MCQ 11 Mark
The point on the curve $y^2 = x$ where tangent makes $45^\circ $ angle with $x-$axis is:
- A
$\Big(\frac{1}{2},\frac{1}{4}\Big)$
- ✓
$\Big(\frac{1}{4},\frac{1}{2}\Big)$
- C
$(4, 2)$
- D
$(1, 1)$
AnswerCorrect option: B. $\Big(\frac{1}{4},\frac{1}{2}\Big)$
Let the required point be $(\text{x}_1,\text{y}_2).$
The tangent makes an angle of $45^\circ$ with the $x-$axis.
Slope of the tangent $=\tan 45^\circ=1$
Since, the point lies on the curve.
Hence, $\text{y}_{1}^2=\text{x}_1$
Now,
$\text{y}^2=\text{x}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\approx\frac{1}{2\text{y}}$
Slope of the tangent $=\bigg(\frac{\text{dy}}{\text{dx}}\bigg)_{(\text{x}_1,\text{y}_1)}=\frac{1}{2\text{y}_1}$
Given:
$\frac{1}{2\text{y}_1}=1$
$\Rightarrow2\text{y}_1=1$
$\Rightarrow\text{y}_1=\frac{1}{2}$
Now,
$\text{x}_1=\text{y}^2_1-\big(\frac{1}{2}\big)^2-\frac{1}{4}$
$\therefore(\text{x}_1,\text{y}_1)=\Big(\frac{1}{4},\frac{1}{2}\Big)$
View full question & answer→MCQ 21 Mark
The point on the curve $y^2 = x$ where tangent makes $45°$ angle with $x-$axis is:
- A
$\Big(\frac{1}{2},\frac{1}{4}\Big)$
- ✓
$\Big(\frac{1}{4},\frac{1}{2}\Big)$
- C
$(4,2)$
- D
$(1,1)$
AnswerCorrect option: B. $\Big(\frac{1}{4},\frac{1}{2}\Big)$
Let the required point be $(x_1, y_1)$.
The tangent makes an angle of $45^\circ$ with the $x-$axis,
$\therefore$ Slope of the tangen $= \tan 45^\circ = 1$
Since, the point lies on the curve.
Hence, $\text{y}^2=\text{x}_1$
Now, $\text{y}^2=\text{x}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=1$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}}$
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=\frac{1}{2\text{y}_1}$
Given:
$\frac{1}{2\text{y}_1}=1$
$\Rightarrow2\text{y}_1=1$
$\Rightarrow\text{y}_1=\frac{1}{2}$
Now,
$\text{x}_1=\text{y}_1^2=\Big(\frac{1}{2}\Big)^2=\frac{1}{4}$
$\therefore(\text{x}_1,\text{y}^2_1)=\Big(\frac{1}{4},\frac{1}{2}\Big)$
View full question & answer→MCQ 31 Mark
The point on the curve $y = x^2 - 3x + 2$ where tangent is perpendicular to $y = x$ is:
- A
$(0, 2)$
- ✓
$(1, 0)$
- C
$(-1, 6)$
- D
$(2, -2)$
AnswerCorrect option: B. $(1, 0)$
$\text{y}=\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$
Let $(x_2,y_2)$ be the required point.
Since, the point lies on the curve,
Hence, $\text{y}_1=\text{x}_1^2-3\text{x}_1+2$
Now, $\text{y}=\text{x}^2-3\text{x}+2$
$\therefore\frac{\text{dy}}{\text{dx}}=2\text{x}-3$
Slope of the perpendicular to this line.
$\therefore$ Slope of the tangent $=\frac{-1}{\text{slope of the line}}=\frac{-1}{1}=-1$
Now,
$2\text{x}_1-3=-1$
$\Rightarrow 2\text{x}_1=2$
$\Rightarrow\text{x}_1=2$
and
$\text{y}_1=\text{x}_1^2-3\text{x}_1+2=1-3+2=0 $
$\therefore(\text{x}_1,\text{y}_1)=(1,0)$
View full question & answer→MCQ 41 Mark
At what point the slope of the tangent to the curve $x^2 + y^2 - 2x - 3 = 0$ is zero:
- A
$(3, 0), (-1, 0)$
- B
$(3, 0), (1, 2)$
- C
$(-1, 0), (1, 2)$
- ✓
$(1, 2), (1, -2)$
AnswerCorrect option: D. $(1, 2), (1, -2)$
Let $(x_1, y_1)$ be the required point.
