If 0(01 (3x + 5)t dx)(1/t) = (alpha/5e)((8/5))(2/3), then alpha is equal to

Q68 jee_main_2025_24_jan_evening Number of Real Solutions of Equations

The number of real solution(s) of the equation

x²+3x+2= |x-3|, |x+2|

is:

A. $2$ $2$
B. $0$ $0$
C. $3$ $3$
D. $1$ $1$

Solution

Related Formula

The function $f(x), g(x)$ $\min\{f(x), g(x)\}$ chooses the lower vertical path between the two curves at any coordinate $x$ $x$.

Core Logic

Analyze the conditions for the right-hand function $|x-3|, |x+2|$ $\min\{|x-3|, |x+2|\}$:

The intersection of $|x-3| = |x+2|$ $|x-3| = |x+2|$ happens at $x - 3 = -(x + 2) \Rightarrow 2x = 1 \Rightarrow x = 0.5$ .
For $x \leq 0.5$ $x \le 0.5$, $|x+2| \leq |x-3| \Rightarrow = |x+2|$ .
For $x > 0.5$ $x > 0.5$, $|x-3| \leq |x+2| \Rightarrow = |x-3|$ .

Min function intersection graph for Q68 - JEE Main 2025 Evening

Step 1: Check Interval $x \leq -2$ $x \le -2$

Here, $|x+2| = -(x+2) = -x-2$ $|x+2| = -(x+2) = -x-2$:

x² + 3x + 2 = -x - 2 \Rightarrow x² + 4x + 4 = 0

(x+2)² = 0 \Rightarrow x = -2

This is a valid solution as it lies precisely within the interval condition boundary.

Step 2: Check Interval $-2 < x \leq 0.5$ $-2 < x \le 0.5$

Here, $|x+2| = x+2$ $|x+2| = x+2$:

x² + 3x + 2 = x + 2 \Rightarrow x² + 2x = 0

x(x+2) = 0 \Rightarrow x = 0 or x = -2

Only $x = 0$ $x = 0$ fits inside this interval.

Step 3: Check Interval $x > 0.5$ $x > 0.5$

Here, $= |x-3| = 3-x$ $\min = |x-3| = 3-x$:

x² + 3x + 2 = 3 - x \Rightarrow x² + 4x - 1 = 0

x = -4 \pm \sqrt(16 - 4(1)(-1))2 = -2 \pm \sqrt(5)

Evaluating values: $-2 + \sqrt(5) \approx 0.236$ $-2 + \sqrt{5} \approx 0.236$, which does not satisfy $x > 0.5$ $x > 0.5$. Thus, no real roots occur in this span.

Combining valid points, we find exactly 2 distinct real solutions ( $x = -2, 0$ $x = -2, 0$).

Pattern Recognition

Sketching a rough visualization showing the parabola crossing below the sharp wedge of the combined absolute values makes it visually clear that there are exactly two crossing points, confirming the algebraic count.

Chapter Mix

Class 11 Mathematics: Quadratic Equations Class 12 Mathematics: Limits, Continuity and Differentiability

Q57 jee_main_2025_24_jan_morning Limits of Algebraic and Trigonometric Functions

x \to 0 x (\sqrt(2 ² x + 3 x) - \sqrt( ² x + x + 4))

is equal to :

A. $0$ $0$
B. $12\sqrt(5)$ $\frac{1}{2\sqrt{5}}$
C. $1\sqrt(15)$ $\frac{1}{\sqrt{15}}$
D. $- 12\sqrt(5)$ $-\frac{1}{2\sqrt{5}}$

Solution

Related Formula

To evaluate limits of indeterminate types containing radical forms, rationalize the numerator directly by multiplying by its conjugate element matching:

(\sqrt(A) - \sqrt(B))(\sqrt(A) + \sqrt(B)) = A - B

Core Logic

Rewrite the expression as a fraction with $x$ $\sin x$ in the denominator and rationalize the numerator:

x \to 0 (2 ² x + 3 x) - ( ² x + x + 4) x \cdot (\sqrt(2 ² x + 3 x) + \sqrt( ² x + x + 4))

= x \to 0 ² x + 3 x - x - 4 x \cdot (\sqrt(2 ² x + 3 x) + \sqrt( ² x + x + 4))

Step 1: Simplify Numerator and Group Terms

Express the numerator terms to isolate algebraic patterns:

