NIMCET Mathematics Previous Year Questions (PYQs) – Page 7 of 95
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If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, then
$|\vec{a}-\vec{b}|^2 + |\vec{b}-\vec{c}|^2 + |\vec{c}-\vec{a}|^2$ does not exceed:
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Solution
Expand
\[|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} = 2 - 2\vec{a}\cdot\vec{b}\]
\[|\vec{b}-\vec{c}|^2 = 2 - 2\vec{b}\cdot\vec{c}\]
\[|\vec{c}-\vec{a}|^2 = 2 - 2\vec{c}\cdot\vec{a}\]
Add:
\[= 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\]
Find Maximum :
\[|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0\]
\[3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq 0\]
\[\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} \geq -\frac{3}{2}\]
Final Answer:
\[6 - 2 \times \left(-\frac{3}{2}\right) = 6 + 3 = \boxed{9}\]
✅ Expression does not exceed 9
If $H$ is the harmonic mean between $P$ and $Q$, then $\dfrac{H}{P} + \dfrac{H}{Q}$ is
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Solution
Harmonic mean of $P$ and $Q$ is
$H = \dfrac{2PQ}{P+Q}$
Now,
$\dfrac{H}{P} + \dfrac{H}{Q} = H\left(\dfrac{1}{P} + \dfrac{1}{Q}\right)
= H \cdot \dfrac{P+Q}{PQ}$
Substitute
$H = \dfrac{2PQ}{P+Q}$:
$\dfrac{H}{P} + \dfrac{H}{Q}
= \dfrac{2PQ}{P+Q} \cdot \dfrac{P+Q}{PQ}
= 2$
So the correct Answer is $2$.
Probability a blade is defective $=0.002$, packet of $10$ blades.
Find packets with no defective blades in $10000$ packets.
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Solution
$P(\text{good})=0.998$
Probability all 10 good $=(0.998)^{10}\approx 0.9802$
So packets $=10000\times0.9802\approx9802$
Let $f(x) = \lfloor x^2 - 3 \rfloor$ where $\lfloor \cdot \rfloor$ is the greatest integer function.
Number of points in $(1,2)$ where $f$ is discontinuous:
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Solution
Discontinuity occurs when
$x^2 - 3$ is an integer
Let
$x^2 - 3 = k \Rightarrow x = \sqrt{k+3}$
We need $1 < x < 2$ → square both sides:
$1 < \sqrt{k+3} < 2$
$\Rightarrow 1 < k+3 < 4$
$\Rightarrow -2 < k < 1$
Possible integer values: $k = -1, 0$
So number of discontinuities = $2$
The number of values of $k$ for which the system of equations
$(k+1)x + 8y = 4k$ and $kx + (k+3)y = 3k-1$ has infinitely many solutions, is
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Solution
$ (k+1)x + 8y = 4k $
$ kx + (k+3)y = 3k - 1 $
Since the system has infinitely many solutions,
$ \dfrac{k+1}{k} = \dfrac{8}{k+3} = \dfrac{4k}{3k-1} $
Taking 1st and 3rd ratio:
$ (k+1)(3k-1) = 4k^2 $
$ 3k^2 + 2k - 1 = 4k^2 $
$ k^2 - 2k + 1 = 0 $
$ (k-1)^2 = 0 $
$ \therefore k = 1 $
Regression lines:
$3x+2y=26$, $6x+y=31$
Correlation between $x,y$ is
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Solution
Slope of line of $y$ on $x$:
$3x+2y=26 \Rightarrow y=-\dfrac{3}{2}x+13$
Slope of line of $x$ on $y$:
$6x+y=31 \Rightarrow x=-\dfrac{1}{6}y+\dfrac{31}{6}$
Product of slopes $=r^{2}$ with sign of slopes:
$r=\sqrt{(-3/2)\cdot(-1/6)}$
$=\sqrt{1/4}$
$=1/2$
Both slopes negative ⇒ $r$ negative.
Both slopes negative ⇒ $r$ negative.
If $a + b + c \neq 0$, the system of equations:
$(b+c)(y+z) - ax = b - c$
$(c+a)(z+x) - by = c - a$
$(a+b)(x+y) - cz = a - b$
has:
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Solution
This is a symmetric linear system.
Solving gives the unique solution:
$x = y = z = 0$
Since $a+b+c \neq 0$, determinant $\neq 0$ → unique solution.
The sum
$^{20}C_8 + ^{20}C_9 + ^{21}C_{10} + ^{22}C_{11} - ^{23}C_{11}$
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Solution
Use Pascal identity:
$^{n}C_r + ^{n}C_{r+1} = ^{n+1}C_{r+1}$
So,
$^{20}C_8 + ^{20}C_9 = ^{21}C_9$
Then,
$^{21}C_9 + ^{21}C_{10} = ^{22}C_{10}$
Next,
$^{22}C_{10} + ^{22}C_{11} = ^{23}C_{11}$
Thus the expression becomes:
$^{23}C_{11} - ^{23}C_{11} = 0$
$\therefore$ the answer is 0.
Car travels half distance with $v_1$, half with $v_2$.
Average speed is
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Solution
For equal distances:
Average velocity $=\dfrac{2v_1v_2}{v_1+v_2}$
If $y = f(x)$ is odd and differentiable on $(-\infty,\infty)$ such that
$f'(3) = -2$, then $f'(-3)$ equals:
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Solution
If $f(x)$ is odd, then:
$f(-x) = -f(x)$
Differentiate both sides:
$f'(-x)(-1) = -f'(x)$
$f'(-x) = f'(x)$
So derivative is even.
Therefore:
$f'(-3) = f'(3) = -2$
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