Since, the point lie on the curve.
Hence, $\text{x}^2_1+\text{y}_1^2-2\text{x}_1-3=0 \ ...(1)$
Now, $\text{x}^2+\text{y}^2-2\text{x}-3=0$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}-2=0$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2-2\text{x}}{2\text{y}}=\frac{1-\text{x}}{\text{y}}$
Now,
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=\frac{1-\text{x}_1}{\text{y}_1}$
Slope of tangent $= 0 ($Given$)$
$\therefore\frac{1-\text{x}_1}{\text{y}_1}=0$
$\Rightarrow1-\text{x}_1=0$
$\Rightarrow\text{x}_1=1$
From $(1),$ we get
$\text{x}_1^2+\text{y}_1^2-2\text{x}_1-3=0$
$\Rightarrow1+\text{y}_1^2-4=0$
$\Rightarrow\text{y}_1=\pm2$
So, the points are $(1, 2)$ and $(1, -2).$
View full question & answer→MCQ 51 Mark
The normal at the point $(1, 1)$ on the curve $2y + x^2 = 3$ is:
- A
$x + y = 0$
- ✓
$x - y = 0$
- C
$x + y + 1 = 0$
- D
$x - y = 1$
AnswerCorrect option: B. $x - y = 0$
$2y + x^2 = 3$
$2\frac{\text{dy}}{\text{dx}}+2\text{x}=0$
$\frac{\text{dy}}{\text{dx}}=-\text{x}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=-1$
Slope of the normal $= 1$
Equation of the normal
$y - 1 = x - 1$
$y = x$
$x - y = 0$
View full question & answer→MCQ 61 Mark
Any tangent to the curve $y = 2x^7 + 3x + 5$:
AnswerCorrect option: C. Makes an acute angle with $x-$axis.
We have, $y = 2x^7 + 3x + 5$
$\frac{\text{dy}}{\text{dx}}=14\text{x}^6+3$
$\Rightarrow\frac{\text{dy}}{\text{dx}}>3$
$(\because x^6$ is always positive for any real value of $x)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}>0$
So, $\tan\theta>0$
Hence, $\theta$ lies in first quadrant.
Thus, the tangent to the curve makes an acute angle with $x-$axis.
View full question & answer→MCQ 71 Mark
The slope of the tangent to the curve $x = t^2 + 3 t - 8, y = 2t^2 - 2t - 5$ at point $(2, -1)$ is:
- A
$\frac{22}{7}$
- ✓
$\frac{6}{7}$
- C
$-6$
- D
$\text{None of these}$
AnswerCorrect option: B. $\frac{6}{7}$
$\text{x}=\text{t}^2+3\text{t}-$ and $\text{y}=2\text{t}^2-2\text{t}^2-2\text{t}-5$
$\frac{\text{dx}}{\text{dt}}=2\text{t}+3$ and $\frac{\text{dy}}{\text{dt}}=4\text{t}-2$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{4\text{t}-2}{2\text{t}+3}$
The given point is $(2, -1).$
$\therefore\text{x}=2$ and $\text{y=}-1$
Now,
$\text{t}^2+3\text{t}-8=2 $ and $2\text{t}^2-2\text{t}-5=-1$
Let u solve one of these to get the value of $t.$
$\text{t}^2+3\text{t}-10=0$ and $2\text{t}^2-2\text{t}-4=0$
$\Rightarrow(\text{t}+5)(\text{t}-2)=0$ and $(2\text{t}-2)(\text{t}-2)=0$
$\Rightarrow\text{t}=-5\text{ or }\text{t}=2$
These two have $t = 2$ as a comman solution.
$\therefore$ Slope of the tangent $= \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=2}=\frac{8-2}{4+3}=\frac{6}{7}$
View full question & answer→MCQ 81 Mark
The equation of the normal to the curve $3x^2 - y^2 = 8$ which is parallel to $x + 3y = 8$ is:
- A
$x + 3y = 8$
- B
$x + 3y + 8 = 0$
- ✓
$x + 3y ± 8 = 0$
- D
$x + 3y = 0$
AnswerCorrect option: C. $x + 3y ± 8 = 0$
Since the normal is parallel to the given line, the equation of normal will be of the given form.
