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Day 246: JEE Mains Maths

5 questions. Try each one first, then open its answer.

  1. Question 1

    Application of Integrals. Asked once: 7 Apr 2025 (same type asked 5 times) See all 5 of this type

    66
    If the area of the region bounded by the curves y=4−x24y= 4 -\frac{x^{2}}{4} and y=x−42y=\frac{x- 4}{2} is equal to α\alpha, then 6α6\alpha equals.

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  2. Question 2

    Limits and Continuity. Asked once: 28 Jun 2022 (same type asked 5 times) See all 5 of this type

    156
    Let f,g:R→Rf,g:R \rightarrow R be functions defined by f(x)={[x],x<0∣1−x∣,x≥0f(x) =\left\{ \begin{matrix} \lbrack x\rbrack, & x < 0 \\ |1 - x|, & x \geq 0 \end{matrix} \right. and
    g(x)={ex−xx<0(x−1)2−1,x≥0g(x) =\left\{ \begin{matrix} e^{x}- x & x < 0 \\ (x - 1)^{2}- 1, & x \geq 0 \end{matrix} \right.
    where [x]\lbrack x\rbrack denote the greatest integer less than or equal to xx. Then, the function fog is discontinuous at exactly:

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  3. Question 3

    Relations and Functions. Asked once: 24 Jan 2025 (same type asked 5 times) See all 5 of this type

    127
    The function f:(−∞,∞)→(−∞,1)f:( - \infty,\infty) \rightarrow ( - \infty,1), defined by f(x)=2x−2−x2x+2−xf(x) =\frac{2^{x}-2^{-x}}{2^{x}+2^{-x}} is :

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  4. Question 4

    Three Dimensional Geometry. Asked once: 23 Jan 2025 (same type asked 5 times) See all 5 of this type

    59
    Let P be the foot of the perpendicular from the point Q(10,−3,−1)Q(10, - 3, - 1) on the line x−37=y−2−1=z+1−2\frac{x - 3}{7}=\frac{y - 2}{- 1}=\frac{z + 1}{- 2}. Then the area of the right angled triangle PQRPQR, where RR is the point (3,−2,1)(3, - 2,1), is

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  5. Question 5

    Application of Integrals. Asked once: 30 Jan 2024 (same type asked 5 times) See all 5 of this type

    68
    The area (in square units) of the region bounded by the parabola y2=4(x−2)y^{2}= 4(x- 2) and the line y=2x−8y= 2x- 8 is

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