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Kerala State Board New Syllabus Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 8 Application of Integrals.

Kerala Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 8 Application of Integrals

Plus Two Maths Application of Integrals 4 Marks Important Questions

Question 1.
(i) The area bounded by the curve y = f(x) x-axis and the line x= a and x= b is……
(ii) Find the area enclosed between the Parabola y = x² and the straight line 2x – y + 3 = 0 (March – 2010)
Answer:

Question 2.
Find the area enclosed between the curve x² = 4y and the line x = 4y – 2 (March -2011)
Answer:

Question 3.
(i) Area of the shaded portion in the figure is equal to

(ii) Consider the curves y = x², x = 0, y = 1, y = 4.
Draw a rough sketch and shade the region bounded by these curves, Find area of the shaded region.
Answer:

Question 4.
Consider the following figure:

(i) Find the point of Intersection P of the circle x² + y² = 32 and the line y = x.
(ii) Find the area of the shaded region. (EDUCATE – 2017; March – 2013; March – 2014)
Answer:

Question 5.
(a) The area bounded by the curve, above the x-axis, between x = a and x = b is

(b) Find the area of the circle x² + y² = 4 using integration. (March – 2016)
Answer:

Question 6.
(i) The area bounded by y = 2cosx , the x-axis from x = 0 to x = \(\frac{\pi}{2}\) is
(a) 0
(b) 1
(c) 2
(d) -1
(ii) Find the area of the region bounded by the y² = 4ax and x² = 4ay, a > 0 (March – 2017)
Answer:

When x = 4a, y = 4a and x = 0, y = 0.
Therefore the point is (0, 0) and (4a, 4a).

Plus Two Maths Application of Integrals 6 Marks Important Questions

Question 1.
Consider the circle x² + y² = 16 and the straight line \(y=\sqrt{3} x\) as shown ¡n the figure

(i) Find the points A and B as shown in the figure.
(ii) Find the area of the shaded region in the figure using definite integral. (May -2010)
Answer:
(i) The point of intersection of x² + y² = l6 and

Question 2.
(i) Draw the rough sketch of \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)
(ii) Find the area bounded by the above curve using integration. (May – 2011)
Answer:
(i) The curve is an ellipse.

Question 3.
(i) Find the area enclosed between the curve y² = x, x = 1, x = 4 and x-axis.
(ii) Using ntegration, find the area of the region bounded by the triangle whose vertices are (1, 0), (2, 2) and (3, 1). (March-2012)
Answer:

Required area ΔABC = Area ΔAB2
+ Area 2BC3 – Area ΔAC3
Equation AC is y = 2(x – 1)
Equation BC is y = 4 – x
Equation AB is y = 1/2 (x – 1)

Question 4.

Using the above figure
Find the equation of AB.
Findthe point P.
Find the area of the shaded region by integration. (May – 2013)
Answer:
(i) The equation of a line passing through (0,2)

Question 5.
Consider the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\) and the line \(\frac{x}{3}+\frac{y}{2}=1\).
(a) Find the points where the line intersects the ellipse?
(b) Shade the smaller region bounded by the ellipse and the line.
(c) Find the area of the shaded region. (May – 2014)
Answer:

Question 6.
Consider the function f(x) = |x| + 1; g(x) = 1 – |x|
(a) Sketch the graph and shade the enclosed region between them.
(b) Find the area of the shaded region. (March – 2015)
Answer:

(b) The equation of the line through AB is

Question 7.
Using the given figure answer the following questions.

Define the equation of the given curve.
Find the area of the enclosed region.
Find the area when a = lo and b = 5. (March – 2011; May – 2015; March – 2017)
Answer:
(i) The figure represents an ellipse with equation

Ellipse is symmetric w.r.t coordinate axles. Therefore the area of the enclosed region is same as four times area enclosed by the curve in first quadrant.

Kerala State Board New Syllabus Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 7 Integrals.

Kerala Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 7 Integrals

Plus Two Maths Application of Derivatives 3 Marks Important Questions

Question 1.
Find the following integrals. (May -2011)
\(\begin{array}{l}
\text { (i) } \int x^{2} e^{2 x} d x \\
\text { (ii) } \int e^{x} \sin x d x
\end{array}\)
Answer:

Question 2.
(i) \(\int e^{x} \sec x(1+\tan x) d x=\ldots \ldots\)
(a) ex cosx + c (b) ex sec x + c
(C) ex tanx + c (d) ex sin x + c
(ii) Find \(\int \sin 2 x \cos 3 x d x\) (March – 2014)
Answer:

Question 3.
Find the following integrals.
(i) \(\begin{array}{l}
\text { (i) } \int \frac{1}{(x+1)(x+2)} d x \\
\text { (ii) } \int \frac{2 x-1}{(x-1)(x+2)^{2}} d x
\end{array}\) (March – 2014)
Answer:

Plus Two Maths Application of Derivatives 4 Marks Important Questions

Question 1.
Consider the integral \(I=\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x\)
(i) Express \(I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x\)
(ii) Show that \(I=\frac{\pi^{2}}{4}\) (March – 2012)
Answer:

Question 2.
(i) Evaluate: \(\int_{2}^{3} \frac{x}{x^{2}+1} d x\)
(ii) Evaluate: \(\int_{0}^{\pi} \frac{x}{1+\sin x} d x\) (March – 2014)
Answer:

Question 3.
(a) What is the value of \(\int_{0}^{1} x(1-x)^{9} d x\) If the
\(\begin{array}{llll}
\text { (i) } \frac{1}{10} & \text { (ii) } \frac{1}{11} & \text { (iii) } \frac{1}{90} & \text { (iv) } \frac{1}{110}
\end{array}\)
(b) Find \(\int_{0}^{1}(2 x+3) d x\) of a sum. (March – 2015)
Answer:

Question 4.
Evaluate \(\int_{0}^{x} \log (1+\cos x) d x\)
Answer:

Question 5.
Find \(\int_{0}^{5}(x+1) d x \text { as limit of a sum. }\)
Answer:

Question 6.
Evaluate \(\int_{0}^{4} x^{2} d x\) as the limit of a sum. (March – 2017)
Answer:

Plus Two Maths Application of Derivatives 6 Marks Important Questions

Question 1.
(i) Fill in the blanks; \(\int \frac{1}{x} d x=\)_____
(ii) Evaluate \(\int \frac{5 x+1}{x^{2}-2 x-35} d x\)
(iii) Integrate with respect to x. \(\sqrt{x^{2}+4 x+8}\) (March – 2010)
Answer:

Question 2.
(i) Evaluate \(\int-\frac{\cos e c^{2} x}{\sqrt{\cot ^{2} x+9}} d x\)
(ii) Evaluate \(\int\left(\cos ^{-1} x\right)^{2} d x\) (May -2010)
Answer:

Question 3.
(i) Evaluate \(\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x\)
(ii) Evaluate \(\int_{0}^{2} e^{x} d x \text { as limit of a sum. }\) (May -2010)
Answer:

Question 4.
(i) Fill in the blanks \(\int \cot x d x=\)_____
(ii) Evaluate the integrals
\(\begin{array}{l}
\text { (a) } \int \sin 2 x \cos 4 x d x \\
\text { (b) } \int \frac{x}{(x+1)(x+2)} d x
\end{array}\) (March -2011)
Answer:

Question 5.
(i) Evaluate \(\int_{0}^{1} x d x\) as the limit of a sum.
(ii) Evaluate \(\int_{0}^{1} x(1-x)^{n} d x\) (March – 2011)
Answer:

Question 6.
(i) Evaluate \(\int_{1}^{2} \frac{1}{x(1+\log x)^{2}} d x\)
(ii) Evaluate \(\int_{0}^{3}\left(2 x^{2}+3\right) d x\) as the limit of a sum. (May – 2011)
Answer:

Question 7.
(i) What is \(\int \frac{1}{9+x^{2}} d x=?\)
(ii) Evaluate the integrals \(\int \frac{1}{1+x+x^{2}+x^{3}} d x\) (May – 2012)
Answer:

Question 8.
(i) Evaluate \(\int_{0}^{3} f(x) d x\)
where \(f(x)=\left\{\begin{array}{ll}
x+3, & 0 \leq x \leq 2 \\
3 x, & 2 \leq x \leq 3
\end{array}\right.\)

(ii) Prove that \(\int_{0}^{1} \log \left(\frac{x}{1-x}\right) d x=\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x\) Find the value of \(\int_{0}^{1} \log \left(\frac{x}{1-x}\right) d x\) (May – 2012)
Answer:

Question 9.
(i) Find \(\int \cot x d x=\ldots \ldots\)
(ii) Find \(\int x \log x d x\)
(iii) Find \(\int \frac{x-1}{(x-2)(x-3)} d x\) (March – 2013)
Answer:

Question 10.
Evaluate
\(\text { (i) } \int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x\)
\(\text { (ii) } \int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}\) (May – 2013)
Answer:

Question 11.
Evaluate
\(\begin{array}{l}
\text { (i) } \int x^{2} \tan ^{-1} x d x \\
\text { (ii) } \int_{-1}^{2} x^{3}-x d x
\end{array}\) (May – 2013)
Answer:

Question 12.
Evaluate \(\int_{0}^{\pi / 4} \log (1+\tan x) d x\) (March – 2013)
Answer:

Question 13.
(a) The value of \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x d x\) (May – 2014)
(b) Prove that \(\int_{0}^{\pi} \frac{x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} d x=\frac{\pi^{2}}{2 a b}\)
Answer:

Question 14.
(a) \(\int \frac{1}{x^{2}+a^{2}} d x=\)
(b) Find \(\int \frac{1}{9 x^{2}+6 x+5} d x\)
(c) Find \(\int \frac{x}{(x-1)^{2}(x+2)} d x\) (May – 2014)
Answer:

Question 15.
Integrate the following
\(\begin{array}{l}
\text { (a) } \frac{x-1}{x+1} \\
\text { (b) } \frac{\sin x}{\sin (x-a)} \\
\text { (c) } \frac{1}{\sqrt{3-2 x-x^{2}}}
\end{array}\) (March – 2015)
Answer:

Question 16.
(a) Prove that \(\int \cos ^{2} x d x=\frac{x}{2}+\frac{\sin 2 x}{4}+c\)
(b)Find \(\int \frac{1}{\sqrt{2 x-x^{2}}} d x\)
(c) Find \(\int x \cos x d x\) (May – 2015)
Answer:

Question 17.
Find the following:
\(\begin{array}{l}
\text { (i) } \int \frac{1}{x\left(x^{7}+1\right)} d x \\
\text { (ii) } \int_{1}^{4}|x-2| d x
\end{array}\)
Answer:

Question 18.
Find \(\int_{0}^{\frac{\pi}{2}} \log \sin x d x\)
Answer:

Question 19.
Find the following: \(\begin{array}{l}
\text { (i) } \int \cot x \log (\sin x) d x \\
\text { (ii) } \int \frac{1}{x^{2}+2 x+2} d x \\
\text { (iii) } \int x e^{9 x} d x
\end{array}\) (May – 2017)
Answer:

Kerala State Board New Syllabus Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 6 Application of Derivatives.

Kerala Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 6 Application of Derivatives

Plus Two Maths Application of Derivatives 4 Marks Important Questions

Question 1.
(a) Find the equation of the tangent to the curve \(x^{\frac{2}{3}}+y^{\frac{2}{3}}=2\) at (1,1).
(b) Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. (May – 2015)
Answer:

Question 2.
(a) The slope of the tangent to the curve given
\(x=1-\cos \theta, y=\theta-\sin \theta \text { by at } \theta=\frac{\pi}{2}\)
(i) 0
(ii) – 1
(iii) 1
(iv) Not defined.

(b) Find the intervals in which the function f(x) = x² – 4x + 6 is strictly decreasing.
(C) Find the minimum and maximum value, if any, of the function f(x) = (2x – 1)² + 3 (March – 2016)
Answer:
(a) (iii) 1
(b) Given; f(x) = x² – 4x + 6 ⇒ f’(x) = 2x – 4
For turning points; f’(x) = 2x – 4 0 ⇒ x = 2
So volurn.,e is niaxirnum when h = 2r
The intervals are (- ∞, 2); (2, ∞)
f’(0) = 2 x 0 – 4 = -4
Therefore f(x) is decreasing in (- ∞, 2)
(c) f(x) = (2 x 1)² + 3
⇒ f’(x) 2(2x – 1) x 2 f”(x) = 8
For tuming points; f’(x) = 8x – 4 = 0 ⇒ x = 1/2
f(x) has minimum value at x = 1/2 minimum value is \(f\left(\frac{1}{2}\right)=3\)
2)

Question 3.
(a) Which of the following function has neither local maxima nor local minima?
(i) f(x) = x² + x
(ii) f(x) = logx
(iii) f(x) = x³ – 3x + 3
(iv) f(x) = 3 + |x|
(b) Find the equation of the tangent to the curve y = 3x² at (1,1). (March – 2016)
Answer:

Question 4.
(i) The slope of the normal to the curve, y = x³ – x² at (1, -1) is
(a) 1
(b) – 1
(c) 2
(d)0

