1. If r=b eθ cota, where a, b are constants then ‘s’ is represented by which one of the following equation?
a) s=r sec(a)+c
b) s=r cos(a)+c
c) s=r+c
d) s=r tan(a)+c
Explanation: r=b eθ cota
\(\frac{dr}{dθ} = b e^{θ cota}(cot a)=r cot(a) \)
w.k.t \(\frac{ds}{dr} = \sqrt{1+r^2 (\frac{dθ}{dr})^2} \,where\, \frac{dθ}{dr}=\frac{1}{r} \,tana\)
\(\frac{ds}{dr} = \sqrt{1+tan^2 a} = sec \,a \rightarrow s = \int sec \,a \,dr \, = r sec(a)+c,\) where c is constant of integration.
2. What is the maximum area of the rectangle with perimeter 620 mm?
a) 24,025 mm2
b) 22,725 mm2
c) 24,000 mm2
d) 24,075 mm2
Explanation: Let x be the length of the rectangle and y be the width of the rectangle. Then, Area A is,
A=x*y …………………………………………………. (1)
Given: Perimeter of the rectangle is 620 mm. Therefore,
P=2(x+y)
620=2(x+y)
x+y=310
y=310-x
We can now substitute the value of y in (1)
A=x*(310-x)
A=310x-x2
To find maximum value we need derivative of A,
dA/dx=310-2x
To find maximum value, \(\frac{dA}{dx}=0 \)
310-2x=0
2x=310
x=155 mm
Therefore, when the value of x=155 mm and the value of y=310-155=155 mm, the area of the rectangle is maximum, i.e., A=155*155=24,025 mm2
3. Power Rule of the derivative states that, \(\frac{d(x^n)}{dx}=nx^{n-1}.\)
a) True
b) False
Explanation: Power Rule is given by, \(\frac{d(x^n)}{dx}=nx^{n-1}.\)
For example, \(\frac{d(x^4)}{dx}=4x^3. \)
4. Which of the following trigonometric function derivatives is correct?
a) \(\frac{d(sinx)}{dx}=-cosx\)
b) \(\frac{d(secx)}{dx}=secxtanx\)
c) \(\frac{d(tanx)}{dx}=cot^2 x\)
d) \(\frac{d(cosx)}{dx}=sinx\)
Explanation: Correct forms of Trigonometric Derivative Functions
- \(\frac{d(sinx)}{dx}=cosx \)
- \(\frac{d(cosx)}{dx}=-sinx\)
- \(\frac{d(secx)}{dx}=secxtanx\)
- \(\frac{d(tanx)}{dx}=sec^2 x \)
5. Product Rule of the derivative is given by ________
a) \(\frac{d(fg)}{dx}=\frac{d(g)}{dx}*\frac{d(f)}{dx}\)
b) \(\frac{df}{dx}*\frac{dg}{dx}=\frac{d(g)}{dx}+\frac{d(f)}{dx}\)
c) \(\frac{d(fg)}{dx}=f \frac{d(g)}{dx}*g \frac{d(f)}{dx}\)
d) \(\frac{d(fg)}{dx}=f \frac{d(g)}{dx}+g \frac{d(f)}{dx}\)
Explanation: Product Rule of derivative is given by,
\( \frac{d(fg)}{dx}=f \frac{d(g)}{dx}+g \frac{d(f)}{dx}\)
6. What is the derivative of \(\frac{3x^2}{(1-sinx)}\)?
a) \(\frac{6xsinx-3x^2 cosx}{(1-sinx)^2} \)
b) \(\frac{6x}{1-sinx} \)
c) \(\frac{6xsinx-3x^2 cosx-6x}{(1-sinx)^2} \)
d) \(\frac{6xsinx-3cosx-6}{(1-sinx)^2} \)
Explanation: Given: y = \(\frac{3x^2}{(1-sinx)}\)
\(\frac{dy}{dx}=\frac{3x^2(\frac{d(1-sinx)}{dx})-(\frac{d(3x^2)}{dx})(1-sinx)}{(1-sinx)^2}\)
\(\frac{dy}{dx}=\frac{3x^2(-cosx)-6x(1-sinx)}{(1-sinx)^2}\)
\(\frac{dy}{dx}=\frac{6xsinx-3x^2 cosx-6x}{(1-sinx)^2}\)
7. Which of the following is a type of Iterative method of solving non-linear equations?
a) Graphical method
b) Interpolation method
c) Trial and Error methods
d) Direct Analytical methods
Explanation: There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods).
8. The half-interval method in numerical analysis is also known as __________
a) Newton-Raphson method
b) Regula Falsi method
c) Taylor’s method
d) Bisection method
Explanation: The Bisection method, also known as binary chopping or half-interval method, is a starting method which is used, where applicable, for few iterations, to obtain a good initial value.
9. Given \(∫_0^8 x^{\frac{1}{3}} dx,\) find the error in approximating the integral using Simpson’s \(\frac{1}{3}\) Rule with n=4.
a) 1.8
b) 2.9
c) 0.3
d) 0.35
Explanation:Given: \(∫_0^8 x^{\frac{1}{3}} dx,\), n = 3,
Let \(f(x)= x^{\frac{1}{3}},\)
\(∆x = \frac{b-a}{2}=\frac{8-0}{2}=4\)………………since b=8, a=0 (limits of the given integral)
Hence endpoints xi have coordinates {0, 2, 4, 6, 8}.
Calculating the function values at xi, we get,
\(f(0)= 0^{\frac{1}{3}}=0\)
\(f(2)= 2^{\frac{1}{3}}\)
\(f(4)= 4^{\frac{1}{3}}\)
\(f(6)= 6^{\frac{1}{3}}\)
\(f(8)= 8^{\frac{1}{3}} =2\)
Substituting these values in the formula,
\(∫_0^8x^{\frac{1}{3}} dx ≈ \frac{∆x}{3} [f(0)+4f(2)+2f(4)+4f(6)+f(8)]\)
\(≈ \frac{2}{3}[0+4(2^{\frac{1}{3}})+ 2(4^{\frac{1}{3}})+ 4(6^{\frac{1}{3}})+2] ≈ 11.65\)
Actual integral value,
\(∫_0^8x^{\frac{1}{3}}dx = \left(\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right)_0^8=12\)
Error in approximating the integral = 12 – 11.65 = 0.35
10. The curvature of a function f(x) is zero. Which of the following functions could be f(x)?
a) ax + b
b) ax2 + bx + c
c) sin(x)
d) cos(x)
Explanation: The expression for curvature is
k=\(\left |\frac{f”(x)}{(1+[f'(X)]^2)^{\frac{3}{2}}} \right |\)
Given that k = 0 we have
f” (x) = 0
f(x) = a + b.