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AMU MCA Mathematics PYQ

Qus : 1 AMU MCA PYQ

Let $f(x,y,z) = e^{\sqrt{1 - x^2 - y^2}}$. Then the range of $f$ is

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Solution

For real values:

$1 - x^2 - y^2 \ge 0$

$x^2 + y^2 \le 1$

Minimum value of $\sqrt{1 - x^2 - y^2}$ is $0$ (when $x^2+y^2=1$)

Maximum value is $1$ (when $x=y=0$)

So exponent varies from $0$ to $1$

$e^0 = 1$

$e^1 = e$

Hence range is $[1, e]$

Qus : 2 AMU MCA PYQ

Let $G$ be a group having elements $a$ and $b$ such that $O(a)=4$, $O(b)=2$ and $a^3 b = ba$. Then $O(ab)$ is

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Solution

Given:

$a^4 = e$

$b^2 = e$

Also $a^3 b = ba$

Multiply by $a$:

$a^4 b = baa$

$b = baa$

$ab = a(baa) = (ab)^{-1}$

Hence $(ab)^2 = e$

Therefore order of $ab = 2$

Qus : 3 AMU MCA PYQ

The value of $\iint_R y\sin(ny)\,dA$, where $R=[1,2]\times[0,\pi]$ is

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Solution

$\iint_R y\sin(ny)\,dA=\int_{1}^{2}\!\!dx\int_{0}^{\pi} y\sin(ny)\,dy$

$=\int_{0}^{\pi} y\sin(ny)\,dy$

Let $u=y,\; dv=\sin(ny)\,dy \Rightarrow v=-\frac{\cos(ny)}{n}$

$\int_{0}^{\pi} y\sin(ny)\,dy=\left[-\frac{y\cos(ny)}{n}\right]_{0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\cos(ny)\,dy$

$=-\frac{\pi\cos(n\pi)}{n}+\frac{1}{n^2}\left[\sin(ny)\right]_{0}^{\pi}=-\frac{\pi(-1)^n}{n}$

Qus : 4 AMU MCA PYQ

The volume of the solid in the first octant bounded by the cylinder $z=9-y^2$ and the plane $x=2$ is

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Solution

First octant: $0\le x\le 2$, $0\le y\le 3$, $0\le z\le 9-y^2$

$V=\int_{0}^{2}\!\!dx\int_{0}^{3}(9-y^2)\,dy=2\left[9y-\frac{y^3}{3}\right]_{0}^{3}=2(27-9)=36$

Qus : 5 AMU MCA PYQ

The volume of the solid that lies under the paraboloid $z=x^2+y^2$ above the $xy$-plane, and inside the cylinder $x^2+y^2=2x$ is

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Solution

Use polar: $x=r\cos\theta,\; y=r\sin\theta$

Cylinder: $r^2=2r\cos\theta \Rightarrow r=2\cos\theta,\; -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$

$V=\iint_D (x^2+y^2)\,dA=\int_{-\pi/2}^{\pi/2}\int_{0}^{2\cos\theta} r^2\cdot r\,dr\,d\theta$

$=\int_{-\pi/2}^{\pi/2}\left[\frac{r^4}{4}\right]_{0}^{2\cos\theta}d\theta=\int_{-\pi/2}^{\pi/2}4\cos^4\theta\,d\theta$

$=8\int_{0}^{\pi/2}\cos^4\theta\,d\theta=8\cdot\frac{3\pi}{16}=\frac{3\pi}{2}$

Qus : 6 AMU MCA PYQ

Consider the series $x_{n+1}=\frac{x_n}{2}+\frac{9}{8x_n}$, $x_0=0.5$ obtained from the Newton-Raphson method. The series converges to

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Solution

$x_{n+1}=\frac{1}{2}\left(x_n+\frac{9}{4x_n}\right)$ is Newton iteration for $\sqrt{\frac{9}{4}}$

$\sqrt{\frac{9}{4}}=\frac{3}{2}=1.5$

Qus : 7 AMU MCA PYQ

$\lim_{(x,y)\to(0,0)} f(x,y)$ where $f(x,y)=\tan^{-1}\left(\frac{|x|+|y|}{x^2+y^2}\right)$ is

