Mathematics for Information Science 1. Question Papers, ktu 2024 scheme questions grouped by module and year, with worked solutions and links to the original papers..

Module 1 · Limits, continuity and derivatives

2024 · Q1 · 3 marks

Question. Find where $g(x)=\begin{cases}(x^2-x-6)/(x-3),&x\ne3\\5,&x=3\end{cases}$ is continuous.

Solution. For $x\ne3$, factor $x^2-x-6=(x-3)(x+2)$, so $g(x)=x+2$. Its limit at $3$ is $5=g(3)$. Thus the removable discontinuity is filled, and $g$ is continuous for every real $x$.

2024 · Q2 · 3 marks

Question. Find the first and second derivatives of $(x^3+7)/x$.

Solution. Write $f=x^2+7x^{-1}$. Then $f'=2x-7x^{-2}$ and $f''=2+14x^{-3}$, for $x\ne0$.

2024 · Q9(a) · 3 marks

Question. Evaluate $\lim_{x\to0}\left(\frac1{x-1}+\frac1{x+1}\right)/x$.

Solution. The numerator is $2x/(x^2-1)$, so the quotient is $2/(x^2-1)$. The limit is $-2$.

2024 · Q9(b) · 3 marks

Question. Find the linearization of $f(x)=\cos x$ at $x=\pi/2$.

Solution. $f(a)=0$ and $f'(a)=-1$ at $a=\pi/2$. Hence $L(x)=-(x-\pi/2)=\pi/2-x$.

2024 · Q9(c) · 3 marks

Question. Determine the concavity of $x^3-6x^2+9x+1$.

Solution. $f''(x)=6x-12$. The graph is concave down for $x<2$, concave up for $x>2$, with inflection point $(2,3)$.

2024 · Q10(a) · 6 marks

Question. Show $y=|x|$ is differentiable on either side of zero but not at zero.

Solution. For $x<0$, $|x|=-x$ and $y'=-1$; for $x>0$, $|x|=x$ and $y'=1$. At zero, the left and right derivatives differ, so $y'(0)$ does not exist.

2024 · Q10(b) · 3 marks

Question. If $A=\pi D^2/4$, find $dA/dD$ at $D=10\,\mathrm m$.

Solution. $dA/dD=\pi D/2$; at $D=10$, the area changes at $5\pi\,\mathrm m^2/\mathrm m$ per metre increase in diameter.

2025 · Q9(a) · 4 marks

Question. Find the first and second derivatives of $y=\frac{(x+1)(x^2+x+1)}{x^3}$.

Solution. Expand and write $y=1+2x^{-1}+2x^{-2}+x^{-3}$. Thus $\displaystyle y'=-\frac2{x^2}-\frac4{x^3}-\frac3{x^4}$ and $\displaystyle y''=\frac4{x^3}+\frac{12}{x^4}+\frac{12}{x^5}$, for $x\ne0$.

2025 · Q9(b) · 5 marks

Question. Verify that $(2,4)$ lies on $x^3+y^3-9xy=0$, then find the tangent and normal there.

Solution. Substitution gives $8+64-72=0$. Implicit differentiation gives $3x^2+3y^2y'-9(y+xy')=0$. At $(2,4)$, $30y'=24$, so $y'=4/5$. The tangent is $y-4=\frac45(x-2)$; the normal is $y-4=-\frac54(x-2)$.

2025 · Q10(a) · 2 marks

Question. Find the linearization of $f(x)=x+\frac1x$ at $x=1$.

Solution. Since $f(1)=2$ and $f'(x)=1-x^{-2}$ gives $f'(1)=0$, the linearization is $L(x)=2$.

2025 · Q10(b) · 4 marks

Question. Determine the concavity and inflection point(s) of $f(x)=x^3-3x^2+2$.

Solution. $f''=6x-6$ changes sign at $x=1$, from negative to positive. The inflection point is $(1,0)$; the graph is concave down for $x<1$ and up for $x>1$.

2025 · Q10(c) · 3 marks

Question. Find $dy/dx$ if $x^2+xy+y^2=1.2$.

