TS Intermediate Mathematics 1st Year Model Paper 2024

SECTION-A

Very Short Answer Type Questions.

(i) Answer ALL questions.

(ii) Each question carries TWO marks.

Q1: Show that the points (-5, 1), (5, 5), and (10, 7) are collinear.

Answer:
To prove the points are collinear, calculate the slopes between two pairs of points:

• Slope between (-5, 1) and (5, 5):
  
  $$\text{slope}_1 = \frac{5 – 1}{5 – (-5)} = \frac{4}{10} = \frac{2}{5}.$$

   

   

   

  slope1​=5−(−5)5−1​=104​=52​.

• Slope between (5, 5) and (10, 7):
  
  $$\text{slope}_2 = \frac{7 – 5}{10 – 5} = \frac{2}{5}.$$

   

   

   

  slope2​=10−57−5​=52​.

Since the slopes are equal, the points are collinear.

---

Q2: Find the distance between the parallel lines:

$$5x – 3y – 4 = 0, \quad 10x – 6y – 9 = 0.$$

5x−3y−4=0,10x−6y−9=0.

Answer:
Using the formula for distance between two parallel lines:

$$\text{Distance} = \frac{|C_2 – C_1|}{\sqrt{A^2 + B^2}}.$$

Distance=A2+B2​∣C2​−C1​∣​.

For the given lines, the distance is:

$$\text{Distance} = \frac{1}{2\sqrt{34}}.$$

Distance=234​1​.

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Q3: Find the fourth vertex of the parallelogram whose consecutive vertices are
$(2, 4, -1)$

 

 

 

$(2,4,-1)$,
$(3, 6, -1)$

 

 

 

$(3,6,-1)$, and
$(4, 5, 1)$

 

 

 

$(4,5,1)$.

Answer:
In a parallelogram, the diagonals bisect each other. Using the midpoint property, the fourth vertex is

$(5, 7, 1)$

(5,7,1).

---

Q4: Find the angle between the planes:

$$x + 2y + 1 = 0, \quad 5x + 3y – 8 = 0.$$

x+2y+1=0,5x+3y−8=0.

Answer:
The angle between two planes is given by:

$$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}.$$

cosθ=∣n1​​∣∣n2​​∣∣n1​​⋅n2​​∣​.

Substituting the normal vectors of the planes:

$$\theta = \cos^{-1} \left( \frac{11}{\sqrt{170}} \right).$$

θ=cos−1(170​11​).

---

Q5: Compute
$\lim_{x \to 0} \frac{\pi^x – 1}{\sqrt{1 + x} – 1}$

 

 

 

limx→0​1+x​−1πx−1​.

Answer:
Using L’Hopital’s Rule:

$$\lim_{x \to 0} \frac{\pi^x – 1}{\sqrt{1 + x} – 1} = 2 \ln \pi.$$

x→0lim​1+x​−1πx−1​=2lnπ.

Q6: Compute
$\lim_{x \to \frac{\pi}{2}} \frac{\cos(x)}{x – \frac{\pi}{2}}$

 

 

 

limx→2π​​x−2π​cos(x)​.

Answer: This is an indeterminate form

$\frac{0}{0}$

00​, so apply L’Hopital’s Rule.

Differentiate the numerator and denominator:

$$\frac{d}{dx} (\cos(x)) = -\sin(x), \quad \frac{d}{dx} \left( x – \frac{\pi}{2} \right) = 1.$$

dxd​(cos(x))=−sin(x),dxd​(x−2π​)=1.

Now, evaluate the limit:

$$\lim_{x \to \frac{\pi}{2}} \frac{-\sin(x)}{1} = -\sin\left( \frac{\pi}{2} \right) = -1.$$

x→2π​lim​1−sin(x)​=−sin(2π​)=−1.

Thus, the limit is

$\boxed{-1}$

−1​.

---

Q7: If
$f(x) = 2x^2 + 3x – 5$

 

 

 

f(x)=2x2+3x−5, then prove that
$f'(0) + 3f'(-1) = 0$

 

 

 

f′(0)+3f′(−1)=0.

Answer: First, find

$f'(x)$

f′(x):

$$f'(x) = 4x + 3.$$

f′(x)=4x+3.

Now, evaluate at

$x = 0$

x=0 and

$x = -1$

x=−1:

$$f'(0) = 3, \quad f'(-1) = -1.$$

f′(0)=3,f′(−1)=−1.

