B.Tech 1st Year Mathematics-I (BT-1002) Solved Question Paper June 2022

Question 114 marks

Q1(a)Not in current syllabus7 marks

Find the tangent at a point '

t

' on the curve

x = a cosh t, y = b sinh t

Solution

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{b cosh t}{a sinh t} .

Tangent: $y - b sinh t = \frac{b cosh t}{a sinh t} (x - a cosh t)$ . Cross-multiplying,

b x cosh t - a y sinh t = ab (cosh^{2} t - sinh^{2} t) = ab .

\frac{x}{a} cosh t - \frac{y}{b} sinh t = 1

Q1(b)Unit 1 · Rolle's & Mean Value theorems7 marks

Verify Rolle's theorem for the function

f (x) = x^{3} - 6 x^{2} + 11 x - 6

Solution

No interval is printed, so we take the interval between two zeros of $f$ . Factorising,

f (x) = (x - 1) (x - 2) (x - 3) \Rightarrow f (1) = f (3) = 0.

Take $[1, 3]$ . A polynomial is continuous on $[1, 3]$ and differentiable in $(1, 3)$ , and $f (1) = f (3)$ . So all three conditions hold.

f^{'} (x) = 3 x^{2} - 12 x + 11 = 0 \Rightarrow x = \frac{12 \pm 144 - 132}{6} = 2 \pm \frac{1}{3} .

$2 - \frac{1}{3} \approx 1.42$ and $2 + \frac{1}{3} \approx 2.58$ both lie in $(1, 3)$ .

c = 2 \pm \frac{1}{3} \in (1, 3)

; Rolle's theorem is verified. (On

[1, 2]

take

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

; on

[2, 3]

take

c = 2 + \frac{1}{3}

Question 214 marks

Q2(a)Not in current syllabus7 marks

Evaluate: (i)

\int \frac{x ^{3}}{1 + x ^{8}} d x

(ii)

\int x sin^{- 1} x d x

Solution

(i) Put $x^{4} = t, 4 x^{3} d x = d t$ : $\frac{1}{4} \int \frac{d t}{1 + t ^{2}} = \frac{1}{4} tan^{- 1} (x^{4}) + C$ .

(ii) By parts with $u = sin^{- 1} x$ :

\int x sin^{- 1} x d x = \frac{x ^{2}}{2} sin^{- 1} x - \frac{1}{2} \int \frac{x ^{2}}{1 - x ^{2}} d x = \frac{x ^{2}}{2} sin^{- 1} x - \frac{1}{4} (sin^{- 1} x - x 1 - x^{2}) + C .

(i)

\frac{1}{4} tan^{- 1} x^{4} + C

(ii)

\frac{2 x ^{2} - 1}{4} sin^{- 1} x + \frac{x}{4} 1 - x^{2} + C

Q2(b)Not in current syllabus7 marks

Evaluate: (i)

D^{n} (e^{a x})

(ii)

D^{n} (x^{n})

Solution

(i) $D (e^{a x}) = a e^{a x}, D^{2} (e^{a x}) = a^{2} e^{a x}, \dots \Rightarrow D^{n} (e^{a x}) = a^{n} e^{a x}$ .

(ii) $D (x^{n}) = n x^{n - 1}, D^{2} (x^{n}) = n (n - 1) x^{n - 2}, \dots, D^{n} (x^{n}) = n (n - 1) \dots 1 \cdot x^{0}$ .

(i)

a^{n} e^{a x}

(ii)

n!

Question 314 marks

Q3(a)Unit 1 · Taylor & Maclaurin (one variable)7 marks

Expand the function

f (x) = e^{x}

by Maclaurin's theorem.

Solution

Maclaurin's series: $f (x) = f (0) + x f^{'} (0) + \frac{x ^{2}}{2 !} f^{''} (0) + \frac{x ^{3}}{3 !} f^{'''} (0) + \dots$

For $f (x) = e^{x}$ : $f^{(n)} (x) = e^{x}$ for every $n$ , so $f^{(n)} (0) = 1$ .

e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots + \frac{x ^{n}}{n !} + \dots = n = 0 \sum \infty \frac{x ^{n}}{n !} .

