1.Radius of Curvature Cartesian Form-B.sc. Math Solutions

Radius of Curvature Cartesian Form-B.sc. Math Solutions

In this article Radius of Curvature Cartesian Form-B.sc. Math Solutions,we will read specific questions of radius of curvature and solved them to understand easily.
“Before diving into these solutions, make sure you understand the foundational concept of “Radius of Curvature in UG Mathematics” and have reviewed the essential “Parametric and Cartesian Formula for Curvature-B.sc. Math“.This will make following today’s step-by-step solutions much easier.”

2.Derivation of Cartesian Radius of Curvature Formula

Let the curve be y = f(x) \cdots (1)
Then we know \frac{dy}{dx} = \tan\psi \cdots (1)
Differentiating w.r.t. x, we get
\frac{d^2y}{dx^2} = \sec^2\psi \frac{d\psi}{dx} = \sec^2\psi \frac{d\psi}{ds} \cdot \frac{ds}{dx} \\ = (1 + \tan^2\psi) \cdot \frac{1}{\rho} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \quad \left[\because \frac{d\psi}{ds} = \frac{1}{\rho}\right] \\ = \left[1 + \left(\frac{dy}{dx}\right)^2\right] \cdot \frac{1}{\rho} \cdot \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \quad \left[\because \tan\psi = \frac{dy}{dx}\right] \\ \frac{d^2y}{dx^2} = \frac{1}{\rho} \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}} \\ \Rightarrow \frac{1}{\rho} = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}
Note: From definition of radius of curvature it is clear that the value of \rho depends on the curve and not on axes. So interchanging the axes of x and y, we have
\rho = \frac{\left[1 + \left(\frac{dx}{dy}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2x}{dy^2}}
and this form is used when \frac{dy}{dx} is infinite.

Also Read This Article:- Radius of curvature

3.Working Rule to Solve Cartesian Curvature Problems

Radius of Curvature Cartesian Form-B.sc. Math Solutions

Step : 1. Find the first derivative from Given Equation y = f(x)
y' = \frac{dy}{dx}
Step : 2. Find the second derivative
y'' = \frac{d^2y}{dx^2}
Step : 3. Substitute the values into the formula of radius curvature
\rho = \frac{\left[1 + (y')^2\right]^{\frac{3}{2}}}{y''}
Step : 4. Calculate the radius of curvature at the given point (x_1, y_1)

4.Radius of Curvature Cartesian Form

Find the radius of curvature at the point t on the following curve:
Example:1.x = c \sin 2\theta (1 + \cos 2\theta), y = c \cos 2\theta (1 - \cos 2\theta)
Solution: x = c \sin 2\theta (1 + \cos 2\theta), y = c \cos 2\theta (1 - \cos 2\theta) \text{ Write the parametric equations as } \\ x = c\left(\sin 2\theta + \frac{1}{2}\sin 4\theta\right) and \\ y = c\left(\cos 2\theta - \frac{1+\cos 4\theta}{2}\right) \\ \text{ Differentiating w.r.t. } \theta, \text{ we get } \\ \dot{x} = 2c(\cos 2\theta + \cos 4\theta) \\ \Rightarrow \dot{x} = 4c \cos 3\theta \cos \theta \\ \dot{y} = 2c(\sin 4\theta - \sin 2\theta) \\ \Rightarrow \dot{y} = 4c \cos 3\theta \sin \theta \\ \text{ Again differentiating w.r.t. } \theta, \text{ we get } \\ \ddot{x} = -4c(3\sin 3\theta \cos \theta + \cos 3\theta \sin \theta) \\ \ddot{y} = 4c(-3\sin 3\theta \sin \theta + \cos 3\theta \cos \theta) \\ \text{Radius of Curvature Formula: } \\ \rho = \frac{(x'^2 + y'^2)^{\frac{3}{2}}}{x' y'' - y' x''} \\ = \frac{ [(16c^2 \cos^2 3\theta \cos^2 \theta + 16c^2 (\cos 3\theta \sin \theta)^2]^{\frac{3}{2}}}{\begin{array}{c}4c \cos 3\theta \cos \theta \times 4c(-3\sin 3\theta \sin \theta  + \cos 3\theta \cos \theta)  \\- 4c \cos 3\theta \sin \theta (-4c)(3\sin 3\theta \cos \theta + \cos 3\theta \sin \theta)\end{array}} \\ = \frac{[16c^2 \cos^2 3\theta (\cos^2 \theta + \sin^2 \theta)]^{\frac{3}{2}}}{\begin{array}{c}-48c^2 \cos 3\theta \cos \theta \sin 3\theta \sin \theta + 16c^2 \cos^2 3\theta \cos^2 \theta\\ + 48c^2 \cos 3\theta \cos \theta \sin 3\theta \sin \theta + 16c^2 \cos^2 3\theta \sin^2 \theta\end{array}}\\ = \frac{(16c^2 \cos^2 3\theta)^{\frac{3}{2}}}{16c^2 \cos^2 3\theta (\cos^2 \theta + \sin^2 \theta)} \\ = \frac{64c^3 \cos^3 3\theta}{16c^2 \cos^2 3\theta} \\ \Rightarrow \rho = 4c \cos 3\theta
Example:2.Prove that the radius of curvature of the catenary
y = \frac{1}{2}a \left(e^{\frac{x}{a}} + e^{-\frac{x}{a}}\right) is \frac{y^2}{a}, and that of the catenary of uniform strength y = c \log \sec \left(\frac{x}{c}\right)  \text{ is } c \sec \left(\frac{x}{c}\right)
[Jodhpur, 80 (S)]
Solution: y = \frac{1}{2}a \left(e^{\frac{x}{a}} + e^{-\frac{x}{a}}\right) \\ = a \cosh \left(\frac{x}{a}\right) \dots (1) \\ \text{ Differentiating w.r.t. x, we get } \\ \frac{dy}{dx} = a \sinh \left(\frac{x}{a}\right) \cdot \frac{1}{a} = \sinh \left(\frac{x}{a}\right) \\ \text{ Again differentiating w.r.t. x, we get } \\ \frac{d^2y}{dx^2} = \frac{1}{a} \cosh \left(\frac{x}{a}\right) \\ \text{ Radius of curvature formula } \\ \rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} \\ = \frac{\left[1 + \sinh^2 \frac{x}{a}\right]^{\frac{3}{2}}}{\frac{1}{a} \cosh \left(\frac{x}{a}\right)} \\ = \frac{a \left[\cosh^2 \frac{x}{a}\right]^{\frac{3}{2}}}{\cosh \left(\frac{x}{a}\right)} \quad \left[\because 1 + \sinh^2 \frac{x}{a} = \cosh^2 \frac{x}{a}\right] \\ = \frac{a \cosh^3 \frac{x}{a}}{\cosh \frac{x}{a}} = a \cosh^2 \frac{x}{a} \\ \therefore \rho = \frac{y^2}{a} \quad [\text{from (1)}] \\ \text{ Again } y = c \log \sec \left(\frac{x}{c}\right) \\ \text{ Differentiating w.r.t. x, we get } \\ \frac{dy}{dx} = \frac{c}{\sec \left(\frac{x}{c}\right)} \cdot \sec \left(\frac{x}{c}\right) \tan \left(\frac{x}{c}\right) \cdot \frac{1}{c} \\ \Rightarrow \frac{dy}{dx} = \tan \left(\frac{x}{c}\right) \\ \text{ Again differentiating w.r.t. x, we get } \\ \frac{d^2y}{dx^2} = \sec^2 \left(\frac{x}{c}\right) \cdot \frac{1}{c} \\ \Rightarrow \frac{d^2y}{dx^2} = \frac{1}{c} \sec^2 \left(\frac{x}{c}\right) \\ \text{ Radius of Curvature Formula } \\ \rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} \\ = \frac{\left(1 + \tan^2 \frac{x}{c}\right)^{\frac{3}{2}}}{\frac{1}{c} \sec^2 \frac{x}{c}} \\ = \frac{c \left(\sec^2 \frac{x}{c}\right)^{\frac{3}{2}}}{\sec^2 \frac{x}{c}} \\ = \frac{c \sec^3 \frac{x}{c}}{\sec^2 \frac{x}{c}} \\ \Rightarrow \rho = c \sec \left(\frac{x}{c}\right) \\ \text{ Hence proved. }
Example:3.Prove that for the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \\ \rho = \frac{a^2b^2}{p^3}
p being the perpendicular from the centre upon the tangent at (x,y)
[Jodhpur BE III,79]
Solution: \text{ Equation of ellipse } \\ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \cdots (1) \\ \text{ Parametric equation of ellipse } \\ x = a \cos \theta, y = b \sin \theta \\ \text{ Differentiating w.r.t.} \theta , \text{ we get } \\ x' = -a \sin \theta, y' = b \cos \theta \\ \text{ Again differentiating w.r.t. x, we get } \\ x'' = -a \cos \theta, y'' = -b \sin \theta \\ \text{ Radius of Curvature Formula } \\ \rho = \frac{(x'^2 + y'^2)^{\frac{3}{2}}}{x' y'' - y' x''} \\ = \frac{(a^2 \sin^2 \theta + b^2 \cos^2 \theta)^{\frac{3}{2}}}{(-a \sin \theta)(-b \sin \theta) - (b \cos \theta)(-a \cos \theta)} \\ = \frac{(a^2 \sin^2 \theta + b^2 \cos^2 \theta)^{\frac{3}{2}}}{ab (\sin^2 \theta + \cos^2 \theta)} \\ \Rightarrow \rho = \frac{(a^2 \sin^2 \theta + b^2 \cos^2 \theta)^{\frac{3}{2}}}{ab} \\ \text{ Equation of tangent on ellipse at } (a \cos \theta, b \sin \theta) \\ \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 \\ \text{ Perpendicular on tangent from centre (0, 0) of ellipse } \\ p = \frac{1}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} \\ \Rightarrow p = \frac{ab}{\sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}} \\ \Rightarrow (a^2 \sin^2 \theta + b^2 \cos^2 \theta)^{\frac{3}{2}} = \frac{a^3 b^3}{p^3} \dots (3) \\ \text{ From equation (2) and (3), we get } \\ \rho = \frac{a^3 b^3}{p^3} \times \frac{1}{ab} \\ \Rightarrow \rho = \frac{a^2 b^2}{p^3}
Example:4.If CP,CD be a pair of conjugate semi-diameters of an ellipse,prove that the radius of curvature \rho is \frac{CD^3}{ab} a and b being the lengths of the semi-axes of the ellipse.
[Raj. 69,78]
Solution: \text{ Parametric equation of ellipse } \\ x = a \cos \theta, y = b \sin \theta \\ \text{ Then Radius of curvature by question no. 3, equation (2) } \\ \rho = \frac{(a^2 \sin^2 \theta + b^2 \cos^2 \theta)^{\frac{3}{2}}}{ab} \dots (1) \\  \text{ We know that by coordinate geometry if CP and CD are } \\ \text{conjugate diameter of ellipse then their difference eccentric angles are} \frac{\pi}{2} \\ \text{ So coordinate of D will be } \left[a \cos\left(\frac{\pi}{2}+\theta\right), b \sin\left(\frac{\pi}{2}+\theta\right)\right] \\ or (-a \sin \theta, b \cos \theta). \\ CD^2 = (-a \sin \theta - 0)^2 + (b \cos \theta - 0)^2 \\ \Rightarrow CD^2 = a^2 \sin^2 \theta + b^2 \cos^2 \theta \dots (2) \\ \text{ From equation (1) and (2), we get } \\ \rho = \frac{(CD^2)^{\frac{3}{2}}}{ab} \\ \Rightarrow \rho = \frac{CD^3}{ab}

