Top 25 Repeated JEE Mains Questions to Score 90+ in Mathematics

High-weightage Chapters in JEE Main Maths 2027

Some chapters contribute to a large portion of the questions asked every year in JEE Main. Students can study these high weightage chapters from the entire JEE Main Maths syllabus to score maximum marks. Students can start with NCERT to get the basic idea about the type of questions covered in these chapters. Given below are 6 chapters based on previous year question analysis that are most important and frequently asked in JEE Main.

Most Repeated Topics in JEE Main Maths 2027

To score maximum marks it is important to focus more on the specific topics that are asked in JEE Main.These topics are those from which JEE Main Mathematics high scoring questions are asked, instead of studying the entire chapter prioritizing the topics mentioned below in the form of table:

Top 25 Repeated JEE Main Maths Questions with Solutions 2027

The JEE Main Mathematics PYQs most repeated are based on concepts that appear repeatedly in previous-year exams. These questions are highly recommended and mostly based on NCERT that helps students to score good marks in JEE main. Refer to the JEE Main Maths questions for 90+ marks given below:

Question 1: Let PQ be a focal chord of the parabola $y^2=36 x$ of length 100, making an acute angle with the positive x-axis. Let the ordinate of P be positive and M be the point on the line segment PQ such that $\mathrm{PM}: \mathrm{MQ}=3: 1$. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line PQ?

(1) $(3,33)$

(2) $(6,29)$

(3) $(-6,45)$

(4) $(-3,43)$

Answer:

$9\left(t+\frac{1}{t}\right)^2=100$

$t=3$

$\Rightarrow \mathrm{P}(81,54) \& Q(1,-6)$

$\mathrm{M}(21,9)$

$\Rightarrow L$ is $(y-9)=\frac{-4}{3}(x-21)$

$3 y-27=-4 x+84$

$4 x+3 y=111$

Hence, the answer is option 4.

Question 2: Let the locus of the center $(\alpha, \beta), \beta>0$,, of the circle which touches the circle $x^2+(y-1)^2=1$ externally and also touches the x-axis be L. Then the area bounded by L and the line $y=4$ is:

(1) $\frac{32 \sqrt{2}}{3}$

(2) $\frac{40 \sqrt{2}}{3}$

(3) $\frac{64}{3}$

(4) $\frac{32}{3}$

Answer:

$(\alpha-0)^2+(\beta-1)^2=(\beta+1)^2$

$\alpha^2=4 \beta$

$x^2=4 y$

$A=2 \int_0^4\left(4-\frac{x^2}{4}\right) d x$

$=2\left[4 x-\frac{x^3}{12}\right]_0^4$

$=\frac{64}{3}$

Hence, the answer is the option (3).

Question 3: Let the abscissae of the two points P and Q on a circle be the roots of $x^2-4 x-6=0$ and the ordinates of P and Q be the roots of $y^2+2 y-7=0$. If PQ is the diameter of the circle $x^2+y^2+2 a x+2 b y+c=0$, then the value of $(a+b-c)$ is

(1) 12

(2) 13

(3) 14

(4) 16

Answer:

The equation of the circle with PQ as the diameter is

$x^2-4 x-6+y^2+2 y-7=0$

$\Rightarrow x^2+y^2-4 x+2 y-13=0$

$\therefore a=-2, b=1, c=-13$

$\therefore a+b-c=12$

Hence, the answer is the option (1).

Question 4: If the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the line $\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$ on the y-axis and the line $\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$ on the y-axis, then the eccentricity of the ellipse is

(1) $\frac{5}{7}$

(2) $\frac{2 \sqrt{6}}{7}$

(3) $\frac{3}{7}$

(4) $\frac{2 \sqrt{5}}{7}$

Answer:

Line 1 and ellipse cut x-axis at x=7 and line 2 and ellipse cut y-axis at $\mathrm{y}=2 \sqrt{6}$.

$\therefore \quad \mathrm{a}=7, \quad \mathrm{~b}=2 \sqrt{6}$

$\mathrm{e}^2=1-\frac{\mathrm{b}^2}{\mathrm{a}^2}=1-\frac{24}{49}=\frac{25}{49}$

$\Rightarrow \mathrm{e}=\frac{5}{7}$

Hence, the answer is the option (1).

