3 Marks - Maths STD 12 Science Questions - Vidyadip

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Question 13 Marks

Find the slope of the normal at the point 't' on the curve $\text{x}=\frac{1}{\text{t}},\text{y}=\text{t}.$

Answer

Here,
$\text{x}=\frac{1}{\text{t}}\text{ and }\text{y}=\text{t}$
$\frac{\text{dx}}{\text{dt}}=\frac{-1}{\text{t}^2}\text{ and }\frac{\text{dy}}{\text{dt}}=1$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{1}{\Big(\frac{-1}{\text{t}^2}\Big)}=-\text{t}^2$
Now,
Slope of the tangent = $\Big(\frac{\text{dy}}{\text{dx}}\Big)=-\text{t}^2$
Slope of the normal $=\frac{-1}{\text{slope of the tangent}}=\frac{-1}{-\text{t}^2}=\frac{1}{\text{t}^2}$

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Question 23 Marks

Find the point on the curve y = x² where the slope of the tangent is equal to the x-coordinate of the point.

Answer

The given equation of the curve is
y = x²...(1)
$\therefore$ slope of tangent to (i) is
$\frac{\text{dy}}{\text{dx}}=2\text{x}\ ...(2)$
According to the question
$\frac{\text{dy}}{\text{dx}}=\text{x}\ ...(3)$
From (2) and (3)
2x = x
⇒ x = 0 & y = 0
Thus, the required point is (0, 0)

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Question 33 Marks

Write the coordinates of the point at which the tangent to the curve y = 2x² - x + 1 is parallel to the line y = 3x + 9.

Answer

Let (x₁, y₁) be the required point.
slope of the given line = 3
Since, the point lies on the curve.
Hence, $\text{y}_1=2\text{x}_1{^2}-\text{x}_1+1\ ...(1)$
Now, $\text{y}=2\text{x}^2-\text{x}+1$
$\therefore\frac{\text{dy}}{\text{dx}}=4\text{x}-1$
Now,
slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=4\text{x}_1-1$
Given:
slope of the tangent = slope of the given line
$\therefore4\text{x}_1-1=3$
$\Rightarrow\text{x}_1=1$
From 1 we get
$\text{y}_1=2-1+1=2$
$\therefore(\text{x}_1,\text{y}_1)=(1,2)$

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Question 43 Marks

Write the angle between the curves y = e^-x and y = e^x at their point of intersections.

Answer

The given equation of curve are,
y = e^-x...(1)
y = e^x...(2)
Solving (1) and (2)
$\text{y}=\frac{1}{\text{e}^\text{x}}=\frac{1}{\text{y}}$
$\Rightarrow\text{y}^2=1$
$\Rightarrow\text{y}=\pm1$
From (2)
$\pm1=\text{e}^\text{x}$
$\Rightarrow\text{x}=0$
So, the point of intersection is P = (0, 1) and Q = (0, -1).
Slope of (2)
$\text{m}_2=\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$\therefore\text{m}_2\text{ at }\text{P}=1\text{ and }\text{m}_2\text{ at }\text{Q}=1$
$\therefore\text{m}_1\times\text{m}_2=-1\times1=-1$
The angle between the curves is 90º

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Question 53 Marks

Show that the tangents to the curve y = 7x³ + 11 at the points x = 2 and x = -2 are parallel.

Answer

The equaation of the given curve is y = 7x³ + 11
$\therefore\frac{\text{dy}}{\text{dx}}=21\text{x}^2$
The slope of the tangent to a curve at $(\text{x}_0,\text{y}_0)$ is $\frac{\text{dy}}{\text{dx}}\Big]_{(\text{x}_0,\text{ y}_0)}$
Therefore, the slope of the tangent at the point where x = 2 is given by, $\frac{\text{dy}}{\text{dx}}\Big]_{(\text{x}=2)}=21(2)^2=84$
it is observed that the slope of the tangents at the points where x = 2 and x = -2 are equal.
Hence, the two tangents are parallel.

