NIMCET Mathematics Previous Year Questions (PYQs) – Page 93 of 95
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 950 questions
A harbour lies in a direction 60° South of West from a fort and at a distance 30 km from it, a ship sets out from the harbour at noon and sails due East at 10 km an hour. The time at which the ship will be 70 km from the fort is
1
2
3
4
Solution
If $\overrightarrow{{a}}$ and $\overrightarrow{{b}}$ are vectors in space, given by $\overrightarrow{{a}}=\frac{\hat{i}-2\hat{j}}{\sqrt[]{5}}$ and $\overrightarrow{{b}}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt[]{14}}$, then the value of$(2\vec{a} + \vec{b}).[(\vec{a} × \vec{b}) × (\vec{a} – 2\vec{b})]$ is
1
2
3
4
Solution
Area of the greatest rectangle that can be inscribed in the ellipse is
1
2
3
4
Solution
Given ellipse: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
**Setting up the Rectangle:**
Let $P(a\cos\theta,\ b\sin\theta)$ be a point on the ellipse
Then the rectangle has:
Length $= 2a\cos\theta$
Breadth $= 2b\sin\theta$
**Area of Rectangle:**
$A = 2a\cos\theta \times 2b\sin\theta$
$A = 4ab\sin\theta\cos\theta$
$A = 2ab\sin 2\theta$
**Maximizing the Area:**
$A$ is maximum when $\sin 2\theta$ is maximum
$\Rightarrow \sin 2\theta = 1$
$\Rightarrow 2\theta = 90^\circ$
$\Rightarrow \theta = 45^\circ$
**Maximum Area:**
$A_{max} = 2ab \times 1$
$\therefore \boxed{A_{max} = 2ab}$
**Setting up the Rectangle:**
Let $P(a\cos\theta,\ b\sin\theta)$ be a point on the ellipse
Then the rectangle has:
Length $= 2a\cos\theta$
Breadth $= 2b\sin\theta$
**Area of Rectangle:**
$A = 2a\cos\theta \times 2b\sin\theta$
$A = 4ab\sin\theta\cos\theta$
$A = 2ab\sin 2\theta$
**Maximizing the Area:**
$A$ is maximum when $\sin 2\theta$ is maximum
$\Rightarrow \sin 2\theta = 1$
$\Rightarrow 2\theta = 90^\circ$
$\Rightarrow \theta = 45^\circ$
**Maximum Area:**
$A_{max} = 2ab \times 1$
$\therefore \boxed{A_{max} = 2ab}$
Given that $\cos 6x=a\cos^6x+b\cos^4x+c\cos^2x+d$ for any real number $x$, find the value of $a+b+c$.
1
2
3
4
Solution
Given:
$\cos 6x=a\cos^6x+b\cos^4x+c\cos^2x+d$
Put $x=0$
$\cos 0=a\cos^6 0+b\cos^4 0+c\cos^2 0+d$
$1=a+b+c+d$ ...(1)
Put $x=\frac{\pi}{2}$
$\cos 3\pi=a\cos^6\frac{\pi}{2}+b\cos^4\frac{\pi}{2}+c\cos^2\frac{\pi}{2}+d$
$-1=d$ ...(2)
From (1),
$a+b+c+d=1$
$a+b+c-1=1$
$a+b+c=2$
Answer: $2$
If $\frac{tanx}{2}=\frac{tanx}{3}=\frac{tanx}{5}$ and x + y + z = π, then the
value of tan2x + tan2y + tan2z is
1
2
3
4
Solution
If $x, y, z$ are three consecutive positive integers, then $log (1 + xz)$ is
1
2
3
4
Solution
Let $\vec{A} = 2\hat{i} + \hat{j} – 2\hat{k}$ and $\vec{B} = \hat{i} + \hat{j}$, If $\vec{C}$ is a vector such that $|\vec{C} – \vec{A}| = 3$ and the angle between A × B and C is ${30^{\circ}}$, then $|(\vec{A} × \vec{B}) × \vec{C}|$ = 3 then the value of $\vec{A}.\vec{C}$ is equal to
1
2
3
4
Solution
. Two common tangents to the circles $x^2 + y^2 = 2a^2$ and parabola $y^2 = 8ax$ are
1
2
3
4
Solution
Given circle: $x^2 + y^2 = 2a^2$
Given parabola: $y^2 = 8ax$
**Equation of tangent to parabola:**
For parabola $y^2 = 8ax$, tangent is:
$y = mx + \dfrac{2a}{m}$ ...(i)
**Condition for tangent to circle:**
For line $y = mx + \dfrac{2a}{m}$ to be tangent to circle $x^2 + y^2 = 2a^2$,
distance from centre $(0,0)$ = radius $= \sqrt{2}\ a$
$\Rightarrow \dfrac{\left|\dfrac{2a}{m}\right|}{\sqrt{1 + m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2a}{|m|\sqrt{1+m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2}{|m|\sqrt{1+m^2}} = \sqrt{2}$
$\Rightarrow |m|\sqrt{1+m^2} = \sqrt{2}$
Squaring both sides:
$\Rightarrow m^2(1+m^2) = 2$
$\Rightarrow m^4 + m^2 - 2 = 0$
$\Rightarrow (m^2 + 2)(m^2 - 1) = 0$
$\Rightarrow m^2 = 1 \quad (\because m^2 = -2 \text{ is not possible})$
$\Rightarrow m = \pm 1$
**Substituting in (i):**
When $m = 1$: $y = x + 2a$
When $m = -1$: $y = -x - 2a$
$\therefore$ The two common tangents are:
$\therefore \boxed{y = x + 2a \quad \text{and} \quad y = -(x + 2a)}$
Given parabola: $y^2 = 8ax$
**Equation of tangent to parabola:**
For parabola $y^2 = 8ax$, tangent is:
$y = mx + \dfrac{2a}{m}$ ...(i)
**Condition for tangent to circle:**
For line $y = mx + \dfrac{2a}{m}$ to be tangent to circle $x^2 + y^2 = 2a^2$,
distance from centre $(0,0)$ = radius $= \sqrt{2}\ a$
$\Rightarrow \dfrac{\left|\dfrac{2a}{m}\right|}{\sqrt{1 + m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2a}{|m|\sqrt{1+m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2}{|m|\sqrt{1+m^2}} = \sqrt{2}$
$\Rightarrow |m|\sqrt{1+m^2} = \sqrt{2}$
Squaring both sides:
$\Rightarrow m^2(1+m^2) = 2$
$\Rightarrow m^4 + m^2 - 2 = 0$
$\Rightarrow (m^2 + 2)(m^2 - 1) = 0$
$\Rightarrow m^2 = 1 \quad (\because m^2 = -2 \text{ is not possible})$
$\Rightarrow m = \pm 1$
**Substituting in (i):**
When $m = 1$: $y = x + 2a$
When $m = -1$: $y = -x - 2a$
$\therefore$ The two common tangents are:
$\therefore \boxed{y = x + 2a \quad \text{and} \quad y = -(x + 2a)}$
Find the area of the triangle formed in the right half-plane by the lines $x-y=0$ and $x+y=0$, and a tangent to the hyperbola $x^2-y^2=a^2$, where $a$ is a non-zero number.
1
2
3
4
Solution
$x^2-y^2=a^2$.
For a hyperbola, the area of the triangle formed by any tangent and its two asymptotes is constant.
Area $=a^2$
NIMCET
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
NIMCET
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
