If $S_n$ denotes the sum of $n$ terms of an A.P. with common difference $d$, then $\lim_n \to \infty \fracS_n - n S_1n(n-1)$ is equal to: (A) $d/2$ (B) $d$ (C) $2d$ (D) $0$
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. tata mcgraw hill mathematics for iit jee
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 $S_n$ denotes the sum of $n$ terms of an A
Hint: Differentiate twice to eliminate constants. Let $f(x) = \int_0^x |t-1| dt$
If $\alpha, \beta$ are roots of $x^2 - 3x + 4 = 0$, then the value of $\alpha^3 + \beta^3 + \frac1\alpha^3 + \frac1\beta^3$ is equal to: (A) 0 (B) $63/64$ (C) $-63/64$ (D) 1
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) .
This is one of the most pirated books in India. Counterfeit copies have poor print quality, missing pages, and wrong answer keys. Always buy from a trusted seller (Amazon, Flipkart Retail, or a physical bookstore) and check for McGraw Hill’s hologram.