The ratio test gives $\frac{x ^{n + 1} / ( n + 1 )!}{x ^{n} / n !} = \frac{∣ x ∣}{n + 1} \to 0$ , so the series is valid for all $x$ .

Q3(b)Unit 1 · Maxima & minima7 marks

Find the maximum or minimum value of the function

f = x y + a^{3} {\frac{1}{x} + \frac{1}{y}}

Solution

Rule: solve $p = f_{x} = 0, q = f_{y} = 0$ ; at each point find $r = f_{xx}, s = f_{x y}, t = f_{y y}$ . $r t - s^{2} > 0, r < 0 \Rightarrow$ max; $r t - s^{2} > 0, r > 0 \Rightarrow$ min; $r t - s^{2} < 0 \Rightarrow$ saddle.

p = y - \frac{a ^{3}}{x ^{2}} = 0, q = x - \frac{a ^{3}}{y ^{2}} = 0 \Rightarrow x^{2} y = a^{3}, x y^{2} = a^{3} .

Dividing: $x = y$ , then $x^{3} = a^{3} \Rightarrow x = y = a$ .

r = \frac{2 a ^{3}}{x ^{3}} = 2, s = 1, t = \frac{2 a ^{3}}{y ^{3}} = 2 at (a, a) .

$r t - s^{2} = 4 - 1 = 3 > 0$ and $r = 2 > 0$ ⇒ minimum.

f (a, a) = a^{2} + a^{3} (\frac{1}{a} + \frac{1}{a}) = 3 a^{2} .

Minimum value

= 3 a^{2}

x = y = a

Question 414 marks

Q4(a)Unit 1 · Maxima & minima7 marks

Find the maximum and minimum value of the function

x^{3} + y^{3} - 3 a x y

Also asked in: Dec 2024 Q1(a), Dec 2025 Q8(b)

Solution

$p = 3 x^{2} - 3 a y, q = 3 y^{2} - 3 a x$ .

$p = 0 \Rightarrow y = \frac{x ^{2}}{a}$ . Put in $q = 0$ : $\frac{x ^{4}}{a ^{2}} = a x \Rightarrow x (x^{3} - a^{3}) = 0 \Rightarrow x = 0$ or $x = a$ .

Stationary points: $(0, 0)$ and $(a, a)$ . Also $r = 6 x, s = - 3 a, t = 6 y$ .

Point	$r$	$s$	$t$	$r t - s^{2}$	Nature
$(0, 0)$	0	$- 3 a$	0	$- 9 a^{2} < 0$	saddle point
$(a, a)$	$6 a$	$- 3 a$	$6 a$	$27 a^{2} > 0$	min if $a > 0$ , max if $a < 0$

f (a, a) = a^{3} + a^{3} - 3 a^{3} = - a^{3} .

For

a > 0

: minimum value

- a^{3}

(a, a)

. For

a < 0

: maximum value

- a^{3}

(a, a)

(0, 0)

is a saddle point (no extremum).

Q4(b)Unit 1 · Partial differentiation7 marks

x^{x} y^{y} z^{z} = c

, then show that at

x = y = z

\frac{\partial ^{2} z}{\partial x \partial y} = - (x lo g e x)^{- 1} .

Solution

Taking logs: $x lo g x + y lo g y + z lo g z = lo g c$ , where $z$ is a function of $x, y$ . Differentiate partially w.r.t. $x$ :

(1 + lo g x) + (1 + lo g z) \frac{\partial z}{\partial x} = 0 \Rightarrow \frac{\partial z}{\partial x} = - \frac{1 + lo g x}{1 + lo g z} .

