Maths M.C.Q (1 Marks)

3. $10\sqrt{3}\text{cm}^2\text{/s}$

Solution:

Let the side of an equilateral triangle be x cm,

$\therefore$ Area of equilateral triangle, $\text{A}=\frac{\sqrt{3}}{4}\text{x}^2\ \ \dots(\text{i})$

Also, $\frac{\text{dx}}{\text{dt}}=2\text{cm/s}$

On differentiating Eq. (i) w.r.t. t, we get

$\frac{\text{dA}}{\text{dt}}=\frac{\sqrt{3}}{2}\cdot2\text{x}\cdot\frac{\text{dx}}{\text{dt}}$

$=\frac{\sqrt{3}}{4}\cdot2\cdot10\cdot2$ $\Big[\because\ \text{x}=10\text{ and }\frac{\text{dx}}{\text{dt}}=2\Big]$

$=10\sqrt{3}\text{cm}^2\text{/s}$

1. A vertical tangent (parallel to y-axis).

Solution:

We are given that $\text{y}=\text{x}^{\frac{1}{5}}$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{5}\text{x}^{\frac{1}{3}-1}$ $\Big[\because\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{n}-1}\Big]$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{5}\text{x}^{\frac{-4}{5}}$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{5\text{x}^{\frac{4}{5}}}$

$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,0)}=\frac{1}{5(0)^{\frac{4}{5}}}=\infty$

So, the curve $\text{y}=\text{x}^{\frac{1}{5}}$ has a vertical tangent at (0, 0), which is parallel to Y-axis.

2. $\text{Decreasing in }\Big(\frac{\pi}{2},\pi\Big)$

Solution:

We have, $\text{f(x)}=4\sin^3\text{x}-6\sin^2\text{x}+12\sin\text{x}+100$

$\therefore\ \text{f(x)}=12\sin^2\text{x}\cdot\cos\text{x}-12\sin\text{x}\cdot\cos\text{x}+12\cos\text{x}$

$=12\cos\text{x}\big[\sin^2\text{x}-\sin\text{x}+1\big]$

$=12\cos\text{x}\big[\sin^2\text{x}+(1-\sin\text{x}\big]$

Now, $1-\sin\text{x}\geq0\text{ and }\sin^2\text{x}\geq0$

$\therefore\ \sin^2\text{x}+\text{f}-\sin\text{x}>0$

Hence, f'(x) > 0, when $\cos\text{x}>0\text{ i.e., x}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$

So, f(x) is increasing when $\text{x}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$

and f'(x) < 0, when $\cos\text{x}<0\text{ i.e., x}\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)$

Hence, f(x) is decreasing when $\text{x}\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)$

Hence, f(x) is decreasing in $\Big(\frac{\pi}{2},\pi\Big)$

3. $\frac{\pi}{2}$

Solution:

We have, x³ - 3xy² + 2 = 0 and 3x²y - y³ - 2 = 0

$\Rightarrow\ 3\text{x}^2-3\Big[\text{x}\cdot2\text{y}\frac{\text{dy}}{\text{dx}}+\text{y}^2\cdot1\Big]+0=0$ and $3\Big[\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{y}\cdot2\text{x}\Big]-3\text{y}^2\frac{\text{dy}}{\text{dx}}-0=0$

$\Rightarrow\ 6\text{xy}\frac{\text{dy}}{\text{dx}}+3\text{y}^2=3\text{x}^2$ and $3\text{y}^2\frac{\text{dy}}{\text{dx}}=3\text{x}^2\frac{\text{dy}}{\text{dx}}+6\text{xy}$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{3\text{x}^2-3\text{y}^2}{6\text{xy}}$ and $\frac{\text{dy}}{\text{dx}}=\frac{6\text{xy}}{3\text{y}^2-3\text{x}^2}$

$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{3(\text{x}^2-\text{y}^2)}{6\text{xy}}$ and $\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{-6\text{xy}}{3(\text{x}^2-\text{y}^2)}$

$\Rightarrow\ \text{m}_1=\frac{\text{x}^2-\text{y}^2}{2\text{xy}}$ and $\text{m}_2=\frac{-2\text{xy}}{\text{x}^2-\text{y}^2}$

$\therefore\ \text{m}_1\text{m}_2=\frac{\text{x}^2-\text{y}^2}{2\text{xy}}\cdot\frac{-(2\text{xy})}{\text{x}^2-\text{y}^2}=-1$

Since, the product of the slopes is -1.