Thus,

$$f'(0) + 3f'(-1) = 3 + 3(-1) = 3 – 3 = 0.$$

f′(0)+3f′(−1)=3+3(−1)=3−3=0.

So, the equation holds true.

---

Q8: Find the derivative of
$\sin^{-1}\left( \frac{2x}{1 + x^2} \right)$

 

 

 

sin−1(1+x22x​).

Answer: Let

$u = \frac{2x}{1 + x^2}$

u=1+x22x​.

The derivative of

$\sin^{-1}(u)$

sin−1(u) is:

$$\frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 – u^2}} \cdot \frac{du}{dx}.$$

dxd​sin−1(u)=1−u2​1​⋅dxdu​.

Now, differentiate

$u$

u:

$$\frac{du}{dx} = \frac{2 – 2x^2}{(1 + x^2)^2}.$$

dxdu​=(1+x2)22−2x2​.

Thus, the derivative is:

$$\frac{dy}{dx} = \frac{2 – 2x^2}{(1 + x^2)^2 \sqrt{1 – \left( \frac{2x}{1 + x^2} \right)^2}}.$$

dxdy​=(1+x2)21−(1+x22x​)2​2−2x2​.

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Q9: Find the approximate value of
$\sqrt{65}$

 

 

 

65​.

Answer: Since

$\sqrt{64} = 8$

64​=8, use the approximation formula:

$$\sqrt{65} \approx 8 + \frac{65 – 64}{2 \times 8} = 8 + \frac{1}{16} = 8.0625.$$

65​≈8+2×865−64​=8+161​=8.0625.

So, the approximate value of

$\sqrt{65}$

65​ is

$\boxed{8.0625}$

8.0625​.

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Q10: Verify Rolle’s Theorem for the function
$y = f(x) = x^2 + 4$

 

 

 

y=f(x)=x2+4 in
$[-3, 3]$

 

 

 

$[-3,3]$.

Answer:

• 
  
  $f(x) = x^2 + 4$

   

   

   

  f(x)=x2+4 is continuous and differentiable.

• 
  
  $f(-3) = f(3) = 13$

   

   

   

  f(−3)=f(3)=13, so

  $f(a) = f(b)$

   

   

  f(a)=f(b).

Now, differentiate:

$$f'(x) = 2x.$$

f′(x)=2x.

Set

$f'(x) = 0$

f′(x)=0:

$$2x = 0 \quad \Rightarrow \quad x = 0.$$

2x=0⇒x=0.

Since

$0$

0 is in

$[-3, 3]$

[−3,3], Rolle’s Theorem is verified, and

$x = 0$

x=0 is the point where

$f'(x) = 0$

f′(x)=0.

SECTION-B

II. Short answer type questions:

(i) Attempt ANY FIVE questions.

(ii) Each question carries FOUR marks.

Q11: If the distance from
$P$

 

 

 

$P$ to the points
$(2, 3)$

 

 

 

$(2,3)$ and
$(2, -3)$

 

 

 

$(2,-3)$ are in the ratio 2:3, then find the equation of the locus of
$P$

 

 

 

$P$.

Answer: Let the coordinates of

$P$

P be

$(x, y)$

(x,y).

Distance from

$P$

P to

$(2, 3)$

(2,3) =

$\sqrt{(x – 2)^2 + (y – 3)^2}$

(x−2)2+(y−3)2​
Distance from

$P$

P to

$(2, -3)$

(2,−3) =

$\sqrt{(x – 2)^2 + (y + 3)^2}$

(x−2)2+(y+3)2​

According to the given ratio,

$$\frac{\text{Distance from } P \text{ to } (2, 3)}{\text{Distance from } P \text{ to } (2, -3)} = \frac{2}{3}.$$

Distance from P to (2,−3)Distance from P to (2,3)​=32​.

Squaring both sides:

$$\frac{(x – 2)^2 + (y – 3)^2}{(x – 2)^2 + (y + 3)^2} = \frac{4}{9}.$$

(x−2)2+(y+3)2(x−2)2+(y−3)2​=94​.

After simplifying, the equation of the locus of

$P$

P is:

$$9\left( (x – 2)^2 + (y – 3)^2 \right) = 4\left( (x – 2)^2 + (y + 3)^2 \right).$$

9((x−2)2+(y−3)2)=4((x−2)2+(y+3)2).

This simplifies to the equation of the locus in standard form.

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Q12: When the axes are rotated through an angle
$\alpha$

 

 

 

$\alpha$, find the transformed equation of
$x \cos \alpha + y \sin \alpha = p$

 

 

 

$x\cos \alpha +y\sin \alpha =p$.

Answer: Let the new coordinates after rotation be

$x’$

x′ and

$y’$

y′. The transformation formulas are:

$$x = x’ \cos \alpha – y’ \sin \alpha$$

x=x′cosα−y′sinα

$$y = x’ \sin \alpha + y’ \cos \alpha$$

y=x′sinα+y′cosα

Substitute these into the given equation:

$$(x’ \cos \alpha – y’ \sin \alpha) \cos \alpha + (x’ \sin \alpha + y’ \cos \alpha) \sin \alpha = p.$$

(x′cosα−y′sinα)cosα+(x′sinα+y′cosα)sinα=p.

Simplifying:

$$x’ (\cos^2 \alpha + \sin^2 \alpha) + y’ (\cos \alpha \sin \alpha – \sin \alpha \cos \alpha) = p.$$

x′(cos2α+sin2α)+y′(cosαsinα−sinαcosα)=p.

$$x’ = p \quad (\text{since } \cos^2 \alpha + \sin^2 \alpha = 1),$$

x′=p(since cos2α+sin2α=1),

Thus, the transformed equation is:

$$x’ = p.$$

x′=p.

---

Q13: Find the value of
$K$

 

 

 

$K$, if the angle between the straight lines
$4x – y + 7 = 0$

 

 

 

$4x-y+7=0$ and
$Kx – 5y – 9 = 0$

 

 

 

$Kx-5y-9=0$ is 45°.

Answer: The angle between two lines is given by the formula:

$$\tan \theta = \left| \frac{m_1 – m_2}{1 + m_1 m_2} \right|.$$

tanθ=​1+m1​m2​m1​−m2​​​.

Where

$m_1$

m1​ and

$m_2$

m2​ are the slopes of the lines.

For the first line

$4x – y + 7 = 0$

4x−y+7=0, the slope

$m_1$

m1​ is:

$$m_1 = \frac{4}{1} = 4.$$

m1​=14​=4.

For the second line

$Kx – 5y – 9 = 0$

Kx−5y−9=0, the slope

$m_2$

m2​ is:

$$m_2 = \frac{K}{5}.$$

m2​=5K​.

Given that the angle between the lines is 45°, so

$\tan 45^\circ = 1$

tan45∘=1.

Thus,

$$\left| \frac{4 – \frac{K}{5}}{1 + 4 \times \frac{K}{5}} \right| = 1.$$

​1+4×5K​4−5K​​​=1.

Simplifying the equation:

$$\frac{4 – \frac{K}{5}}{1 + \frac{4K}{5}} = 1.$$

1+54K​4−5K​​=1.

Multiplying both sides by

$5(5 + 4K)$

5(5+4K), solving for

$K$

K:

$$K = 10.$$

K=10.

---

Q14: Compute
$\lim_{x \to 0} \frac{\sin x}{x^2}$

 

 

 

limx→0​x2sinx​.

Answer: As

$x \to 0$

x→0, the numerator

$\sin x$

sinx approaches 0, and the denominator

$x^2$

x2 also approaches 0. This is an indeterminate form

$\frac{0}{0}$

00​.

To solve this, apply L’Hopital’s Rule:

$$\lim_{x \to 0} \frac{\sin x}{x^2} = \lim_{x \to 0} \frac{\cos x}{2x}.$$

x→0lim​x2sinx​=x→0lim​2xcosx​.

Evaluating the limit:

$$\lim_{x \to 0} \frac{\cos x}{2x} = \frac{1}{0},$$

x→0lim​2xcosx​=01​,

which tends to infinity.

Thus, the limit is

$\boxed{\infty}$

∞​.

---

Q15: Find the derivative of the function
$\cot x$

 

 

 

$\cot x$ from the first principle.

Answer: Using the first principle of differentiation:

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}.$$

f′(x)=h→0lim​hf(x+h)−f(x)​.

Let

$f(x) = \cot x$

f(x)=cotx, so:

$$f'(x) = \lim_{h \to 0} \frac{\cot(x + h) – \cot(x)}{h}.$$

f′(x)=h→0lim​hcot(x+h)−cot(x)​.

Using the identity for the difference of cotangents:

$$\cot(x + h) – \cot(x) = \frac{\sin h}{\sin(x + h)\sin x}.$$

cot(x+h)−cot(x)=sin(x+h)sinxsinh​.

Simplifying, the derivative of

$\cot x$

cotx is:

$$f'(x) = -\csc^2 x.$$

f′(x)=−csc2x.

---

Q16: Find the lengths of sub-tangent and sub-normal at a point on the curve
$y = b \sin\left( \frac{\pi x}{a} \right)$

 

 

 

y=bsin(aπx​).

Answer: For the curve

$y = b \sin\left( \frac{\pi x}{a} \right)$

y=bsin(aπx​), the length of the sub-tangent is given by:

$$T = \frac{y}{y’} = \frac{b \sin\left( \frac{\pi x}{a} \right)}{b \cdot \frac{\pi}{a} \cos\left( \frac{\pi x}{a} \right)} = \frac{a \sin\left( \frac{\pi x}{a} \right)}{\pi \cos\left( \frac{\pi x}{a} \right)}.$$

T=y′y​=b⋅aπ​cos(aπx​)bsin(aπx​)​=πcos(aπx​)asin(aπx​)​.

The length of the sub-normal is given by:

$$N = \frac{y’}{y”} = \frac{b \cdot \frac{\pi}{a} \cos\left( \frac{\pi x}{a} \right)}{-b \cdot \left( \frac{\pi}{a} \right)^2 \sin\left( \frac{\pi x}{a} \right)} = -\frac{a \cos\left( \frac{\pi x}{a} \right)}{\pi \sin\left( \frac{\pi x}{a} \right)}.$$

N=y′′y′​=−b⋅(aπ​)2sin(aπx​)b⋅aπ​cos(aπx​)​=−πsin(aπx​)acos(aπx​)​.

---

Q17: The volume of a cube is increasing at the rate of 8 cm³/sec. How fast is the surface area increasing when the length of an edge is 12 cm?

Answer: Let

$V = s^3$

V=s3 be the volume of the cube and

$A = 6s^2$

A=6s2 be the surface area, where

$s$

s is the side length of the cube.

Given that

$\frac{dV}{dt} = 8$

dtdV​=8 cm³/sec, we need to find

$\frac{dA}{dt}$

dtdA​ when

$s = 12$

s=12.

First, differentiate the volume with respect to time:

$$\frac{dV}{dt} = 3s^2 \frac{ds}{dt}.$$

dtdV​=3s2dtds​.

Substitute

$\frac{dV}{dt} = 8$

dtdV​=8:

$$8 = 3 \times 12^2 \times \frac{ds}{dt}.$$

8=3×122×dtds​.

Solve for

$\frac{ds}{dt}$

dtds​:

$$8 = 3 \times 144 \times \frac{ds}{dt}, \quad \frac{ds}{dt} = \frac{8}{432} = \frac{1}{54}.$$

8=3×144×dtds​,dtds​=4328​=541​.

Now, differentiate the surface area:

$$\frac{dA}{dt} = 12s \frac{ds}{dt}.$$

dtdA​=12sdtds​.

Substitute

$s = 12$

s=12 and

$\frac{ds}{dt} = \frac{1}{54}$

dtds​=541​:

$$\frac{dA}{dt} = 12 \times 12 \times \frac{1}{54} = \frac{144}{54} = \frac{24}{9} = 2.67 \text{ cm}^2/\text{sec}.$$

dtdA​=12×12×541​=54144​=924​=2.67 cm2/sec.

Thus, the surface area is increasing at a rate of

$\boxed{2.67}$

2.67​ cm²/sec.

SECTION-C

III. Long answer type questions :

(i) Attempt ANY FIVE questions.
(ii) Each question carries SEVEN marks.

18. Find the equation of the straight line parallel to the line 3x + 4y = 7 and passing through the point of intersection of the lines x – 2y – 3 = 0 and x + 3y – 6 = 0.

Solution:

1. First, find the point of intersection of the lines
   $x – 2y – 3 = 0$

   and
   $x + 3y – 6 = 0$

   x+3y−6=0. Solve the system of equations:

$$x – 2y = 3 \quad \text{(Equation 1)}$$

$$x + 3y = 6 \quad \text{(Equation 2)}$$

x+3y=6(Equation 2)

From Equation 1,
$x = 2y + 3$

x=2y+3. Substitute this in Equation 2:

$$(2y + 3) + 3y = 6$$

$$2y + 3 + 3y = 6 \quad \Rightarrow \quad 5y = 3 \quad \Rightarrow \quad y = \frac{3}{5}$$

2y+3+3y=6⇒5y=3⇒y=53​

Substitute
$y = \frac{3}{5}$

into
$x = 2y + 3$

x=2y+3:

$$x = 2 \times \frac{3}{5} + 3 = \frac{6}{5} + \frac{15}{5} = \frac{21}{5}$$

x=2×53​+3=56​+515​=521​

Thus, the point of intersection is
$\left(\frac{21}{5}, \frac{3}{5}\right)$

(521​,53​).

2. Now, the given line is
   $3x + 4y = 7$

   3x+4y=7. The slope of this line is:

$$\text{slope} = -\frac{3}{4}$$

slope=−43​

Since the required line is parallel, it will have the same slope. The equation of the line with slope
$-\frac{3}{4}$

passing through
$\left(\frac{21}{5}, \frac{3}{5}\right)$

(521​,53​) is:

$$y – \frac{3}{5} = -\frac{3}{4}\left(x – \frac{21}{5}\right)$$

y−53​=−43​(x−521​)

Multiply through by 20 to eliminate fractions:

$$20\left(y – \frac{3}{5}\right) = -15\left(x – \frac{21}{5}\right)$$

$$20y – 12 = -15x + 63$$

$$15x + 20y = 75$$

15x+20y=75

Thus, the equation of the required line is:

$$15x + 20y = 75$$

15x+20y=75

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19. Show that the lines represented by
$(lx + my)^2 – 3(mx – ly)^2 = 0$

and
$lx + my + n = 0$

form an equilateral triangle with area
$\sqrt{3} \frac{(l^2 + m^2)}{n^2}$

3​n2(l2+m2)​ square units.

Solution:

1. Expand the first equation
   $(lx + my)^2 – 3(mx – ly)^2 = 0$

   (lx+my)2−3(mx−ly)2=0:

$$(lx + my)^2 = 3(mx – ly)^2$$

$$l^2x^2 + 2lmxy + m^2y^2 = 3(m^2x^2 – 2lmxy + l^2y^2)$$

$$l^2x^2 + 2lmxy + m^2y^2 = 3m^2x^2 – 6lmxy + 3l^2y^2$$

$$l^2x^2 + 2lmxy + m^2y^2 – 3m^2x^2 + 6lmxy – 3l^2y^2 = 0$$

$$(l^2 – 3m^2)x^2 + (m^2 – 3l^2)y^2 + 8lmxy = 0$$

(l2−3m2)x2+(m2−3l2)y2+8lmxy=0

2. The general form of an equation representing two lines is
   $A x^2 + B xy + C y^2 = 0$

   Ax2+Bxy+Cy2=0, where the area of the triangle formed by these lines is:

$$\text{Area} = \frac{\sqrt{3}}{4} \frac{(l^2 + m^2)}{n^2}$$

Area=43​​n2(l2+m2)​

Hence, we conclude that the lines form an equilateral triangle with area
$\sqrt{3} \frac{(l^2 + m^2)}{n^2}$

3​n2(l2+m2)​.

---

20. Find the condition for the chord
$lx + my = 1$

of the circle
$x^2 + y^2 = a^2$

x2+y2=a2 (whose center is the origin) to subtend a right angle at the origin.

Solution:

1. The equation of the circle is
   $x^2 + y^2 = a^2$

   x2+y2=a2.

2. The equation of the chord is
   $lx + my = 1$

   lx+my=1.

3. For the chord to subtend a right angle at the origin, the perpendicular distance from the center of the circle (origin) to the chord must be equal to the radius of the circle.

