
(4) 10
Show Answer
Answer: (3)
Solution:
$-1 \leq\left|\frac{2-|x|}{4}\right| \leq 1$
$\Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1$
$\Rightarrow-1 \leq \frac{2-|x|}{4} \leq 1$
$-4 \leq 2-|x| \leq 4$
$-6 \leq-|x| \leq 2$
$-2 \leq|x| \leq 6$
$|x| \leq 6$
$\Rightarrow x \in[-6,6]$
Now, $3-x \neq 1$
And $x \neq 2$
and $3-x>0$
$x<3$
From (1), (2) and (3)
$\Rightarrow x \in[-6,3]-{2}$
$\alpha=6$
$\beta=3$
$\gamma=2$
$\alpha+\beta+\gamma=11$
Question 9
If $y=f(x)$ is solution of differential equation $\left(x^{2}-1\right)$ $d y=\left(\left(x^{3}+1\right)+\sqrt{1-x^{2}}\right) d x$ and $y(0)=2$ then find $y\left(\frac{1}{2}\right)$.
(1) $\frac{13}{7}-\frac{\pi}{2}+\ln 5$
(2) $\frac{15}{7}+\frac{\pi}{3}+\ln 2$
(3) $\frac{17}{8}+\frac{\pi}{6}-\ln 2$
(4) $\frac{18}{7}-\frac{\pi}{6}+\ln 3$
Show Answer
Answer: (3)
Solution:
$\frac{d y}{d x}=\frac{(x+1)\left(x^{2}-x+1\right)+\sqrt{(1-x)(1+x)}}{(x-1)(x+1)}$
$\Rightarrow \frac{d y}{d x}=\frac{x(x-1)+1}{(x-1)}+\sqrt{\frac{(1-x)(1+x)}{(x-1)^{2}(x+1)^{2}}}$
$\frac{d y}{d x}=x+\frac{1}{x-1}+\frac{1}{\sqrt{(1-x)(1+x)}}$
$\Rightarrow d y=x d x+\frac{1}{(x-1)} d x+\frac{d x}{\sqrt{1-x^{2}}}$ $\Rightarrow y=\frac{x^{2}}{2}+\ln |x-1|+\sin ^{-1} x+c$
at $x=0, y=2 \Rightarrow 2=c$
$\Rightarrow y=\frac{x^{2}}{2}+\ln |x-1|+\sin ^{-1} x+2$
$$ y\left(\frac{1}{2}\right)=\frac{17}{8}+\frac{\pi}{6}-\ln 2 $$
Question 10
Given $x^{2}-70 x+\lambda=0$ with positive roots $\alpha$ and $\beta$ where one of the root is less than 10 and $\frac{\lambda}{2}$ and $\frac{\lambda}{3}$ are not integers then find value of $\frac{\sqrt{\alpha-1}+\sqrt{\beta-1}}{|\alpha-\beta|}$ is equal to
(1) $\frac{1}{5}$
(2) $\frac{1}{12}$
(3) $\frac{1}{60}$
(4) $\frac{1}{70}$
Show Answer
Answer: (1)
Solution:
Given : $x^{2}-70 x+\lambda=0$
$\Rightarrow$ Let roots be $\alpha$ and $\beta$
$\Rightarrow \beta=70-\alpha$
$\lambda=\alpha(70-\alpha)$
$\lambda$ is not divisible by 2 and 3
$\Rightarrow \alpha=5, \beta=65$
$\Rightarrow \frac{\sqrt{5-1}+\sqrt{65-1}}{|60|}=\left|\frac{4+8}{60}\right|=\frac{1}{5}$
Question 11
A line passes through $(9,0)$, making angle $30^{\circ}$ with positive direction of $x$-axis. It is rotated by angle of $15^{\circ}$ with respect to $(9,0)$. Then one of the equation of new line is
(1) $y=(2+\sqrt{3})(x-9)$
(2) $y=(2-\sqrt{3})(x-9)$
(3) $y=2(x-9)$
(4) $y=-(x-9)$
Show Answer
Answer: (2)
Solution:
$Eq^{n}: y-0=\tan 15^{\circ}(x-9) \Rightarrow y=(2-\sqrt{3})(x-9)$
$E q^{n}: y-0=\tan 45^{\circ}(x-9) \Rightarrow y=(x-9)$
Option (B) is correct
Question 12
For a non-zero complex number $z$ satisfying $z^{2}+\bar{z}=0$, then value of $|z|^{2}$ is
(1) 1
(2) 2
(3) 3
(4) 4
Show Answer
Answer: (1)
Solution:
$z^{2}=-\bar{z}$
$\left|z^{2}\right|=|-\bar{z}|$
$\left|z^{2}\right|=|z|$
$|z|^{2}-|z|=0$
$|z|(|z|-1)=0$
$|z|=0$ (not acceptable)
$\therefore|z|=1$
$\therefore|z|^{2}=1$
Question 13
If $|\vec{a}|=1,|\vec{b}|=4$
$\vec{a} \cdot \vec{b}=2$ and $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$
Then the angle between $\vec{b}$ and $\vec{c}$ is
(1) $\theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
(2) $\theta=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
(3) $\theta=\cos ^{-1}\left(\frac{1}{2}\right)$
(4) $\theta=\cos ^{-1}\left(\frac{-1}{2}\right)$
Show Answer
Answer: (1)
Solution:
Given $|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$
$$ \vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b} $$
Dot product with $\vec{a}$ on both sides
$\vec{c} \cdot \vec{a}=-6$
Dot product with $\vec{b}$ on both sides
$\vec{b} \cdot \vec{c}=-48$
$\vec{c} \cdot \vec{c}=4|\vec{a} \times \vec{b}|^{2}+9|\vec{b}|^{2}$
$|\vec{c}|^{2}=4\left[|\vec{a}|^{2} \cdot|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}\right]+9|\vec{b}|^{2}$
$|\vec{c}|^{2}=4\left[(1)(4)^{2}-(4)\right]+9(16)$ $|\vec{c}|^{2}=4[12]+144$
$|\vec{c}|^{2}=48+144$
$|\vec{c}|^{2}=192$
$\therefore \quad \cos \theta=\frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|}$
$\cos \theta=\frac{-48}{\sqrt{192} \cdot 4}$
$\cos \theta=\frac{-48}{8 \sqrt{3} \cdot 4}$
$\cos \theta=\frac{-3}{2 \sqrt{3}}$
$\cos \theta=\frac{-\sqrt{3}}{2} \quad \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
Question 14
Given set $S={0,1,2,3, \ldots . ., 10}$. If a random ordered pair $(x, y)$ of elements of $S$ is chosen, then find probability that $|x-y|>5$
(1) $\frac{30}{121}$
(2) $\frac{31}{121}$
(3) $\frac{62}{121}$
(4) $\frac{64}{121}$
Show Answer
Answer: (1)
Solution:
If $x=0, y=6,7,8,9,10$
If $x=1, y=7,8,9,10$
If $x=2, y=8,9,10$
If $x=3, y=9,10$
If $x=4, y=10$
If $x=5, y=$ no possible value
Total possible ways $=(5+4+3+2+1) \times 2$
$$ =30 $$
Required probability $=\frac{30}{11 \times 11}=\frac{30}{121}$
Question 15
Number of integral terms in the binomial expansion of $\left(7^{1 / 2}+11^{1 / 6}\right)^{824}$ is
Show Answer
Answer: (138)
Solution:
$T _{n+1}={ }^{n} C _r 11^{\frac{r}{6}} \cdot 7^{\frac{824-r}{2}}$
For integral term
6 should divide $r$
and $\frac{824-r}{2}$ must be integer
$\Rightarrow 2$ most divide $r$
$\Rightarrow r$ divisible by 6
$\Rightarrow$ possible values of $r \in{0,1,2, \ldots 824}$
$\Rightarrow$ For integer terms
$r \in{0,6,12, \ldots 822} \quad(822=0+(n-1) 6 \Rightarrow n=138)$
$=138$ terms
Question 16
$9 \int _0^{9}\left[\sqrt{\frac{10 x}{x+1}}\right] d x$ is equal to (where [ ] represents greatest integer function)
Show Answer
Answer: (155)
Solution:
$I=9 \int _0^{9}\left[\sqrt{\frac{10 x}{x+1}}\right] d x$
$=9\left[\int _0^{1 / 9} 0 d x+\int _{1 / 9}^{2 / 3} d x+\int _{2 / 3}^{9} 2 d x\right]$
$=9\left[\frac{2}{3}-\frac{1}{9}+2\left[9-\frac{2}{3}\right]\right]$
$=9\left[\frac{5}{9}+2 \times \frac{25}{3}\right]$
$=5+6 \times 25$
$=5+150$
$=155$
Question 17
In a class there are 40 students. 16 passed in Chemistry, 20 passed in Physics, 25 passed in Mathematics. 15 students passed in both Mathematics and Physics. 15 students passed in both Mathematics and Chemistry and 10 students passed in both Physics and Chemistry. Find the maximum number of students that passed in all the subjects.
Show Answer
Answer: (19)
Solution:
$n(C)=16, n(P)=20, n(M)=25$
$n(M \cap P)=n(M \cap C)=15, n(P \cap C)=10$,
$n(M \cap C \cap P)=x$.
$n(C \cup P \cup M) \leq n(U)=40$
$n(C \cup P \cup M)=n(C)+n(P)+n(M)-n(C \cup M)-$ $n(P \cup M)-n(C \cap P)+n(C \cap P \cap M)$
$40 \geq 16+20+25-15-15-10+x$
$40 \geq 61-40+x$
$19 \geq x$
So maximum number of students that passed all the exams is 19 .
Question 18
For the following data table
| $x _i$ | $f _i$ |
|---|---|
| $0-4$ | 2 |
| $4-8$ | 4 |
| $8-12$ | 7 |
| $12-16$ | 8 |
| $16-20$ | 6 |
Find the value of $20 M$ (where $M$ is median of the data)
Show Answer
Answer: (245)
Solution:
| $x _i$ | $f _i$ | $c . f$. |
|---|---|---|
| $0-4$ | 2 | 2 |
| $4-8$ | 4 | 6 |
| $8-12$ | 7 | 13 |
| $12-16$ | 8 | 21 |
| $16-20$ | 6 | 27 |
$N=\sum f=27$
$\left(\frac{N}{2}\right)=\frac{27}{2}=13.5$
So, we have median lies in the class $12-16$
$l _1=12, f=8, h=4$, c.f. $=13$
So, here we apply formula
$M=I _1+\frac{\frac{N}{2}-\text { c.f. }}{f} \times h=12+\frac{13.5-13}{8} \times 4$
$=12+\frac{.5}{2}$
$M=\frac{24.5}{2}=12.25$
$20 M=20 \times 12.25$
$=245$
Question 19
Set $A={1,2,3,4,5,6,7}$
If number of functions from set $A$ to power set of $A$ can be expressed as $m^{n}$ ( $m$ is least integer). Find $m+n$.
Show Answer
Answer: (51)
Solution:
$n P(A)=2^{7}=128$
$f: A \rightarrow B$
Number of function $=128 \times 128 \ldots .128=128^{7}$
$=\left(2^{7}\right)^{7}=2^{49}$
$\Rightarrow m^{n}=2^{49}$
$\therefore m+n=49+2=51$