$\text{x}+3\text{y}=\text{k}$
$3\text{x}^2-\text{y}^2=8$
Let $(x_1, y_1)$ be the point of intersection of the two curves.
Then,
$\text{x}_1+3\text{y}_1=\text{k} \ ...(1)$
$3\text{x}_1^2-\text{y}_1^2=8 \ ...(2)$
Now,$$ $3\text{x}^2-\text{y}^2=8$
On diffierentiating both sides $\text{w.r.t.x,}$ we get
$6\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{6\text{x}}{2\text{y}}=\frac{3\text{x}}{\text{y}}$
Slope of the normal, $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=\frac{3\text{x}_1}{\text{y}_1}$
Slope of the normal, $\text{m}=\frac{-1}{\Big(\frac{3\text{x}}{\text{y}}\Big)}=\frac{-\text{y}_1}{3\text{x}_1}$
Given:
Slope of the normal $=$ Slope of the given line
$\Rightarrow\frac{-\text{y}_1}{3\text{x}_1}=\frac{-1}{3}$
$\Rightarrow\text{y}_1=\text{x}_1\ ...(3)$
From $(2),$ we get
$3\text{x}_1{^2}-\text{x}_1{^2}=8$
$\Rightarrow2\text{x}_1{^2}=8$
$\Rightarrow\text{x}_1{^2}=4$
$\Rightarrow\text{x}_1=\pm2$
Case $1$
when $\text{x}_1=2$
From (3), we get
$\text{y}_1=\text{x}_1=2$
From (3), we get
$2 + 3 (2) = k$
$\Rightarrow 2 + 6 = k$
$\Rightarrow k = 8$
$\therefore$ Equation of the normal from $(1)$
$\Rightarrow x + 3y = 8$
$\Rightarrow x + 3y - 8 = 0$
$\Rightarrow -2 - 6 = k$
$\Rightarrow k = -8$
$\therefore$ Equation of the normal from $(1)$
$\Rightarrow x + 3y = -8$
$\Rightarrow x + 3y - 8 = 0$
From both the case, we get the equation of the normal as:
$\text{x}+3\text{y}\pm8=0$
View full question & answer→MCQ 91 Mark
The point on the curve $9y^2 = x^3$, where the normal to the curve makes equal intercepts with the axes is:
- ✓
$\Big(4,\frac{8}{3}\Big)$
- B
$\Big(-4,\frac{8}{3}\Big)$
- C
$\Big(4,-\frac{8}{2}\Big)$
- D
$\text{None of these}.$
AnswerCorrect option: A. $\Big(4,\frac{8}{3}\Big)$
$9y^2 = x^3$
$\Rightarrow18\text{y}\frac{\text{dy}}{\text{dx}}=3\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{6\text{y}}=\text{slope of tangent}$
Slope of normal $=\frac{-6\text{y}}{\text{x}^2}$
Given that normal makes equal intercept on axes.
$\Rightarrow$ Slope of normal $=\pm1$
$\frac{-6\text{y}}{\text{x}^2}=1$ or $\frac{-6\text{y}}{\text{x}^2=}=-1$
$\Rightarrow\text{y}=\frac{-\text{x}^2}{6}$ or $\text{y}=\frac{\text{x}^2}{6}$
$\text{y}=\frac{-\text{x}^2}{6}$
$9\text{y}^2=\text{x}^3$
$9\Big(\frac{-\text{x}^2}{6}\Big)^2=\text{x}^3$
$\Rightarrow\text{x}=0\text{ or }4$
$\Rightarrow\text{y}=0,\pm\frac{8}{3}$
Point are $\Big(4,\frac{8}{3}\Big)$ and $\Big(4,\frac{-8}{3}\Big)$
View full question & answer→MCQ 101 Mark
The equation to the normal to the curve $\text{y}=\sin\text{x}$ at $(0, 0)$ is:
- A
$x = 0$
- B
$y = 0$
- ✓
$x + y = 0$
- D
$x - y = 0$
AnswerCorrect option: C. $x + y = 0$
Given:
$\text{y}=\sin\text{x}$
On differentiating both sides $w.r.t.x,$ we get
$\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,0)}=\cos=1$
Slope of the normal, $\text{m}=\frac{-1}{1}=-1$
Given:
$(x_1, y_1) = (0, 0)$
$\therefore$ equation of the normal
$= y - y_1 = m (x_1, y_1)$
$\Rightarrow y - 0 = -1 (x - 0)$
$\Rightarrow y = -x$
$\Rightarrow x + y = 0$
View full question & answer→MCQ 111 Mark
The equation of the normal to the curve $\text{y}=\text{x}+\sin\text{x}\cos\text{x}\text{ at }\text{x}=\frac{\pi}{2}$ is
