JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 14 of 275
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If $\displaystyle \int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2),dx = a\tan^{-1}(2) - b\log_e(5)$, where $a,b \in \mathbb{N}$, then $(2a + b)$ is equal to ______.
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Solution
Let $I = \int_{0}^{1} \cot^{-1}(1 - 2x + 4x^2),dx$
$I = \int_{0}^{1} \left(\cot^{-1}(2x-1) - \cot^{-1}(2x)\right),dx \quad ...(1)$
Applying King
$I = \int_{0}^{1} \left(-\cot^{-1}(2x-1) + \cot^{-1}(2x-2)\right),dx \quad ...(2)$
From (1) & (2)
$2I = \int_{0}^{1} \left(\cot^{-1}(2x-2) - \cot^{-1}(2x)\right),dx$
$= \int_{0}^{1} \cot^{-1}(2x-2),dx - \int_{0}^{1} \cot^{-1}(2x),dx$
Applying King
$= \int_{0}^{1} \cot^{-1}(-2x),dx - \int_{0}^{1} \cot^{-1}(2x),dx$
$= \int_{0}^{1} (\pi - \cot^{-1}(2x)),dx - \int_{0}^{1} \cot^{-1}(2x),dx$
$= \int_{0}^{1} \pi,dx - 2\int_{0}^{1} \cot^{-1}(2x),dx$
$= \pi - 2\int_{0}^{1} \cot^{-1}(2x),dx$
By parts
$I = \pi - 2\left[x\cot^{-1}(2x)\right]{0}^{1} + \int{0}^{1} \frac{2x}{1+4x^2},dx$
Let $1 + 4x^2 = t$
$8x,dx = dt$
$I = \pi - 2\cot^{-1}(2) + \frac{1}{4}\int_{1}^{5} \frac{dt}{t}$
$= \pi - 2\cot^{-1}(2) + \frac{1}{4}\ln 5$
$\Rightarrow 2I = 2\pi - 4\cot^{-1}(2) + \frac{1}{2}\ln 5$
Given $\int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2),dx = 4I$
$= 2\left[2\pi - 4\cot^{-1}(2) + \frac{1}{2}\ln 5\right]$
$= 4\pi - 8\cot^{-1}(2) + \ln 5$
$\Rightarrow 2a + b = 8 + 1 = 9$
Let m and M be respectively the minimum and maximum values of
\[
\left|
\begin{array}{ccc}
\cos^{2}x & 1+\sin^{2}x & \sin 2x\\
1+\cos^{2}x & \sin^{2}x & \sin 2x\\
\cos^{2}x & \sin^{2}x & 1+\sin 2x
\end{array}
\right|.
\]
Then the ordered pair (m, M) is equal to :
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Solution
Let $m$ and $M$ be respectively the minimum and maximum values of
\[
\left|\begin{matrix}
\cos^2 x & 1+\sin^2 x & \sin 2x\\
1+\cos^2 x & \sin^2 x & \sin 2x\\
\cos^2 x & \sin^2 x & 1+\sin 2x
\end{matrix}\right|.
\]
Then the ordered pair $(m, M)$ is equal to :
Solution:
Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$
\[
\Delta =
\left|\begin{matrix}
\cos^2 x & 1+\sin^2 x & \sin 2x\\
1 & -1 & 0\\
0 & -1 & 1
\end{matrix}\right|
\]
Expanding along the first row,
\[
\Delta = \cos^2 x(-1) - (1+\sin^2 x)(1) - \sin 2x(1)
\]
\[
\Delta = -(\cos^2 x + \sin^2 x + 1 + \sin 2x)
\]
\[
\Delta = -2 - \sin 2x
\]
Since $\sin 2x \in [-1,1]$,
\[
\Delta \in [-3, -1]
\]
Hence, $m = -3$, $M = -1$
Therefore, the ordered pair is $\boxed{(-3, -1)}$.
If the coefficients of x7 in ${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$ and x$-$7 in ${\left( {{x} - {1 \over {bx^2}}} \right)^{11}}$, b $\ne$ 0, are equal, then the value of b is equal to :
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Solution
Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^{2}-4x-6=0$
and the ordinates of $P$ and $Q$ be the roots of $y^{2}+2y-7=0$.
If $PQ$ is a diameter of the circle $x^{2}+y^{2}+2ax+2by+c=0$, then the value of $(a+b-c)$ is _________.
(A)
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Solution
Let X = ℝ × ℝ. Define a relation R on X by
(a₁,b₁) R (a₂,b₂) ⇔ b₁ = b₂.
Statement I: R is an equivalence relation.
Statement II: For some (a,b) ∈ X, the set S = { (x,y) ∈ X : (x,y) R (a,b) } represents a line parallel to y = x.
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Solution
The centre of a circle $C$ is at the centre of the ellipse $E:\ \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1,\ a>b$. Let $C$ pass through the foci $F_{1}$ and $F_{2}$ of $E$ such that the circle $C$ and the ellipse $E$ intersect at four points. Let $P$ be one of these four points. If the area of the triangle $PF_{1}F_{2}$ is $30$ and the length of the major axis of $E$ is $17$, then the distance between the foci of $E$ is
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Solution
If $\displaystyle \sum_{r=0}^{25} \left\{ {^{50}C_{r}} \cdot {^{\,50-r}C_{\,25-r}} \right\} = K \binom{50}{25}$, then $K$ is equal to :
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Solution
Let z $ \in $ C with Im(z) = 10 and it satisfies ${{2z - n} \over {2z + n}}$ = 2i - 1 for some natural number n. Then :
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Solution
Let $PS$ be the median of the triangle with vertices $P(2,2)$, $Q(6,-1)$ and
$R(7,3)$. The equation of the line passing through $(1,-1)$ and parallel to
$PS$ is :
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Solution
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