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Question 63 Marks

Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{y}=\sqrt{\text{x}}\ \text{at}\ \text{x}=9$

Answer

We know that the slope of the tangent to the curve y = f(x) is
$\frac{\text{dy}}{\text{dx}}=\text{f}'(\text{x})\ ...(1)$
And the slope of the normal is
$\frac{-1}{\frac{\text{dy}}{\text{dx}}}=\frac{-1}{\text{f}'(\text{x})}\ ...(2)$
$\text{y}=\sqrt{\text{x}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$
$\therefore$ slope of tangent at x = 9
$\therefore\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=9}=\frac{1}{2\sqrt{9}}=\frac{1}{6}$
Also, the slope of normal is
$\frac{-1}{\frac{\text{dy}}{\text{dx}}}=\frac{-1}{\text{f}'(\text{x})}=-6$

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Question 73 Marks

Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{y}=\sqrt{\text{x}^3}\ \text{at}\text{ x}=4$

Answer

$\text{y}=\sqrt{\text{x}^3}=\text{x}^\frac{3}{2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3}{2}\text{x}^{\frac{1}{2}}=\frac{3}{2}\sqrt{\text{x}}$
When x = a, $\text{y}=\sqrt{\text{x}^3}=\sqrt{64}=8$
Now,
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(4.8)}=\frac{3}{2}\sqrt{4}=3$
Slope of the normals $=\frac{-1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(4.8)}}=\frac{-1}{3}$

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Question 83 Marks

Find the equation of the tangent to the curve $\sqrt{\text{x}}+\sqrt{\text{y}}=\text{a},$ at the point $\Big(\frac{\text{a}^2}{4},\frac{\text{a}^2}{4}\Big).$

Answer

$\sqrt{\text{x}}+\sqrt{\text{y}}=\text{a},$
Differentiating both sides w.r.t. x,
$\Rightarrow\frac{1}{2\sqrt{\text{x}}}+\frac{1}{2\sqrt{\text{y}}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\sqrt{\text{y}}}{\sqrt{\text{x}}}$
Given $(\text{x}_1,\text{y}_1)=\Big(\frac{\text{a}^2}{4},\frac{\text{a}^2}{4}\Big)$
Slope of tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\Big(\frac{\text{a}^2}{4},\frac{\text{a}^2}{4}\Big)}=\frac{-\sqrt{\frac{\text{a}^2}{4}}}{\sqrt{\frac{\text{a}^2}{4}}}$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\frac{\text{a}^2}{4}=-1\big(\text{x}-\frac{\text{a}^2}{4}\big)$
$\Rightarrow\text{y}-\frac{\text{a}^2}{4}=-\text{x}+\frac{\text{a}^2}{4}$
$\Rightarrow\text{x}+\text{y}=\frac{\text{a}^2}{2}$

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Question 93 Marks

Find the equation of the normal to y = 2x³ - x² + 3 at (1, 4).

Answer

The equation of the curve is
$\text{y}=2\text{x}^3-\text{x}^3+3\ ...(1)$
Slope $=\text{m}=\frac{\text{dy}}{\text{dx}}=6\text{x}^2-2\text{x}$
$\therefore\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,\ 4)}=4$
Now,
The equation of the normal is (1) is
$\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)$
$\Rightarrow(\text{y}-4)=\frac{-1}{4}(\text{x}-1)$
$\Rightarrow\text{x}+4\text{y}=16+1$
$\Rightarrow\text{x}+4\text{y}=17$

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Question 103 Marks

Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{xy}=6\ \text{at}\ (1,6)$

Answer

$\text{xy}=6$
On differentiating both sides w.r.t. x, we get
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{\text{x}}$
Now,
slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,6)}=\frac{-\text{y}}{\text{x}}=\frac{-6}{1}=-6$
slope of the normal $=\frac{-1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,6)}}=\frac{-1}{-6}=\frac{1}{6}$

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Question 113 Marks

Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{y}=\text{x}^3-\text{x}\ \text{at}\ \text{x}=2$

Answer

$\text{y}=\text{x}^3-\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=3\text{x}^2-1$
When $\text{x}=2,\text{y}=\text{x}^3-\text{x}=2^3-2=6$
Now,
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,\ 6)}=3(2 )^2-1=11$
Slope of the normal $=\frac{-1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,\ 6)}}=\frac{-1}{11}$

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Question 123 Marks

Write the equation of the normal to the curve $\text{y}=\text{x}+\sin\text{x}\cos\text{x}\text{ at }\text{x}=\frac{\pi}{2}.$

