Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) – f(x)}}{{x – 1}} = 44$ is _ _ _ _ .
Solution
Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable.
Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n – 1}}f(1) – f'(x) = 44$
$\therefore nf(1) – f'(1) = 44$
$\therefore n.9 – ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$
$ \Rightarrow 9n – 19 = 44$
$\Rightarrow n=7$
$cos^{-1}(cos(-5))+sin^{-1}(sin(6))-tan^{-1}(tan(12)) $ is equal to:
(The inverse trigonometric functions take the principal values)
(A) $3\pi – 11$
(B) $3\pi + 1 $
(C) $4\pi – 11$
(D) $4\pi – 9$
Solution
The given expression can be rewritten as,
$cos^{-1}cos5+sin^{-1}sin 6-tan^{-1} tan12 $
Considering the principal values the given expression can be further rewritten as,
$cos^{-1}cos(2\pi – 5)+sin^{-1}sin (-(2\pi- 6))-tan^{-1} tan(-(4\pi -12)) $
Or $(2\pi – 5)+ (-(2\pi- 6))-(-(4\pi-12)) = 4\pi -11 $
Answer: (C)
A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ .
Solution
For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics.
$tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $
$\Rightarrow \frac {3-r}{4} = \frac {r}{2}$
$\Rightarrow r=1 $
So, $R \equiv ( – 1,0)$
Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$
Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane P. Then which of the following points lies on P?
(A) $(0,2,-4)$
(B) $(4,0,-2)$
(C) $(-2,0,-\frac {1}{2})$
(D) $(3,1,-\frac {1}{2})$
Solution
Bisectors are given by,
$\frac{{x – 2y – 2z + 1}}{3} = \pm \frac{{2x – 3y – 6z + 1}}{7}$
$ \Rightarrow 7x – 14y – 14z + 7 = \pm (6x – 9y – 18z + 3)$
Hence, $x-5y+4z+4=0$ & $13x-23y-32z+10=0$
Let $\theta $ be the angle between $x-2y-2z+1=0$ & $x-5y+4z+4=0$
$\cos \theta = \frac{{|1 + 10 – 8|}}{{3 \times \sqrt {42} }} = \frac{1}{{\sqrt {42} }} < \frac{1}{{\sqrt 2 }}$
$\theta > 45^\circ $
So, $x-5y+4z+4=0$ is the obtuse angle bisector.
$\therefore $ Acute angle bisector $P \equiv 13x – 23y – 32z + 10 = 0$
The point $(-2,0,-\frac {1}{2})$ satisfies P.
Answer: (C)
80 g of copper sulphate $CuSO_4 .5H_2 O $ is dissolved in deionised water to make 5 L of solution. The concentration of the copper sulphate solution is $x \times 10^{-3} mol L^{-1} $. The value of x is _ _ _ _ . (Nearest integer)
[Atomic masses – Cu:63.54 u, S:32 u, O: 16 u, H: 1 u]
Solution
$Molarity = \frac {n}{V} = \frac {w}{M_0 L} $
$\therefore Molarity = \frac {80}{249.54 \times 5 } \approx 64 \times 10^{-3} mol L^{-1} $
$\therefore x = 64 $
For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g) $, when $\Delta S = -176.0 JK^{-1} $ and $\Delta H = -57.8 kJ mol^{-1} $, the magnitude of $\Delta G $ at 298 K for the reaction is _ _ _ _ $kJ mol^{-1} $. (Nearest integer)
Solution
We have, $\Delta G = \Delta H – T \Delta S $
$\Delta S = -176.0 JK^{-1} = -0.176 kJ K^{-1} $
$\Delta G = -57.8 – 298 \times (-0.176) \approx -57.8+52.5 = -5.3 $
$\therefore |\Delta G | = 5.3 kJ mol^{-1} $
Ans: 5
The molar solubility of $Zn(OH)_2 $ in 0.1 M NaOH solution is $x \times 10^{-18} M$. The value of x is _ _ _ _ . (Nearest integer)
(Given: The solubility product of $Zn(OH)_2 $ is $2 \times 10^{-20} $)
Solution
$Zn{(OH)_2}\rightleftharpoons \mathop {Z{n^{2 + }}}\limits_s + 2\mathop {O{H^ – }}\limits_{2s + 0.1} $
${K_{sp}} = [Z{n^{2 + }}]{[O{H^ – }]^2} = s.{(2s + 0.1)^2}$
$\therefore 2 \times {10^{ – 20}} = s.{(2s + 0.1)^2} \approx {0.1^2} \times s$
$ \Rightarrow s = 2 \times {10^{ – 18}}$
$\therefore x = 2 $