- A
$\text{x}= 2$
- B
$\text{x}=\pi$
- ✓
$\text{x}+\text{y}=0$
- D
$2\text{x}=\pi$
AnswerCorrect option: C. $\text{x}+\text{y}=0$
Given:
$\text{y}=\sin\text{x}$
On differentinating both sides $w.r.t. x,$
We get,
$\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,0)}=\cos0=1$
Slope of the normal $\text{m,}=\frac{-1}{1}=-1$
Given:
$(\text{x}_1,\text{y}_1)=(0,0)$
$\therefore$ Equation of the normal
$=\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-0=-1(\text{x}-0)$
$\Rightarrow\text{y}=-\text{x}$
$\Rightarrow\text{x}+\text{y}=0$
View full question & answer→MCQ 121 Mark
The angle of intersection of the curves $xy = a^2$ and $x^2 - y^2 = 2a^2$ is:
- A
$0^\circ$
- B
$45^\circ$
- ✓
$90^\circ$
- D
AnswerCorrect option: C. $90^\circ$
$\text{x}\text{y}=\text{a}^2$ and $\text{x}^2-\text{y}^2=2\text{a}^2$
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$ and $2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{\text{x}}$ and $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
We can see dearly that product of the slopes of $\tan$ gents is $-1$
Hence, angle between two tangents is $90^\circ .$
View full question & answer→MCQ 131 Mark
The slope of the tangent to the curve $x = 3t^2 + 1, y = t^3 -1$ at $x = 1$ is:
- A
$\frac{1}{2}$
- ✓
$0$
- C
$-2$
- D
$\infty$
Answer$\text{x}=3\text{t}^2+1$ and $\text{y}=\text{t}^3-1$
$\frac{\text{dx}}{\text{dt}}=6\text{t}$ and $\frac{\text{dy}}{\text{dt}}=3\text{t}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{t}}{2} \ ...(1)$
But,
$\text{x}=1$
$\Rightarrow3\text{t}^2+1=1$
$\Rightarrow\text{t}=0$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{t}}{2}=0 \ (\because\text{From(i))}$
View full question & answer→MCQ 141 Mark
If the curve $ay + x^2 = 7$ and $x^3 = y$ cut orthogonally at $(1, 1)$, then a is equal to:
AnswerGiven:
$\text{ay}+\text{x}^2=7 \ ...(1)$
$\text{x}^3=\text{y} \ ...(2)$
Point $= (1, 1)$
On differentiatuing $(1) w.r.t.x,$ we get
$\text{a}\frac{\text{dy}}{\text{dx}}+2\text{x}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-2\text{x}}{\text{a}}$
$\Rightarrow\text{m}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=\frac{-2}{\text{a}}$
Again, on differetiating $(2) w.r.t.x,$ we get
$3\text{x}^2=\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=3$
It is given that curves are orthogonal at the given point,
$\therefore\text{m}_1\times\text{m}_2=-1$
$\Rightarrow\frac{-2}{\text{a}}\times3=-1$
$\Rightarrow\text{a}=6$
View full question & answer→MCQ 151 Mark
The line $y = mx + 1$ is a tangent to the curve $y^2 = 4x$, if the value of m is:
- ✓
$1$
- B
$2$
- C
$3$
- D
$\frac{1}{2}$
AnswerLet $(x_1, y_1)$ be the required point.
The slope of the given line is $m.$
we have
$\text{y}^2=4\text{x}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=4$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{2\text{y}}=\frac{2}{\text{y}}$
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=\frac{2}{\text{y}_1}$
Given:
Slope of tangent $= m$
Now,
$\frac{2}{\text{y}_1}=\text{m}\ ...(1)$
Because the given line is a tangent to the given curve at point $(x_1, y_1)$ this point lies on both the line and the curve.
$\therefore\text{y}_1=\text{mx}+1$ and $\text{y}_1^2=4\text{x}_1$
$\Rightarrow\text{x}_1=\frac{\text{y}_1-1}{\text{m}}$ and $\text{x}_1=\frac{\text{y}_1^2}{4}$
So,
$\frac{\text{y}_1-1}{\text{m}}=\frac{\text{y}_1^2}{4}$
$\Rightarrow\frac{\text{y}_1-1}{\Big(\frac{2}{\text{y}_1}\Big)}=\frac{\text{y}_1{^2}}{4}\Big[\text{From} (1)$
$\Rightarrow\frac{\text{y}_1(\text{y}_1-1)}{2}=\frac{\text{y}_1^2}{4}$
$\Rightarrow2\text{y}_1^2-2\text{y}_1=\text{y}_1^2$
$\Rightarrow\text{y}_1^2-2\text{y}_1=0$
$\text{y}_1(\text{y}_1-2)=0$
$\Rightarrow\text{y}_1=0,2$
So, for $\text{y}_1=0,\text{m}=\frac{2}{0}=\infty$
For $\text{y}_1=2,\text{m}=\frac{2}{2}=1$
View full question & answer→MCQ 161 Mark
The point on the curve $y = 6x - x^2$ at which the tangent to the curve is inclined at $\frac{\pi}{4}$ to the line $x + y = 0$ is:
AnswerCorrect option: B. $(3,9)$
Let $(x_1, y_1)$ be the point where the given curve intersect the given line at the given angle.
Since, the point lie on the curve.
Hence, $\text{y}_1=6\text{x}_1-\text{x}_1^2$
Now, $\text{y}=6\text{x}-\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=6-2\text{x}$
$\Rightarrow\text{m}_ 1=6-2\text{x}_1$
and
$x + y = 0$
$\Rightarrow1+\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$
$\Rightarrow\text{m}_2=-1$
it is given that the angles between them is $\frac{\pi}{4}$
$\therefore\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|$
$\Rightarrow\tan\frac{\pi}{4}=\Big|\frac{6-2\text{x}_1+1}{1-6+2\text{x}_1}\Big|$
$\Rightarrow1=\Big|\frac{7-2\text{x}_1}{2\text{x}_1-5}\Big|$
$\Rightarrow\Big|\frac{7-2\text{x}_1}{2\text{x}_1-5}\Big|=1\text{ or }\Big|\frac{7-2\text{x}_1}{2\text{x}_1-5}\Big|=-1$
$\Rightarrow7-2\text{x}_1=2\text{x}_1-5\text{ or }7-2\text{x}_1=-2\text{x}_1+5$
$\Rightarrow4\text{x}_1=12\text{ or }2=0$
$\Rightarrow\text{x}_1=3$
and
$\text{y}_1=6\text{x}_1-\text{x}_1^2=18-9=9$
$\therefore(\text{x}_1,\text{y}_1)=(3,9)$
View full question & answer→MCQ 171 Mark
The point at the curve $y = 12x - x^2$ where the slope of the tangent is zero will be:
- A
$(0, 0)$
- B
$(2, 16)$
- C
$(3, 9)$
- ✓
Answer$\text{y}=12\text{x}-\text{x}^2$
Slope of the tangent $= 0$
$\frac{\text{dy}}{\text{dx}}=0$
$12-2\text{x}=0$
$\Rightarrow\text{x}=6$
$\Rightarrow\text{y}=36$
Point on corve is $(6, 36)$.