Question 5: If for all real triplets $(a, b, c), f(x)=a+b x+c x^2$ then $\int_0^1 f(x) d x$ is equal to:

(1) $2\left\{3 f(1)+2 f\left(\frac{1}{2}\right)\right\}$

(2) $\frac{1}{3}\left\{f(0)+f\left(\frac{1}{2}\right)\right\}$

(3) $\frac{1}{2}\left\{f(1)+3 f\left(\frac{1}{2}\right)\right\}$

(4) $\frac{1}{6}\left\{f(0)+f(1)+4 f\left(\frac{1}{2}\right)\right\}$

Answer:

$\int_0^1\left(a+b x+c x^2\right) d x=a x+\frac{b x^2}{2}+\left.\frac{c x^3}{3}\right|_0 ^1=a+\frac{b}{2}+\frac{c}{3}$

f(1) = a + b + c

f(0) = a

$f\left(\frac{1}{2}\right)=a+\frac{b}{2}+\frac{c}{4}$

Hence, the answer is option 4.

Question 6: The integral $\int \frac{d x}{(x+4)^{8 / 7}(x-3)^{6 / 7}}$ is equal to :

(where C is a constant of integration)

(1) $-\left(\frac{x-3}{x+4}\right)^{-1 / 7}+C$

(2) $\frac{1}{2}\left(\frac{x-3}{x+4}\right)^{3 / 7}+C$

(3) $\left(\frac{x-3}{x+4}\right)^{1 / 7}+C$

(4) $-\frac{1}{13}\left(\frac{x-3}{x+4}\right)^{-13 / 7}+C$

Answer:

$\int\left(\frac{x-3}{x+4}\right)^{\frac{-6}{7}} \frac{1}{(x+4)^2} d x$ Let $\frac{x-3}{x+4}=t^7 \frac{7}{(x+4)^2} d x=7 t^6 d t \int t^{-6} t^6 d t=t+c\left(\frac{x-3}{x+4}\right)^{\frac{1}{7}}+c$

Hence, the answer is option 3.

Question 7: Let $f(x)=\int \frac{\sqrt{x}}{(1+x)^2} d x(x \geqslant 0)$. Then $f(3)-f(1)$ is equal to:

(1) $-\frac{\pi}{12}+\frac{1}{2}+\frac{\sqrt{3}}{4}$

(2) $\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$

(3) $-\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$

(4) $\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$

Answer:

$f(x)=\int_1^3 \frac{\sqrt{x} d x}{(1+x)^2}=\int_1^{\sqrt{3}} \frac{t \cdot 2 t d t}{\left(1+t^2\right)^2}($ put $\sqrt{x}=t)$

$=\left(-\frac{t}{1+t^2}\right)_t^{\sqrt{3}}+\left(\tan ^{-1} t\right)_1^{\sqrt{3}}[$ Applying by parts $]$

$=-\left(\frac{\sqrt{3}}{4}-\frac{1}{2}\right)+\frac{\pi}{3}-\frac{\pi}{4}$

$=\frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{\pi}{12}$

Hence, the answer is option 4.

Question 8: The value of $\sum_{n=1}^{100} \int_{n-1}^n e^{x-[x]} d x$ where [x] is the greatest integer $\leq x$, is :

(1) $100(e-1)$

(2) 100e

(3) 100(1-e)

(4) 100(1+e)

Answer:

$\sum_{n=1}^{100} \int_{n-1}^n e^{\{x\}} d x$, period of $\{x\}=1 \sum_{n=1}^{100} \int_0^1 e^{\{x\}} d x=\sum_{n=1}^{100} \int_0^1 e^x d x \sum_{n=1}^{100}(e-1)=100(e-1)$

Note that:

$x-[x]=\{x\}$, where $[\mathrm{x}]$ is the greatest integer $\leq x$ and $\{x\}$ fractional part of $x$.

Hence, the answer is option 1.