Solution. Differentiate implicitly: $2x+y+x y'+2y y'=0$. Therefore $\displaystyle y'=-\frac{2x+y}{x+2y}$ when $x+2y\ne0$.

Module 2 · Partial derivatives and multivariable limits

2024 · Q3 · 3 marks

Question. Evaluate $\lim_{(x,y)\to(0,0)}(1-\cos(xy))/(xy)$.

Solution. Let $t=xy\to0$. Since $1-\cos t\sim t^2/2$, the ratio is asymptotic to $t/2$ and tends to $0$.

2024 · Q4 · 3 marks

Question. Show $w_{xy}=w_{yx}$ for $w=e^x+x\ln y+y\ln x$.

Solution. $w_x=e^x+\ln y+y/x$, so $w_{xy}=1/y+1/x$. Also $w_y=x/y+\ln x$, so $w_{yx}=1/y+1/x$.

2024 · Q11(a) · 3 marks

Question. Find $w_{xy}$ for $w=xy+e^y/(y^2+1)$.

Solution. $w_x=y$; differentiating with respect to $y$ gives $w_{xy}=1$.

2024 · Q11(b) · 3 marks

Question. Evaluate $\lim_{(x,y)\to(0,0)}(x^2-xy)/(\sqrt{x}-\sqrt{y})$.

Solution. On the real domain $x,y\ge0$, factor the numerator as $x(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})$. The quotient is $x(\sqrt{x}+\sqrt{y})\to0$; the limit is $0$.

2024 · Q11(c) · 3 marks

Question. Find the first partial derivatives of $f=e^{x/y}+e^{z/y}$.

Solution. $f_x=e^{x/y}/y$, $f_z=e^{z/y}/y$, and $\displaystyle f_y=-\frac{x e^{x/y}+z e^{z/y}}{y^2}$, for $y\ne0$.

2024 · Q12(a) · 3 marks

Question. Find the domain and range of $g(x,y)=\ln(x^2+y^2)$.

Solution. The logarithm requires $x^2+y^2>0$, so the domain is $\mathbb R^2\setminus\{(0,0)\}$. Since $x^2+y^2$ can be any positive real number, the range is $\mathbb R$.

2024 · Q12(b) · 6 marks

Question. If $w=e^{x^2y}$, $x=\sqrt{uv}$, $y=1/v$, find $w_u,w_v$ at $(u,v)=(2,2)$.

Solution. $x^2y=(uv)(1/v)=u$, hence $w=e^u$. Thus $(w_u,w_v)=(e^u,0)$ and at $(2,2)$ the answer is $(e^2,0)$.

2025 · Q11(a) · 5 marks

Question. For $f=x\cos y+ye^x$, find $f_{xx},f_{xy},f_{yx},f_{yy}$.

Solution. $f_x=\cos y+ye^x$ and $f_y=-x\sin y+e^x$. Therefore $f_{xx}=ye^x$, $f_{xy}=e^x-\sin y$, $f_{yx}=e^x-\sin y$, and $f_{yy}=-x\cos y$.

2025 · Q11(b) · 4 marks

Question. Show $f=xy^3/(x^2+y^6)$ has no limit at $(0,0)$.

Solution. Along $x=0$, $f=0$. Along $x=y^3$, $f=1/2$ for $y\ne0$. Since these path limits differ, the limit does not exist.

2025 · Q12(a) · 5 marks

Question. For $w=4e^x\ln y$, $x=\ln(r\cos\theta)$, $y=r\sin\theta$, find $w_r,w_\theta$.

Solution. Substitute to get $w=4r\cos\theta\ln(r\sin\theta)$. Hence $\displaystyle w_r=4\cos\theta[\ln(r\sin\theta)+1]$ and $\displaystyle w_\theta=4r[-\sin\theta\ln(r\sin\theta)+\cos\theta\cot\theta]$.

2025 · Q12(b) · 4 marks

Question. Find the partial derivatives of $f=x^3y+2xy^3$ and verify the mixed-derivative theorem.

Solution. $f_x=3x^2y+2y^3$, $f_y=x^3+6xy^2$. Then $f_{xy}=3x^2+6y^2=f_{yx}$.