Answer

$\text{y}=\text{x}+\sin\text{x}\cos\text{x}$
On differentiating both sides w.r.t.x,
$\frac{\text{dy}}{\text{dx}}=1+\cos^2\text{x}-\sin^2\text{x}$
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{2}}=1+\cos^2\frac{\pi}{2}=1-1=0$
Where $\text{x}=\frac{\pi}{2},\text{y}=\frac{\pi}{2}+\sin\frac{\pi}{2}\cos\frac{\pi}{2}=\frac{\pi}{2}$
$\therefore(\text{x}_1,\text{y}_1)=\Big(\frac{\pi}{2},\frac{\pi}{2}\Big)$
Equation of the normal
$=\text{y}-\text{y}_1=\frac{-1}{\text{slope of the tangent}}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\frac{\pi}{2}=\frac{-1}{0}\Big(\text{x}-\frac{\pi}{2}\Big)$
$\Rightarrow\text{x}=\frac{\pi}{2}$
$\Rightarrow2\text{x}=\pi$

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Question 133 Marks

Write the equation of the normal to the curve $\text{y}=\cos\text{x}$ at (0, 1).

Answer

The given equation of curves are

y = 2x² - x + 1 ...(1)

y = 3x + 9 ...(2)

Slope of (1)

$\text{m}_1=\frac{\text{dy}}{\text{dx}}=4\text{x}-1$

Slope of (2)

$\text{m}_2=\frac{\text{dy}}{\text{dx}}=3$

Given that tangent to (1) is parallel to (2)

$\therefore\text{m}_1=\text{m}_2$

⇒ 4x - 1 = 3

⇒ x = 1

From (1)

y = 2 × 1² - 1 + 1 = 2

thus, the required point p = (1, 2).

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Question 143 Marks

Find the slope of the tangent to the curve x = t² + 3t - 8, y = 2t² - 2t - 5 at t = 2.

Answer

We have,
x = t² + 3t - 8, y = 2t² - 2t - 5
$\therefore\frac{\text{dy}}{\text{dt}}=2\text{t}+3,\frac{\text{dy}}{\text{dt}}=4\text{t}-2$
$\Big(\frac{\text{dy}}{\text{dt}}\Big)_{\text{t}-2}=7,\Big(\frac{\text{dy}}{\text{dt}}\Big)_{\text{t}-2}=6$
$\text{m}=\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dy}}{\text{dt}}}=\frac{6}{7}$
$\therefore\text{m}=\frac{6}{7}$

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Question 153 Marks

Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{y}=2\text{x}^2+3\sin\text{x}\ \text{at}\text{ x}=0$

Answer

$\text{y}=2\text{x}^2+3\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\text{x}+3\cos\text{x}$
when $\text{x}=0,\text{y}=2\text{x}^2+3\sin\text{x}=2(0)^2+3\sin0=0$
Now, slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,\ 0)}=4(0)+3\cos0=3$
slope of the normal $=\frac{-1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,\ 0)}}=\frac{-1}{3}$

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Question 163 Marks

Write the coordinates of the point on the curve $\text{x}=\text{e}^\text{t}\cos\text{t},\text{y}=\text{e}^\text{t}$ where the tangent line makes an angle $\frac{\pi}{4}$ with x-axis.

Answer

Here,
$\text{x}=\text{e}^\text{t}\cos\text{t}\text{ and }\text{y}=\text{e}^{\text{t}}\sin\text{t}$
$\frac{\text{dx}}{\text{dt}}=\text{e}^{\text{t}}\cos\text{t}-\text{e}^{\text{t}}\sin\text{t}\text{ and }\frac{\text{dy}}{\text{dt}}=\text{e}^{\text{t}}\sin\text{t}+\text{e}^{\text{t}}\cos\text{t}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{e}^\text{t}\sin\text{t}+\text{e}^\text{t}\cos\text{t}}{\text{e}^\text{t}\sin\text{t}+\text{e}^\text{t}\cos\text{t}}=\frac{\sin\text{t}+\cos\text{t}}{\cos\text{t}-\sin\text{t}}$
Now,
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=\frac{\pi}{4}}=\frac{\sin\frac{\pi}{4}+\cos\frac{\pi}{4}}{\cos\frac{\pi}{4}-\sin\frac{\pi}{4}}=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}=\frac{\frac{2}{\sqrt{2}}}{0}=\infty$
Let $\theta$ be the angle made by the tangent with the x-axis.
$\therefore\tan\theta=\infty$
$\Rightarrow\theta=\frac{\pi}{2}$