Question 9: Let $f: R \rightarrow R$ defined as $f(x)=\left\{\begin{array}{c}x^5 \sin \left(\frac{1}{x}+5 x^2\right), x<0 \\ 0, x=0 \\ x^5 \cos \left(\frac{1}{x}+\lambda z^2, x<0\right)\end{array}\right.$ The value of $\lambda$ for which $f^{\prime \prime}(x)$ exists is.

(1) 2

(2) 3

(3) 5

(4) 6

Answer:

$f^{\prime}(x)=\left\{\begin{array}{c}5 x^4 \sin \left(\frac{1}{x}\right)-x^3 \cos \left(\frac{1}{x}\right)+10 x, x<0 \\ 0, \quad x=0 \\ 5 x^4 \cos \left(\frac{1}{x}\right)+x^3 \sin \left(\frac{1}{x}\right)+2 \lambda, x>0\end{array}\right.$

$f^{\prime \prime}(x)=\left\{\begin{array}{c}\text { term having } \mathrm{x} \text { in muliplication }+10, x<0 \\ 0, \quad x=0 \\ \text { term having } \mathrm{x} \text { in muliplication }+2 \lambda, x>0\end{array}\right.$

L.H.L = R.H.L.

$2 \lambda=10 \lambda=5$

Hence, the answer is option (3).

Question 10: Let f be a twice differentiable function on (1,6). If $(2)=8, f^{\prime}(2)=5, f^{\prime}(x) \geq 1$ and $f^{\prime \prime}(x) \geq 4$, for all $x \epsilon(1,6)$ then:

(1) $f(5)+f^{\prime}(5) \leq 26$

(2) $f(5)+f^{\prime}(5) \geq 28$

(3) $f(5)+f^{\prime \prime}(5) \leq 20$

(4) $f(5) \leq 10$

Answer:

$\mathrm{f}(2)=8, \mathrm{f}^{\prime}(2)=5, \mathrm{f}^{\prime}(\mathrm{x}) \geq 1, \mathrm{f}^{\prime \prime}(\mathrm{x}) \geq 4, \forall \mathrm{x} \in(1,6)$

$f^{\prime \prime}(x)=\frac{f^{\prime}(5)-f^{\prime}(2)}{5-2} \geq 4 \Rightarrow f^{\prime}(5) \geq 17$

$f^{\prime}(x)=\frac{f(5)-f(2)}{5-2} \geq 1 \Rightarrow f(5) \geq 11$

$f^{\prime}(5)+f(5) \geq 28$

Hence, the answer is option 2.

Question 11: The population P=P(t) at time 't' of ascertain species follows the differential equation. Then the time at which the population become zero:

(1) $\log _e 18$

(2) $\frac{1}{2} \log _e 18$

(3) $\log _e 9$

(4) $2 \log _e 18$

Answer:

$\frac{\mathrm{dP}(\mathrm{t})}{\mathrm{dt}}=\frac{\mathrm{P}(\mathrm{t})-900}{2}$

$\int_0^t \frac{\mathrm{dP}(\mathrm{t})}{\mathrm{P}(\mathrm{t})-900}=\int_0^{\mathrm{t}} \frac{\mathrm{dt}}{2}$

$\{\ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|\}_0^{\mathrm{t}}=\left\{\frac{\mathrm{t}}{2}\right\}_0^{\mathrm{t}}$

$\ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|-\ell \mathrm{n}|\mathrm{P}(0)-900|=\frac{\mathrm{t}}{2}$

$\ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|-\ln 50=\frac{\mathrm{t}}{2}$

Let at $t=t_1, P(t)=0$ hence

$\ln |\mathrm{P}(\mathrm{t})-900|-\ln 50=\frac{\mathrm{t}_1}{2}$

$\mathrm{t}_1=2 \ell \mathrm{n} 18$

Hence, the answer is option 4.