(ii) Find the intervals in which the function f(x) = 2x³ – 24x + 25 is increasing or decreasing. (May – 2016)
Answer:
(i) (b) – 1
(ii) f(x) = 2x³ – 24x + 25
f’(x) = 6x² – 24
f’(x) = O
⇒ 6x² – 24 = 0 ⇒ x² = 4 ⇒ x = – 2,2
Therefore the intervals are (-∞, -2); (-2, 2); (2, ∞)
f(x) is increasing in the intervals (-∞, -2); (2, ∞)
f(x) is decreasing in the intervals (-2, 2)

Question 5.
(i) The slope of the normal to the curve, y² – 4x at (1,2) is
(a) 1
(b) 1/2
(c) 2
(d) – 1

(ii) Find the intervals in which the function 2x³ + 9x² + 12x – 1 is strictly increasing. (March – 2017)
Answer:
(i) (b) – 1
(ii) f(x) = 2x³ + 9x² + 12x – 1
f’(x) = 6x² + 18x + 12
= 6(x² + 3x + 2) = 6(x + 1) (x + 2)
f’(x) = O
⇒ 6(x + 1)(x + 2) = 0 ⇒ x = – 1 – 2
Therefore the intervals are
(- ∞, – 2); (- 2, – 1); (- 1, ∞)
In the ¡nterval (- ∞, – 2)
f’( – 3) = 6(- 3 + 1) (- 3 + 2) > 0
Therefore increasing In the interval (- 2, – 1)
f’(- 1.5) = 6(- 1.5 + 1)(- 1.5 + 2) < 0
Therefore decreasing In the interval (- 1, ∞)
f’(0) = 6(0 + 1)(0 + 2) > 0
Therefore increasing

Question 6.
Find two positive numbers whose sum is 16 and sum of whose cubes is minimum. (March – 2017)
Answer:
Let the numbers be x and 16 – x. Then,
S = x³ + (16 – x)³
= S’ = 3x² + 3(16 – x)²(- 1)
⇒ S” = 6x + 6(16 – x)………..(1)
For turning points S’ = 0 ⇒ 3 x² – 3(16 – x)² = 0
⇒ x² – 16 + 32x – x² =0
⇒ – 16² + 32x = 0 = x² = \(\frac{16 \times 16}{32}\) =8
(1) ⇒ S” = 6(8) + 6(16 – 8) > 0
TherefocemrnimumM x = 8
Thusthe numbers are8 and 16 – 8 = 8

Plus Two Maths Application of Derivatives 6 Marks Important Questions

Question 1.
(i) Show that the function x³ – 6x² + 15x + 4 is strictly increasing in R.
(ii) Find the approximate change in volume of a cube of side x meters caused by an increase in the side by 3%.
(iii) Find the equation of the tangent and normal at the point (1,2) on the parabola y² = 4x. (March – 2010)
Answer:
(i) Given; f(x) = x³ – 6x² + 15x + 4
f’(x) = 3x² – 12x + 15 = 3(x² – 4x +5)
= 3(x² – 4x + 4 + 1) = 3(x – 2)² + 1) > 0
For any value of x, f(x) is a strKly ¡ncreasing.

(ii) We have; V = x³ and Δx = 3% of x = 0.03x
\(d V=\frac{d V}{d x} \Delta x=3 x^{2} \Delta x\)
= 3x² x 0.03x = 0.09x³ = 0.09V
\(\Rightarrow \frac{d V}{V}=0.09\)

Therefore 9% is the approximate increase In volume.

(iii) Given; y²….4x ⇒ 2y \(\frac{d y}{d x}\) = 4 ⇒ \(\frac{d y}{d x}=\frac{2}{y}\)
Slope at (1,2) = \(\frac{2}{2}\) = 1
Equation of tangent at (1,2) is; y – 2 = 1(x – 1)
⇒ x – y + 1 = 0
Equation of normal at (1,2) is; y – 2 = – 1(x – 1)
⇒ x + y – 3 = 0

Question 2.
Consider the parametric forms
x = 1 + \(\frac{1}{t}\) – and y = t – \(\frac{1}{t}\) ofa curve
(i) Find \(\frac{d y}{d x}\)
(ii) Find the equation of the tangent at t = 2.
(iii) Find the equation of the normal at t = 2. (May – 2010)
Answer:

Question 3.
(i) The radius of a circle is increasing at the rate of 2cmls. Find the rate at which area of the circle is increasing when radius is
6cm.
(ii) Prove that the function f(x) = log sin x is strictly increasing in \(\left(0, \frac{\pi}{2}\right)\) and strictly decreasing in \(\left(\frac{\pi}{2}, \pi\right)\)
(iii) Find the maximum and minimum value of the function f(x) = x³ – 6x² + 9x + 15. (March – 2011)
Answer:

Question 4.
(i) Find the approximate value of (82)^1/4 up to three places of decimals using differentiation.
(ii) Find two positive numbers such that Their sum is 8 and the sum of their squares is minimum. (May – 2011)
Answer:

(ii) Let the numbers be x and 8 – x. Then,
S = x² + (8 – x)²
⇒ S’ = 2x + 2(8 – x)( – 1)
⇒ S” = 2 + 2 = 4 ………..(1)
For turning points S’ = 0 = 2x – 2(8 – x) = 0
⇒ 4x – 16 = 0 ⇒ x = 4
(1) ⇒ S” = 4 > 0
Therefore minimum at x = 4
Thus the numbers are 4 and 8 – 4 = 4.

Question 5.
(i) The slope of the tangent to the curve y = x³ – 1 at x = 2 is ……….
(ii) Use differentiation to approximate \(\sqrt{36.6}\)
(iii) Find two numbers whose sum is 24 and whose product as large as possible. (March – 2012, March – 2016)
Answer:

Therefore minimum at x =12
Thus the numbers are 12 and 24 – 12 = 12.

Question 6.
(i) Show that the function x³ – 3x² + 6x – 5 is strictly increasing on R.
(ii) Find the interval in which the function f(x) = sin x + cosx; 0 < x < 2π is strictly increasing or strictly decreasing. (May – 2012)
Answer:
(i) Given; f(x) = x³ – 3x² + 6x – 5
f’(x) = 3x² – 6x + 6 = 3(x² – 2x +2)
= 3(x² – 2x + 1 + 1) 3(x – 1)² + 1) > 0
For any value cit x, f(x) is a strictly increasing.

Question 7.
(i) Find the slope of the normal to the curve y = sinθ at θ = π/4
(ii) Show that the function f(x) = x³ – 6x² + 15x + 4 is strictly increasing in R.
(iii) Show that all rectangles with a given perimeter, the square has the maximum area. (March – 2013)
Answer:

(ii) f(x) = x³ – 6x² + 15x + 4
Differentiating w.r.t x;
f(x) = 3x² – 12x + 15 = 3(x² – 4x + 5)
= 3 (x² – 4x + 4 + 1)
= 3 ((x – 2)² + 1) > 0, ∀x∈R
Therefore fis strictly increasing in R.