² x + 3 x - 4 - x = ( x - 1)( x + 4) - x

Substitute this back into our rationalized limit format:

= x \to 0 ( x - 1)( x + 4) - x x \cdot (\sqrt(2 ² x + 3 x) + \sqrt( ² x + x + 4))

Step 2: Distribute $x$ $\sin x$ in Denominator

Split the limit across the two separated numerator expressions:

= x \to 0 [ ( x - 1)/( x) \cdot ( x + 4) - 1 ] \cdot 1\sqrt(2(1)+3) + \sqrt(1+0+4)

Evaluate the limit component values:

x \to 0 ( x - 1)/( x) = x \to 0 (-2 ²(x/2))/(2 (x/2) (x/2)) = x \to 0 [- (x/2)] = 0

Substituting this zero value simplifies the numerator expression directly:

= [ 0 \cdot (1 + 4) - 1 ] \cdot 1\sqrt(5) + \sqrt(5) = -12\sqrt(5)

Pattern Recognition

Recognizing that $( x - 1)/( x) \to 0$ $\frac{\cos x - 1}{\sin x} \to 0$ as $x \to 0$ $x \to 0$ isolates the non-vanishing trigonometric components without needing full multi-stage application of L'Hôpital's rule.

Chapter Mix

Class 11 Mathematics: Limits and Derivatives

Q72 jee_main_2025_28_jan_evening Limits of Trigonometric Functions

Let

f(x)= narrow \inftyΣr=0ⁿ( (x/2r+1)+ ³(x/2r+1)1- ²(x/2r+1)).

Then

xarrow0 ex-ef(x)(x-f(x))

is equal to

Numerical Answer. Answer: 1 to 1

Solution

Related Formula

Trigonometric identity:

( θ + ³θ)/(1- ²θ) = θ ( (1+ ²θ)/(1- ²θ) ) = ( θ)/( 2θ)

Also note standard telescopic identity:

2φ - φ = ( φ)/( 2φ)

Core Logic

Let $θ = x2r+1$ $\theta = \frac{x}{2^{r+1}}$. The term inside the summation simplifies to:

((x)/(2^r)) - ( x2r+1)

Now, evaluating the summation:

Σr=0ⁿ [ ((x)/(2^r)) - ( x2r+1) ] = x - ( x2ⁿ⁺¹)

Taking the limit as $n \to \infty$ $n \to \infty$, $( x2ⁿ⁺¹) \to (0) = 0$ . Therefore, $f(x) = x$ $f(x) = \tan x$.

Step 1: Evaluate the Limit

We need to find:

xarrow0 ex-exx- x

Factor out $ex$ $e^{\tan x}$ from the numerator:

xarrow0 ex \cdot [ ex- x - 1x- x ]

Let $u = x - x$ $u = x - \tan x$. As $x \to 0$ $x \to 0$, $u \to 0$ $u \to 0$. The limit becomes:

uarrow0 e⁰ \cdot [ (e^u - 1)/(u) ] = 1 \times 1 = 1

Pattern Recognition

Standard limit substitution $y \to 0 (e^y - 1)/(y) = 1$ $\lim_{y \to 0} \frac{e^y - 1}{y} = 1$ applies cleanly whenever the argument in the exponent matches the entire denominator layout.

Chapter Mix

Class 11 Mathematics: Trigonometry Class 12 Mathematics: Limits, Continuity and Differentiability

Q jee_main_2025_29_jan_morning Limits of Special Series

The value of

n\to \infty(ΣK = 1ⁿ(k³ + 6k² + 11k + 5)/((k + 3)!))

is:

A. $(4)/(3)$ $\frac{4}{3}$
B. 2
C. $(7)/(3)$ $\frac{7}{3}$
D. $(5)/(3)$ $\frac{5}{3}$

Solution

Related Formula

Σk=1\infty ( (1)/(k!) - (1)/((k+3)!) ) Telescoping Series simplification

Core Logic

Rewrite the numerator polynomial to establish factor terms matching the factorial expansion base $(k+3)$ $(k+3)$:

k³ + 6k² + 11k + 5 = (k³ + 6k² + 11k + 6) - 1 = (k+1)(k+2)(k+3) - 1

Step 1: Simplify General Term

Tk = ((k+1)(k+2)(k+3))/((k+3)!) - (1)/((k+3)!)