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Solution

As $(x,y)\to(0,0)$, $|x|+|y|\sim r$ and $x^2+y^2=r^2$ so $\frac{|x|+|y|}{x^2+y^2}\sim\frac{1}{r}\to\infty$

Hence $\tan^{-1}(\cdot)\to\frac{\pi}{2}$

Qus : 8 AMU MCA PYQ

The value of $\int_C (y^2\,dx + x^2\,dy)$, where $C$ is the triangle given by $x=0$, $x+y=1$, $y=0$ is

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Solution

Using Green’s theorem:

$\oint_C (M\,dx+N\,dy)=\iint_D\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA$

Here $M=y^2,\; N=x^2$ so $\frac{\partial N}{\partial x}=2x,\; \frac{\partial M}{\partial y}=2y$

Integral $=\iint_D 2(x-y)\,dA$ over triangle with vertices $(0,0),(1,0),(0,1)$

By symmetry $\iint_D x\,dA=\iint_D y\,dA$ so $\iint_D (x-y)\,dA=0$

Hence value $=0$

Qus : 9 AMU MCA PYQ

The solution of the linear difference equation $y_{k+1}-ay_k=0$, $a\ne 1$ is

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Solution

$y_{k+1}=ay_k \Rightarrow y_k=ca^k$

Qus : 10 AMU MCA PYQ

The general solution of the differential equation $(D^2 - a^2)^3 y = e^{ax}$, where $D=\frac{d}{dx}$ is

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Solution

Characteristic equation:

$(m^2-a^2)^3=0$

Roots: $m=\pm a$ each of multiplicity 3.

Hence complementary function:

$(c_1+c_2x+c_3x^2)e^{ax}+(c_4+c_5x+c_6x^2)e^{-ax}$

Since RHS is $e^{ax}$ and $m=a$ has multiplicity 3,

Particular integral = $x^3\frac{e^{ax}}{3!\,(2a)^3} =\frac{x^2e^{ax}}{8a^2}$

Qus : 11 AMU MCA PYQ

The shortest distance from the point $(1,0,-2)$ to the plane $x+2y+z=4$ is

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Solution

Distance formula:

$d=\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}$

Plane: $x+2y+z-4=0$

$d=\frac{|1(1)+2(0)+1(-2)-4|}{\sqrt{1^2+2^2+1^2}}$

$=\frac{|1-2-4|}{\sqrt6}=\frac{5}{\sqrt6} =\frac{5\sqrt6}{6}$

Qus : 12 AMU MCA PYQ

Let $f(x)$ be continuous whose values are known at $-2,-1,1,2$. If the Lagrange interpolation formula $f(x)=L_1f(-2)+L_2f(-1)+L_3f(1)+L_4f(2)$ is used to approximate $f(0)$, then $L_3$ is

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Solution

Lagrange basis for node $x_3=1$:

$L_3(0)=\frac{(0+2)(0+1)(0-2)}{(1+2)(1+1)(1-2)}$

$=\frac{(2)(1)(-2)}{(3)(2)(-1)} =\frac{-4}{-6}=\frac{2}{3}$

Qus : 13 AMU MCA PYQ

In the analysis of data of a randomized block design with 5 blocks and 4 treatments, the error degrees of freedom is:

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Solution

For Randomized Block Design:

Error d.f. = $(t-1)(b-1)$

Here $t=4$, $b=5$

Error d.f. = $(4-1)(5-1)=3 \times 4 = 12$

Qus : 14 AMU MCA PYQ

The following LPP

Maximize $z = x_1 + x_2$

Subject to

$x_1 + x_2 \le 1$

$-3x_1 + x_2 \ge 3$

$x_1, x_2 \ge 0$

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Solution

From $-3x_1 + x_2 \ge 3$

$x_2 \ge 3 + 3x_1$

But from $x_1 + x_2 \le 1$

$x_2 \le 1 - x_1$

So we need:

$3 + 3x_1 \le 1 - x_1$

$4x_1 \le -2$

$x_1 \le -\frac{1}{2}$

This contradicts $x_1 \ge 0$

Hence problem is infeasible.

Qus : 15 AMU MCA PYQ

If $x \ge 1$ is the critical region for testing $H_0:\theta=2$ against $H_1:\theta=1$, on the basis of a single observation from the population

$f(x;\theta)=\theta e^{-\theta x},\quad x \ge 0$

then the value of the level of significance is

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Solution

Level of significance:

$\alpha = P(X \ge 1 \mid \theta=2)$

$= \int_{1}^{\infty} 2e^{-2x} dx$

$= \left[-e^{-2x}\right]_{1}^{\infty}$

$= e^{-2}$

$= \frac{1}{e^2}$

Qus : 16 AMU MCA PYQ

The standard deviation of sample means is called

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Solution

Standard deviation of sample mean distribution is called Standard Error.

Qus : 17 AMU MCA PYQ

If $X$ has an exponential distribution with mean $3$, then the variance of $X$ is

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Solution

For exponential distribution:

Mean $= \frac{1}{\lambda}$

Variance $= \frac{1}{\lambda^2}$

Given mean $=3 \Rightarrow \lambda=\frac{1}{3}$

Variance $=\frac{1}{(1/3)^2}=9$

Qus : 18 AMU MCA PYQ

A lot consists of 20 defective and 80 non-defective items. Two items are drawn at random without replacement. What is the probability that both items are defective?

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Solution

Total items = 100

Probability both defective:

$=\frac{20}{100}\times\frac{19}{99}$

$=\frac{380}{9900}=\frac{19}{495}$

Qus : 19 AMU MCA PYQ

Suppose random variable $X$ has possible values $1,2,3,\dots$ and $P(X=j)=\frac{1}{2^j}$, $j=1,2,3,\dots$ then $P(X \text{ is even})$ equals

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Solution

Even values: $2,4,6,\dots$

$P(\text{even})=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\dots$

$=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots$

This is geometric series:

$=\frac{1/4}{1-1/4}=\frac{1/4}{3/4}=\frac{1}{3}$

Qus : 20 AMU MCA PYQ

Coefficient of correlation is equal to

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Solution

Correlation coefficient $r=\pm\sqrt{b_{xy}b_{yx}}$

Hence it is geometric mean of two regression coefficients.

Qus : 21 AMU MCA PYQ

If $X$ has discrete uniform distribution with pmf $f(x)=\frac{1}{8}$ for $x=1,2,\dots,8$, then the mean of the distribution is

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Solution

Mean of uniform integers $1$ to $8$:

$=\frac{1+8}{2}=4.5$

Qus : 22 AMU MCA PYQ

The equation of the plane through the intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to $X$-axis is

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Solution

Qus : 23 AMU MCA PYQ

If $P(X=0)=1-P(X=1)$ and $E(X)=3V(X)$, the value of $P(X=0)$ is

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Solution

Let $p=P(X=1)$

Then $P(X=0)=1-p$

$E(X)=p$

$V(X)=p(1-p)$

Given $p=3p(1-p)$

$1=3(1-p)$

$1=3-3p$

$3p=2$

$p=\frac{2}{3}$

Therefore $P(X=0)=1-p=\frac{1}{3}$

Qus : 24 AMU MCA PYQ

If the point $(a, a)$ falls between the lines $|x + y| = 4$, then

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Solution

Qus : 25 AMU MCA PYQ

If $X$ is a discrete random variable taking positive integer values and possesses memory-less property, then the distribution of $X$

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If each observation is multiplied by a constant $k$, then new S.D. = $|k| \times$ old S.D.

Here $k=2$

New S.D. = $2 \times 8 = 16$

Qus : 34 AMU MCA PYQ

The standard deviation of a variable $x$ is $10$. Then the standard deviation of $50 + 5x$ is

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Solution

Qus : 35 AMU MCA PYQ

The standard deviation of first $n$ natural numbers is

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Solution

For numbers $1,2,\dots,n$:

Mean $=\frac{n+1}{2}$

Ratio $=\frac{m}{m_0}=3$

Qus : 46 AMU MCA PYQ

A dictionary is printed consisting of $7$ letters words only that can be made with letters of word $CRICKET$. If the words are printed at the alphabetical order as in an ordinary dictionary, then the number of words before the word $CRICKET$ is

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Solution

Qus : 47 AMU MCA PYQ

Which of the following potentials does not satisfy Laplace’s equation?

