Available Questions 601 found Page 13 of 31
Standalone Questions
#1247
Mathematics
Definite Integrals
VSA
UNDERSTAND
2024
AISSCE(Board Exam)
KNOWLEDGE
2 Marks
Evaluate: $\int_{0}^{\pi/2}sin~2x~cos~3x~dx$
Key:
Sol:
Sol:
#1246
Mathematics
Continuity and Differentiability
VSA
REMEMBER
2024
AISSCE(Board Exam)
KNOWLEDGE
2 Marks
Check the differentiability of $f(x)=\begin{cases}x^{2}+1,&0\le x<1\\ 3-x,&1\le x\le2\end{cases}$ at $x=1.$
Key:
Sol:
Sol:
#1245
Mathematics
Derivatives
VSA
UNDERSTAND
2024
AISSCE(Board Exam)
KNOWLEDGE
2 Marks
If $x=e^{x/y}$, prove that $\frac{dy}{dx}=\frac{log~x-1}{(log~x)^{2}}$
Key:
Sol:
Sol:
#1244
Mathematics
Inverse Trigonometric Functions
VSA
APPLY
2024
AISSCE(Board Exam)
KNOWLEDGE
2 Marks
Evaluate : $sec^{2}(tan^{-1}\frac{1}{2})+cosec^{2}(cot^{-1}\frac{1}{3})$
Key:
Sol:
Sol:
#985
Physics
Nuclei
VSA
APPLY
2025
KNOWLEDGE
2 Marks
State two important properties of the nuclear force.
Key:
Sol:
Sol:
hkdcdj ds
#982
Physics
Alternating Current
graph plot and analyze
#977
Mathematics
Relations and Functions
ASSERTION_REASON
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
Assertion (A): Let $f(x) = e^{x}$ and $g(x) = \log x$. Then $(f + g)x = e^{x} + \log x$ where domain of $(f + g)$ is $\mathbb{R}$.
Reason (R): $\text{Dom}(f + g) = \text{Dom}(f) \cap \text{Dom}(g)$.
Key: D
Sol:
Sol:
#976
Mathematics
Vector Algebra
ASSERTION_REASON
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
Assertion (A) : If $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=256$ and $|\vec{b}|=8$, then $|\vec{a}|=2$.
Reason (R) : $\sin ^{2} \theta+\cos ^{2} \theta=1$ and $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$ and $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$.
Key: A
Sol:
Sol:
#975
Mathematics
Inverse Trigonometric Functions
ASSERTION_REASON
REMEMBER
2025
AISSCE(Board Exam)
Competency
1 Marks
Assertion (A) : Set of values of $\sec^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is a null set.
Reason (R) : $\sec^{-1}$ x is defined for $x \in \mathbb{R}-(-1, 1)$.
Key: A
Sol:
Sol:
#974
Mathematics
Continuity and Differentiability
ASSERTION_REASON
REMEMBER
2025
AISSCE(Board Exam)
KNOWLEDGE
1 Marks
Assertion (A): $f(x) = \begin{cases} x\sin\frac{1}{x}, & x\neq 0 \\ 0, & x=0 \end{cases}$ is continuous at $x=0$.
Reason (R): When $x \to 0$, $\sin\frac{1}{x}$ is a finite value between $-1$ and $1$.
Key:
Sol:
Sol:
#971
Mathematics
Linear Programming
MCQ_SINGLE
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
In a Linear Programming Problem (LPP), the objective function $Z = 2x + 5y$ is to be maximised under the following constraints :$x + y \leq 4$, $3x + 3y \geq 18$, $x, y \geq 0$
Study the graph and select the correct option.
(A) lies in the shaded unbounded region.
(B) lies in $\triangle AOB$.
(C) does not exist.
(D) lies in the combined region of $\triangle AOB$ and unbounded shaded region.
Key: C
Sol:
Sol:
The feasible region of a Linear Programming Problem is the set of points that satisfy all the given constraints simultaneously.Constraint 1 requires the points to be in the region where the sum of $x$ and $y$ is less than or equal to 4.Constraint 2 requires the points to be in the region where the sum of $x$ and $y$ is greater than or equal to 6.Mathematically, a number cannot be both $\le 4$ and $\ge 6$ at the same time. Visually, looking at the graph, there is a clear gap between the shaded region $\Delta AOB$ and the shaded unbounded region above $PQ$. The two regions do not overlap.3. ConclusionSince there is no common region that satisfies all constraints, the feasible region is an empty set.Without a feasible region, there are no valid values for $x$ and $y$ to substitute into the objective function $Z$. Therefore, an optimal solution (maximum value) cannot be found.Answer:The correct option is (C) does not exist.
#970
Mathematics
Linear Programming
MCQ_SINGLE
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
For a Linear Programming Problem (LPP), the given objective function is $Z = x + 2y$. The feasible region PQRS determined by the set of constraints is shown as a shaded region in the graph.Which of the following statements is correct ?