Module 3 · Gradients and extrema

2024 · Q5 · 3 marks

Question. Find the directions of zero change in $f=(x^2+y^2)/2$ at $(1,1)$.

Solution. $\nabla f=(x,y)$, so at $(1,1)$ a direction $\mathbf u$ has zero directional derivative when $(1,1)\cdot\mathbf u=0$. The unit directions are $\pm(1,-1)/\sqrt2$.

2024 · Q6 · 3 marks

Question. Find the critical point(s) of $f=5xy-7x^2+3x-6y+2$.

Solution. Set $f_x=5y-14x+3=0$ and $f_y=5x-6=0$. This gives $(x,y)=(6/5,69/25)$, the only critical point.

2024 · Q13(a) · 5 marks

Question. For $w=f(x-y,y-z,z-x)$, show $w_x+w_y+w_z=0$.

Solution. Let the three arguments be $u,v,t$. By the chain rule, $w_x=f_u-f_t$, $w_y=-f_u+f_v$, and $w_z=-f_v+f_t$. Their sum is zero.

2024 · Q13(b) · 4 marks

Question. Find the directional derivative of $g=(x-y)/(xy+2)$ at $(1,-1)$ in the direction $12\mathbf i+5\mathbf j$.

Solution. At $(1,-1)$, $\nabla g=(3,-3)$. Normalize the direction to $(12/13,5/13)$. The directional derivative is $\nabla g\cdot\mathbf u=21/13$.

2024 · Q14(a) · 6 marks

Question. Find the local extrema of $f=10xy e^{-(x^2+y^2)}$.

Solution. $f_x=10y e^{-r^2}(1-2x^2)$ and $f_y=10x e^{-r^2}(1-2y^2)$. Critical points are $(0,0)$ and $x^2=y^2=1/2$. The origin is a saddle. At $(\pm1/\sqrt2,\pm1/\sqrt2)$ with equal signs, $f=5/e$ (local maxima); with opposite signs, $f=-5/e$ (local minima).

2024 · Q14(b) · 3 marks

Question. Find $\nabla f$ for $f=x^2-xy+y^2-y$ at $(1,-1)$.

Solution. $\nabla f=(2x-y,-x+2y-1)$, so $\nabla f(1,-1)=(3,-4)$.

2025 · Q13(a) · 5 marks

Question. Locate the relative extrema and saddle points of $f=x^3+3xy^2-15x^2-15y^2+72x$.

Solution. The critical points from $f_x=3x^2+3y^2-30x+72=0$, $f_y=6y(x-5)=0$ are $(4,0),(6,0),(5,1),(5,-1)$. The Hessian determinant is positive with negative diagonal at $(4,0)$: local maximum $f=112$. It is positive with positive diagonal at $(6,0)$: local minimum $f=108$. At $(5,\pm1)$ the determinant is negative: saddle points.

2025 · Q13(b) · 4 marks

Question. Find the absolute extrema of $f=2x^2-4x+y^2-4y+1$ on the triangle bounded by $x=0$, $y=2$, and $y=2x$ in the first quadrant.

Solution. The vertices are $(0,0),(0,2),(1,2)$. Since $f=2(x-1)^2+(y-2)^2-5$, the minimum is $-5$ at $(1,2)$. Vertex values are $1,-3,-5$, so the absolute maximum is $1$ at $(0,0)$.

2025 · Q14(a) · 4 marks

Question. Find the directional derivative of $f=xe^y$ in direction $\mathbf i-\mathbf j$.

Solution. $\nabla f=(e^y,xe^y)$. The unit direction is $(1,-1)/\sqrt2$, so $D_{\mathbf u}f=e^y(1-x)/\sqrt2$ at the point of evaluation.

2025 · Q14(b) · 5 marks

Question. Find $z_x,z_y$ at the origin if $x^3+z^2+ye^{xz}+z\cos y=0$.

Solution. Let $F(x,y,z)=0$. Then $z_x=-F_x/F_z$ and $z_y=-F_y/F_z$. At the origin, $F_x=0$, $F_y=1$, and $F_z=1$, so $z_x=0$ and $z_y=-1$.