Question 12: The function $f(x)=\frac{4 x^3-3 x^3}{6}-2 \sin x+(2 x-1) \cos x$

(1) Decreases in $\left(-\infty, \frac{1}{2}\right)$

(2) Increases in $\left(-\infty, \frac{1}{2}\right)$

(3) Increases in $\left(\frac{1}{2}, \infty\right)$

(4) Decreases in $\left(\frac{1}{2}, \infty\right)$

Answer:

$f(x)=\frac{4 x^3-3 x^2}{6}-2 \sin x+(2 x-1) \cos x$

$f^{\prime}(x)=(2 x-1)(x-\sin x)$

$\Rightarrow \mathrm{f}(\mathrm{x}) \geq 0$ in $\mathrm{x} \in\left[\frac{1}{2}, \infty\right)$

and $f(x) \leq 0$ in $x \in\left(-\infty, \frac{1}{2}\right]$

Hence, the answer is option 3.

Question 13: Imaginary part of $(3+2 \sqrt{-54})^{\frac{1}{2}}-(3-2 \sqrt{-54})^{\frac{1}{2}}$ can be:

(1) 6

(2) $\sqrt{6}$

(3) $-\sqrt{6}$

(4) $-2 \sqrt{6}$

Answer:

$(3+2 \sqrt{-54})^{1 / 2}-(3-2 \sqrt{-54})^{1 / 2}$

$(3+2 \cdot 3 \sqrt{6} i)^{1 / 2}-(3-2 \cdot 3 \sqrt{6} i)^{1 / 2}$

$\left[3^2+(\sqrt{6} i)^2+2 \cdot 3 \sqrt{6} i\right]^{1 / 2}-\left[3^2+(\sqrt{6} i)^2-2 \cdot 3 \sqrt{6} i\right]^{1 / 2}$

$|3+\sqrt{6} i|-|3-\sqrt{6}|$

$b_1= \pm \sqrt{6}$ and $b_2= \pm \sqrt{6}$

So the value of $b_1+b_2=2 \sqrt{6} \quad$ or $\quad-2 \sqrt{6} \quad$ or $\quad 0$

Hence, the answer is option 4.

Question 14: Let $\alpha$ and $\beta$ be the roots of $x^2-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^2-6 x+q=0$. If $\alpha, \beta, \gamma, \delta$. forms a geometric progression. Then the ratio $(2 q+p):(2 q-p)$ is :

(1) 3:1

(2) 9:7

(3) 5:3

(4) 33:31

Answer: The general term or $\mathrm{n}^{\text {th }}$ term of a geometric progression is $a_n=a r^{n-1}$

Now,

$x^2-3 x+p=0$ roots are $\alpha \& \beta \alpha+\beta=3 \alpha \beta=p$

$x^2-6 x+q=0$ roots are $\gamma \& \delta \gamma+\delta=6 \gamma \delta=q$

$\alpha, \beta, \gamma \& \delta$ are in GPLet, $\alpha=a, \beta=a r, \gamma=a r^2$ and $\delta=a r^3$

$a+a r=3 \quad \ldots(1) a r^2+a r^3=6 \quad \ldots(2)(2) \div(1) r^2=2$

So, $\frac{2 q+p}{2 q-p}=\frac{2 r^5+r}{2 r^5-r}=\frac{2 r^4+1}{2 r^4-1}=\frac{9}{7}$

Hence, the answer is option 2.

Question 15: Let $\lambda \neq 0$ be in R. If $\alpha$ and $\beta$ are the roots of the equation, $x^2-x+2 \lambda=0$ and $\alpha$ and $\gamma$ are the roots of the equation, $3 x^2-10 x+27 \lambda=0$, then $\frac{\beta \gamma}{\lambda}$ is equal to:

(1) 25

(2) 18

(3) 9

(4) 36

Answer:

$\alpha+\beta=1, \alpha \beta=2 \lambda \alpha+\beta=\frac{10}{3}, \quad \alpha \gamma=\frac{27 \lambda}{3}=9 \lambda \gamma-\beta=\frac{7}{3}$

$\frac{\gamma}{\beta}=\frac{9}{2} \Rightarrow \gamma=\frac{9}{2} \beta=\frac{9}{2} \times \frac{2}{3} \Rightarrow \gamma=3 \frac{9}{2} \beta-\beta=\frac{7}{3} \frac{9}{2} \beta=\frac{7}{3} \Rightarrow \beta=\frac{2}{3}$

$\alpha=1-\frac{2}{3}=\frac{1}{3} 2 \lambda=\frac{2}{9} \Rightarrow \lambda=\frac{1}{9} \frac{\beta \gamma}{\lambda}=\frac{\frac{2}{3} \times 3}{\frac{1}{9}}=18$

Hence, the answer is option (2).