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Solution

Laplace equation: $\nabla^2\phi=0$

(a) $\nabla^2=2+10-12=0$ ✔

(b) $\nabla^2=0$ ✔

(d) $\nabla^2=4-8+2=-2 \ne 0$ ✘

$x_1 \le 3$

$x_1, x_2 \ge 0$

has optimal value as

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Solution

Qus : 76 AMU MCA PYQ

The function $f(x)=\dfrac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ is not defined at $x=\pi$. The value of $f(\pi)$, so that $f(x)$ is continuous at $x=\pi$, is

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Solution

Qus : 77 AMU MCA PYQ

The following LPP:

Minimize $Z = x - y$

Subject to:

$2x + 3y \le 6$

$0 \le x \le 4$

Solution:

Standardized regression coefficient:

$\beta = b \left(\frac{S_x}{S_y}\right)$

From the figure,

$ BC=\sqrt{AB^2-AC^2}=\sqrt{1-x^2} $

$ \Rightarrow \cos\alpha=\sqrt{1-x^2},\ \sin\alpha=x $

$ \Rightarrow \cos\beta=x,\ \sin\beta=\sqrt{1-x^2} $

$ \Rightarrow \sin^{-1}x=\alpha,\ \cos^{-1}x=\beta $

In right triangle,

$ \alpha+\beta=\frac{\pi}{2} $

$ \Rightarrow \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2} $

$ \Rightarrow \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x $

Qus : 171 AMU MCA PYQ

Let $f(x)$ be a function defined by $f(x)=\begin{cases}4x-5,\ x\le2 \ x-\lambda,\ x>2 \end{cases}$ If $\lim_{x\to2} f(x)$ exists, then the value of $\lambda$ is

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Solution

For limit to exist,

$ \lim_{x\to2^-} f(x)=\lim_{x\to2^+} f(x) $

$ \lim_{x\to2^-} f(x)=4(2)-5=3 $

$ \lim_{x\to2^+} f(x)=2-\lambda $

Equating,

$ 3=2-\lambda $

$ \Rightarrow \lambda=-1 $

Qus : 172 AMU MCA PYQ

The tangent of the angle between the lines whose intercepts on the axes are respectively $a,-b$ and $b,-a$ is

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Solution

Equations of lines:

$ \frac{x}{a}-\frac{y}{b}=1 $

$ \frac{x}{b}-\frac{y}{a}=1 $

Slopes:

$ m_1=\frac{b}{a},\quad m_2=\frac{a}{b} $

Angle between lines:

$ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| $

$ =\left|\frac{\frac{b}{a}-\frac{a}{b}}{1+\frac{b}{a}\cdot\frac{a}{b}}\right| $

$ =\left|\frac{\frac{b^2-a^2}{ab}}{2}\right| $

$ =\left|\frac{b^2-a^2}{2ab}\right| $

has a unique solution then

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Solution

Determinant $\ne 0$ $\left|\begin{matrix}2 & a & 6 \ 1 & 2 & b \ 1 & 1 & 3\end{matrix}\right| \ne 0$ $=2(6-b)-a(3-b)+6(1-2)$ $=12-2b-3a+ab-6$ $=ab-3a-2b+6$ $=(a-2)(3-b)\ne 0$ $\Rightarrow a\ne2$ or $b\ne3$

Qus : 220 AMU MCA PYQ

A discrete random variable $X$ taking non-negative values has the following moment generating function $M_X(t) = e^{2(e^t-1)},; -\infty < t < \infty$ Then, the value of $P(X \le 1)$ is:

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Solution

Given MGF corresponds to Poisson distribution with $\lambda = 2$. $P(X \le 1) = P(0)+P(1)$ $= e^{-2} + 2e^{-2} = 3e^{-2}$

Qus : 221 AMU MCA PYQ

If $a, b, c$ are roots of the equation $x^3 + px + q = 0$, then the value of $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$ is

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Solution

Qus : 222 AMU MCA PYQ

Max $z = 6x_1 - x_2$

Subject to

$2x_1 - x_2 \le 2$

$Mode = 3Median − 2Mean$

Rearranging:

$P(X \le k) = (k/N)^n$

$P(X \le k-1) = ((k-1)/N)^n$

So,

$P(X = k) = P(X \le k) - P(X \le k-1)$

Solution

Using Chebyshev inequality:

$P(|X-\mu|\ge m)\le \frac{Var(X)}{m^2}$

$1-\frac{9}{m^2}\ge 0.99$

$\frac{9}{m^2}\le 0.01$

$m^2\ge 900$

$m=30$

Qus : 318 AMU MCA PYQ

If $f(x)=\frac{x}{2}-1$, then on the interval $[0,\pi]$

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Solution

$f(x)=0$ at $x=2$ ⇒ $\frac{1}{f(x)}$ discontinuous $\tan[f(x)]$ continuous (no $\frac{\pi}{2}$ hit) Final Answer: $\boxed{\text{tan continuous, 1/f discontinuous}}$

Qus : 319 AMU MCA PYQ

If the mean and variance of a binomial variate $X$ are 8 and 4 respectively, then $P(X<3)$ equals:

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Solution

For binomial distribution:

Mean $=np=8$

Variance $=np(1-p)=4$

So,

$8(1-p)=4$

$1-p=\frac12$

$p=\frac12$

Then $n=16$

Now,

$P(X<3)=P(0)+P(1)+P(2)$

$= \frac{\binom{16}{0}+\binom{16}{1}+\binom{16}{2}}{2^{16}}$

$= \frac{1+16+120}{2^{16}}$

$= \frac{137}{2^{16}}$

Qus : 320 AMU MCA PYQ

If $y = \tan^{-1}(\sqrt{1+x^2} - x)$, then $\frac{dy}{dx}$ equals

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Solution

Let $u=\sqrt{1+x^2}-x$ Use identity ⇒ $u = \frac{1}{\sqrt{1+x^2}+x}$ So $y=\tan^{-1}u$ Differentiate ⇒ $\frac{dy}{dx} = -\frac{1}{2(1+x^2)}$ Final Answer: $\boxed{-\frac{1}{2(1+x^2)}}$

Qus : 321 AMU MCA PYQ

When 40% of a number is added to 42, the result is the number itself. The number is:

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Solution

Let the number $= x$ $0.4x + 42 = x$ $x - 0.4x = 42$ $0.6x = 42$ $x = 70$

Qus : 322 AMU MCA PYQ

For a normal distribution, the coefficient of Kurtosis $\beta_2$ and $\gamma_2$ are:

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Solution

$ \beta_2=3 $

Bowley’s coefficient uses quartiles: $Sk = \dfrac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1}$

Qus : 340 AMU MCA PYQ

A root of the equation $17x^2 + 17x \tan\left(2\tan^{-1}\frac{1}{5} - \frac{\pi}{4}\right) - 10 = 0$ is

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Skewness of binomial:

$\gamma = \dfrac{1-2p}{\sqrt{np(1-p)}}$

Zero when:

$1-2p=0$

Position at $t=2$:

$\vec{r}=5\hat{i}-8\hat{j}$

Velocity:

$\vec{v}=\frac{d\vec{r}}{dt}=-4t\hat{j}$

At $t=2$:

$\vec{v}=-8\hat{j}$

Momentum:

$\vec{p}=m\vec{v}=2(-8\hat{j})=-16\hat{j}$

Angular momentum:

$\vec{L}=\vec{r}\times\vec{p}$

$=(5\hat{i}-8\hat{j})\times(-16\hat{j})$

From first:

$x = \frac{u}{2}$

From second:

$y = x - v = \frac{u}{2} - v$

From third:

$z = w - 5x - 4y$

$z = w - \frac{5u}{2} - 4\left(\frac{u}{2} - v\right)$

$z = w - \frac{5u}{2} - 2u + 4v$

$z = w - \frac{9u}{2} + 4v$

Thus inverse:

$T^{-1}(x,y,z) = \left(\frac{x}{2}, \frac{x - 2y}{2}, \frac{-9x + 8y + 2z}{2}\right)$

