JEE MAIN 2026 Previous Year Questions (PYQs) – Page 21 of 25
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 242 questions
Let $ f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}} $ be such that $ f(0) = -26 + 24\log_e 2 $. If $ f(1) = a + b\log_e 3 $, where $ a, b \in \mathbb{Z} $, then $ a + b $ is equal to
1
2
3
4
Solution
$ f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}} $
Put $ x = t^6 \Rightarrow dx = 6t^5 dt $
$ = \int \frac{6t^5 dt}{t^4 + 2t^3} = 6\int \frac{t^2 - 4}{t + 2} dt $
$ = 6\left[ \frac{t^2}{2} - 2t + 4\ln(t + 2) \right] + C $
$ = 3x^{1/3} - 12x^{1/6} + 24\ln(x^{1/6} + 2) + C $
$ f(0) = 24\ln 2 + C = -26 + 24\ln 2 \Rightarrow C = -26 $
Now
$ f(1) = -35 + 24\ln 3 = a + b\ln 3 $
$ \Rightarrow a = -35,; b = 24 $
$ \Rightarrow a + b = -11 $
$ \frac{6}{3^{26}} + \frac{10.1}{3^{25}} + \frac{10.2}{3^{24}} + \frac{10.2^2}{3^{23}} + \ldots + \frac{10.2^{24}}{3} $ is equal to
1
2
3
4
Solution
$ S = \frac{6}{3^{26}} + \frac{10}{3^{25}}\left[ \frac{(6)^{25} - 1}{6 - 1} \right] $
$ S = \frac{6}{3^{26}} + \frac{10}{3^{25}} \cdot \frac{6^{25} - 1}{5} $
$ S = \frac{2}{3^{25}} + 2\left[ 2^{25} - \frac{1}{3^{25}} \right] $
$ S = 2^{26} $
$ \sum_{r=1}^{25} \frac{r}{r^2 + r + 1} = \frac{p}{q} $, where $ p $ and $ q $ are integers such that $ \gcd(p, q) = 1 $, then $ p + q $ is equal to
1
2
3
4
Solution
$ S = \sum \frac{r}{(r^2 + r + 1)} $
$ = \frac{1}{2} \sum \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) $
$ = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \ldots + \left(\frac{1}{601} - \frac{1}{651}\right) \right] $
$ = \frac{1}{2}\left(1 - \frac{1}{651}\right) = \frac{325}{651} $
$ \Rightarrow p = 325,; q = 651 \Rightarrow p + q = 976 $
Three persons enter in a lift at the ground floor. The lift will go upto $ 10^{th} $ floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to ______
1
2
3
4
Solution
$ S = \sum \frac{r}{(r^2 + r + 1)} $
$ = \frac{1}{2} \sum \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) $
$ = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \ldots + \left(\frac{1}{601} - \frac{1}{651}\right) \right] $
$ = \frac{1}{2}\left(1 - \frac{1}{651}\right) = \frac{325}{651} $
$ \Rightarrow p = 325,; q = 651 \Rightarrow p + q = 976 $
Let $ f $ be a differentiable function satisfying
$ f(x) = 1 - 2x + e^{-x} \int_0^x e^t f(t),dt,; x \in \mathbb{R} $
and let
$ g(x) = \int_0^x (f(t) + 2)^5 (t - 4)^6 (t + 12)^{17} dt,; x \in \mathbb{R} $.
If $ p $ and $ q $ are respectively the points of local minima and local maxima of $ g $, then the value of $ p + q $ is equal to ______
1
2
3
4
Solution
$ f'(x) - 2f(x) = 2x - 3 $
$ \frac{dy}{dx} - 2y = 2x - 3 $
$ y e^{-2x} = \int e^{-2x}(2x - 3)dx $
On solving we get
$ f(x) = 1 - x $
$ g'(x) = (3 - x)^5 (x - 4)^6 (x + 12)^{17} $
Local maxima $ \Rightarrow q = 3 $
Local minima $ \Rightarrow p = -12 $
$ \Rightarrow p + q = 9 $
If the distance of the point $ P(43, \alpha, \beta) $ from the line
$ \vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k}),; \mu \in \mathbb{R} $ along a line with direction ratios $ 3, -1, 0 $ is $ 13\sqrt{10} $, then $ \alpha^2 + \beta^2 $ is equal to ______
1
2
3
4
Solution
$ \frac{x - 43}{3} = \frac{y - \alpha}{-1} = \frac{z - \beta}{0} $
$ \Rightarrow P(43 + 3\lambda,; \alpha - \lambda,; \beta) $
$ \Rightarrow \alpha = 13,; \beta = 1 $
$ \Rightarrow \alpha^2 + \beta^2 = 170 $
Let $ A = \left[ \matrix{ 3 & -4 \cr 1 & -1 } \right] $ and $ B $ be two matrices such that
$ A^{100} = 100B + I $. Then the sum of all the elements of $ B^{100} $ is ______
1
2
3
4
Solution
$ A = I + \left[ \matrix{ 2 & -4 \cr 1 & -2 } \right] $, let $ M = \left[ \matrix{ 2 & -4 \cr 1 & -2 } \right] $
$ M^2 = \left[ \matrix{ 0 & 0 \cr 0 & 0 } \right] \Rightarrow M^3 = M^4 = \ldots = M^{100} $
$ A^{100} = (I + M)^{100} = I + 100M $
$ A^{100} = I + 100B \Rightarrow M = B $
$ \Rightarrow B^{100} = \left[ \matrix{ 0 & 0 \cr 0 & 0 } \right] $
Sum of all elements $ = 0 $
If the domain of the function
$ f(x) = \cos^{-1}\left(\frac{2x-5}{11-3x}\right) + \sin^{-1}(2x^2 - 3x + 1) $
is the interval $[\alpha, \beta]$, then $ \alpha + 2\beta $ is equal to:
1
2
3
4
Solution
$ f(x) = \cos^{-1}!\left(\frac{2x-5}{11-3x}\right) + \sin^{-1}(2x^2 - 3x + 1) $
For $\cos^{-1}$:
$ -1 \le \frac{2x-5}{11-3x} \le 1 $
For $\sin^{-1}$:
$ -1 \le 2x^2 - 3x + 1 \le 1 $
First solve:
$ -1 \le 2x^2 - 3x + 1 \le 1 $
$ -1 \le 2x^2 - 3x + 1 \Rightarrow 2x^2 - 3x + 2 \ge 0 $
$ 2x^2 - 3x \le 0 $
So,
$ x \in [0, \frac{3}{2}] \quad ...(i) $
Now solve:
$ -1 \le \frac{2x-5}{11-3x} \le 1 $
First part:
$ \frac{2x-5}{11-3x} + 1 \ge 0 $
$ \frac{2x-5 + 11 - 3x}{11-3x} \ge 0 $
$ \frac{6 - x}{11-3x} \ge 0 $
Second part:
$ \frac{2x-5}{11-3x} - 1 \le 0 $
$ \frac{2x-5 - (11-3x)}{11-3x} \le 0 $
$ \frac{5x - 16}{11-3x} \le 0 $
From sign analysis:
$ x \in (-\infty, \frac{16}{5}] \cup (\frac{11}{3}, \infty) $
and
$ x \in (-\infty, \frac{11}{3}) \cup [6, \infty) $
Intersection:
$ x \in (-\infty, \frac{16}{5}] \cup [6, \infty) \quad ...(ii) $
Final intersection of (i) and (ii):
$ x \in [0, \frac{3}{2}] $
So,
$ \alpha = 0,\ \beta = \frac{3}{2} $
$ \alpha + 2\beta = 0 + 2 \cdot \frac{3}{2} = 3 $
The area of the region, inside the ellipse $x^2 + 4y^2 = 4$ and outside the region bounded by the curves $y = |x| - 1$ and $y = 1 - |x|$, is:
1
2
3
4
Solution
The number of relations, defined on the set ${a,b,c,d}$, which are both reflexive and symmetric, is equal to:
1
2
3
4
Solution
Set has 4 elements → total pairs = $4 \times 4 = 16$
Reflexive condition:
All $(a,a), (b,b), (c,c), (d,d)$ must be present → fixed
Remaining pairs = $16 - 4 = 12$
Symmetric condition:
Pairs form symmetric pairs like:
$(a,b)$ with $(b,a)$ → 1 choice pair
Total such independent pairs = $\frac{12}{2} = 6$
Each pair → 2 choices (present or not)
Total relations:
$ = 2^6 = 64 $
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
