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NCERT Exemplar · Class 12 Mathematics Application of Integrals

34 questions · 34 still being checked

EXERCISE 8.3 31–34 (part 4 of 4)

  1. Choose the correct answer from the given four options in each of the Exercises $\displaystyle 24$ to 34.

    Exercise 31

    The area of the region bounded by the ellipse x225+y216=1\displaystyle \frac{x^2}{25}+\frac{y^2}{16}=1 is
    (A)
    20π\displaystyle 20 \pi sq units
    (B)
    20π2\displaystyle 20 \pi^2 sq units
    (C)
    16π2\displaystyle 16 \pi^2 sq units
    (D)
    25π\displaystyle 25 \pi sq units

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    NCERT’s answer
    A
    (A) \(\displaystyle 20\pi\) sq units. The ellipse is symmetric about both axes, so \(\displaystyle A = 4\times\) (first-quadrant area).NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q31\[y = \frac{4}{5}\sqrt{25 - x^2} \quad (y \ge 0) \]\[A = 4\int_0^5 \frac{4}{5}\sqrt{25 - x^2}\,dx = \frac{16}{5}\int_0^5 \sqrt{5^2 - x^2}\,dx \]\[= \frac{16}{5}\left[\frac{x}{2}\sqrt{25 - x^2} + \frac{25}{2}\sin^{-1}\frac{x}{5}\right]_0^5 \]\[= \frac{16}{5}\cdot\frac{25}{2}\cdot\frac{\pi}{2} = 20\pi \]

  2. Exercise 32

    The area of the region bounded by the circle x2+y2=1\displaystyle x^2+y^2=1 is
    (A)
    2π\displaystyle 2 \pi sq units
    (B)
    π\displaystyle \pi sq units
    (C)
    3π\displaystyle 3 \pi sq units
    (D)
    4π\displaystyle 4 \pi sq units

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    NCERT’s answer
    B
    (B) \(\displaystyle \pi\) sq units. The circle is symmetric about both axes, so \(\displaystyle A = 4\times\) (first-quadrant area).NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q32\[A = 4\int_0^1 \sqrt{1 - x^2}\,dx \]\[= 4\left[\frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}x\right]_0^1 \]\[= 4\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \pi \]

  3. Exercise 33

    The area of the region bounded by the curve y=x+1\displaystyle y=x+1 and the lines x=2\displaystyle x=2 and x=3\displaystyle x=3 is
    (A)
    72\displaystyle \frac{7}{2} sq units
    (B)
    92\displaystyle \frac{9}{2} sq units
    (C)
    112\displaystyle \frac{11}{2} sq units
    (D)
    132\displaystyle \frac{13}{2} sq units

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    NCERT’s answer
    A
    (A) \(\displaystyle \frac{7}{2}\) sq units. The line lies above the \(\displaystyle x\)-axis for \(\displaystyle 2\le x\le 3\), so the area is the plain integral.NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q33\[A=\int_2^3 (x+1)\,dx \] \[=\left[\frac{x^2}{2}+x\right]_2^3 \] \[=\left(\frac{9}{2}+3\right)-(2+2)=\frac{7}{2} \]

  4. Exercise 34

    The area of the region bounded by the curve x=2y+3\displaystyle x=2 y+3 and the y\displaystyle y lines. y=1\displaystyle y=1 and y=−1\displaystyle y=-1 is
    (A)
    4\displaystyle 4 sq units
    (B)
    32\displaystyle \frac{3}{2} sq units
    (C)
    6\displaystyle 6 sq units
    (D)
    8\displaystyle 8 sq units

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    NCERT’s answer
    C
    (C) \(\displaystyle 6\) sq units. The region lies between the curve and the \(\displaystyle y\)-axis, since \(\displaystyle x = 2y + 3 > 0\) for \(\displaystyle -1 \le y \le 1\):NCERT_Solution_Class12_Maths_Exemplar_Ch8_Ex8-3_Q34\[A = \int_{-1}^{1} x\,dy = \int_{-1}^{1} (2y+3)\,dy \]\[= \left[y^2 + 3y\right]_{-1}^{1} \]\[= (1 + 3) - (1 - 3) = 6 \]