If A is the area in the first quadrant enclosed by the curve \(C: 2x^2 – y + 1 = 0\), the tangent to C at the point (1, 3) and the line x + y = 1, then the value of 60A is ________
Correct Answer: 16
Solution and Explanation
Shaded Area Calculation
The equation of the curve is given as:
\( y = 2x^2 + 1 \)
The tangent at the point \( P(1, 3) \) is:
\( y = 4x - 1 \)
Step 1: Solve the Integral
We compute the integral of the curve from \( x = 0 \) to \( x = 1 \):
\[ \int_0^1 (2x^2 + 1) \, dx = \left[ \frac{2x^3}{3} + x \right]_0^1 \]
Evaluating the integral:
\[ = \left( \frac{2(1)^3}{3} + (1) \right) - \left( \frac{2(0)^3}{3} + 0 \right) = \frac{2}{3} + 1 = \frac{5}{3} \]
Step 2: Compute the Areas of the Triangles
1. Area of Triangle \( \triangle QOT \)
The area of \( \triangle QOT \) is:
\[ \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} \]
2. Area of Triangle \( \triangle PQR \)
The vertices of \( \triangle PQR \) are \( P(1,3), Q(1,0), R\left( \frac{1}{4}, 0 \right) \).
We compute the area using the formula for the area of a triangle with given vertices:
\[ \text{Area} = \frac{1}{2} \left| 1(0-0) + 1(0-3) + \frac{1}{4}(3-0) \right| = \frac{1}{2} \left| 0 - 3 + \frac{3}{4} \right| = \frac{1}{2} \cdot \frac{9}{4} = \frac{9}{8} \]
3. Area of Triangle \( \triangle QRS \)
The vertices of \( \triangle QRS \) are \( Q(1,0), R\left( \frac{1}{4}, 0 \right), S\left( \frac{2}{5}, \frac{13}{5} \right) \).
We compute the area similarly:
\[ \text{Area} = \frac{1}{2} \left| 1(0 - \frac{13}{5}) + \frac{1}{4}(\frac{13}{5} - 0) + \frac{2}{5}(0 - 0) \right| = \frac{1}{2} \left| -\frac{13}{5} + \frac{13}{20} \right| \]
Simplifying the expression:
\[ = \frac{1}{2} \cdot \frac{39}{40} = \frac{39}{40} \]
Step 3: Combine the Results
Now, we combine all the computed areas:
\[ A = \frac{5}{3} - \frac{1}{2} - \frac{9}{8} + \frac{9}{40} \]
To simplify, we find the common denominator (120):
\[ A = \frac{200}{120} - \frac{60}{120} - \frac{135}{120} + \frac{27}{120} = \frac{200 - 60 - 135 + 27}{120} = \frac{32}{120} = \frac{8}{30} = \frac{16}{60} \]
Final Answer
The shaded area is \( A = \frac{16}{60} \).
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Tangents and Normals Questions
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is:
- A line passing through the point $ A(-2, 0) $, touches the parabola $ P: y^2 = x - 2 $ at the point $ B $ in the first quadrant. The area of the region bounded by the line $ AB $, parabola $ P $, and the x-axis is:
Let a circle $C_1$ be obtained on rolling the circle $x^2+y^2-4 x-6 y+11=0$ upwards 4 units on the tangent $T$ to it at the point $(3,2)$ Let $C_2$ be the image of $C_1$ in $T$ Let $A$ and $B$ be the centers of circles $C_1$ and $C_2$ respectively, and $M$ and $N$ be respectively the feet of perpendiculars drawn from $A$ and $B$ on the $x$-axis. Then the area of the trapezium AMNB is:
- Let a tangent to the curve $9 x^2+16 y^2=144$ intersect the coordinate axes at the points $A$ and $B$. Then, the minimum length of the line segment $AB$ is
- The tangents at the point $A (1,3)$ and $B (1,-1)$ on the parabola $y^2-2 x-2 y=1$ meet at the point $P$ Then the area (in unit $^2$ ) of $\triangle PAB$ is :-
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.