Similarly $\frac{\partial z}{\partial y} = - \frac{1 + lo g y}{1 + lo g z}$ . Now differentiate this w.r.t. $x$ ( $y$ constant):

\frac{\partial ^{2} z}{\partial x \partial y} = (1 + lo g y) \cdot \frac{1}{( 1 + lo g z ) ^{2}} \cdot \frac{1}{z} \frac{\partial z}{\partial x} = - \frac{( 1 + lo g x ) ( 1 + lo g y )}{z ( 1 + lo g z ) ^{3}} .

At $x = y = z$ :

\frac{\partial ^{2} z}{\partial x \partial y} = - \frac{( 1 + lo g x ) ^{2}}{x ( 1 + lo g x ) ^{3}} = - \frac{1}{x ( 1 + lo g x )} = - \frac{1}{x ( lo g e + lo g x )} = - \frac{1}{x lo g e x} .

\frac{\partial ^{2} z}{\partial x \partial y} = - (x lo g e x)^{- 1}

. Hence proved.

Question 514 marks

Q5(a)Unit 1 · Partial differentiation7 marks

The radius of a sphere is found to be 20 cm with a possible error of 0.02 cm. Find the relative error in calculating the volume.

Solution

$V = \frac{4}{3} π r^{3} \Rightarrow lo g V = lo g \frac{4}{3} π + 3 lo g r$ . Taking differentials:

\frac{δ V}{V} = 3 \frac{δ r}{r} = 3 \times \frac{0.02}{20} = 0.003.

Relative error

= 0.003

(percentage error

= 0.3%

Q5(b)Unit 1 · Partial differentiation7 marks

x^{y} + y^{x} = c

, then find

\frac{d y}{d x}

Solution

Let $f (x, y) = x^{y} + y^{x} - c = 0$ . For an implicit function, $\frac{d y}{d x} = - \frac{f _{x}}{f _{y}}$ .

f_{x} = y x^{y - 1} + y^{x} lo g y, f_{y} = x^{y} lo g x + x y^{x - 1} .

\frac{d y}{d x} = - \frac{y x ^{y - 1} + y ^{x} lo g y}{x ^{y} lo g x + x y ^{x - 1}}

Question 614 marks

Q6(a)Unit 1 · Partial differentiation7 marks

x^{x} y^{y} z^{z} = c

, then show that at

x = y = z

\frac{\partial ^{2} z}{\partial x \partial y} = - (x lo g e x)^{- 1} .

Solution

This question also appears in this paper as Q4(b); the solution is repeated here.

Taking logs: $x lo g x + y lo g y + z lo g z = lo g c$ , where $z$ is a function of $x, y$ . Differentiate partially w.r.t. $x$ :

(1 + lo g x) + (1 + lo g z) \frac{\partial z}{\partial x} = 0 \Rightarrow \frac{\partial z}{\partial x} = - \frac{1 + lo g x}{1 + lo g z} .

Similarly $\frac{\partial z}{\partial y} = - \frac{1 + lo g y}{1 + lo g z}$ . Now differentiate this w.r.t. $x$ ( $y$ constant):

\frac{\partial ^{2} z}{\partial x \partial y} = (1 + lo g y) \cdot \frac{1}{( 1 + lo g z ) ^{2}} \cdot \frac{1}{z} \frac{\partial z}{\partial x} = - \frac{( 1 + lo g x ) ( 1 + lo g y )}{z ( 1 + lo g z ) ^{3}} .

At $x = y = z$ :

\frac{\partial ^{2} z}{\partial x \partial y} = - \frac{( 1 + lo g x ) ^{2}}{x ( 1 + lo g x ) ^{3}} = - \frac{1}{x ( 1 + lo g x )} = - \frac{1}{x ( lo g e + lo g x )} = - \frac{1}{x lo g e x} .

\frac{\partial ^{2} z}{\partial x \partial y} = - (x lo g e x)^{- 1}

. Hence proved.

Q6(b)Unit 2 · Limit of a sum7 marks

Evaluate

n \to \infty lim {\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{2 n}}

Solution

Rule: $n \to \infty lim \frac{1}{n} r = 1 \sum n f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$ .