Tk = (1)/(k!) - (1)/((k+3)!)

This creates a clean telescoping layout format structure.

Step 2: Sum the Series

Writing out expanded \partial sums up to infinity:

S = ( (1)/(1!) + (1)/(2!) + (1)/(3!) + (1)/(4!) + ) - ( (1)/(4!) + (1)/(5!) + (1)/(6!) + )

All higher terms cancel out systematically, leaving exactly the leading remaining fragments:

S = (1)/(1!) + (1)/(2!) + (1)/(3!) = 1 + (1)/(2) + (1)/(6) = (10)/(6) = (5)/(3)

Pattern Recognition

Whenever factorials dominate fraction denominators, manipulate structural terms to align components via Telescoping sums ( $Vₙ - Vn-k$ $V_n - V_{n-k}$).

Chapter Mix

Class 11 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Sequences and Series

Q73 jee_main_2025_29_jan_morning Sandwich Theorem and Greatest Integer Function

Let [t] be the greatest integer less than or equal to t. Then the least value of

p in N

for which

x \to 0⁺ ( x ( [ 1x ] + [ 2x ] + + [ px ] ) - x² ( [ 1x² ] + [ 2²x² ] + + [ 9²x² ] ) ) \geq 1

is equal to

Numerical Answer. Answer: 24

Solution

Related Formula

x \to 0^+ x [ (k)/(x) ] = k

Σk=1ⁿ k = (n(n+1))/(2), Σk=1ⁿ k² = (n(n+1)(2n+1))/(6)

Core Logic

Using properties of Greatest Integer Function limits, as $x \to 0^+$ $x \to 0^+$, fraction values diverge cleanly to continuous variable distributions:

x \to 0^+ x [ (k)/(x) ] = k Σk=1p k = (p(p+1))/(2)

Similarly, for the second block component part:

x \to 0^+ x² [ (k²)/(x²) ] = k² Σk=1⁹ k² = (9 \times 10 \times 19)/(6) = 285

Step 1: Setup Inequality Formulation

Combine evaluated limits component parts:

(p(p+1))/(2) - 285 \geq 1

(p(p+1))/(2) \geq 286 p(p+1) \geq 572

Step 2: Solve for least natural number

Evaluate product bounds of adjacent integers: If $p = 23 23 \times 24 = 552$ $p = 23 \implies 23 \times 24 = 552$ (False) If $p = 24 24 \times 25 = 600$ $p = 24 \implies 24 \times 25 = 600$ (True) Therefore, the least natural value of $p$ $p$ is $24$ $24$.

Pattern Recognition

Greatest Integer fractions simplify directly to standard scalar values inside limits evaluated at infinity or zero, letting you drop brackets and treat them as arithmetic sequences.

Chapter Mix

Class 11 Mathematics: Limits

If t→ 0(∫₀¹ (3x + 5)^t dx)(1)/(t) = (α)/(5e)((8)/(5))(2)/(3)$\lim_{t\to 0}\left(\int_0^1 (3x + 5)^t dx\right)^{\frac{1}{t}} = \frac{\alpha}{5e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$, then α$\alpha$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Apply Taylor Series or L'Hopital's Theorem

Step 2: Complete the Final Form Match

Pattern Recognition

Chapter Mix

Reference Study Guides

More Limits Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Check Interval x ≤ -2$x \le -2$

Step 2: Check Interval -2 < x ≤ 0.5$-2 < x \le 0.5$

Step 3: Check Interval x > 0.5$x > 0.5$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Numerator and Group Terms

Step 2: Distribute x$\sin x$ in Denominator

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate the Limit

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify General Term

Step 2: Sum the Series

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setup Inequality Formulation

Step 2: Solve for least natural number

Pattern Recognition

Chapter Mix

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If $t\to 0(\int₀¹ (3x + 5)^t dx)(1)/(t) = (α)/(5e)((8)/(5))(2)/(3)$ , then $α$ $\alpha$ is equal to

Step 1: Check Interval $x \leq -2$ $x \le -2$

Step 2: Check Interval $-2 < x \leq 0.5$ $-2 < x \le 0.5$

Step 3: Check Interval $x > 0.5$ $x > 0.5$

Step 2: Distribute $x$ $\sin x$ in Denominator