Qus : 363 AMU MCA PYQ

If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ satisfies the matrix equation $A^2 - kA + 2I = 0$, then the value of $k$ is:

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Solution

Trace of $A = 3 + (-2) = 1$

Determinant of $A = (3)(-2) - (-8) = -6 + 8 = 2$

Characteristic equation of $A$ is:

$\lambda^2 - (\text{trace})\lambda + \det = 0$

$\lambda^2 - \lambda + 2 = 0$

By Cayley–Hamilton theorem:

$A^2 - A + 2I = 0$

Comparing with $A^2 - kA + 2I = 0$

$k = 1$

Qus : 364 AMU MCA PYQ

The general solution of $y = 2x \frac{dy}{dx} + y\left(\frac{dy}{dx}\right)^2$ is

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Put $p = \frac{dy}{dx}$

$y = 2xp + yp^2$

$y(1 - p^2) = 2xp$

$y = \frac{2xp}{1 - p^2}$

This is Clairaut’s form. Hence general solution is

$2xc - y + c^2 = 0$

Qus : 365 AMU MCA PYQ

The equation of the cone reciprocal to $x^2 + 2y^2 + 3z^2 = 0$ is

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Solution

Reciprocal cone coefficients are reciprocals of given coefficients.

So equation becomes

$\frac{x^2}{1} + \frac{y^2}{1/2} + \frac{z^2}{1/3} = 0$

Multiplying gives

$6x^2 + 3y^2 + 2z^2 = 0$

Qus : 366 AMU MCA PYQ

The direction cosines of the line which is equally inclined to the axes is

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Solution

If equally inclined then $l = m = n$

$l^2 + m^2 + n^2 = 1$

$3l^2 = 1$

$l = \pm\frac{1}{\sqrt{3}}$

Qus : 367 AMU MCA PYQ

Let $T : \mathbb{R}^2 \to \mathbb{R}^3$ defined by $T(x,y) = (-x-y, 3x+8y, 9x-11y)$. Then the rank and nullity of $T$ are respectively

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Matrix form:

$\begin{bmatrix} -1 & -1 \\ 3 & 8 \\ 9 & -11 \end{bmatrix}$

Columns are independent ⇒ Rank = 2

Nullity = 2 - 2 = 0

Qus : 368 AMU MCA PYQ

If PSP' and QSQ' be two perpendicular focal chords of parabola $y^2 = 4x$, then $\frac{1}{SP \cdot SP'} + \frac{1}{SQ \cdot SQ'}$ equals

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

For perpendicular focal chords of parabola, required sum = $\frac{1}{4}$

Qus : 369 AMU MCA PYQ

The equation of axis of the conic $\sqrt{ax} + \sqrt{by} = 1$ is

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Solution

Qus : 370 AMU MCA PYQ

If the Trapezoidal rule with interval $[0,1]$ is exact for approximating the integral $\int_{0}^{1} (x^3 - cx^2)\,dx$, then the value of $c$ is

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Actual integral:

$\int_{0}^{1} (x^3 - cx^2)\,dx = \left[\frac{x^4}{4} - \frac{cx^3}{3}\right]_0^1 = \frac{1}{4} - \frac{c}{3}$

Trapezoidal rule with one interval:

$T = \frac{1}{2}[f(0) + f(1)]$

$f(0) = 0$

$f(1) = 1 - c$

$T = \frac{1}{2}(1 - c)$

Since exact:

$\frac{1}{4} - \frac{c}{3} = \frac{1}{2}(1 - c)$

Solving:

$\frac{1}{4} - \frac{c}{3} = \frac{1}{2} - \frac{c}{2}$

Multiply by 12:

$3 - 4c = 6 - 6c$

$2c = 3$

$c = 1.5$

Qus : 371 AMU MCA PYQ

If $S$ is the set of all real numbers except $-1$ and $*$ is defined by $a * b = a + b + ab$, then the inverse of $2 * 3$ is

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

First compute:

$2 * 3 = 2 + 3 + (2)(3) = 11$

Inverse element $x$ satisfies:

$11 * x = 0$ (identity element is $0$)

$11 + x + 11x = 0$

$x(12) = -11$

$x = -\frac{11}{12}$

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