JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 5 of 275
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Let $A, B$ and $C$ be sets such that $\varnothing \ne A\cap B \subseteq C$. Which of the following statements is not true?
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Solution
The area of the region described by
$A=\{(x,y):x^{2}+y^{2}\le 1 \text{ and } y^{2}\le 1-x\}$ is :
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Solution
For the matrices
$A = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} -29 & 49 \ -13 & 18 \end{bmatrix}$
if $(A^{15} + B)\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$, then among the following which one is true?
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Solution
Here
$A^n = \begin{bmatrix} 2n+1 & -4n \ n & -2n+1 \end{bmatrix}$
$\Rightarrow A^{15} = \begin{bmatrix} 31 & -60 \ 15 & -29 \end{bmatrix}$
$A^{15} + B = \begin{bmatrix} 2 & -11 \ 2 & -11 \end{bmatrix}$
Now
$\begin{bmatrix} 2 & -11 \ 2 & -11 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$
$\Rightarrow 2x - 11y = 0$
$\Rightarrow x = 11,; y = 2$
The position of a moving car at time t is given by f(t) = at2 + bt + c, t > 0, where a, b and c are realnumbers greater than 1. Then the average speed of the car over the time interval [t1, t2] isattained at the point :
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Solution
If ${}^n{P_r} = {}^n{P_{r + 1}}$ and ${}^n{C_r} = {}^n{C_{r - 1}}$, then the value of r is equal to :
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Solution
If $z = x+iy$ satisfies $|z|-2=0$ and $|z-i|-|z+5i|=0$, then :
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Solution
If the orthocentre of the triangle whose vertices are $(1,2)$, $(2,3)$ and $(3,1)$ is $(\alpha,\beta)$, then the quadratic equation whose roots are $\alpha+4\beta$ and $4\alpha+\beta$ is:
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Solution
If the coefficients of $x^4$, $x^5$, and $x^6$ in the expansion of $(1+x)^n$ are in arithmetic progression,
then the maximum value of $n$ is:
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Solution
Let the area of $\triangle$ PQR with vertices P(5,4), Q(-2,4) and R(a,b) be 35 square units. If its orthocenter and centroid are $O\!\left(2,\dfrac{14}{5}\right)$ and C(c,d) respectively, then c+2d is equal to:
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Solution
Given $P(5,4),\ Q(-2,4)$ and orthocentre $O\left(2,\frac{14}{5}\right)$
$PQ$ is horizontal ⇒ altitude from $R$ is vertical ⇒ $x=2$
So $a=2$
Area $=\frac{1}{2}\times |PQ|\times \text{height}$
$=\frac{1}{2}\times 7 \times |b-4|=35$
$\Rightarrow |b-4|=10 \Rightarrow b=14 \text{ or } -6$
Using orthocentre condition (slopes), valid point is $R(2,-6)$
Centroid:
$c=\frac{5+(-2)+2}{3}=\frac{5}{3},\quad d=\frac{4+4-6}{3}=\frac{2}{3}$
$c+2d=\frac{5}{3}+\frac{4}{3}=3$
$\boxed{3}$
$PQ$ is horizontal ⇒ altitude from $R$ is vertical ⇒ $x=2$
So $a=2$
Area $=\frac{1}{2}\times |PQ|\times \text{height}$
$=\frac{1}{2}\times 7 \times |b-4|=35$
$\Rightarrow |b-4|=10 \Rightarrow b=14 \text{ or } -6$
Using orthocentre condition (slopes), valid point is $R(2,-6)$
Centroid:
$c=\frac{5+(-2)+2}{3}=\frac{5}{3},\quad d=\frac{4+4-6}{3}=\frac{2}{3}$
$c+2d=\frac{5}{3}+\frac{4}{3}=3$
$\boxed{3}$
Let the domains of the functions $f(x)=\log_{4}\big(\log_{3}\big(\log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)\big)\big)$ and $g(x)=\sin^{-1}\left(\dfrac{7x+10}{x-2}\right)$ be $(\alpha,\beta)$ and $[\gamma,\delta]$, respectively. Then $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}$ is equal to
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Solution
For $f(x)=\log_{4}\big(\log_{3}\big(\log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)\big)\big)$
Condition:
$\log_{3}(\cdot)>0 \Rightarrow \log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)>1$
$\Rightarrow 8-\log_{2}(x^{2}+4x+5)>7$
$\Rightarrow \log_{2}(x^{2}+4x+5)<1$
$\Rightarrow x^{2}+4x+5<2$
$\Rightarrow (x+2)^2<1$
$\Rightarrow -3$
So $(\alpha,\beta)=(-3,-1)$
For $g(x)=\sin^{-1}\left(\frac{7x+10}{x-2}\right)$
Condition: $-1\le \frac{7x+10}{x-2}\le 1$
Solve:
$\frac{7x+10}{x-2}\le 1 \Rightarrow \frac{6x+12}{x-2}\le 0 \Rightarrow \frac{x+2}{x-2}\le 0$
$\Rightarrow -2\le x<2$
$\frac{7x+10}{x-2}\ge -1 \Rightarrow \frac{8x+8}{x-2}\ge 0 \Rightarrow \frac{x+1}{x-2}\ge 0$
$\Rightarrow x\le -1 \text{ or } x>2$
Intersection: $[-2,-1]$
So $(\gamma,\delta)=(-2,-1)$
Now:
$\alpha^2+\beta^2+\gamma^2+\delta^2=9+1+4+1=15$
$\boxed{15}$
Condition:
$\log_{3}(\cdot)>0 \Rightarrow \log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)>1$
$\Rightarrow 8-\log_{2}(x^{2}+4x+5)>7$
$\Rightarrow \log_{2}(x^{2}+4x+5)<1$
$\Rightarrow x^{2}+4x+5<2$
$\Rightarrow (x+2)^2<1$
$\Rightarrow -3$
So $(\alpha,\beta)=(-3,-1)$
For $g(x)=\sin^{-1}\left(\frac{7x+10}{x-2}\right)$
Condition: $-1\le \frac{7x+10}{x-2}\le 1$
Solve:
$\frac{7x+10}{x-2}\le 1 \Rightarrow \frac{6x+12}{x-2}\le 0 \Rightarrow \frac{x+2}{x-2}\le 0$
$\Rightarrow -2\le x<2$
$\frac{7x+10}{x-2}\ge -1 \Rightarrow \frac{8x+8}{x-2}\ge 0 \Rightarrow \frac{x+1}{x-2}\ge 0$
$\Rightarrow x\le -1 \text{ or } x>2$
Intersection: $[-2,-1]$
So $(\gamma,\delta)=(-2,-1)$
Now:
$\alpha^2+\beta^2+\gamma^2+\delta^2=9+1+4+1=15$
$\boxed{15}$
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