View full question & answer→MCQ 181 Mark
The angle of intersection of the curves $\text{y}=2\sin^2\text{x}$ and $\text{y}=\cos2\text{x}\text{ at }\text{x}=\frac{\pi}{6}$ is:
- A
$\frac{\pi}{4}$
- B
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{3}$
- D
$\text{None of these.}$
AnswerCorrect option: C. $\frac{\pi}{3}$
$\text{y}=2\sin^2\text{x}\Rightarrow\frac{\text{dy}}{\text{dx}}=4\sin\text{x}\cos\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}-\frac{\pi}{6}}=\sqrt{3}=\text{m}_1$
$\text{y}=\cos2\text{x}\Rightarrow\frac{\text{dy}}{\text{dx}}=-2\sin2\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}-\frac{\pi}{6}}=-\sqrt{3}=\text{m}_1$
$\tan\theta\Big|\frac{\sqrt{3}+\sqrt{3}}{1-3}\Big|=\sqrt{3}$
$\Rightarrow\theta=\tan^{-1}\sqrt{3}=\frac{\pi}{3}$
View full question & answer→MCQ 191 Mark
The equations of tangent at those points where the curve $y = x^2 - 3x + 2$ meets $x-$axis are:
AnswerCorrect option: B. $x + y - 1 = 0 = x - y - 2$
$\text{y}=\text{x}^2-3\text{x}+2$
Slope of tangent
$\frac{\text{dy}}{\text{dx}}=2\text{x}-3$
Tangent meets at $x-$axis hence $y = 0.$
$\text{x}^2-3\text{x}+2=0$
$(\text{x}-2)(\text{x}-1)=0$
$\text{x}=2$ or $\text{x}=1 $
For $\text{x}=2\Rightarrow\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,0)}=1$
Equation of tangent $m = -1$, point $(1, 0)$
$\text{y}-0-1(\text{x}-1)$
$\Rightarrow\text{x}+\text{y}-1=0$
View full question & answer→MCQ 201 Mark
If the tangent to the curve $x = at^2, y = 2$ at is perpendicular to $x-$axis, then its point of contact is:
- A
$(a, a)$
- B
$(0, a)$
- ✓
$(0, 0)$
- D
$(a, 0)$
AnswerCorrect option: C. $(0, 0)$
$x = at^2$ and $y = 2at$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{at}$ and $\frac{\text{dy}}{\text{dt}}=2\text{a}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{a}}{2\text{at}}=\frac{2\text{a}}{\text{y}}$
Slope of tangent $=\frac{2\text{a}}{\text{y}}$
Tangent is perpendicular to $y-$axis.
$\Rightarrow$Tangent is parallel to $x-$axis
$\frac{2\text{a}}{\text{y}}=0$
$\text{a}=0$
$\Rightarrow\text{x}=0$ and $\text{y}=0$
Point is $(0, 0)$.
View full question & answer→MCQ 211 Mark
The normal to the curve $x^2 = 4y$ passing through $(1, 2)$ is:
- ✓
$x + y = 3$
- B
$x − y = 3$
- C
$x + y = 1$
- D
$x − y = 1$
AnswerCorrect option: A. $x + y = 3$
$x^2 = 4y$
$2\text{x}=4\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,2)}=\frac{1}{2}=\text{m}$
Slope of normal $=\frac{-1}{\frac{\text{dy}}{\text{dx}}}=\frac{-1}{\frac{\text{x}}{\text{2}}}=\frac{-2}{\text{x}}$
Let $(X, Y)$ be the point where normal and curve intersect
$\therefore $ Slope of normal at $(X, Y) =\frac{-2}{\text{X}}$
Equation of normal passing through $(X, Y)$ with slope $\frac{-2}{\text{X}}$ is
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\text{y}-\text{Y}=\frac{-2}{\text{h}}(\text{x}-\text{X})$
Since normal passes through $(1, 2)$ it will satisfy its equation
$\text{Y}=2+\frac{-2}{\text{h}}(\text{1}-\text{X})\ \dots(1)$
Since $(X, Y)$ lies on the curve $x^2 = 4y$
$X^2 = 4Y .......(2)$
Using $(1)$ and $(2)$
$\Rightarrow\ 2+\frac{-2}{\text{h}}(\text{1}-\text{X})=\frac{\text{X}^2}{4}$
$2+\frac{2}{\text{X}}-2=\frac{\text{X}^2}{4}$
$\frac{2}{\text{X}}=\frac{\text{X}^2}{4}$
$\text{X}^3=8$
$\text{X}=2$
Putting $X = 2$ in $(2)$
$\text{Y}=\frac{\text{X}^2}{4}=\frac{(2)^2}{4}=\frac{4}{4}=1$
Hence, $X = 2, Y = 1$