(A) Z is minimum at $(\frac{18}{7}, \frac{2}{7})$
(B) Z is maximum at R$(\frac{7}{2}, \frac{3}{4})$
(C) (Value of Z at P) > (Value of Z at Q)
(D) (Value of Z at Q) < (Value of Z at R)
Key:
Sol:
Sol:
#969
Mathematics
Inverse Trigonometric Functions
MCQ_SINGLE
REMEMBER
2025
AISSCE(Board Exam)
KNOWLEDGE
1 Marks
The following graph is a combination of : (A) $y = \sin^{-1} x$ and $y = \cos^{-1} x$
(B) $y = \cos^{-1} x$ and $y = \cos x$
(C) $y = \sin^{-1} x$ and $y = \sin x$
(D) $y = \cos^{-1} x$ and $y = \sin x$
Key: C
Sol:
Sol:
#968
Mathematics
Inverse Trigonometric Functions
MCQ_SINGLE
ANALYZE
2025
AISSCE(Board Exam)
Competency
1 Marks
The given graph illustrates : (A) $y = \tan^{-1}x$
(B) $y = \csc^{-1}x$
(C) $y = \cot^{-1}x$
(D) $y = \sec^{-1}x$
Key: A
Sol:
Sol:
#959
Mathematics
Relations and Functions
MCQ_SINGLE
APPLY
2025
JEE Main 2025
Competency
0 Marks
Let A = { (α, β) ∈ R x R : |α - 1| ≤ 4 and |β - 5| ≤ 6} and B = { (α, β) ∈ R × R: 16(α-2)²+9(β-6)² ≤ 144}. Then
(A) A ⊂ B
(B) B ⊂ A
(C) neither A ⊂ B nor B ⊂ A
(D) A ∪ B = {(x, y) : -4 ≤ x ≤ 4, -1 ≤ y ≤ 11}
Key: B
Sol:
Sol:
#945
Mathematics
Inverse Trigonometric Functions
VSA
APPLY
2024
KNOWLEDGE
2 Marks
Express \(\tan^{-1}(\frac{\cos~x}{1-\sin~x})\) where \(\frac{-\pi}{2}\lt x\lt \frac{\pi}{2}\) in the simplest form.
Key:
Sol:
Sol:
Let $y = \tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)$.First, use half-angle identities to rewrite the numerator and denominator:$$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)$$$$1 - \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2$$Substitute these back into the expression:$$\frac{\cos x}{1-\sin x} = \frac{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2}$$Cancel the common term $\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)$:$$= \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$$Divide the numerator and denominator by $\cos\frac{x}{2}$:$$= \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}$$Since $\tan\frac{\pi}{4} = 1$, we can rewrite this as:$$= \frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}}$$Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:$$= \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$$Therefore:$$y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right)$$Since $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, the angle lies within the principal range, so:$$y = \frac{\pi}{4} + \frac{x}{2}$$
#944
Mathematics
Inverse Trigonometric Functions
VSA
APPLY
2024
KNOWLEDGE
2 Marks
Find the value of \(\tan^{-1}(-\frac{1}{\sqrt{3}})+\cot^{-1}(\frac{1}{\sqrt{3}})+\tan^{-1}[\sin(-\frac{\pi}{2})].\)
Key:
Sol:
Sol:
Let the given expression be $E$.$$E = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right]$$Evaluate each term separately using principal values:$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$(Since $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\tan^{-1}(-x) = -\tan^{-1}x$)$\cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}$(Since $\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}$)For the third term, first evaluate the sine function:$\sin\left(-\frac{\pi}{2}\right) = -1$So, $\tan^{-1}(-1) = -\frac{\pi}{4}$Substitute these values back into the expression:$$E = -\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4}$$Find a common denominator (which is 12):$$E = \frac{-2\pi + 4\pi - 3\pi}{12}$$$$E = \frac{-\pi}{12}$$
#943
Mathematics
Inverse Trigonometric Functions
VSA
APPLY
2024
KNOWLEDGE
2 Marks
Find the domain of the function \(f(x)=\sin^{-1}(x^{2}-4).\) Also, find its range.
Key:
Sol:
Sol:
Domain
We need the argument of the arcsine to be in $[-1, 1]$:$$-1 \le x^2 - 4 \le 1$$Add 4 to all parts:$$3 \le x^2 \le 5$$Taking the square root, we get two intervals:$$\sqrt{3} \le |x| \le \sqrt{5}$$This splits into positive and negative regions:$$x \in [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$$
Range
From the inequality above, we know that as $x$ varies over the domain, the term $(x^2 - 4)$ covers the entire interval $[-1, 1]$.Therefore, $f(x)$ covers all possible values of $\sin^{-1}(u)$:$$y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$
#942
Mathematics
Inverse Trigonometric Functions
VSA
APPLY
2024
KNOWLEDGE
2 Marks
Find the principal value of \(\tan^{-1}(1)+\cos^{-1}(-\frac{1}{2})+\sin^{-1}(-\frac{1}{\sqrt{2}}).\)
Key:
Sol:
Sol:
Let the expression be $E$.$$E = \tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)$$
Evaluate each term using principal value branches:
$\tan^{-1}(1) = \frac{\pi}{4}$(Since $\tan\frac{\pi}{4} = 1$)
$\cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$(Since $\cos^{-1}(-x) = \pi - \cos^{-1}x$)
$\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}$(Since $\sin^{-1}(-x) = -\sin^{-1}x$)
Substitute these values back into $E$:$$E = \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{4}$$Cancel the $\frac{\pi}{4}$ terms:$$E = \frac{2\pi}{3}$$
#941
Mathematics
Inverse Trigonometric Functions
VSA
APPLY
2024
KNOWLEDGE
2 Marks
Evaluate: \(\sec^{2}(\tan^{-1}\frac{1}{2})+cosec^{2}(\cot^{-1}\frac{1}{3})\)
Key:
Sol:
Sol: