JEE MAIN 2026 Previous Year Questions (PYQs) – Page 19 of 25
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Solution
Solution
$ |PR| = 9 \Rightarrow a^2 + b^2 + 16 = 81 \Rightarrow a^2 + b^2 = 65 $
Solution
Solution
Solution
Coordinates of centroid of triangle $ ABC $ are
$ \frac{2}{3}(t_1^2 + t_2^2 + 1) = \frac{7}{3} \Rightarrow t_1^2 + t_2^2 = \frac{5}{2} $
$ \frac{4}{3}(t_1 + t_2) = \frac{4}{3} \Rightarrow t_1 + t_2 = 1 $
$ (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2 \Rightarrow t_1 t_2 = -\frac{3}{4} $
$ (t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1 t_2 = 4 $
$ (BC)^2 = 4(t_1 - t_2)^2 + 16(t_1 - t_2)^2 $
$ \Rightarrow (BC)^2 = 80 $
If $ E(X) = \frac{263}{15} $, then $ P(X < 20) $ is equal to :
Solution
$ P(X < 20) = \sum_{X=14}^{19} P(X) = \frac{11}{15} $
Solution
$ f(x) = \begin{cases} \cos \pi x, & x \to 1^- \ -\frac{\sin(x-1)}{(x-1)}, & x \to 1^+ \end{cases} $
RHL $ = \lim_{x \to 1} -\frac{\sin(x-1)}{(x-1)} = -1 $
LHL $ = \lim_{x \to 1} \cos \pi x = -1,; f(1) = -1 $
$ f(x) $ is continuous at $ x = 1 $
$ f(x) = \begin{cases} -\frac{\sin(x-1)}{(x-1)}, & x \to -1^- \ \cos \pi x, & x \to -1^+ \end{cases} $
RHL $ = \lim_{x \to -1} \cos \pi x = -1 $
LHL $ = \lim_{x \to -1} -\frac{\sin(x-1)}{(x-1)} = \frac{\sin 2}{2} $
$ f(x) $ is discontinuous at $ x = -1 $
Considering the principal values of inverse trigonometric functions, the value of the expression
$ \tan \left( 2\sin^{-1} \left( \frac{2}{\sqrt{13}} \right) - 2\cos^{-1} \left( \frac{3}{\sqrt{10}} \right) \right) $ is equal to :
Solution
Solution
Solution
JEE MAIN
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