(iii) Let x and ybe the length and breadth of a rectangle with area A and perimeter P.

Question 8.
A right circular cylinder is inscribed in a given cone of radius R cm and height H cm as shown in the figure.

(i) Find the Surface Area S of the circular cylinder as a function of x.
(ii) Find a relation connecting x and R when S is a maximum. (May – 2013)
Answer:
(i) There are two similar triangles ΔDJB and ΔDHF

Question 9.
(i) Which of the following function is Increasing for all values of x in its domain?
(a) sin x
(b) log x
(c) x²
(d) |x|

(ii) Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2,0) and (4,4).
(iii) Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 24x – 6x². (March – 2014)
Answer:
(i) (b) log x
(ii) Given; y = (x – 2)² ⇒ \(\frac{d y}{d x}\) = 2(x – 2)
Slope of the chord = \(\frac{4-0}{4-2}=2\)
\(\Rightarrow 2=2(x-2) \Rightarrow x=3 \Rightarrow y=(3-2)^{2}=1\)
Therefore the required point is (3, 1)

(iii) Given; p(x) = 41 – 24x – 6x²
p’(x) = – 24 – 12x
p”(x) = – 12
For turning points p’(x) = – 24 – 12x = 0
⇒ x = -2
Since p”(x) = – 12 always maximum Therefore maximum value p(- 2) = 41 – 24(- 2) 6(- 2)² = 65

Question 10.
(a) Find the slope of the tangent to the parabola y² = 4ax at (at², 2at).
(b) Find the intervals in which the function x² – 2x + 5 is strictly increasing.
(c) A spherical bubble volume at the rate of which the diminishing when the is decreasing in 2cm³/sec. Find the surface area is radius is 3cm. (May – 2014)
Answer:

Question 11.
(a) Which of the following function is always increasing?
(i) x + sin 2x
(ii) x – sin 2x
(ill) 2x + sin 3x
(iv) 2x – sin 2x
(b) The radius of a cylinder is increasing at a rate of 1cm/s and its height decreasing at a rate of 1cm/s. Find the rate of change of its volume when the radius is 5cm and the height is 5cm.
(c) Write the equation of tangent at (1,1) on the curve 2x² + 3y² = 5. (March – 2015)
Answer:

HSE Kerala Board Syllabus HSSLive Plus Two Computer Application Chapter Wise Previous Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala Plus Two Chapter Wise Previous Questions and Answers. Here HSSLive.Guru have given Higher Secondary Kerala Plus Two Computer Application Chapter Wise Previous Year Important Questions and Answers based on CBSE NCERT syllabus.

Kerala Plus Two Computer Application Chapter Wise Previous Year Questions and Answers

• Chapter 1 Review of C++ Programming
• Chapter 2 Arrays
• Chapter 3 Functions
• Chapter 4 Web Technology
• Chapter 5 Web Designing Using HTML
• Chapter 6 Client-Side Scripting Using Java Script
• Chapter 7 Web Hosting
• Chapter 8 Database Management System
• Chapter 9 Structured Query Language
• Chapter 10 Enterprise Resource Planning
• Chapter 11 Trends and Issues in ICT

We hope the given HSE Kerala Board Syllabus HSSLive Plus Two Computer Application Chapter Wise Previous Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus Two Computer Application Chapter Wise Previous Year Important Questions and Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

Kerala State Board New Syllabus Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 5 Continuity and Differentiability.

Kerala Plus Two Maths Chapter Wise Previous Questions Chapter 5 Continuity and Differentiability

Plus Two Maths Continuity and Differentiability 3 Marks Important Questions

Question 1.
Consider \(f(x)=\left\{\begin{array}{ll}
\frac{x^{2}-x-6}{x+2}, & x \neq-2 \\
-5, & x=-2
\end{array}\right.\)

(i) Find f(-2)
(ii) Check whether the function f(x) is continuous at x= -2. (March – 2009)
Answer:

Question 2.
If f(x) = sin(Log x), prove that x² y₂ + xy₁ + y = 0 (May -2009)
Answer:
Given; y sin(Iogx)
Differentiating with respect to X;
\(y_{1}=\cos (\log x) \frac{1}{x} \Rightarrow x y_{1}=\cos (\log x)\)
Again differentiating with respect to x
\(\begin{array}{l}
\Rightarrow x y_{2}+y_{1}=-\sin (\log x) \frac{1}{x} \\
\Rightarrow x^{2} y_{2}+x y_{1}=-y \Rightarrow x^{2} y_{2}+x y_{1}+y=0
\end{array}\)

Question 3.
(i) Establish that g(x) =1 – x + |x| is continuous at origin.
(ii) Check whether h(x) = |l – x + |x|| is continuous at origin. (March – 2010)
Answer:
(i) Given; g(x) = 1 – x + |x| ⇒ g(x) (1 – x) + |x|
Here g(x) is the sum of two functions continuous functions hence continuous.
(ii) We have;
\(\begin{array}{l}
f o g(x)=f(g(x)) \\
=\quad f(1-x+|x|)=|1-x+| x \mid=h(x)
\end{array}\)
The composition of two continuous functions is again continuous. Therefore h(x) continuous.

Question 4.
Find \(\frac{d y}{d x}\) of the following
\(\begin{array}{l}
\text { (i) } x=\sqrt{a^{\sin ^{4} 4}} \quad y=\sqrt{a^{\cos ^{-1} t}} \\
\text { (ii) } y=\cos ^{-1} \frac{\left(1-x^{2}\right)}{\left(1+x^{2}\right)}, 0<x<1 \\
\text { (iii) } y=\sin ^{-1} 2 x \sqrt{1-x^{2}}, y_{\sqrt{2}}<x<y_{\sqrt{2}}
\end{array}\)
Answer:

Question 5.
Find \(\frac{d y}{d x} \text { if } x^{3}+2 x^{2} y+3 x y^{2}+4 y^{3}=5\) (March – 2015)
Answer:

Question 6.
Find all points of discontinuity of f where f is defined by \(f(x)=\left\{\begin{array}{ll}
2 x+3, & x \leq 2 \\
2 x-3, & x>2
\end{array}\right.\) (March – 2016)
Answer:
In both the intervals x \(\leq[latex] 2 and x > 2 the function f(x) is a polynomial so continuous. So we havetocheckthe continuity at x = 2.

Question 7.
If e^x-y = x^y, then prove that [latex]\frac{d y}{d x}=\frac{\log x}{[\log \operatorname{ex}]}\) (May – 2014; March – 2016)
Answer:

Plus Two Maths Continuity and Differentiability 4 Marks Important Questions

Question 1.
Find \(\frac{d y}{d x}\) of the following (March – 2009)
\(\begin{array}{l}
\text { (i) } y=\sin ^{-1}\left(3 x-4 x^{3}\right)+\cos ^{-1}\left(4 x^{3}-3 x\right) \\
\text { (ii) } y=\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)
\end{array}\)
Answer:

Question 2.
Consider the function f(x) = |x| x ∈ R
(i) Draw the graph of f(x) =|x|
(ii) Show that the function is continuous at x = 0. (March – 2010)
Answer:

f(0^–) f(0^–) = 0. therefore continuous at x = 0.
AIso from the figure we can see that the graph does not have a break or jump.