Question 16: Let p and q are two positive numbers such that $p+q=2$ and $p^4+q^4=272$. Then p and q are root of equation:

(1) $x^2-2 x+8=0$

(2) $x^2-2 x+136=0$

(3) $x^2-2 x+16=0$

(4) $x^2-2 x+2=0$

Answer: $\left(p^2+q^2\right)^2-2 p^2 q^2=272\left((p+q)^2-2 p q\right)^2-2 p^2 q^2=27216-16 p q+2 p^2 q^2=272(p q)^2-8 p q-128=0 p q=\frac{8 \pm 24}{2}=16,-8 p q=16$

Hence, the answer is option 3.

Question 17: Let $\mathrm{S}=\left\{\left(\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right) ; \mathrm{a}, \mathrm{b} \in\{1,2,3, \ldots 100\}\right\}$ and let $\mathrm{T}_{\mathrm{n}}=\left\{\mathrm{A} \in \mathrm{S}: \mathrm{A}^{\mathrm{n}(\mathrm{n}+1)}=\mathrm{I}\right\}$. Then the number of elements in $\cap_{n=1}^{100} T_n$ is:

Answer:

$\mathrm{S}=\left[\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right] ; \quad \mathrm{a}, \mathrm{b} \in\{1,2,3, \ldots 100\}$

$\mathrm{T}_{\mathrm{n}}=\left\{\mathrm{A} \in \mathrm{s}: \mathrm{A}^{\mathrm{n}(\mathrm{n}+1)}=\mathrm{I}\right\}$

$\mathrm{A}^2=\left[\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right]\left[\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right]=\left[\begin{array}{cc}1 & -a+a b \\ 0 & b^2\end{array}\right]$

b should be 1

$A^2=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ hence a can take any value

Total number of elements = 100

Hence, the answer is 100.

Question 18: Let $A=\left(\begin{array}{ll}2 & -2 \\ 1 & -1\end{array}\right)$ and $B=\left(\begin{array}{ll}-1 & 2 \\ -1 & 2\end{array}\right)$. Then the number of elements in the set $\left\{(\mathrm{n}, \mathrm{m}): \mathrm{n}, \mathrm{m} \in\{1,2, \ldots \ldots, 10\}\right.$ and $\left.\mathrm{nA}^{\mathrm{n}}+\mathrm{mB}^{\mathrm{m}}=\mathrm{I}\right\}$ is ________.

Answer: $\mathrm{A}^2=\mathrm{A}$ and $\mathrm{B}^2=\mathrm{B}$

Therefore equation $\mathrm{nA}^{\mathrm{n}}+\mathrm{mB}^{\mathrm{m}}=\mathrm{I}$ becomes $\mathrm{nA}+\mathrm{mB}=\mathrm{I}$, which gives $\mathrm{m}=\mathrm{n}=\mathrm{I}$

So, only one element i.e. (1,1) is possible.

Hence, the answer is 1.

Question 19: Let the eccentricity of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is reciprocal to that of the hyperbola $2 x^2-2 y^2=1$. If the ellipse

intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____:

Answer: $\mathrm{E}: \frac{\mathrm{x}^4}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{b^2}=1 \rightarrow \mathrm{e}$

$H: x^2-y^2=\frac{1}{2} \Rightarrow e^{\prime}=\sqrt{2}$

$\mathrm{e}=\frac{1}{\sqrt{2}}$

$\because e^2=\frac{1}{2}$

$1-\frac{b^2}{a^2}=\frac{1}{2} \Rightarrow \frac{b^2}{a^2}=\frac{1}{2}$$a^2=2 b^2$

$\mathrm{E} \& \mathrm{H}$ are at a right angle they are confocal Focus of Hyperbola = focus of ellipse

$\left( \pm \frac{1}{\sqrt{2}} \cdot \sqrt{2}, 0\right)=\left( \pm \frac{\mathrm{a}}{\sqrt{2}}, 0\right)$

$\mathrm{a}=\sqrt{2}$

$\because a^2=2 b^2 \Rightarrow b^2=1$

Length of $L R=\frac{2 b^2}{a}=\frac{2(1)}{\sqrt{2}}$

$=\sqrt{2}$Square of LR $=2$

Hence, the answer is the 2.

Question 20: If the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the line $\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$ on the x-axis and the line $\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$ on the y-axis, then the eccentricity of the ellipse is

Answer:

Line 1 and ellipse cut x-axis at x=7 and line 2 and ellipse cut y-axis at $\mathrm{y}=2 \sqrt{6}$.

$\therefore \quad \mathrm{a}=7, \quad \mathrm{~b}=2 \sqrt{6}$

$\mathrm{e}^2=1-\frac{\mathrm{b}^2}{\mathrm{a}^2}=1-\frac{24}{49}=\frac{25}{49}$

$\Rightarrow \mathrm{e}=\frac{5}{7}$

Hence, the answer is $\frac{5}{7}$

Question 21: If the system of equations

$
\begin{aligned}
& (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\
& \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\
& (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9
\end{aligned}
$

has infinitely many solutions, then $\lambda^2+\lambda$ is equal to:

(1) 10

(2) 12

(3) 6

(4) 20

Answer:

$
\begin{aligned}
& (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\
& \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\
& (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9
\end{aligned}
$

For infinitely many solutions

$
\begin{aligned}
& \mathrm{D}=\left|\begin{array}{ccc}
\lambda-1 & \lambda-4 & \lambda \\
\lambda & \lambda-1 & \lambda-4 \\
\lambda+1 & \lambda+2 & -(\lambda+2)
\end{array}\right|=0 \\\\
& (\lambda-3)(2 \lambda+1)=0 \\\\
& \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc}
5 & \lambda-4 & \lambda \\
7 & \lambda-1 & \lambda-4 \\
9 & \lambda+2 & -(\lambda+2)
\end{array}\right|=0 \\\\
& 2(3-\lambda)(23-2 \lambda)=0 \\\\
& \lambda=3 \\\\
& \therefore \lambda^2+\lambda=9+3=12
\end{aligned}
$

Hence, the correct answer is option (2).

Question 22: Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then, which of the following points lies on the line of the shortest distance between $\mathrm{L}_1$ and $\mathrm{L}_2$?

(1) $\left(-\frac{5}{3},-7,1\right)$

(2) $\left(2,3, \frac{1}{3}\right)$

(3) $\left(\frac{8}{3},-1, \frac{1}{3}\right)$

(4) $\left(\frac{14}{3},-3, \frac{22}{3}\right)$

Answer:

$\begin{aligned} & \mathrm{P}(2 \lambda+1,3 \lambda+2,4 \lambda+3) \text { on } \mathrm{L}_1 \\\\& \mathrm{Q}(3 \mu+2,4 \mu+4,5 \mu+5) \text { on } \mathrm{L}_2 \\\\& \text{Direction ratio's of}\ \overrightarrow{P Q}=3 \mu-2 \lambda+1,4 \mu-3 \lambda+2,5 \mu-4 \lambda+2 \\\\& \overrightarrow{P Q} \perp \mathrm{L}_1\end{aligned}$
$
\begin{aligned}
& \Rightarrow(3 \mu-2 \lambda+1) 2+(4 \mu-3 \lambda+2) 3+(5 \mu-4 \underline{\lambda}+ 2) 4=0 \\\\
& ⇒38 \mu-29 \lambda+16=0.........(1)\\\\
& \mathrm{PQ} \perp \mathrm{~L}_2 \\\\
& \Rightarrow(3 \mu-2 \lambda+1) 3+(4 \mu-3 \lambda+2) 4+(5 \mu-4 \underline{\lambda}+ 2) 5=0 \\\\
& ⇒50 \mu-38 \lambda+21=0.......(2)\\\\
& \text {Solving equation 1 and 2, we get, }\\\\
& \lambda=\frac{1}{3} ; \mu=\frac{-1}{6} \\\\
& \therefore \mathrm{P}\left(\frac{5}{3}, 3, \frac{13}{3}\right), \mathrm{Q}\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)
\end{aligned}
$
The vector $\overrightarrow{P Q}=Q-P=\left(-\frac{1}{6}, \frac{1}{3}, -\frac{1}{6}\right)$.

