Tata Mcgraw Hill Mathematics For Iit Jee -

The number of ways in which 5 boys and 3 girls can be seated in a row so that no two girls sit together is divisible by 144. If the number of ways is $N$, find the value of $N/144$.

Phase 2 (Class 12 + revision): Return to earlier chapters (e.g., Trigonometry) and attempt Exercise 2. Mark difficult problems with a red pen for future revision.

Hint: Intersection at $(0,0)$ and $(1,1)$. Integral $\int_0^1 (\sqrtx - x^2) dx$. tata mcgraw hill mathematics for iit jee

This paper is structured as , consisting of Single Correct MCQs, Multiple Correct MCQs, and Integer Type questions.

Hint: Put $x = \tan\theta$. The expression simplifies to $\theta/2$. $y = \frac12\tan^-1x$. The number of ways in which 5 boys

If you can solve every “Difficult” problem in the Calculus and Coordinate Geometry sections of this book blindfolded, you are not just prepared for JEE Mathematics—you are ready to teach it.

| Feature | Tata McGraw Hill (RD Sharma) | Cengage (G. Tewani) | Arihant (SK Goyal) | | :--- | :--- | :--- | :--- | | | Enormous solved examples | Concept application & illustrations | Topic-wise past papers | | Theory Depth | Moderate | High (with conceptual notes) | Moderate to Low | | Problem Difficulty Gradient | Gradual, well-graded | Steep, jumps quickly | Mixed, sometimes erratic | | Best for | Building speed & pattern recognition | Deep conceptual clarity | Exam-focused revision | | Price-to-Value | Excellent (highly economical) | Expensive | Moderate | Mark difficult problems with a red pen for future revision

Let $f(x) = \int_0^x |t-1| dt$. Then: (A) $f(x)$ is continuous everywhere. (B) $f(x)$ is differentiable everywhere. (C) $f(x)$ has a local minimum at $x=1$. (D) $f'(1)$ does not exist.

If the distance between the plane $Ax-2y+z = d$ and the plane containing the lines $\fracx-12=\fracy-23=\fracz-34$ and $\fracx-23=\fracy-34=\fracz-45$ is $\sqrt6$, find the value of $|d|$.

The book is typically organized into , covering the entire JEE syllabus (both Main and Advanced). The structure follows a logical flow:

Hint: $D = \beginvmatrix 1 & a & 0 \ 0 & 1 & a \ a & 0 & 1 \endvmatrix = 1(1) - a(0-a) + 0 = 1 + a^2$. For infinite solutions, $D=0 \implies a^2 = -1$. Real value? Wait. The system is $x+ay=0$, $az+y=0$, $ax+z=0$. From eq 1: $x = -ay$. Eq 2: $y = -az$. Eq 3: $z = -ax$. Substitute: $x = -a(-az) = a^2z = a^2(-ax) = -a^3x$. $x(1+a^3) = 0$. So infinite solutions if $1+a^3 = 0 \implies a=-1$. Is $a=0$ possible? $x=0, y=0, z=0$. Unique solution $(0,0,0)$. So not C. Answer is (B) . Wait , if $a=0$, solution is $x=0, y=0, z=0$. It is unique, not infinite. If $a=-1$, $x=y, y=z, z=x$. $x=y=z$. Infinite solutions (any point on line $x=y=z$). Correct Answer: (B) .