Module 4 · Constrained optimization and linear programming

2024 · Q7 · 3 marks

Question. Find the extreme values of $xy$ subject to $x+y=16$.

Solution. Substitute $y=16-x$: $f=x(16-x)=64-(x-8)^2$. The maximum is $64$ at $(8,8)$; there is no minimum over the real line because $f\to-\infty$ as $|x|\to\infty$.

2024 · Q8 · 3 marks

Question. Form an LPP for normal and super painkiller bottles with profits $11,15$ and resource limits A:1400, B:800, C:1800.

Solution. Let $x,y$ be weekly normal and super bottles. Maximize $Z=11x+15y$ subject to $2x+y\le1400$, $x+y\le800$, $x+3y\le1800$, $x,y\ge0$.

2024 · Q15(a) · 6 marks

Question. Find the point on $2x+y-z=5$ nearest the origin.

Solution. The nearest point lies along the plane’s normal $(2,1,-1)$. Projecting onto the plane gives $P=\frac5{2^2+1^2+(-1)^2}(2,1,-1)=(5/3,5/6,-5/6)$.

2024 · Q15(b) · 3 marks

Question. Run two steepest-descent steps for $f=x^2+y^2$ from $(1,1)$ with fixed step $\alpha=0.1$.

Solution. $\nabla f=(2x,2y)$, so each step multiplies both coordinates by $1-2\alpha=0.8$. The iterates are $(0.8,0.8)$ then $(0.64,0.64)$.

2024 · Q16(a) · 5 marks

Question. Find the largest and smallest $xy$ on $x^2/8+y^2/2=1$.

Solution. Let $x=\sqrt8\cos t$, $y=\sqrt2\sin t$. Then $xy=2\sin(2t)$; maximum $2$, minimum $-2$.

2024 · Q16(b) · 4 marks

Question. Maximize $z=6x_1+11x_2$ subject to $2x_1+x_2\le104$, $x_1+2x_2\le76$, $x_1,x_2\ge0$.

Solution. Feasible vertices are $(0,0),(52,0),(44,16),(0,38)$. Their objective values are $0,312,440,418$. The maximum is $440$ at $(44,16)$.

2025 · Q15(a) · 5 marks

Question. Minimize $z=20x+10y$ subject to $x+2y\le40$, $3x+y\ge30$, $4x+3y\ge60$, $x,y\ge0$.

Solution. Feasible corner points are $(15,0),(40,0),(6,12),(4,18)$. Their costs are $300,800,240,260$, so the minimum is $240$ at $(6,12)$.

2025 · Q15(b) · 4 marks

Question. Find the extreme values of $x^2+y^2+z^2$ subject to $x+y+z=1$, $x-y=0$.

Solution. Set $x=y=t$, $z=1-2t$. Then $f=6t^2-4t+1$, minimized at $t=1/3$, giving $(1/3,1/3,1/3)$ and $f=1/3$. There is no maximum because the constraint line is unbounded and $f\to\infty$.

2025 · Q16(a) · 5 marks

Question. Take three steepest-descent steps for $f=3x^2+4y^2$ from $(1,1)$ with fixed $\alpha=0.01$.

Solution. The update is $(x,y)\mapsto(0.94x,0.92y)$. The iterates are $(0.94,0.92)$, $(0.8836,0.8464)$, and $(0.830584,0.778688)$; the final value is approximately $f=4.4950$.

2025 · Q16(b) · 4 marks

Question. Formulate the three-product production LPP. Profits A,B,C are Rs.3,2,4; machine times are $(4,3,5)$ and $(2,2,4)$ minutes, with capacities 2000 and 2500. Produce at least 100 A, 200 B, 50 C, and no more than 150 A.

Solution. Let $a,b,c$ be production quantities. Maximize $Z=3a+2b+4c$ subject to $4a+3b+5c\le2000$, $2a+2b+4c\le2500$, $100\le a\le150$, $b\ge200$, $c\ge50$.

Source note: The transcribed questions are selected items from the linked papers; consult the PDFs for the complete paper and original layout.