Hence, both the curves are intersecting at right angle i.e., making $\frac{\pi}{2}$ with each other.

2. 12

Solution:

We have, y = -x³ + 3x² + 9x - 27

$\therefore\ \frac{\text{dy}}{\text{dx}}=-3\text{x}^2+6\text{x}+9=$ Slope of the curve

And $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-6\text{x}+6=-(\text{x}-1)$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=0\Rightarrow-6(\text{x}-1)=0\Rightarrow\text{x}=1$

Now,$\frac{\text{d}^3\text{y}}{\text{dx}^3}=-6<0$

So, the slope is maximum when x = 1

$\therefore\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}=1)}=-3.1^2+6.1+9=12$

2.  $[-2,-1]$

Solution:

We have, f(x) = 2x³ + 9x² + 12x - 1

$\therefore$ f'(x) = 6x² + 18x + 12

= 6(x² + 3x + 2) = 6(x + 2)(x + 1)

So, $\text{f}'(\text{x})\leq0,$ for decreasing.

On drawing number lines as below.

We see that f'(x) is decreasing in [-2, -1].

3. $\text{x} + 3\text{y} \pm8 = 0$

Solution:

we have, the equation of the curve is 3x² - y² = 8 ...(i)

Also, the given equation of the line is x + 3y = 8

$\Rightarrow\ 3\text{y}=8-\text{x}$

$\Rightarrow\ \text{y}=-\frac{\text{x}}{3}+\frac{8}{3}$

Thus, slope of the line is $-\frac{1}{3}$ which should be equal to slope of the equation of normal to the curve.

On differentiating Eq. (i) 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 curve

Now, slope of normal to the curve $=-\frac{1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)}$

$=-\frac{1}{\Big(\frac{3\text{x}}{\text{y}}\Big)}=-\frac{\text{y}}{3\text{x}}$

$\therefore\ -\Big(\frac{\text{y}}{3\text{x}}\Big)=-\frac{1}{3}$

$\Rightarrow\ 3\text{y}=-3\text{x}$

$\Rightarrow\ \text{y}=\text{x}$

On substituting the value of the given equation of the curve, we get

$3\text{x}^2-\text{x}^2=8$

$\Rightarrow\ \text{x}^2=\frac{8}{2}$

$\Rightarrow\ \text{x}=\pm2$

For $\text{x}=2,3(2)^2-\text{y}^2=8$

$\Rightarrow\ \text{y}^2=4$

$\Rightarrow\ \text{y}=\pm2$

and for $\text{x}=-2,3(-2)^2-\text{y}^2=8$

$\Rightarrow\ \text{y}=\pm2$

So, the points at which normal are parallel to the given line are $(\pm2,\pm2).$

Hence, the equation of normal at $(\pm2,\pm2)$ is

$\text{y}-(\pm2)=-\frac{1}{3}[\text{x}-(\pm2)]$

$\Rightarrow\ 3[\text{y}-(\pm2)]=-[\text{x}-(\pm2)]$

$\Rightarrow\ \text{x}+3\text{y}\pm8=0$

1. x + 5y = 2

Solution:

We have, equation of the curve y(1 + x²) = 2 - x ...(i)

It is given that the curve crosses x-axis

Putting y = 0 in equation (i), we get

$\therefore$ 0(1 + x²) = 2 - x

⇒ x = 2

So, the curve passes through the point (2, 0).

Now differentiating equation (i) w.r.t. x, we get

$\therefore\ \text{y}\times(0+2\text{x})+(1+\text{x}^2)\cdot\frac{\text{dy}}{\text{dx}}=0-1$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-1-2\text{xy}}{1+\text{x}^2}$

$\therefore\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,0)}=\frac{-1-2\times0}{1+2^2}=-\frac{1}{5}=$ slope of tangent to the curve

$\therefore$ Equation of tangent of the curve passing through (2, 0) is

$\text{y}-0=-\frac{1}{5}(\text{x}-2)$

or $5\text{y}+\text{x}=2$

2. $\Big(\frac{-1}{2},0\Big)$

Solution:

We have, y = e^2x

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}$

Since, it passes through the point (0, 1).

$\therefore\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,1)}=2\cdot\text{e}^{2\cdot0}=2=$ Slope of tangent to the curve

The equation of tangent is given by

y - 1 = 2(x - 0)

⇒ y - 1 = 2x

⇒ y = 2x + 1

We are given that, the tangent to curve y = e^2x at the point (0, 1) meets X-axis i.e., y = 0.

$\therefore\ 0=2\text{x}+1$

$\Rightarrow\ \text{x}=-\frac{1}{2}$

Thus, the required point is $\Big(\frac{-1}{2},0\Big).$

4. (2, 2), (-2, 34)

Solution:

The given equation of curve is

y = x³ - 12x + 18

$\therefore\ \frac{\text{dy}}{\text{dx}}=3\text{x}^2-12$ [on differenttiating w.r.t.x]

So, the slope of line parallel to the X-axis

$\therefore\ \Big(\frac{\text{dy}}{\text{dx}}\Big)=0$

$\Rightarrow\ 3\text{x}^2-12=0$

$\Rightarrow\ \text{x}^2=\frac{12}{3}=4$

$\therefore\ \text{x}=\pm2$

For x = 2, y = 2³ - 12 × 2 + 18 = 2

and for x = -2, y = (-2)³ -12(-2) + 18 = 34

So, the points are (2, 2) and (-2, 34).

2.  $\frac{1}{20}\text{radian/}\sec$

Solution:

Let the angle between floor and the ladder be $\theta.$

Let at any time 't' AB = x cm and BC = y cm

$\therefore\ \sin\theta=\frac{\text{x}}{500}$ and $\cos\theta=\frac{\text{y}}{500}$

$\Rightarrow\ \text{x}=500\sin\theta$ and $\text{y}=500\cos\theta$

Also it is given that $\frac{\text{dx}}{\text{dt}}=10\text{cm/s}$

$\Rightarrow\ 500\cdot\cos\theta\cdot\frac{\text{d}\theta}{\text{dt}}=10\text{cm}/ \text{s}$

$\Rightarrow\ \frac{\text{d}\theta}{\text{dt}}=\frac{10}{500\cos\theta}=\frac{1}{50\cos\theta}$

For $\text{y}=2\text{m}=20\text{cm},$

$\frac{\text{d}\theta}{\text{dt}}=\frac{1}{50\cdot\frac{\text{y}}{500}}=\frac{10}{\text{y}}$ $=\frac{10}{200}=\frac{1}{20}\text{radian/}\sec$ 

1. Always increases.

Solution:

We have, $\text{f(x)}=\tan\text{x}-\text{x}$

$\therefore\ \text{f}'(\text{x})=\sec^2\text{x}-1$

Since, $\text{f}'(\text{x})>0,\forall\text{ x}\in\text{R}$

Hence, f(x) always increases.

1. $1<\text{x}<3$

Solution:

We have, y = x(x - 3)²

$\therefore\ \frac{\text{dy}}{\text{dx}}$ =2(x - 3).1 + (x - 3)².1

= 2x² - 6x + x² + 9 = 6x = 3x² - 12x + 9

= 3(x² - 3x - x + 3) = 3(x - 3)(x - 1)

So, y = x(x - 3)² decreases for (1, 3).

[since, y' < 0 for all x $\in$ (1, 3), hence y is decreasing on (1, 3)]

4. Neither maximum nor minimum.

Solution:

We have, $\text{f(x)}=2\sin3\text{x}+3\cos3\text{x}$

$\therefore\ \text{f}'(\text{x})=2\cdot\cos3\text{x}3+3(-\sin3\text{x})\cdot3$

$\Rightarrow\ \text{f}'(\text{x})=6\cos3\text{x}-9\sin3\text{x}\ \ \dots(\text{i})$

Now, $\text{f}''(\text{x})=-18\sin3\text{x}-27\cos3\text{x}$

$=-9(2\sin3\text{x}+3\cos3\text{x})$

$\therefore\ \text{f}'\Big(\frac{5\pi}{6}\Big)=6\cos\Big(3\cdot\frac{5\pi}{6}\Big)-9\sin\Big(3\cdot\frac{5\pi}{6}\Big)$

$=6\cos\frac{5\pi}{2}-9\sin\frac{5\pi}{2}$

$=6\cos\Big(2\pi+\frac{\pi}{2}\Big)-9\sin\Big(2\pi+\frac{\pi}{2}\Big)$

$=-9\neq0$

So, $\text{x}=\frac{5\pi}{6}$ cannot be point of maxima or minima.