$\text{y}-\text{Y}=\frac{-2(\text{x}-\text{X})}{\text{X}}$
$\text{y}-\text{1}=\frac{-2(\text{x}-\text{2})}{\text{2}}$
$\text{y}-1=-1(\text{x}-2)$
$\text{y}-1=-\text{x}+2$
$\text{x}+\text{y}=2+1$
$\text{x}+\text{y}=3$
View full question & answer→MCQ 221 Mark
The angle between the curves $y^2 = x$ and $x^2= y$ at $(1, 1)$ is:
- A
$\tan^{-1}\frac{4}{3}$
- ✓
$\tan^{-1}\frac{3}{4}$
- C
$90^\circ$
- D
$45^\circ$
AnswerCorrect option: B. $\tan^{-1}\frac{3}{4}$
$\tan^{-1}\frac{3}{4}$
Given:
$\text{y}^2=\text{x} \ ...(1)$
$\text{x}^2=\text{y} \ ...(2)$
Point $= (1, 1)$
On diffierentiating $\text{(1) w.r.t. x,}$ we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}}$
$\Rightarrow\text{m}_1=\frac{1}{2}$
On differentiating $\text{(2) w.r.t. x,}$ we get
$2\text{x}=\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{m}_2(1)=2$
Now,
$\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{\frac{1}{2}-2}{1+\frac{1}{2}\times2}\Big|=\frac{3}{4}$
$\Rightarrow\theta=\tan^{-1}\big(\frac{3}{4}\big)$
View full question & answer→MCQ 231 Mark
The angle of intersection of the parabolas $y^2 = 4 ax$ and $x^2 = 4ay$ at the origin is:
- A
$\frac{\pi}{6}$
- B
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{2}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: C. $\frac{\pi}{2}$
Given:
$y^2 = 4ax ...(1)$
$x^2 = 4ay ...(2)$
Point $= (0, 0)$
On differentiating $\text{(1) w.r.t. x}$,
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\approx\frac{2\text{a}}{\text{y}}$
$\Rightarrow\text{m}_1\approx\infty$
Now, on differentiating $\text{(2) w.r.t. x}$,
$2\text{x}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\approx\frac{\text{x}}{2\text{u}}=0$
$\therefore\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{\infty}{1+0}\Big|=\infty$
$\Rightarrow\theta=\tan^{-1}\infty=\frac{\pi}{2}$
View full question & answer→MCQ 241 Mark
The slope of the tangent to the curve $x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is:
- A
$\frac{22}{7}$
- ✓
$\frac{6}{7}$
- C
$\frac{7}{6}$
- D
$-\frac{6}{7}$
AnswerCorrect option: B. $\frac{6}{7}$
$x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$
$\frac{\text{dx}}{\text{dt}}=2\text{t}+3,\frac{\text{dy}}{\text{dt}}=4\text{t}-2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{4\text{t}-2}{2\text{t}-3}$
Point is given by $(2, -1)$
$x = 2$ and $y = -1$
$t^2 + 3t - 8 = 2$ and $2t^2 - 2t - 5 = -1$
$t^2 + 3t - 10 = 0$ and $2t^2 - 2t - 4 = 0$
$(t + 5) (t - 2) = 0$ and $(2 + 2) (t + 2) = 0$
For $t = 2$ both equation has common solution.
$\Rightarrow $ Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}-2}=\frac{4\times2-2}{2\times2+3}=\frac{6}{7}$
View full question & answer→MCQ 251 Mark
The curves $y = ae^x$ and $y = be^{−x}$ cut orthogonally, if:
- A
$a = b$
- B
$a = -b$
- ✓
$ab = 1$
- D
$ab = 2$
AnswerCorrect option: C. $ab = 1$
Given:
$\text{y}=\text{ae}^\text{x} \ ...(1)$
$\text{y}=\text{be}^{\text{-x}} \ ...(2)$
Let the point of intersection of these two curves be $(x_1, y_1)$.
Now,
On differentiating (1) w.r.t. x, we get
$\frac{\text{dy}}{\text{dx}}=\text{ae}{^\text{x}}$
$=\text{m}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=\text{ae}^{\text{x}^{1}}$
Again, on diffirentiating (2)w.r.t. x, we get
$\frac{\text{dy}}{\text{dx}}=-\text{be}^{\text{-x}}$
$\Rightarrow\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=-\text{be}^{-\text{x}_1}$
It is given that the curves cut orthaonally.