Radius of Curvature Cartesian Form-B.sc. Math Solutions

Example:5. In the curve y = ae^{\frac{x}{a}}, prove that \rho = a \sec^2 \theta \cosec \theta \text{ where } \theta = \tan^{-1}\left(\frac{y}{a}\right).
Solution: y = ae^{\frac{x}{a}} \\ \text{ Differentiating w.r.t. x, we get } \\ \frac{dy}{dx} = ae^{\frac{x}{a}} \cdot \frac{1}{a} = e^{\frac{x}{a}} \\ \text{ Again differentiating w.r.t. x, we get } \\ \frac{d^2y}{dx^2} = \frac{1}{a} e^{\frac{x}{a}} \\ \text{Radius of Curvature Formula } \rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} \\ = \frac{\left[1 + e^{\frac{2x}{a}}\right]^{\frac{3}{2}}}{\frac{1}{a} e^{\frac{x}{a}}} \\ = \frac{a \left[1 + e^{\frac{2x}{a}}\right]^{\frac{3}{2}}}{e^{\frac{x}{a}}} \\ = \frac{a \left[1 + \frac{y^2}{a^2}\right]^{\frac{3}{2}}}{\frac{y}{a}} \quad \left[\because y = ae^{\frac{x}{a}}\right] \\ \text{ put } \frac{y}{a} = \tan \theta \\ = \frac{a \left[1 + \tan^2 \theta\right]^{\frac{3}{2}}}{\tan \theta} \\ = \frac{a \left(\sec^2 \theta\right)^{\frac{3}{2}}}{\tan \theta} \\ = a \sec^3 \theta \cdot \cot \theta \\ = a \sec^3 \theta \cdot \frac{\cos \theta}{\sin \theta} \\ \Rightarrow \rho = a \sec^2 \theta \cosec \theta
Example:6. For the parabola y^2 = 4ax, prove that \rho = \frac{2(SP)^{\frac{3}{2}}}{\sqrt{a}} where \rho denotes the radius of curvature and SP is the focal distance of the point P(x_1, y_1) on the parabola.
Solution:\text{ Equation of parabola } y^2 = 4ax \\ \text{ Let } P(at^2, 2at) \text{ be any point on parabola then parametric equation of parabola. } \\ x = at^2, y = 2at. \\ \text{ Differentiating w.r.t. t, we get } \\ x' = 2at, y' = 2a \\ \text{ Again differentiating w.r.t. t, we get } \\ x'' = 2a, y'' = 0 \\ \text{ Radius of curvature formula} \\ \rho = \frac{(x'^2 + y'^2)^{\frac{3}{2}}}{x' y'' - y' x''} \\ = \frac{\left[(2at)^2 + (2a)^2\right]^{\frac{3}{2}}}{2at \times 0 - 2a \times 2a} \\ = \frac{(4a^2t^2 + 4a^2)^{\frac{3}{2}}}{-4a^2} \\ = \frac{(4a^2)^{\frac{3}{2}} (1 + t^2)^{\frac{3}{2}}}{-4a^2} \\ \Rightarrow \rho = 2a (1 + t^2)^{\frac{3}{2}} \quad [\text{Numerically}] \dots (1) \\ \text{ We know that by coordinate geometry } \\ \text{ SP =Distance between P and focus S = a + x } \\ \Rightarrow SP = a + at^2 = a(1 + t^2) \cdots (2) \\ \text{ From (1) and (2) } \\ \Rightarrow \rho = 2a \left(\frac{SP}{a}\right)^{\frac{3}{2}} \\ \Rightarrow \rho = \frac{2(SP)^{\frac{3}{2}}}{\sqrt{a}}
Example:7. Show that the radius of curvature for the curve y = a\cosh\left(\frac{x}{a}\right) is equal to the portion of the normal intercepted between the curve and the x-axis and that it varies as the square of the ordinate.
Solution: \text{ The curve is } y = a\cosh\left(\frac{x}{a}\right) \cdots (1) \\ \text{ Differentiating w.r.t. x, we get } \\ \frac{dy}{dx} = a\sinh\left(\frac{x}{a}\right) \cdot \frac{1}{a} = \sinh\left(\frac{x}{a}\right) \\ \Rightarrow \frac{dy}{dx} = \sinh\left(\frac{x}{a}\right) \\ \text{ Again differentiating w.r.t. x, we get } \frac{d^2y}{dx^2} = \frac{1}{a} \cosh\left(\frac{x}{a}\right) \\ \text{ Radius of curvature Formula } \\ \rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} \\ = \frac{\left[1 + \sinh^2\left(\frac{x}{a}\right)\right]^{\frac{3}{2}}}{\frac{1}{a} \cosh\left(\frac{x}{a}\right)} \\ = \frac{a \left[\cosh^2\left(\frac{x}{a}\right)\right]^{\frac{3}{2}}}{\cosh\left(\frac{x}{a}\right)} \\ = \frac{a \cosh^3\left(\frac{x}{a}\right)}{\cosh\left(\frac{x}{a}\right)} = a \cosh^2\left(\frac{x}{a}\right) \\ \Rightarrow \rho = a \cosh^2\left(\frac{x}{a}\right) \cdots (2) \\ \text{ Length of Normal } = y \sec\psi \\ = y \sqrt{1 + \tan^2\psi} \\ = y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \\ = y \sqrt{1 + \sinh^2\left(\frac{x}{a}\right)} \\ = y \sqrt{\cosh^2\left(\frac{x}{a}\right)} \\ = y \cosh\left(\frac{x}{a}\right) \\ = a \cosh^2\left(\frac{x}{a}\right) \cdots (3) \\ \text{ from (2) and (3) } \\ \rho = \text{Normal between curve and x-axis} \\ \rho \propto y^2
Example:8.If \rho_1, \rho_2 be the radii of curvature at the extremities of a focal chord of a parabola whose latus rectum is l, prove that
(\rho_1)^{-\frac{2}{3}} + (\rho_2)^{-\frac{2}{3}} = (l)^{-\frac{2}{3}}
Solution: \text{ Let the coordinates of P and Q, and extremities } \\ \text{ of a focal chord of a focal chord be } (at_1^2, 2at_1) \text{ and } (at_2^2, 2at_2) \text{ Then } t_1 t_2 = -1 \text{ (By coordinate geometry) } \cdots (1) \\ \text{ Also the parametric equations of the parabola } y^2 = 4ax \text{ are } \\ x = at^2, y = 2at \\ \because x' = 2at, y' = 2a \text{ and } x'' = 2a, y'' = 0 \\ \because \rho \text{ at } at^2 = \frac{(x'^2 + y'^2)^{\frac{3}{2}}}{x' y'' - y' x''} \\ = \frac{[4a^2t^2 + 4a^2]^{\frac{3}{2}}}{2at \cdot 0 - 2a \cdot 2a} = \frac{8a^3(1 + t^2)^{\frac{3}{2}}}{-4a^2} \\ \Rightarrow \rho = 2a(1 + t^2)^{\frac{3}{2}} \text{ (Numerically)} \\ \therefore \rho_1 = \text{radius of curvature at } (at_1^2, 2at_1) \\ = 2a(1 + t_1^2)^{\frac{3}{2}} \\ \text{ Similarly } \rho_2 = 2a(1 + t_2^2)^{\frac{3}{2}} \\ \therefore (\rho_1)^{-\frac{2}{3}} + (\rho_2)^{-\frac{2}{3}} = (2a)^{-\frac{2}{3}} (1 + t_1^2)^{-1} + (2a)^{-\frac{2}{3}} (1 + t_2^2)^{-1} \\ = (2a)^{-\frac{2}{3}} \left[\frac{1}{1 + t_1^2} + \frac{1}{1 + t_2^2}\right] \\ = (2a)^{-\frac{2}{3}} \left[\frac{t_1^2 + t_2^2 + 2}{t_1^2 + t_2^2 + t_1^2 t_2^2 + 1}\right] \\ = (2a)^{-\frac{2}{3}} \left[\frac{t_1^2 + t_2^2 + 2}{t_1^2 + t_2^2 + 2}\right] \quad [\text{from (1)}] \\ = (2a)^{-\frac{2}{3}} \\ \Rightarrow (\rho_1)^{-\frac{2}{3}} + (\rho_2)^{-\frac{2}{3}} = (l)^{-\frac{2}{3}} \quad [\because l = 2a]
Example : 9. In the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, show that the radius of curvature at an end of the major axis is equal to the semi-latus rectum of the ellipse.
Solution: \text{ The equation of ellipse } \\ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \cdots (1) \\ \text{ Differentiating w.r.t. x,we get }, \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \\ \Rightarrow \frac{dy}{dx} = -\frac{b^2 x}{a^2 y} \\ \text{ Again differentiating w.r.t. x,we get } \\ \therefore \frac{d^2y}{dx^2} = -\frac{b^2}{a^2} \left[\frac{y \cdot 1 - x \frac{dy}{dx}}{y^2}\right] \\ = -\frac{b^2}{a^2 y^2} \left[y - x \left(-\frac{b^2 x}{a^2 y}\right)\right] \\ = -\frac{b^2}{a^2 y^2} \left(\frac{a^2 y^2 + b^2 x^2}{a^2 y}\right) = -\frac{b^4}{a^4 y^3} \left[\frac{x^2}{a^2} + \frac{y^2}{b^2}\right] \\ \frac{d^2y}{dx^2} = -\frac{b^4}{a^4 y^3} \cdot 1 \quad [\text{from (1)}] \\ \rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} = \frac{\left[1 + \left(-\frac{b^2 x}{a^2 y}\right)^2\right]^{\frac{3}{2}}}{\frac{b^4}{a^4 y^3}} \\ = \frac{(a^4 y^2 + b^4 x^2)^{\frac{3}{2}}}{a^4 b^4} \quad [\text{(neglecting -ve sign)}] \\ \text{ Now the coordinates of one end of major axis are (a, 0), so }\\ \rho \text{ at } (a, 0) = \left[\frac{a^4 \cdot 0^2 + b^4 a^2}{a^4 b^4}\right]^{\frac{3}{2}} \\ = \frac{b^6 a^3}{a^4 b^4} = \frac{b^2}{a} \\ \Rightarrow \rho \text{ at } (a, 0) = \text{semi-latus rectum of the ellipse} \quad (1)
With the above illustrations,one can understand the Radius of Curvature Cartesian Form-B.sc. Math Solutions.