Hence, f(x) at $\text{x}=\frac{5\pi}{6}$ is neither maximum nor minimum.

3. One maxima and one minima.

Solution:

We have, f(x) = 2x³ - 3x² - 12x + 4

⇒ f'(x) = 6x² - 6x - 12

⇒ f'(x) = 6(x² - x - 2)

⇒ f'(x) = 6(x + 1)(x - 2)

Find the critical points by equating f'(x) to 0.

$\therefore$ f'(x) = 0

⇒ 6(x + 1)(x - 2) = 0

⇒ x = -1 and x = +2

From the above number line, we can conclude that, x = -1 is point of local maxima and x = 2 is point of local minima.

Thus, f(x) has one maxima and one minima.

3. $\text{e}^{\frac{1}{\text{e}}}$

Solution:

Let $\text{y}=\Big(\frac{1}{\text{x}}\Big)^\text{x}$

$\Rightarrow\ \log\text{y}=\text{x}\cdot\log\frac{1}{\text{x}}=-\text{x}\cdot\log\text{x}$

Diffrentiating both sides w.r.t x, we get,

$\frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=-\text{x}\cdot\frac{1}{\text{x}}-\log\text{x}$

$=-1-\log\text{x}$

$\therefore\ \frac{\text{dy}}{\text{dx}}=-\Big(\frac{1}{\text{x}}\Big)^\text{x}(1+\log\text{x})$

Now, $\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\ 1+\log\text{x}=0$

$\Rightarrow\ \text{x}=\frac{1}{\text{e}}$

Sign scheme of f'(x) is as shown in the following figure.

From the figure, $\text{x}=\frac{1}{\text{e}}$ is the point of maxima

Hence, maximum value of y is $\Big(\frac{1}{\frac{1}{\text{e}}}\Big)^{\frac{1}{\text{e}}}=\text{e}^{\frac{1}{\text{e}}}$

2. 0

Solution:

We have, f(x) = x³ - 18x² + 96x

$\therefore$ f'(x) = 3x² - 36x + 96

f'(x) = 0

$\therefore$ 3x² - 36x + 96 = 0

⇒ 3(x² - 12x + 32) = 0

⇒ (x - 8)(x - 4) = 0

⇒ x = 4, 8

For least value of f(x) in [0, 9], we should find the value of function at x = 0,

4, 8, 9

f(0) = 0

f(4) = 4³ - 18 × 4² + 96 × 4 = 64 - 288 + 384 = 160

f(8) = 8³ - 18 × 8² + 96 × 8 = 128

f(9) = 9³ - 18 × 9² + 96 × 9 = 729 - 1458 + 864 = 195

Thus, absolute minimum value of f on [0, 9] is 0 occurring at x = 0.

4. Is an increasing function.

Solution:

We have, $\text{f(x)}=2\text{x}+\cos\text{x}$

$\Rightarrow\ \text{f}'(\text{x})=2+(-\sin\text{x})$

$=2-\sin\text{x}$

Since, the maximum value of $\sin\text{x}$ is 1.

Hence, $\text{f}'(\text{x})>0,\forall\text{ x}$

Thus, f'(x) is an increasing function.

3.  $\cos\text{x}$

Solution:

$​​\text{f}_1(\text{x})=\sin2\text{x},$ increases from '0' to '1' in $\Big(0,\frac{\pi}{2}\Big)$

$\text{f}_2(\text{x})=\tan\text{x}$ is increasing function in each quadrant.