$\therefore\text{m}_1\times\text{m}_2=-1$
$\Rightarrow\text{ae}^{\text{x}_1}\times (-\text{be}^-\text{x}_1)=-1$
$\Rightarrow\text{ab}=1$
View full question & answer→MCQ 261 Mark
If the line $y = x$ touches the curve $y = x^2 + bx + c$ at a point $(1, 1)$ then:
- A
$b = 1, c = 2$
- ✓
$b = -1, c = 1$
- C
$b = 2, c = 1$
- D
$b = -2, c = 1$
AnswerCorrect option: B. $b = -1, c = 1$
We can find the slope of the line by differentiating w.r.t.x,
Slope of the given line $= 1$
Now,
$\text{y}=\text{x}^2+\text{bx}+\text{c} \ ...(1)$
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=2+\text{b}$
Given:
Slope of the tangent $= 1$
$\Rightarrow2+\text{b}=1$
$\text{b}=-1$
On substituting $b = -1, x = 1$ and $y = 1$ in (1), we get
$\Rightarrow1=1-1+\text{c}$
$\text{c}=1$
View full question & answer→MCQ 271 Mark
The equation of the normal to the curve $y = x(2 - x)$ at the point $(2, 0)$ is:
- ✓
$x - 2y = 2$
- B
$x - 2y + 2 = 0$
- C
$2x + y = 4$
- D
$2x + y - 4 = 0$
AnswerCorrect option: A. $x - 2y = 2$
Here,
$\text{y}=\text{x}(2-\text{x})=2\text{x}-\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2-2\text{x}$
Slope of the tangent $=\bigg(\frac{\text{dy}}{\text{dx}}\bigg)_{(2,0)}=2-4=-2$
Slope of the normal $\text{m}=\frac{-1}{-2}=\frac{1}{2}$
Given:
$(\text{x}_1,\text{y}_1)=(2,0)$
$\therefore$ Equation of the normal
$=\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-0=\frac{1}{2}(\text{x}-2)$
$\Rightarrow2\text{y}=\text{x}-2$
$\Rightarrow\text{x}-2\text{y}=2$
View full question & answer→MCQ 281 Mark
If the curves $y = 2e^x$ and $y = ae^{−x}$ intersect orthogonally, then $a =$
- ✓
$\frac{1}{2}$
- B
$\frac{-1}{2}$
- C
$2$
- D
$2\text{e}^2$
AnswerCorrect option: A. $\frac{1}{2}$
$= 2e^x$ and $y = ae^{−x}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_1=2\text{e}^\text{x}$ and $\Big(\frac{\text{dy}}{\text{dx}}\Big)_2=-\text{ae}^\text{x}$
Given that $\Big(\frac{\text{dy}}{\text{dx}}\Big)_1\times\Big(\frac{\text{dy}}{\text{dx}}\Big)_2=-1$
$2\text{e}^\text{x}-\text{ae}^{-\text{x}}=-1$
$2\text{a}=1$
$\text{a}=\frac{1}{2}$
View full question & answer→MCQ 291 Mark
The equation of the normal to the curve $x = a \cos^3 \theta $, $y = a \sin^3 θ$ at the point $\theta=\frac{\pi}{4}$ is:
- A
$x = 0$
- B
$y = 0$
- ✓
$c = y$
- D
$x + y = a$
AnswerCorrect option: C. $c = y$
$\text{x}=\text{a}\cos^3\theta $ and $\text{y}=\text{a}\sin^3\theta$
$\text{x}=\text{a}\cos^3\frac{\pi}{4}$ and $\text{y}=\text{a}\sin^3\frac{\pi}{4}$
$\text{x}=\frac{\text{a}}{2\sqrt{2}}$ and $\frac{\text{a}}{2\sqrt{2}}$
Point is $\Big(\frac{\text{a}}{2\sqrt{2}},\frac{\text{a}}{2\sqrt{2}}\Big)$
$\text{x}=\text{a}\cos^3\theta$ and $\text{y}=\text{a}\sin^3\theta$
$\Rightarrow\cos\theta=3\sqrt{\frac{\text{x}}{\text{a}}}$ and $\sin\theta=3\sqrt{\frac{\text{y}}{\text{a}}}$
$\sin^2\theta+\cos^2\theta=1$
$\Rightarrow\Big(\frac{\text{x}}{\text{a}}\Big)^{\frac{2}{3}}+\Big(\frac{\text{y}}{\text{a}}\Big)^{\frac{2}{3}}=1$
View full question & answer→