Question 3.
(i) Find the derivative of y = x^a + a^x with respect to x.
(ii) If e^y (x + 1) = 1 , showthat \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\) (May – 2011)
Answer:

Question 4.
(i) Check the continuity of the function given by f(x) \(f(x)=\left\{\begin{array}{ll}
x \sin \frac{1}{x}, & x \neq 0 \\
1, & x=0
\end{array}\right.\)

(ii) Verify Mean Value Theorem for the function f(x) = x + 1/x in the interval [1,3]. (May – 2011)
Answer:


Hence Mean Value Theorem ¡s verified.

Question 5.
(i) Determine the value of k so that the function (May – 2012)

Answer:

Question 6.
Consider a fUnction f: R → R defined by
\(f(x)=\left\{\begin{array}{cc}
a+x, & x \leq 2 \\
b-x, & x>2
\end{array}\right.\)

(i) Find a relation between a and b if f is continuous at x = 2.
(ii) Find a and b, if f is continuous at x2 and a + b = 2. (May – 2013)
Answer:
(i) Since fis continuous at x = 2, we have;

(ii) Given a = 2 …(2) Solving (1) and (2) we have;
⇒ 2a = – 2 ⇒ a = – 1
⇒ b = 2 – a = 2 + 1 = 3

Question 7.
(i) Find if x = a(t – sin t) y = a(1 + cos t)
(ii) Verify Rolles theorem for the function f(x) = x2 + 2 in the interval [-2, 2] (March – 2014)
Answer:

Question 8.
(a) Find the relationship between a and b so that the function f defined by
\(f(x)=\left\{\begin{array}{ll}
a x^{2}-1, & x \leq 2 \\
b x+3, & x>2
\end{array}\right.\) is continuous.

(b) Verify mean value theorem for the function f(x) = x2 – 4x -3 ¡n the interval [1, 4]. (May – 2014)
Answer:
(a) Since fis continuous

Hence mean value theorem satisfies for the funcion.

Question 9.
(a) Find ‘a’ and ‘b’ if the function
\(f(x)=\left\{\begin{array}{ll}
\frac{\sin x}{x}, & -2 \leq x \leq 0 \\
a \times 2^{x}, & 0 \leq x \leq 1 \\
b+x, & 1<x \leq 2
\end{array}\right.\) is continous on [-2, 2]

(b) How many of the functions
f(x) = |x|, g(x) = |x|², h(x) = |x|³ are not differentiable at x = 0?
(i) 0
(ii) 1
(iii) 2
(iv) 3 (March – 2015)
Answer:
(a) Since f(x) is continuous on [-2, 2]

Question 10.
(a) Find the relation between ‘a’ and ‘b’ if the function f defined by
\(f(x)=\left\{\begin{array}{l}
a x+1, x \leq 3 \\
b x+3, x>3
\end{array}\right.\) is continuous.
lbx+3.x>3
(b) If e^y (x + 1) = 1, show thats \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\) (May -2015)
Answer:

Question 11.
Find the value of a and b such that the function \(f(x)=\left\{\begin{array}{cc}
5 a & x \leq 0 \\
a \sin x+\cos x & 0<x<\frac{\pi}{2} \\
b-\frac{\pi}{2} & x \geq \frac{\pi}{2}
\end{array}\right.\) is continuous. (March – 2010)
Answer:

Question 12.
(i) Find \(\frac{d y}{d x}, \text { if } x=a \cos ^{2} \theta ; y=b \sin ^{2} \theta\)
(ii) Find the second derivative of the function y = e^x sinx. (May – 2017)
Answer:

Question 13.
Find \(\frac{d y}{d x}\) of the following (4 score each)
(i) y^x = x^y (May – 2015)
(ii) (COSx)^y = (cosy)^x (March – 2017)
Answer:

Plus Two Maths Continuity and Differentiability 6 Marks Important Questions

Question 1.
Find \(\frac{d y}{d x}\) if
(i) sinx + cosy = xy
(ii) x = acos³t, y = asin³t
(iii) y = x^x + (logx)^x (May -2009; May -2011; March -2015)
Answer:
(i) Given; sinx + cosy = xy
Differentiating with respect to x;

Question 2.
(i) Let y =3 cos(log x) + 4 sin (log x)
(a) Find \(\frac{d y}{d x}\)
(b) Prove that x² y₂ + xy₁ + y = 0

(ii) (a) Find the derivative of y = e^2x+logx
(b) Find \(\frac{d y}{d x}\)
if x = a (θ – sinθ), y = a(1 – cosθ) (March – 2009)
Answer:

Question 3.
(i) Show that the function f (x) defined by f(x) = sin (cosx) is a continuous function.
(ii) If \(\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}\), Show that \(\frac{d^{2} y}{d x^{2}}=\frac{-\frac{d^{2} x}{d y^{2}}}{\left(\frac{d x}{d y}\right)^{3}}\) (May -2010)
Answer:
Given; f(x) = sin(cos x)
Let g(x) = sin(x) and h(x) = cos x
Both these function are trigonometric functions hence continuous.
goh(x) = g(h(x)) = g(cos x) = sin(cos x) = f(x)

Since f(x) is the composition of two continuous functions, hence continuous.

Question 4.
(i) Let y = x^{sin x} + (sinx)^x. Find \(\frac{d y}{d x}\)
(ii) Given; \(y=\sqrt{\tan ^{-1} x}\)
(a) \(2\left(1+x^{2}\right) y \frac{d y}{d x}=1\)
(b) \(\left(1+x^{2}\right) y \frac{d^{2} y}{d x^{2}}+\left(1+x^{2}\right)\left(\frac{d y}{d x}\right)^{2}+2 x y \frac{d y}{d x}=0\) (May – 2010; Onam – 2017)
Answer:

Question 5.
(i) The function \(f(x)=\left\{\begin{array}{ll}
5, & x \leq 2 \\
a x+b, 2< & x<10 \text { is } \\
21, & x \geq 10
\end{array}\right.\) continuous. Find a and b
(ii) Find \(\frac{d y}{d x}\) (a) if y = Sin (xsinx)
(iii) If y = ae” + be’; show that \(\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0\) (March – 2011)
Answer:

Question 6.
(i) Match the following.