$\text{f}_3(\text{x})=\cos\text{x},$ decreases from '1' to '0' in $\Big(0,\frac{\pi}{2}\Big) $

$ \text{f}_4(\text{x})=\cos3\text{x},$ decreases if $3\text{x}\in\Big(0,\frac{\pi}{2}\Big)\ \text{or x }\in\Big(0,\frac{\pi}{6}\Big)$ 

3. 1

Solution:

Let f(x) = x² - 8x + 17

$\therefore$ f'(x) = 2x - 8

So, f'(x) = 0, gives x = 4

Now,f''(x) = 2 > 0, $\forall$ x

So, x =4 is the point of local minima.

$\therefore$ Minimum value of f(x) at x = 4,

f(4) = 4 × 4 - 8 × 4 + 17 = 1

4. 6

Solution:

We are given that, ay + x² = 7 and x³ = y cut orthogonally at (1, 1).

On differentiating w.r.t. x, we get

$\text{a}\cdot\frac{\text{dy}}{\text{dx}}+2\text{x}=0$ and $3\text{x}^2=\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\frac{2\text{x}}{\text{a}}$ and $\frac{\text{dy}}{\text{dx}}=3\text{x}^2$

$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=\frac{-2}{\text{a}}=\text{m}_1$ and $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=3.1=3=\text{m}_2$

Since, the curves cut orthogonally at (1, 1)

$\text{m}_1\cdot\text{m}_2=-1$

$\Rightarrow\ \Big(\frac{-2}{\text{a}}\Big)\cdot3=-1$

$\Rightarrow\ \text{a}=6$

4. 0.32

Solution:

We have, y = x⁴ - 10

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=4\text{x}^3$

and $\triangle\text{x}=2.00-1.99=0.01$

$\therefore\ \triangle\text{y}=\frac{\text{dy}}{\text{dx}}\times\triangle\text{x}$

$=4\text{x}^3\times\triangle\text{x}$

$=4\times2^3\times0.01$

$=32\times0.01=0.32$

So, the approximate change in y is 0.32.

2.  $\frac{6}{7}$

Solution:

Given equation of curve is

$\text{x}=\text{t}^2+3\text{t}-8$ and $\text{y}=2\text{t}^2-2\text{t}-5$

$\therefore\ \frac{\text{dx}}{\text{dy}}=2\text{t}+3$ and $\frac{\text{dy}}{\text{dx}}=4\text{t}-2$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{4\text{t-2}}{2\text{t}+3}\ \ \dots(\text{i})$

Since, the curve passes through the point (2, -1), we have

2 = t² + 3t - 8

and -1 = 2t² - 2t - 5

⇒ t² + 3t - 10 = 0

And 2t² - 2r - 4 = 0

⇒ t² + 5t - 2t - 10 = 0

And 2t² + 2t - 4t - 4 = 0

⇒ t(t + 5) - 2(t + 5) = 0

And 2t(t + 1) - 4(t + 1) = 0

⇒ (t - 2)(t + 5) = 0

And (2t - 4)(t + 1) = 0

⇒ t = 2, -5 and t = 1, 2

⇒ t = 2 (common value)

From equation (i),

$\therefore$ Slope of tangent, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=2}=\frac{4\times2=2}{2\times2+3}=\frac{6}{7}$ 

2. $\text{x}=\frac{1}{\text{e}}$

Solution:

We have, f(x) = x^x

Let us suppose y = x^x

Taking logarithm on both sides, we get

$\log\text{y}=\text{x}\log\text{x}$

$\therefore\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\text{x}\cdot\frac{1}{\text{x}}+\log\text{x}\cdot1$ $\big[\because(\text{fg})'=\text{fg}'+\text{gf}'\big]$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(1+\log\text{x})\cdot\text{x}^\text{x}$

Find the critical points by equating $\frac{\text{dy}}{\text{dx}}$ to 0.

$\therefore\ \frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\ (1+\log\text{x})\text{x}^\text{x}=0$

$\Rightarrow\ \log\text{x}=-1$ as $\text{x}^\text{x}\neq0$

$\Rightarrow\ \log\text{x}=\log\text{e}^{-1}$

$\Rightarrow\ \text{x}=\text{e}^{-1}$

$\Rightarrow\ \text{x}=\frac{1}{\text{e}}$

Hence, f(x) has a stationary point at $\text{x}=\frac{1}{\text{e}}.$