The line passing through $P=\left(\frac{5}{3}, 3, \frac{13}{3}\right)$ in the direction $\overrightarrow{P Q}$ has parametric equations:
$x=\frac{5}{3}+t\left(-\frac{1}{6}\right), \quad y=3+t\left(\frac{1}{3}\right), \quad z=\frac{13}{3}+t\left(-\frac{1}{6}\right) .$
Rearranged as a symmetric form:
$\frac{x-\frac{5}{3}}{-\frac{1}{6}}=\frac{y-3}{\frac{1}{3}}=\frac{z-\frac{13}{3}}{-\frac{1}{6}}=t$
$⇒\frac{x-\frac{5}{3}}{\frac{1}{6}}=\frac{y-3}{-\frac{1}{3}}=\frac{z-\frac{13}{3}}{\frac{1}{6}}$
$⇒\frac{x-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{z-\frac{13}{3}}{1}$.

Here, point $\left(\frac{14}{3},-3, \frac{22}{3}\right)$ lies on the line PQ as the point satisfies the line equation.

Hence, the correct answer is option (4).

Question 23: The distance of the point (1, −2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and 2x −2y + z + 12=0, is :

(1) $2 \sqrt{2}$

(2) 2

(3) $\sqrt{2}$

(4) $\frac{1}{\sqrt{2}}$

Answer:

As we learned in

Distance of a point from plane (Cartesian form) -

The length of the perpendicular from $P\left(x_1, y_1, z_1\right)$ to the plane

$a x+b y+c z+d=0$ is given by $\frac{\left[a x_1+b y_1+c z_1+d\right]}{\left|\sqrt{a^2+b^2+c^2}\right|}$

The normal vector of the plane is $\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & -2 & 1\end{array}\right|=3 \hat{i}+3 \hat{j}$

So the equation of the plane is

$3 x+3 y=C$

passes through (1,2,2)

3+6=C

3x+3y=9 => x+y=3

Distance $=\left|\frac{1-2-3}{\sqrt{2}}\right|=\frac{4}{\sqrt{2}}=2 \sqrt{2}$

Hence, the answer is the option 1.

Question 24: If the angle between the lines, $x / 2=y / 2=z / 1$ and $\frac{5-x}{-2}=\frac{7 y-14}{p}=\frac{z-3}{4}$ is $\cos ^{(-1)} \frac{2}{3}$, then p is equal to :

(1) 7/2

(2) 2/7

(3) -7/4

(4) -4/7

Answer:

$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$

$\frac{x-5}{2}=\frac{y-2}{p / 7}=\frac{z-13}{4}$

$\cos \Theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$

$2 / 3=\frac{4+2 p / 7+4}{3 \times \sqrt{\left(20+\frac{p^2}{49}\right)}}$

$\sqrt{20+\frac{p^2}{49}}=4+p / 7$

$p=7 / 2$

Hence, the answer is the option 1.

Question 25: Consider the three planes

$P_1: 3 x+15 y+21 z=9$

$P_2: x-3 y-z=5$ and

$P_3: 2 x+10 y+14 z=5$

Then, which one of the following is true ?

(1) $P_2$ and $P_3$ are parallel

(2) $P_1$ and $P_2$ are parallel

(3) $P_1$ and $P_3$ are parallel

(4) $P_1, P_2$ and $P_3$ all are parallel.

Answer:

Given equation of planes are

$P_1: x+5 y+7 z=3 P_2: x-3 y-z=5 P_3: x+5 y+7 z=\frac{5}{2}$

$P_1$ and $P_3$ are parallel.

Hence, the answer is the option 3.