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Question 173 Marks

Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1-\cos\theta)\text{at}\ \theta=\frac{\pi}{2}$

Answer

$\text{x}=\text{a}(\theta-\sin\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}(1-\cos\theta)$
$\text{y}=\text{a}(1-\cos\theta)$
$\Rightarrow\frac{\text{dy}}{\text{d}\theta}=\text{a}(\sin\theta)$
$\theta\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{a}(\sin\theta)}{\text{a}(1-\cos\theta)}=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}=\cot\frac{\theta}{2}$
Now,
slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\theta=\frac{\pi}{2}}=\cot\Big(\frac{\frac{\pi}{2}}{2}\Big)=\cot\Big(\frac{\pi}{4}\Big)=1$
slope of the normal $=\frac{-1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\theta=\frac{\pi}{2}}}=\frac{-1}{1}=-1$

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Question 183 Marks

Write the coordinates of the point on the curve y² = x where the tangent line makes an angle $\frac{\pi}{4}$ with x-axis.

Answer

We have,
$\text{y}^2=\text{x}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}}=\text{m}_1$
We have given that slopes of the tangent slope of tangent makes an angle.
$\frac{\pi}{4}$ with x-axis.
$\therefore\text{m}_1=\tan\frac{\pi}{4}=1$
$\Rightarrow\frac{1}{2\text{y}}=1$
$\Rightarrow\text{y}=\frac{1}{2}$
$\therefore\text{x}=\frac{1}{4}$
Hence, the required point is $\Big(\frac{1}{4},\frac{1}{2}\Big).$

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Question 193 Marks

Find the coordinates of the point on the curve y² = 3 - 4x where tangent is parallel to the line 2x + y - 2 = 0.

Answer

The given equation of curve is
y² = 3 - 4x ...(1)
2x + y - 2 = 0 ...(2)
Slope of (1)
$\text{m}_1=\frac{\text{dy}}{\text{dx}}=\frac{-4}{2\text{y}}$
Slope of (2)
$\text{m}_2=\frac{\text{dy}}{\text{dx}}=-2$
Given that the tangent to (1) is parallel to (2)
$\therefore\text{m}_1=\text{m}_2$
$\Rightarrow\frac{-4}{2\text{y}}=-2$
$\Rightarrow\text{y}=1$
From (1)
$1=3-4\text{x}$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, the required point is $\Big(\frac{1}{2},1\Big).$

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Question 203 Marks

Write the angle between the curves y² = 4x and x² = 2y - 3 at the point (1, 2).

Answer

Given:
y² = 4x ...(1)
x² = 2y − 3 ...(2)
On differentiating (1) w.r.t.x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=4$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{y}}$
$\Rightarrow\text{m}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,2)}=\frac{2}{2}=1$
On differentiating (2) w.r.t.x, we get
$2\text{x}=2\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}$
$\Rightarrow\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,2)}=1$
Thus, we get
$\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|$
$\Rightarrow\tan\theta=\Big|\frac{1-1}{1+1}\Big|$
$\Rightarrow\tan\theta=0$
$\Rightarrow\theta=0$

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Question 213 Marks

When the curve crosses the y-axis, the point on the curve is of the form (0, y).

Answer

Here,
y = x² - x + 2
⇒ y = 0 - 0 + 2 = 2
so, the point where the curve crosses the y-axis is (0, 2).
Now,
y = x² - x + 2
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{x}-1$
Slope of the tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,2)}=2(0)-1=-1$
$\therefore(\text{x}_1,\text{y}_1)=(0,2)$
and
Equation of tangent
= y - y₁ = m(x - x₁)
⇒ y - 2 = -1(x - 0)
⇒ y - 2 = -x
⇒ x + y - 2 = 0

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3 Marks - Maths STD 12 Science Questions - Vidyadip