The perpendicular distance
$d$

from the center to the chord
$lx + my = 1$

lx+my=1 is given by:

$$d = \frac{|l \cdot 0 + m \cdot 0 – 1|}{\sqrt{l^2 + m^2}} = \frac{1}{\sqrt{l^2 + m^2}}$$

d=l2+m2​∣l⋅0+m⋅0−1∣​=l2+m2​1​

For the chord to subtend a right angle at the origin, we set this distance equal to the radius
$a$

a:

$$\frac{1}{\sqrt{l^2 + m^2}} = a$$

l2+m2​1​=a

Squaring both sides:

$$\frac{1}{l^2 + m^2} = a^2$$

l2+m21​=a2

Thus, the condition is:

$$l^2 + m^2 = \frac{1}{a^2}$$

l2+m2=a21​

---

21. Find the angle between the lines whose direction cosines are given by the equations
$3l + m + 5n = 0$

and
$6mn – 2nl + 5lm = 0$

6mn−2nl+5lm=0.

Solution:

1. The direction cosines of the lines are
   $l, m, n$

   . The angle
   $\theta$

   between two lines with direction cosines
   $(l_1, m_1, n_1)$

   and
   $(l_2, m_2, n_2)$

   (l2​,m2​,n2​) is given by:

$$\cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$$

cosθ=l12​+m12​+n12​​l22​+m22​+n22​​l1​l2​+m1​m2​+n1​n2​​

2. Using the given equations
   $3l + m + 5n = 0$

   and
   $6mn – 2nl + 5lm = 0$

   , solve for the relationship between
   $l$

   ,
   $m$

   , and
   $n$

   n.

3. Calculate the cosine of the angle between the two lines using the formula provided.

---

22. If
$y = x\sqrt{a^2 + x^2} + a^2\log(x + \sqrt{a^2 + x^2})$

, then show that
$\frac{dy}{dx} = 2\sqrt{a^2 + x^2}$

dxdy​=2a2+x2​.

Solution:

Differentiate
$y = x\sqrt{a^2 + x^2} + a^2\log(x + \sqrt{a^2 + x^2})$

y=xa2+x2​+a2log(x+a2+x2​) using the product rule and chain rule.

1. The first term
   $x\sqrt{a^2 + x^2}$

   xa2+x2​ is differentiated using the product rule:

$$\frac{d}{dx}\left(x\sqrt{a^2 + x^2}\right) = \sqrt{a^2 + x^2} + x \cdot \frac{1}{2\sqrt{a^2 + x^2}} \cdot 2x = \sqrt{a^2 + x^2} + \frac{x^2}{\sqrt{a^2 + x^2}}$$

dxd​(xa2+x2​)=a2+x2​+x⋅2a2+x2​1​⋅2x=a2+x2​+a2+x2​x2​

2. The second term
   $a^2\log(x + \sqrt{a^2 + x^2})$

   a2log(x+a2+x2​) is differentiated using the chain rule:

$$\frac{d}{dx}\left(a^2\log(x + \sqrt{a^2 + x^2})\right) = \frac{a^2}{x + \sqrt{a^2 + x^2}} \cdot \left(1 + \frac{x}{\sqrt{a^2 + x^2}}\right)$$

dxd​(a2log(x+a2+x2​))=x+a2+x2​a2​⋅(1+a2+x2​x​)

Combining the terms, you get:

$$\frac{dy}{dx} = 2\sqrt{a^2 + x^2}$$

dxdy​=2a2+x2​

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23. If the tangent at any point on the curve
$x^{2/3} + y^{2/3} = a^{2/3}$

x2/3+y2/3=a2/3 intersects the coordinate axes in A and B, then show that the length AB is constant.

Solution:

1. Find the equation of the tangent to the curve
   $x^{2/3} + y^{2/3} = a^{2/3}$

   x2/3+y2/3=a2/3 at any point. Use the derivative of the curve to find the slope of the tangent.

2. The tangent intersects the coordinate axes, and you can use the intercept form to determine the coordinates of points A and B.

3. Show that the length of the line segment AB remains constant for all points on the curve.

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24. From a rectangular sheet of dimensions 30 cm x 80 cm, four equal squares of side ‘x’ cm are removed at the corners and the sides are then turned up so as to form an open rectangular box. Find the value of x, so that the volume of the box is the greatest.

Solution:

1. The volume
   $V$

   V of the box is given by:

$$V = x(30 – 2x)(80 – 2x)$$

V=x(30−2x)(80−2x)

2. To find the value of
   $x$

   that maximizes the volume, differentiate
   $V$

   with respect to
   $x$

   and set
   $\frac{dV}{dx} = 0$

   dxdV​=0.

3. Solve for
   $x$

   x to find the maximum volume.