JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 7 of 275
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Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is:
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Solution
Given data: $2,3,3,4,5,7,a,b$ (8 observations)
Mean $=4 \Rightarrow$ total sum $=8\times 4=32$
Sum of known terms $=24 \Rightarrow a+b=8$
Variance $=2 \Rightarrow \sum (x-4)^2=8\times 2=16$
Known contribution $=16 \Rightarrow (a-4)^2+(b-4)^2=0$
⇒ $a=b=4$
Data: $2,3,3,4,4,4,5,7$
Mode $=4$
Mean deviation about mode:
$=\frac{1}{8}[|2-4|+|3-4|+|3-4|+0+0+0+|5-4|+|7-4|]$
$=\frac{1}{8}[2+1+1+0+0+0+1+3]=\frac{8}{8}=1$
Mean $=4 \Rightarrow$ total sum $=8\times 4=32$
Sum of known terms $=24 \Rightarrow a+b=8$
Variance $=2 \Rightarrow \sum (x-4)^2=8\times 2=16$
Known contribution $=16 \Rightarrow (a-4)^2+(b-4)^2=0$
⇒ $a=b=4$
Data: $2,3,3,4,4,4,5,7$
Mode $=4$
Mean deviation about mode:
$=\frac{1}{8}[|2-4|+|3-4|+|3-4|+0+0+0+|5-4|+|7-4|]$
$=\frac{1}{8}[2+1+1+0+0+0+1+3]=\frac{8}{8}=1$
The value of $\lambda$ such that the sum of the squares of the roots of the quadratic equation
$x^2 + (3 - \lambda)x + 2 = \lambda$
has the least value, is –
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Solution
The general solution of the differential equation (y2
– x3)dx – xydy = 0 (x $ \ne $ 0) is :
(where c is a constant of integration)
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Solution
If $A$ is a $3\times 3$ non-singular matrix such that $AA' = A'A$ and
$B = A^{-1}A'$, then $BB'$ equals :
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Solution
Let $y = y(x)$ be the solution of differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3\sin x$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $y(0) = -\frac{7}{4}$. Then $y\left(\frac{\pi}{6}\right)$ is equal to:
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Solution
$\frac{dy}{dx} - 2y\cos x = 2\cos x + 3\sin x \cos x$
$I.F. = e^{-2\sin x}$
$e^{-2\sin x}y = \int e^{-2\sin x}(3\sin x \cos x + 2\cos x),dx$
$= e^{-2\sin x}\left(-\frac{3}{2}\sin x - \frac{7}{4}\right) + C$
$\Rightarrow y = -\frac{3}{2}\sin x - \frac{7}{4} + Ce^{2\sin x}$
$y(0) = -\frac{7}{4} \Rightarrow C = 0$
$y\left(\frac{\pi}{6}\right) = -\frac{3}{2}\cdot\frac{1}{2} - \frac{7}{4} = -\frac{5}{2}$
If $\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n$ and $\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na$(n, a > 1) then the standard deviation of n observations x1, x2, ..., xn is :
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Solution
If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and ${{37} \over 4}$ respectively, then (a $-$ b)2 is equal to :
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Solution
Let $P$ and $Q$ be any points on the curves $(x-1)^{2}+(y+1)^{2}=1$ and $y=x^{2}$, respectively.
The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval:
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Solution
Let $9=x_{1} < x_{2} < \ldots < x_{7}$ be in an A.P. with common difference d. If the standard deviation of $x_{1}, x_{2}..., x_{7}$ is 4 and the mean is $\bar{x}$, then $\bar{x}+x_{6}$ is equal to :
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Solution
If the function
$
f(x)=
\begin{cases}
\dfrac{7^{x}-9^{x}-8^{x}+1}{\sqrt{2}-\sqrt{1+\cos^{2}x}}, & x\neq0,\\[6pt]
a\log_{e}2\log_{e}3, & x=0
\end{cases}
$
is continuous at $x=0$, then the value of $a^{2}$ is equal to:
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Solution
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