(ii) If y = sin^-1 x, prove that (1 – x²) y₂ – xy₁ = 0 (March – 2012; May -2017)
Answer:

Question 7.
(i) Consider \(f(x)=\left\{\begin{array}{ll}
3 x-8, & x \leq 5 \\
2 k, & x>5
\end{array}\right.\) Find the value of k if f(x) is continuous at x = 5.
(ii) Find \(\frac{d y}{d x}, \text { if } y=(\sin x)^{\log x}, \sin x>0\)
(iii) If y = (sin-1 x)2, then show that \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=2\). (March -2013)
Answer:

Question 8.
(i) Find, if y = 1ogx, x>0
(ii) Is f(x) = |x| differentiable at x = 0?
(iii) Find if x = sin θ – cos θ and y= sinθ + cosθ (May – 2013)
Answer:

Kerala State Board New Syllabus Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 4 Determinants.

Kerala Plus Two Maths Chapter Wise Previous Questions Chapter 4 Determinants

Plus Two Maths Determinants 3 Marks Important Questions

Question 1.
Prove that \(\begin{array}{|lll|}
1 ! & 2 ! & 3 ! \\
2 ! & 3 ! & 4 ! \\
3 ! & 4 ! & 5 !
\end{array}\) (March – 2015)
Answer:

Question 2.
Using properties of determinants prove the following. (March – 2010; Christmas -2017)
Answer:

Plus Two Maths Determinants 4 Marks Important Questions

Question 1.
Consider the matrix \(A=\left[\begin{array}{ll}
2 & 5 \\
3 & 2
\end{array}\right]\)

(i) Find adj (A)
(ii) Find A1
(iii) Using A^-1 solve the system of linear equations 2x + 5y = 13x + 2y = 7 (March – 2010)
Answer:

Plus Two Maths Determinants 6 Marks Important Questions

Question 1.
Consider the matrix \(A=\left[\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right]\)

(i) Using the column operat,on
C₁ → C₁ + C₂ + C₃,
show that \(|A|=(a+b+c)\left|\begin{array}{ccc}
1 & b & c \\
1 & c & a \\
1 & a & b
\end{array}\right|\)
(ii) Show that |A| = – (a3 + b3 + e3 — 3abc)
(iii) Find A x adj(A) if a = 1,b = 10,c = 100 (May – 2010)
Answer:

Question 2.
(i) (a) If \(A=\left[\begin{array}{ccc}
1 & 1 & 5 \\
0 & 1 & 3 \\
0 & -1 & -2
\end{array}\right]\)

What is the value of |3A|?
(b) Find the equation of the line joining the points (1,2) and (-3,-2) using determinants.
(ii) Show that \(\left|\begin{array}{lll}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{array}\right|=(a-b)(b-c)(c-a)\)
Answer:

(b) Let (x,y) be the coordinate of any point on The line, then (1,2), (-3, -2) and (x, y) are collinear.

Hence the area formed will be zero.

Question 3.
Consider the following system of linear equations; x + y + z = 6, x – y + z = 2, 2x + y + z = 1
(i) Express this system of equations in the Standard form AXB
(ii) Prove that A is non-singular.
(iii) Find the value of x, y and z satisfying the above equation. (May – 2011)
Answer:

Question 4.
(i) lf \(\left|\begin{array}{ll}
x & 3 \\
5 & 2
\end{array}\right|=5\), then x = ………..
(ii) Prove that
\(\left|\begin{array}{ccc}
y+k & y & y \\
y & y+k & y \\
y & y & y+k
\end{array}\right|=k^{2}(3 y+k)\)
(iii) Solve the following system of linear Equations, using matrix method; 5x + 2y = 3, 3x + 2y = 5 (March – 2012)
Answer:

Question 5.
(i) Let B is a square matrix of order 5, then |kB| is equal to ………..
(a) |B|
(b) k|B|
(c) k⁵|B|
(d) 5|B|

(ii) Prove that \(\left|\begin{array}{lll}
1 & x & x^{2} \\
1 & y & y^{2} \\
1 & z & z^{2}
\end{array}\right|=(x-y)(y-z)(z-x)\)
(iii) Check the consistency of the following equations; 2x + 3y + z = 6, x + 2y – z = 2, 7x + y + 2z =10 (May – 2012)
Answer:

Therefore the system is consistent and has unique solutions.

Question 6.
(i) Find the values of x in which \(\left|\begin{array}{ll}
3 & x \\
x & 1
\end{array}\right|=\left|\begin{array}{ll}
3 & 2 \\
4 & 1
\end{array}\right|\)

(ii) Using the property of determinants, show that the points A(a,b + c), B(b,c + a), C(c,a + b) are collinear.
(iii) Examine the consistency of system of following equations: 5x – 6y + 4z = 15, 7x + y – 3z = 19, 2x + y + 6z = 46 (EDUMATE – 2017; March – 2013)
Answer:

Since, the system is consistent and has unique solutions.

Question 7.
Consider a system of linear equations which is given below;
\(\begin{array}{l}
\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4 ; \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1 \\
\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2
\end{array}\)

(i) Express the above equation in the matrix form AX = B.
(ii) Find A^-1, the inverse of A.
(iii) Find x,y and z. (May – 2013)
Answer:

Question 8.
Consider the matrices \(A=\left[\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right]\)

(i) Find A² – 7A – 21 = 0
(ii) Hence find A-1
(iii) Solve the following system of equations using matrix method 2x + 3y = 4; 4x + 5y = 6 (March – 2014)
Answer:

(iii) The given system of equations can be converted into matrix form AX = B

Question 9.
(i) Let A be a square matrix of order 2 x 2 then |KA| is equal to
(a) K|A|
(b) K²|A|
(c) K³|A|
(d) 2K|A|

(ii) Prove that
\(\left|\begin{array}{ccc}
\mathbf{a}-\mathbf{b}-\mathbf{c} & \mathbf{2 a} & 2 \mathbf{a} \\
2 \mathrm{~b} & \mathrm{~b}-\mathrm{c}-\mathrm{a} & 2 \mathrm{~b} \\
2 \mathrm{c} & 2 \mathrm{c} & \mathrm{c}-\mathrm{a}-\mathrm{b}
\end{array}\right|=(\mathrm{a}+\mathrm{b}+\mathrm{c})^{3}\)

(iii) Examine the consistency of the system of Equations. 5x + 3y = 5; 2x + 6y = 8 (May- 2014)
Answer:

(iii) The given system of equation can be written in matrix form as

solution exist and hence it is consistent.