The general term is $\frac{1}{n + r}$ , $r = 1$ to $n$ . Take $\frac{1}{n}$ outside:

n \to \infty lim r = 1 \sum n \frac{1}{n + r} = n \to \infty lim \frac{1}{n} r = 1 \sum n \frac{1}{1 + \frac{r}{n}} = \int_{0}^{1} \frac{d x}{1 + x} = [lo g (1 + x)]_{0}^{1} .

Limit

= lo g 2

Question 714 marks

Q7(a)Unit 2 · Double & triple integrals7 marks

Evaluate

\int_{1}^{2} \int_{1}^{x} \frac{d x d y}{x y}

Solution

The variable limit $x$ belongs to $y$ , so integrate w.r.t. $y$ first:

I = \int_{1}^{2} \frac{1}{x} [lo g y]_{1}^{x} d x = \int_{1}^{2} \frac{lo g x}{x} d x = [\frac{( lo g x ) ^{2}}{2}]_{1}^{2} .

I = \frac{( lo g 2 ) ^{2}}{2} \approx 0.2402

Q7(b)Unit 2 · Double & triple integrals7 marks

Evaluate

\iint_{R} y d x d y

, where

R

is the region bounded by the parabolas

y^{2} = 4 a x

and

x^{2} = 4 a y

Solution

The parabolas meet at $(0, 0)$ and $(4 a, 4 a)$ . For a vertical strip, $y$ runs from $\frac{x ^{2}}{4 a}$ to $2 a x$ , and $x$ from 0 to $4 a$ .

I = \int_{0}^{4 a} [\frac{y ^{2}}{2}]_{x^{2} /4 a}^{2 a x} d x = \frac{1}{2} \int_{0}^{4 a} (4 a x - \frac{x ^{4}}{16 a ^{2}}) d x

= \frac{1}{2} [2 a x^{2} - \frac{x ^{5}}{80 a ^{2}}]_{0}^{4 a} = \frac{1}{2} (32 a^{3} - \frac{1024 a ^{5}}{80 a ^{2}}) = \frac{1}{2} (32 a^{3} - \frac{64 a ^{3}}{5}) .

I = \frac{48}{5} a^{3}

Question 814 marks

Q8(a)Unit 2 · Double & triple integrals7 marks

Evaluate

\int_{0}^{1} \int_{0}^{x^{2}} e^{y / x} d y d x

Solution

\int_{0}^{x^{2}} e^{y / x} d y = x [e^{y / x}]_{0}^{x^{2}} = x (e^{x} - 1) .

I = \int_{0}^{1} (x e^{x} - x) d x = [(x - 1) e^{x}]_{0}^{1} - \frac{1}{2} = (0 + 1) - \frac{1}{2} .

I = \frac{1}{2}

Q8(b)Unit 2 · Double & triple integrals7 marks

Evaluate

\int_{0}^{2} \int_{0}^{x} \int_{0}^{x + y} e^{x} (y + 2 z) d z d y d x

Solution

z-integration: $\int_{0}^{x + y} (y + 2 z) d z = y (x + y) + (x + y)^{2} = (x + y) (x + 2 y)$ .

y-integration:

\int_{0}^{x} (x^{2} + 3 x y + 2 y^{2}) d y = x^{3} + \frac{3}{2} x^{3} + \frac{2}{3} x^{3} = \frac{19}{6} x^{3} .

x-integration (by parts, $\int x^{3} e^{x} d x = e^{x} (x^{3} - 3 x^{2} + 6 x - 6)$ ):

I = \frac{19}{6} \int_{0}^{2} x^{3} e^{x} d x = \frac{19}{6} [e^{x} (x^{3} - 3 x^{2} + 6 x - 6)]_{0}^{2} = \frac{19}{6} (2 e^{2} + 6) .

I = \frac{19}{3} (e^{2} + 3) \approx 65.80