5.Common Pitfalls/Mistakes to Avoid During Exam

Radius of Curvature Cartesian Form-B.sc. Math Solutions

Mistake:1.Sign (+/-) errors while calculating \frac{dy}{dx}
Mistake:2.Negative Radius:
\rho(Radius of Curvature) always represents a magnitude (length);therefore,if the result is negative,take the modulus |.| to express it as a positive value.

6.Cartesian vs Parametric Comparison Table

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7.Frequently Asked Questions Related to Radius of Curvature Cartesian Form-B.sc. Math Solutions

**छात्र-छात्राओं से आज का प्रश्न**

*”एक स्त्री का परिचय देते हुए एक पुरुष ने कहा ‘इसके पिता का एकमात्र पुत्र मेरा पिता है’। वह पुरुष उस स्त्री से कैसे सम्बन्धित है?”*
**Today’s Question to Students**
*”Introducing a woman,a man said,’The only son of her father is my father’.How is the man related to the woman?”*
*पिछली प्रश्नोत्तरी का उत्तर*(D)
\bigcirc \square A \rightarrow  बच्चे टेलिविजन देखें
\triangle ? \& \rightarrow  टेलीविजन और कार्टून
\square \times + \rightarrow   बच्चे खेल खेलें
(1) और (3) सेः
बच्चें \rightarrow \square
अतः तब (3) से ज्ञात करना कठिन हो जाता है कि ‘खेल’ के लिए ‘\times‘ और ‘+’ में से किसका प्रयोग किया गया है।
*Previous Quiz Solution*(D)
\bigcirc \square A \rightarrow  Watch Kids Television
\triangle ? \& \rightarrow  Television and cartoons
\square \times + \rightarrow  Play Kids Games
From (1) and (3):
Children \rightarrow \square
Therefore,then from (3) it becomes difficult to find out which of the ‘\times‘ and ‘+’ is used for ‘game’.
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*