Question 10.
(a) Choose the correct statement related to the matnces \(A=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right], B=\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
\(\begin{array}{l}
\text { (i) } A^{3}=A, B^{3} \neq B \\
\text { (ii) } A^{3} \neq A, B^{3}=B \\
\text { (iii) } A^{3}=A, B^{3}=B \\
\text { (iv) } A^{3} \neq A, B^{3} \neq B
\end{array}\)

(b) lf \(M=\left[\begin{array}{ll}
7 & 5 \\
2 & 3
\end{array}\right]\) then verity the equation M² – 10M + 11 I₂ = O

(c) Inverse of the matrix \(\left[\begin{array}{lll}
0 & 1 & 2 \\
0 & 1 & 1 \\
1 & 0 & 2
\end{array}\right]\) (March – 2015)
Answer:

Question 11.
Solve the system of Linear equations x + 2y + z = 8; 2x + y – z = 1; x – y + z = 2 (March – 2015)
Answer:

Question 12.
(a) If \(\left|\begin{array}{ll}
x & 1 \\
1 & x
\end{array}\right|=15\) then find the value of X.

(b)Solve the following system of equations 3x – 2y + 3z = ?, 2x + y – z = 1 4x – 3y + 2z = 4 (May – 2015)
Answer:

Question 13.
(i)The value of the determinant \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & -1 \\
1 & 1 & -1
\end{array}\right|\) is
(a) -4
(b) 0
(c) 1
(d) 4

(ii) Using matrix method, solve the system of linear equations x + y + 2z = 4; 2x – y + 3z = 9; 3x – y – z = 2 (May – 2016)
Answer:
(i) (d) 4
(ii) Express the given equation in the matrix form as AX = B

Question 14.
(i) If \(A=\left[\begin{array}{ll}
a & 1 \\
1 & 0
\end{array}\right]\) is such that A² = I then a equals
(a) 1
(b) -1
(c) 0
(d) 2

(ii)Solve the system of equations x – y + z = 4; 2x + y – 3z = 0; x + y + z = 2 Using matrix method. (March – 2017)
Answer:

Question 15.
(i) IfA is a 2 x 2 matrix with |A| = 5, then |adjA| is
(a) 5
(b) 25
(c) 1/5
(d) 1/25

(ii) Solve the system of equations using matrix method.
x + y + z = 1; 2x + 3y – z = 6; x – y + z = -1 (May – 2017)
Answer:
(i) (a) 5
(ii) LetA X=B,

Kerala State Board New Syllabus Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 12 Linear Programming.

Kerala Plus Two Maths Chapter Wise Previous Questions Chapter 12 Linear Programming

Plus Two Maths Linear Programming 4 Marks Important Questions

Question 1.
Consider the linear programming problem;
Maximise; Z = x +y , 2x + y – 3< 0, x – 2y + 1 < 0, y < 3, x < 0, y < 0
(i) Draw its feasible region.
(ii) Find the corner points of the feasible region.
(iii) Find the corner at which Z attains its maximum. (March – 2012)
Answer:

(ii) In the figure the shaded region ABC is the feasible region. Here the region is bounded. The corner points are A(1, 1), B(5, 3), C(O, 3).
(iii) Given; Z = x + y

Since maximum value of Z occurs at B, the soluion is Z = (5) + (3) = 8.

Question 2.
Consider the LPP Minimise; Z = 200 k + 500y, x + 2y > 10, 3x + 4y < 24, x > 0, y > 0
(i) Draw the feasible region.
(ii) Find the co-ordinates of the comer points of the feasible region.
(iii) Solve the LPP. (May – 2012)
Answer:

(ii) In the figure the shaded region ABC is the feasible region. Here the region is bounded. The corner points are 4(4, 3), 5(0, 6), C(0, 5)
(iii) Given; Z = 200x + 500y

Since minimum value of Z occurs at A, the soluion is Z = 200(4) + 500(3) = 2300.

Question 3.
Consider the LPP
Maximise; Z = 5x + 3y
Subject to; 3x + 5y < 15, 5x + 2y < 10x, y > 0
(i) Draw the feasible region.
(ii) Find the corner points of the feasible region.
(iii) Find the corner at which Z attains its maximum. (March – 2013)
Answer:
In the figure, the shaded region OABC is the feasible region. Here the region is bounded.
The corner points are

Given; Z = 5x + 3y

Since maximum value of Z occurs at B, the soluion is Z =

Question 4.
Consider the linear programming problem: Minimize Z = 3x + 9y Subject to the constraints: x + 3y < 60 x + y > 10, x < y, x > 0, y > 0
(i) Draw its feasible region.
(ii) Find the vertices of the feasible region
(iii) Find the minimum value of Z subject to the given constraints. (March-2014, SAY-2016)
Answer:

(ii) The feasible region is ABCD.
Solving x + y = 10, x = y we get B(5, 5)
Solving x + 3y = 60, x = y we get C(15, 15)
Hence the comer points are A(0, 10) , B(5, 5), C(15, 15), D(0, 20)
(iii) Given; Z = 3x + 9y

Form the table, minumum value of Z is 6 O at B(5, 5).

Question 5.
Consider the linear inequalities 2x + 3y < 6; 2x + y < 4; x, y < 0
(a) Mark the feasible region.
(b) Maximise the function z = 4x + 5y subject to the given constraints. (March – 2014)
Answer:

(b) 15.2x + 3y = 6

2x + y = 4

Maximum at x = 1.5, y = 1
Maximum value is Z = 11

Question 6.
Consider the linear programming problem: Minimise Z = 4x + 4y Subject to x + 2y < 8; 3x + 2y < 12x, y < 0
(a) Mark its feasible region.
(b) Find the comer points of the feasible region.
(c) Find the corner at which Z attains its minimum. (May – 2014)
Answer:
(a) x + 2y = 8,

3x + 12y = 12

The comer points are 0(0, 0), A(4, 0), B(2, 3), C(0, 4)

(c)

Z attains minimum at (4, 0).

Question 7.
Consider the linear programming problem: Maximum z = 4x + y
Subject to constraints: x + y < 50, 3x + y < 9x, y < 0
(a) Draw the feasible region
(b) Find the corner points of the feasible region
(c) Find the corner at which ‘z’ attains its maximum value. (May – 2015)
Answer:
(a) x + y = 50,

3x + y = 90

(b) Solving the equations we get the points as
O(0, 0) A(30, 0); B(20, 30); C(0, 50)

(c)

Z attains maximum at A(30, 0)

Question 8.
Consider the LPP
Maximise z = 3x + 2y
Subject to the constraints: x + 2y < 10, 3x + y < 15; x, y < 0
(a) Draw its feasible region
(b) Find the corner points of the feasible region
(c) Find the maximum value of Z. (March – 2016)
Answer:

(b) The corner points 0(0,0), A(5,0), B(4,3), C(0, 5)
Z is maximum at B(4, 3), z = 18.

(c)

Question 9.
Consider the linear programming problem: Maximum z = 50x + 40y
Subject to constraints:
x + 2y < 10; 3x + 4y < 24; x, y < 0
(i) Draw the feasible region
(ii) Find the comer points of the feasible region
(iii) Find the maximum value of z. (March – 2017)
Answer:

(ii) In the figure the shaded region ABC is the feasible region. Here the region is bounded. The corner points are A(4, 3), B(0, 6), C(0, 5)
(iii) Given; Z = 50x + 40y

Since minimum value of Z occurs at A, the soluion is Z = 50(4) + 4(3) = 320.

Plus Two Maths Linear Programming 6 Marks Important Questions

Question 1.
A furniture dealer sells only tables and chairs. He has Rs. 12,000 to invest and a space to store 90 pieces. A table costs him Rs. 400 and a chair Rs. 100. He can sell a table at a profit of Rs. 75 and a chair at a profit of Rs. 25. Assume that he can sell all the items. The dealer wants to get maximum profit.
(i) By defining suitable variables, write the objective function.
(ii) Write the constraints.
(iii) Maximise the objective function graphically. (March – 2010)
Answer:
(i) Let x be the number of Tables and y be the number of Chairs. Then; Maximise; z = 75x + 25y
(ii) Furniture constraints x + y < 90
Investment constraint 400x + 100y < 12000
Therefore;Maximise; Z = 75x + 25y, x + y < 90, 4x + y < 120, x<0, y<0
(iii) In the figure the shaded region OABC is the feasible region. Here the region is bounded. The corner points are O(0, 0), A(30, 0) B(10, 80), C(0, 90).

Given; Z = 75x + 25y

Since minimum value of Z occurs at B, the soluion is Z = 2750.

Question 2.
A company produces two types of cricket balls A and B. The production time of one ball of type B is double type A (time in units). The company has the time to produce a maximum of 2000 balls per day. The supply of raw materials is sufficient for the production of 1500 balls (both A and B) per day. The company wants to make maximum profit by making a profit of Rs. 3 from a ball of type A and Rs. 5 from type B.

Then,
(i) By defining suitable variables write the objective function.
(ii) Write the constraints.
(iii) How many balls should be produced in each type per day in order to get maximum profit? (May – 2010)
Answer:
(i) Let x be the number of balls of type A and y be the number of balls of type B. Then Maximise profit is Z = 3x + 5y
(ii) Balls constraints 2x + y < 2000 investment constraint x + y < 1500
Therefore; Maximise; Z = 3x + 5y, 2x + y < 2000, x + y < 1500, x < 0, y < 0

(iii) In the figure the shaded region OABC is the fesible region. Here the region ¡s bounded. The corner points are O(0, 0), A(1000, 0) B(500, 1000), C(0, 1500). Given; Z = 3x + 5y

Since maximum value of Z occurs at C, the soluion is Z = 3(0) + 5(1500) = 7500.

Question 3.
The graph of a linear programming problem is given below. The shaded region is the feasible region. The objective function is Maximise; Z = px + qy

(i) What are the co-ordinates of the corners of the feasible region?
(ii) Write the constraints.
(iii) If the Max. Z occurs at A and B, what ¡s the relation between p and q?
(iv) If q = 1, write the objective function.
(v) Find Max. Z. (March – 2011)
Answer:
(i) From the figure the feasible region is OABC.
Then the comer points are;
A is (5, 0), B is (3, 4), C is (0, 5) and O (0, 0)
(ii) The constraints are 2x + y < 10, x + 3y < 15, x < 0, y < 0
(iii) Given; Z = px + qy

Since maximum at A and B we have;
⇒ 3p + 4q = 5p ⇒ 2p = 4q ⇒ p = 2q
(iv) When q = 1, then p ⇒ 2q ⇒ p = 2
Objective function is; Z = 2x + y
(v) We have; Z px + qy at B Z has maximum ⇒ Z = 2(3) + 4 = 10

Question 4.
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and I hour on machine B to produce a package of bolts. He earns a profit of Rs. 17.50 per package on nuts and Rs. 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he produced each day so as to maximise the profit if he operates his machines for at the most 12 hours a day?
(i) By suitable defining the variables write the objective function of the problem.
(ii) Formulate the problem as a linear programming problem(LPP)
(iii) Solve the LPP graphically and find the number of packages of nuts and bolts to be manufactured. (May -2011)
Answer:
(i) Let x be the number of packages of nuts produced and y be the number of packages of bolts produced. Then;
Maximise profit is; Z = 17. 5x + 7y
(ii) Time constraint for Machine A; x + 3y < 12
Time constraint for Machine B; 3x + y < 12
Therefore; Maximise; Z = 17.5x + 7y, x + 3y < 12, 3x + y < 12, x < 0, y < 0
(iii) In the figure the shaded region OABC is the visible region. Here the region is bounded. The corner points are 0(0,0), A (4, 0) B(3, 3), C(0, 4).

Given; Z = 17.5x + 7y

Since maximum value of Z occurs at B, the soluion is Z = 17.5(3) + 7(3) = 73.5.

Question 5.
A bakery owner makes two types of cakes A and B. Three machines are needed for this purpose. The time (in minutes) required for making each type of cakes in each machine is given below;

Each machine is available for almost 6 hours per day. Assume that all cakes will be sold out every day. The bakery owner wants to make a maximum profit per day by making Rs. 7.5 from type A and Rs. 5 from type B.
(i) Write the objective function by defining suitable variables.
(ii) Write the constraints.
(iii) Find the maximum profit graphically. (May- 2013, EDUMATE – 2017)
Answer:
(i) Number of cake of type A: x
Number of cake of type B: y
Then profit function is Maximise: Z = 75x + 5y

(ii) 12x + 6y < 360; 18x + 0y < 360
6x + 9y < 360; x > 0, y > 0
Simplifying we get;
2x – i – y < 60………..(1)
x < 20………..(2)
2x + 3y < 120………..(3)
x > 0, y > 0

The feasible region is OABCDO
Solving (1) and (2) we get the point B- (2020)
Solving (1) and (3) we get the point C- (15,30)
A-(20,0), O-(0,0), D-(0,40)
Given; Z = x + y

Since maximum value of Z occurs at B, the soluion is Z = 250 (20, 20).

Question 6.
In a factory, there are two machines A and B producing toys. They respectively produce 60 and 80 units in one hour. A can run a maximum of 10 hours and B a maximum of 7 hours a day. The cost of their running per hour respectively amounts to 2,000 and 2,500 rupees. The total duration of working these machines cannot exceed 12 hours a day. If the total cost cannot exceed Rs. 25,000 per day and the total daily production is at least 800 units, then formulate the problem mathematically. (March – 2014)
Answer:
Let x be the running time for machine A and y be the running time for machine B.
Since machines cannot work more than 12 hours x + y < 12
Since maximum production of two machines is 800 units.
60x + 80y < 800
Maximum cost of production is 25000, 2000x + 2500y < 25000
0 < x < 10, 0 < y < 7