NIMCET Mathematics Previous Year Questions (PYQs) – Page 90 of 95
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In ΔABC, if a = 2, b = 4 and ∠C = 60°, then A and B are respectively equal to
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Solution
The sum of infinite terms of decreasing GP is equal to the greatest value of the function $f(x) = x^3
+ 3x – 9$ in the
interval [–2, 3] and difference between the first two terms is f '(0). Then the common ratio of the GP is
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Solution
? GP and Function Relation
Given: \( f(x) = x^3 + 3x - 9 \)
The sum of infinite GP = max value of \( f(x) \) on [−2, 3]
The difference between first two terms = \( f'(0) \)
Step 1: \( f(x) \) is increasing ⇒ Max at \( x = 3 \)
\( f(3) = 27 \Rightarrow \frac{a}{1 - r} = 27 \)
Step 2: \( f'(x) = 3x^2 + 3 \Rightarrow f'(0) = 3 \)
⇒ \( a(1 - r) = 3 \)
Step 3: Solve:
\( a = 27(1 - r) \)
\( \Rightarrow 27(1 - r)^2 = 3 \Rightarrow (1 - r)^2 = \frac{1}{9} \Rightarrow r = \frac{2}{3} \)
✅ Final Answer: \( r = \frac{2}{3} \)
The vertex of the parabola whose focus is (-1,1) and directrix is 4x + 3y - 24 = 0 is
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Solution
Given: Focus $S(-1, 1)$ and directrix $4x + 3y - 24 = 0$
The vertex is the midpoint of focus and foot of perpendicular from focus to directrix
Foot of perpendicular from $S(-1, 1)$ to $4x + 3y - 24 = 0$:
$\dfrac{x + 1}{4} = \dfrac{y - 1}{3} = -\dfrac{4(-1) + 3(1) - 24}{4^2 + 3^2}$
$= -\dfrac{-4 + 3 - 24}{16 + 9}$
$= -\dfrac{-25}{25}$
$= 1$
$\therefore x + 1 = 4 \Rightarrow x = 3$
$\therefore y - 1 = 3 \Rightarrow y = 4$
$\therefore$ Foot of perpendicular $Z = (3, 4)$
Since vertex $V$ is midpoint of $SZ$:
$V = \left(\dfrac{-1 + 3}{2},\ \dfrac{1 + 4}{2}\right)$
$V = \left(\dfrac{2}{2},\ \dfrac{5}{2}\right)$
$\therefore \boxed{V = \left(1,\ \dfrac{5}{2}\right)}$
The vertex is the midpoint of focus and foot of perpendicular from focus to directrix
Foot of perpendicular from $S(-1, 1)$ to $4x + 3y - 24 = 0$:
$\dfrac{x + 1}{4} = \dfrac{y - 1}{3} = -\dfrac{4(-1) + 3(1) - 24}{4^2 + 3^2}$
$= -\dfrac{-4 + 3 - 24}{16 + 9}$
$= -\dfrac{-25}{25}$
$= 1$
$\therefore x + 1 = 4 \Rightarrow x = 3$
$\therefore y - 1 = 3 \Rightarrow y = 4$
$\therefore$ Foot of perpendicular $Z = (3, 4)$
Since vertex $V$ is midpoint of $SZ$:
$V = \left(\dfrac{-1 + 3}{2},\ \dfrac{1 + 4}{2}\right)$
$V = \left(\dfrac{2}{2},\ \dfrac{5}{2}\right)$
$\therefore \boxed{V = \left(1,\ \dfrac{5}{2}\right)}$
Forces of magnitude 5, 3, 1 units act in the directions
6i + 2j + 3k, 3i - 2j + 6k, 2i - 3j - 6k respectively on a particle which is displaced from the
point (2, −1, −3) to (5, −1, 1). The total work done by the force is
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Solution
In a right angled triangle, the hypotenuse is four times the perpendicular drawn to it from the opposite vertex. The value of one of the acute angles is
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Solution
If $ \int \frac{xe^{x}}{\sqrt{1+e^{x}}}=f(x)\sqrt{1+e^{x}}-2log \frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1}+C$ then $f(x)$ is
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Solution
If $f(x)=\lim _{{x}\rightarrow0}\, \frac{{6}^x-{3}^x-{2}^x+1}{\log _e9(1-\cos x)}$ is a real number then $\lim _{{x}\rightarrow0}\, f(x)$
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Solution
The number of points in $(-\infty,\infty)$, for which $x^2 - x\sin x - \cos x = 0$ is
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Solution
Let $f(x) = x^2 - x\sin x - \cos x$
Differentiating with respect to $x$:
$f'(x) = 2x - \sin x - x\cos x + \sin x$
$f'(x) = 2x - x\cos x$
$f'(x) = x(2 - \cos x)$
Since $-1 \leq \cos x \leq 1$ $\Rightarrow$ $2 - \cos x \geq 1 > 0$ always
$\therefore$ Sign of $f'(x)$ depends on sign of $x$
$f'(x) < 0$ for $x < 0$ and $f'(x) > 0$ for $x > 0$
$\therefore f(x)$ has minimum at $x = 0$
Minimum value $= f(0) = 0 - 0 - \cos 0 = -1 < 0$
Also, $f(-x) = (-x)^2 - (-x)\sin(-x) - \cos(-x)$
$= x^2 - x\sin x - \cos x = f(x)$
$\therefore f(x)$ is an even function
As $x \rightarrow \pm\infty$, $f(x) \rightarrow +\infty$
Since $f(0) = -1 < 0$ is minimum and $f(x) \rightarrow +\infty$ on both sides
$\therefore$ Graph of $f(x)$ crosses $x$-axis at exactly two points
One in $(-\infty, 0)$ and one in $(0, \infty)$
$\therefore$ Total number of points $= \boxed{2}$
Differentiating with respect to $x$:
$f'(x) = 2x - \sin x - x\cos x + \sin x$
$f'(x) = 2x - x\cos x$
$f'(x) = x(2 - \cos x)$
Since $-1 \leq \cos x \leq 1$ $\Rightarrow$ $2 - \cos x \geq 1 > 0$ always
$\therefore$ Sign of $f'(x)$ depends on sign of $x$
$f'(x) < 0$ for $x < 0$ and $f'(x) > 0$ for $x > 0$
$\therefore f(x)$ has minimum at $x = 0$
Minimum value $= f(0) = 0 - 0 - \cos 0 = -1 < 0$
Also, $f(-x) = (-x)^2 - (-x)\sin(-x) - \cos(-x)$
$= x^2 - x\sin x - \cos x = f(x)$
$\therefore f(x)$ is an even function
As $x \rightarrow \pm\infty$, $f(x) \rightarrow +\infty$
Since $f(0) = -1 < 0$ is minimum and $f(x) \rightarrow +\infty$ on both sides
$\therefore$ Graph of $f(x)$ crosses $x$-axis at exactly two points
One in $(-\infty, 0)$ and one in $(0, \infty)$
$\therefore$ Total number of points $= \boxed{2}$
The position vectors of points A and B are 
and 
.
Then the position vector of point p dividing AB in
the ratio m : n is
and .
Then the position vector of point p dividing AB in
the ratio m : n is1
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Solution
A is targeting B, B and C are targeting A. Probability of hitting the target by A, B and C are $\frac{2}{3}, \frac{1}{2}$ and $\frac{1}{3}$ respectively. If A is hit then the probability that B hits the target and C does not, is
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Solution
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